simulescumechanics

TECHNICAL UNIVERSITY OF CIVIL ENGINEERING

Mechanics of Materials Student Workbook

with Mathcad Applications

-Volume II-

Eng. Ion S. Simulescu, MPh, PhD, PE Associate Professor

Bucharest 2007

I

PREFACE The present textbook is the second volume from a series of textbooks, titled Mechanics of Materials Student Workbook, intended to familiarize the students enrolled in the first semester of their sophomore year at the Technical University of Bucharest, School of Civil Engineering, with the practical application of the theoretical concepts developed during the weekly lectures. This textbook has in fact a complementary role to the theoretically orientated textbook published under the name of Lectures in Mechanics of Materials, volume I. A number of six chapters are covered by this textbook. Briefly, the organization is: Chapter 5 - Pure Shear, Chapter 6 - Connections of Axially Deformed Members, Chapter 7 - Bending of Plane Linear Beams, Chapter 8 - Bending Deflection of Linear Beams, Chapter 9 - Torsion and Chapter 10 - Plane Stress Transformation. The organization of the individual chapter is kept identical to the organization utilized in the first volume. Each chapter starts with a theoretical section, named “Theoretical Background”, where the most important theoretical aspects are succinctly discussed. This section is followed by the “Solved Problems” section which contains a number of representative in-depth solved problems. Finally, the last section is the “Proposed Problems” section where a relatively large number of problems are proposed to the student for private exercise. With the intent to increase the student appetite towards using the modern capability of the numerical computer, the problems contained in the “Solved Problems” sections are solved using the MATHCAD software capabilities in parallel to the more classical method of the manual calculation. I can not conclude without expressing my appreciation for the help and dedication provided during the tedious period of the text editing and figures preparation by my students Alin Radu, Diana Ene, Irina Lazar and Dan Peticila. It is also my pleasure to acknowledge the help that I received during the initial phase of preparation of this textbook from my younger colleague Eng. Ovidiu Bogdan. Finally, the author expresses his sincere gratitude to Professor Dr. Eng. Dan Cretu, the Chairman of the Strength of Materials Department, for his encouragements and support in the realization of this textbook. Dr. Eng Ion S. Simulescu Bucharest, Romania. September, 2007.

II

TABLE OF CONTENTS

Chapter 5 Pure Shear 1

5.1 Theoretical Background 5.2 Solved Problems 5.3 Proposed Problems

Chapter 6 Connections of Axially Deformed Members 11

6.1 Theoretical Background 6.1.1 Analysis of Connections Comprised of Discrete Connectors 6.1.2 Analysis of Continuous Welded Connections 6.1.3 Analysis of Connected Parts 6.1.4 Connection Analysis. Practical Steps

6.2 Solved Problems 6.3 Proposed Problems

Chapter 7 Bending of Plane Linear Beams 67 - Stress Distribution

7.1 Theoretical Background 7.1.1 Pure Bending 7.1.2 Nonuniform Bending 7.1.3 Strength Condition 7.1.4 Built-Up Beam Connectors


Chapter 8 Bending of Plane Linear Beams 193 - Deflection

8.1 Theoretical Background 8.1.1 Integration of the Moment - Curvature Differential Equation 8.1.2 Integration of the Load-Deflection Differential Equation 8.1.3 Boundary and Continuity Integration Conditions 8.1.4 Superposition Method


III

Appendix 8.1 Deflection Formulae

Chapter 9 Torsion 305

9.1 Theoretical Background 9.1.1 Member with Circular Cross-Section 9.1.2 Member with Solid Rectangular Cross-Section 9.1.3 Member with Solid Elliptical Cross-Section 9.1.4 Member with Closed Thin-Wall Cross-Section 9.1.5 Member with Open Thin-Wall Cross-Section


Chapter 10 Plane Stress Transformation 340

10.1 Theoretical Background 10.1.1 Plane Stress Transformation Equations 10.1.2 Principal Stresses and Directions 10.1.3 Maximum Shear Stresses and Directions 10.1.4 Mohr’s Circle for Plane Stresses


References 375

1

CHAPTER 5 Pure Shear

5.1. Theoretical Background

In accordance with the shear stress definitions the shear stress τ is defined as:

AV

A ∆∆

= →∆ 0limτ (5.1)

where V∆ is the tangential stress vector acting on the infinitesimal area A∆ .

If the particular case of the linear plane beam is considered, the resultant shear

forceV acting on the beam cross-section is obtained by integrating the shear stress

vector over the entire area A of the cross-section:

∫=A

dAV *τ (5.2)

If only the shear force V is considered acting on the cross-section, the cross-section is

subjected to pure shear. The pure shear condition is more a practical desiderate than a

real condition. In general, as indicated by the differential relation (5.3),

)(xVdxdM

= (5.3)

In the absence of the distributed moments )(xm , the existence of the shear force V on

the cross-section is related with the simultaneous existence of the bending moment

)(xM on the same cross-section. Consequently, the “pure shear” is realized only if

the magnitude of the bending moment can be neglected.

In general, the distribution of shear stress τ is not uniform on the cross-section and,

consequently, can not be determined from the integral equation (5.2). In some cases

the average shear stress avgτ may be used in place of the true shear stress τ without

introducing a significant error. The average shear stress avgτ is calculated as:

AV

avg =τ (5.4)

2

The shear force V and the shear stress avgτ are illustrated in Figure 5.1.

Figure 5.1 Pure Shear

(a) Shear Force and (b) Average Shear Stress

The existence of shear stress implies the existence of shear strain, which represents

the change of the right angle. The original right angle shown in Figure 5.2.a is

modified into an acute angle *θ as depicted in Figure 5.2.b.

Figure 5.2 Shear Strain

(a) Undeformed Angle and (b) Deformed angle

The shear strainγ , representing the angular change, and measured in radians, is

calculated as:

*

2θπγ −= (5.5)

If the angular change is a small angle, the shear strain is approximated by its tangent:

s

s

Lδ

θπγ =−≅ )2

tan( * (5.6)

3

For the particular case of the homogeneous linear elastic material, the constitutive

low, the relation between the shear strain and stress, is described by Hook’s law as:

γτ *G= (5.7)

where G is the shear modulus characterizing the material.

The strength verification formula is:

allavg ττ ≤ (5.8)

where allτ is the allowable stress of the material, a value experimentally determined.

5.2. Solved Problems

The following section contains a number of practical situations which match the pure

shear assumptions presented above. The calculation of the connections subjected to

axial deformation, also involving the pure shear assumptions, is presented in the next

chapter.

Problem 5.2.1 The hole in a plate with dimensions shown in Figure 5.2.1 is punched

out by a hydraulic punch press. If the plate has a thickness of t=1.6 mm and is made

of steel with an average punching shear resistance of 262 MPa, what is the required

punching force P?

Figure 5.2.1

4

A. General Observations

A.1- The punching force P induces a constant average shear stress avgτ considered

uniformly distributed on the lateral surface located around the perimeter of the hole.

A.2 - Numerical Application

The following numerical data is provided:

Fv 262MPa:= - punching shear resistance of steel;

t 1.6mm:= - hole thickness;

hr 3mm:= - hole radius;

hl 38mm:= - hole length.

B. Calculations

The perimeter Hp and the lateral area Ls of the hole are:

Hp 2 hl π hr⋅( )+ ⋅:= →+⇒ ]3.0*8.3[*2 π Hp 9.485cm=

Ls Hp t⋅:= →⇒ 16.0*5.9 Ls 1.518cm2=

The punching force P is then calculated as:

P Ls Fv⋅:= →⇒ − Pam 624 10*262*10*518.1 P 3.976 104× N=

P 39.761kN=

Problem 5.2.2 The bearing pad shown in Figure 5.2.2, consisting of two steel plates

bonded to a chloroprene elastomer, is subjected, during a static laboratory test, to a

shear force V=5.35 kN. Considering the lower steel plate rigidly attached to the base,

the relative lateral displacement of the top steel plate measured is 6 mm. What is the

shear modulus of elasticity of the chloroprene elastomer?

5

Figure 5.2.2


A1 - It is assumed that the shear force V is applied at the center of the top steel plate

and, consequently, only a lateral displacement δ=6 mm of the chloroprene elastomer

in X direction is induced. The undeformed and deform condition of the elastomer are

shown in the sketch below:

A2 - The plate dimensions are:

a 125mm:= b 150mm:= t 38mm:=

B. Calculations

The shear strain xyγ is calculated:

γxy atanδt

:= →

⇒ −

8.36.0tan 1 γxy 0.157=

The average shear stress avxy _τ is obtained as:

τxy_avV

Apl:= →⇒ − 24

3

10*5.18710*35.5

mN

τxy_av 2.853 105× Pa=

where the area Apl a b⋅:= →⇒ 15*5.12 Apl 187.5cm2=

The shear modulus of elasticity is:

6

Gτxy_av

γxy:= →⇒

157.010*853.2 5 Pa G 1.822 106× Pa=

5.3 Proposed Problems

Problem 5.3.1 The 500 kN capacity hydraulic punch press, illustrated in Figure 5.3.1,

is used to punch a circular hole in a 10 mm thick aluminum plate. If the average

punching shear resistance of this plate is 250 MPa, what is the maximum diameter of

hole that can be punched?

Figure 5.3.1

Problem 5.3.2 The joint, shown in Figure 5.3.2, located between two concrete slabs, A

and B, is filled with a flexible epoxy. The joint has a height h = 150 mm, length

L = 2.0 m and thickness t = 12 mm. When subjected to vertical loading the slabs

displace vertically relative to each other with a distance d = 0.075 mm. Determine: (a)

the average shear strain in the epoxy and (b) the magnitude of the shear force at the

joint if the epoxy has a shear modulus of 960 GPa.

Figure 5.3.2

7

Problem 5.3.3 The flexible connection consisting of two rubber pads is bonded to

steel plates as illustrated in Figure 5.3.3. The rubber pads are 200 mm long, 150 mm

wide and 12 mm thick. Calculate: (a) the average shear strain occurring in the rubber

if the force P = 15 kN and the shear modulus for the rubber is G = 830 kPa, and (b)

the relative horizontal displacement between the interior plate and the outer plates.

Figure 5.3.3

Problem 5.3.4 The shock absorber constructed from an outer steel tube, a central steel

bar and a hollow rubber cylinder bonded to the steel components is used to support a

delicate instrument transmitting a vertical force P. The pertinent dimensions of the

shock absorber are shown in Figure 5.3.4. Calculate the following: (a) the shear stress

in the rubber at a radial distance r measured from the center of the steel supporting bar

and (b) the vertical displacement of the central steel bar, assuming that the shear

modulus of the rubber is known and the steel components are rigid elements.

Figure 5.3.4

8

Problem 5.3.5 A reinforcing bar is casted in concrete, as shown in Figure 5.3.5, with

the intent to conduct a “pull-out” capacity test. Assuming that the rebar has a diameter

d = 12 mm, an embedment length L = 30 cm and is subjected to a tensile force P =

17.5 kN determine: (a) the average shear stress, the bond stress, developed between

the concrete and the steel if the uniform distribution is considered and (b) the

maximum shear stress if the following distribution of the shear bond is accepted as

)*6**9*4(**4

3233

max xxLLL

+−=ε

τ , where x is measured from the interior end

of the bar toward the concrete surface.

Figure 5.3.5

Problem 5.3.6 A rectangular plate made of plastic, pictured in the undeformed

position with dashed lines, is uniformly deformed into the shape of the parallelogram,

marked in Figure 5.3.6 as a shaded surface. Calculate the average shear strain and

stress in the plate after the deformation considering a shear modulus of elasticity of

G=1GPa.

Figure 5.3.6

9

Problem 5.3.7 The shear stress τ applied on the edges of a rectangular plate, shown

in Figure 5.3.7, induces a shear strain 0012.0=γ . Determine the horizontal

displacement Aδ and the corresponding shear stressτ , if the shear modulus of

elasticity of the material is G=10GPa.

Figure 5.3.7

Problem 5.3.8 The horizontal plate, pictured in Figure 5.3.8, is subjected to an axial

load P=250 kN. The plate is mounted into a C shaped lug using two rubber pads. If

the dimensions of the rubber pads are b=50mm, h= 25mm and w=80mm, respectively,

calculate: (a) the average shear stress avgτ in the rubber pads, (b) the average shear

strain avgγ in the rubber pads considering a shear modulus of elasticity for the rubber

G=0.8MPA, and (c) the relative horizontal displacement between the lugs and the

horizontal plate.

Figure 5.3.8

Problem 5.3.9 The pair of players, pictured in Figure 5.3.9, are held together by a

central bolt with a diameter d = 60mm and made of steel with an allowable shear

10

stress MPashearboltall 80__ =τ . A rigid part C is pressed between the players’ jaws by

applying simultaneously a pair of forces P at location A. Determine the maximum

force which can be applied to part C.

Figure 5.3.9

11

CHAPTER 6 Connections of Axially Deformed Members


This chapter presents the most important aspects of the analysis of the connections of

the axially deformed members. A detailed theoretical presentation is made in Chapter

6 of the book titled Lectures in Mechanics of Materials - Volume I, and only the

most important results, necessary to assist the problem solving process, are reprinted.

The connections are grouped in two main categories: (a) connections with discrete

connectors and (b) connections with continuous connectors. The first category refers

to connections realized with bolts, rivets or nails, while the second category includes

welded or glued connections. In engineering practice one can encounter mixed

connections, but those are not the subject of this chapter.

6.1.1 Analysis of Connections Comprised of Discrete Connectors

In general, a connection is realized from a single connector or from a multitude of

single connectors (multiple connectors). The single connector is a small cylindrical

deformable solid characterized by a tiny cross-section by comparison with its length.

Both types of connections can be treated in a universal way if the following

assumption is accepted:

Assumption 1: It is commonly assumed, when the geometry of the connection is

symmetrical and the load passes through the connection’s centroid, that

the load is equally shared by all connectors.

The mathematical transposition of the above assumption is:

conn

memberconngle n

FP =_sin (6.1)

where: conngleP _sin is the axial force transmitted trough the single connector;

memberF is the axial force in the member;

12

connn is the number of single connectors participating in the axial force

of the member.

The axial force pertinent to the single connector, conngleP _sin , must be less or equal than

the capable force characterizing the single connector connglePcap _sin :

connglegle_conn PcapP _sinsin ≤ (6.2)

The capable force connglePcap _sin developed by a single connector is obtained

considering two failure scenarios: (1) bearing failure and (2) shearing failure.

• Failure Scenario 1 (Bearing Failure):

The capable force resulting from bearing failure, connglebearPcap _sin_ , is calculated

as:

bearallpartpartsbconngle ttdbearPcap __sin *),min(*_ σ←→= (6.3)

where: bd is the diameter of the connector;

←→ partparts tt , are the sum of the thicknesses of the parts moving in

opposite directions assuming the rupture of the connector;

bearall _σ is the allowable bearing stress of the material.

Observations: (1) the bearing takes place at the curved surface of contact between the

bolt and the axially deformed member. Due to the difficulties related

with the exact calculation of the contact area and the variation of the

bearing normal stress on this area a simpler expression,

),min(* ←→ partpartsb ttd , is adopted in the calculation contained in the

above formula (6.3);

(2) the allowable bearing stress bearall _σ , determined from laboratory

tests, is supposed to be known in the calculations and is considered

uniformly distributed on the contact area.

13

(3) the error introduced by using the simpler expression of the contact

area and a uniformly distributed bearing stress is considered during the

evaluation of the allowable bearing stress bearall _σ value assigned by

the codes.

• Failure Scenario 2 (Shearing Failure):

The capable force resulting from shearing failure, conngleshearPcap _sin_ , is obtained

as:

shearallb

shearconngled

nshearPcap _

2

_sin *4

**_ τ

π= (6.4)

where: shearn is the number of the shearing surfaces of the connector and is

equal to the number of the interfaces of the connected components;

bd is the diameter of the connector;

shearall _τ is the allowable shearing stress of the connector’s material.

Observations: (1) the formula (6.4) is not an “exact” formula, but for its simplicity is

adopted by all international codes;

(2) the shear stress shearall _τ , acting in the shearing cross-section of the

connector, is considered uniformly distributed and is obtained from

laboratory tests;

(3) the error induced by the usage of the formula (6.4) is considered at

the evaluation of the shearall _τ value assigned by the codes.

The capable force developed by the individual connector, connglePcap _sin , is calculated

as:

)_,_min( _sin_sin_sin conngleconngleconngle shearPcapbearPcapPcap = (6.5)

14

6.1.2 Analysis of Continuous Welded Connections

In the “Manual of Steel Construction Allowable Stress Design”, published by the

American Institute of Steel Construction (AISC), the welds are categorized in four

basic groups: groove, fillet, slot and plug welds. These categories are illustrated in

Figure 6.1. In engineering practice, the fillet weld is the most commonly used type of

weld, due to its facile fabrication.

The allowable stress on various types of welds is dependent upon the effective area

of the weld. The effective area of the groove or fillet weld is calculated as the product

of the effective throat dimension et and the length of the weld. For fillet welds the

effective throat dimension is defined as the shortest dimension measured from the

weld root to the face of the weld. Assuming that the weld has equal legs of length a

the effective throat is equal to2

a .

Figure 6.1 Basic Weld Types

The calculation of the effective throat, accordingly to AISC, is shown in Figure 6.2.

15

Figure 6.2 Effective Throat Dimension for Fillet Welds

(a) Equal Legs and (b) Unequal Legs

For full penetration groove welds, the AISC defines the effective throat as a function

of the thickness of the parts joined in the connection as pictured in Figures 6.3.a and

6.3.b. In the case of partial penetration welds, the effective throat is a function of the

depth of the preparation as shown in Figures 6.3.c and 6.3.d.

Assumption 2: It is commonly assumed that an individual weld fails in shearing along

the effective area.

The capable force developed by a single weld, weldcapP _ , is calculated:

weldalleffweldweldcap tlP __ ** τ= (6.6)

where: weldl , efft are the effective lengths and throats of the weld;

weldall _τ is the allowable shear stress in the weld.

Observations: (1) the failure surface presumed in formula (6.6) is justified by

laboratory tests;

(2) the allowable bearing stress weldall _τ , determined from laboratory

tests, is supposed known in the calculations and is considered

uniformly distributed on the contact area.

(3) the error introduced by using the simpler expression of the contact

area and an uniformly distributed shearing stress is considered during

the evaluation of the allowable bearing stress weldall _τ value assign by

the codes.

16

Figure 6.3 Groove Weld Effective Throat

The capable force developed by a single weld, weldcapP _ , must be greater or equal to

the axial force corresponding to the specific weld:

weldweldcap FP ≥_ (6.7)

If all the welds comprising a welded connection are geometrically identical, parallel

and equidistant to the longitudinal axis of the member, the axial force of the member

memberF is considered equally distributed to each weld:

weld

memberweld n

FF = (6.8)

where: weldn is the number of welds existing in the connection.

Note: Special cases of eccentrically welds and combination of groove and fillet welds

are presented in the next section.

6.1.3 Analysis of Connected Parts

The axial load existing in the axially deformed member is transferred between the

parts of the connections through the connectors. The regular verification described in

Chapter 4 titled “Axial Deformation” applies. This verification is summarized with

the following equation:

17

allcritical

membertcritcorresmembertcritical A

Fσσ ≤= _sec__

_sec_ (6.9)

where: membertcritical _sec_σ is the normal stress in any critical cross-section;

criticalA is the effective area of the critical cross-section;

membertcritcorresF _sec__ is the corresponding axial force acting in the critical

cross-section considered;

allσ is the allowable normal stress of the material.

6.4 Connection Analysis. Practical Steps

The practical analysis of a connection subjected to axial force is conducted

considering the following main steps:

(1) disregard the existence of the connection and calculate the axial force acting

on the member related with the connection;

(2) identify the connecting parts and the connection type;

(3) draw the “transmission” diagram. This diagram indicates the transfer of the

axial force between the parts comprised in the connection. The “transmission”

diagram is the most suggestive for the connections realized from multiple

discrete connectors. For connections realized with a single bolt or for welded

connections the “transmission” diagram becomes trivial;

(4) Verify the member and all other constituent parts of the connection

considering all critical cross-sections and their corresponding axial forces as

described in section 6.4;

(5) Verify the connection’s capacity.

(6) The capacity of the connection is represented by the minimum capacity of the

parts and of the connection.

These steps are exemplified in the next section.

18


Problem 6.2.1 The flat bar illustrated in Figure 6.2.1 has a width b = 50 mm and

thickness t =6 mm, and is loaded in tension by a force P.

Figure 6.2.1

The bar is attached to a support by a pin of diameter d that passes through a hole of

the same size in the bar. The distance between the hole and the end of the bar is h =

25 mm. The allowable tensile stress on the net cross section of the bar, along the

planes a-b and c-d, is 112 MPa. The allowable shear stress in the pin is 64 MPa, while

the allowable shear stress in the bar along planes e-b and f-c is 43 MPa. Calculate: (a)

the pin diameter for which the load P will be maximum and (b) the corresponding

maximum value of the load.


A.1 – The system contains two structural elements: (a) the axially deformed flat bar

and (b) the bolt with an unknown diameter d . Four failure scenarios are emphasized

and sysP together with its corresponding bolt diameter are identified.


τall_shear_bolt 64 MPa⋅:= - allowable shear stress for the bolt;

τall_shear_bar 43 MPa⋅:= allowable shear stress of the bar;

19

σall 112 MPa⋅:= - allowable tensile stress in the bar;

b 50 mm⋅:= - dimensions of the flat bar’s cross-section;

t 6 mm⋅:=

h 25 mm⋅:= - edge distance.

B. Calculations

B.1 – Flat Bar Capacity Calculation

B.1.a - Scenario 1 – Material failure on cdab − cross-section

P1 d( ) t b d−( )⋅ σall⋅:= Nd 64 10*112*10*)5(*6.0 −−⇒

B.1.b - Scenario 2 – Material failure on fceb − cross-sections

P2 d( ) 2 t hd2

+⋅ τall_shear_bar⋅

⋅:= Nd

+⇒ − 64 10*43*10*)

25.2(*6.0*2

B.2 – Bolt Capacity Calculation

B.2.a - Scenario 3 - Shear failure of the bolt

P3 d( ) 2π d2⋅

4τall_shear_bolt⋅

⋅:= Nd )10*64*10*

4*(*2 64

2−⇒

π

B.2.b - Scenario 4 - Bearing failure of the bolt

P4 d( ) d t⋅ 1.6⋅ σall⋅:= Nd 64 10*112*6.1*10*6.0* −⇒

B.3 – Calculation of the Capacity of the System

Pmin d( ) min P1 d( ) P2 d( ), P3 d( ), P4 d( ),( ):=

The four previously calculated capacities )(),(),( 321 dPdPdP and )(4 dP , dependent

on the bolt diameter d , are plotted in the graphs shown below. The bolt diameter d ,

varies from 1 cm to 4 cm.

20

0.01 0.015 0.02 0.025 0.03 0.0350

5 .104

1 .105

1.5 .105

2 .105

P1 d( )

P2 d( )

P3 d( )

P4 d( )

d

The second graph pictures the capacities )(),(),( 321 dPdPdP and )(4 dP together with

the minimum envelope function )(min dP .

0.01 0.015 0.02 0.025 0.03 0.0350

5 .104

1 .105

1.5 .105

2 .105

P1 d( )

P2 d( )

P3 d( )

P4 d( )

Pmin d( )

d The third graph contains only the variation of the )(min dP . Consequently, the

maximum value of the axial force characterizing the system is identified

as NPsys 18635= . Its corresponding bolt diameter is cmd 245.2= .

21

0.01 0.013 0.016 0.019 0.022 0.025 0.028 0.031 0.034 0.037 0.045000

7500

1 .104

1.25 .104

1.5 .104

1.75 .104

2 .104

Pmin d( )

d

Problem 6.2.2 Calculate the maximum axial load capP which can be applied to the

lap- splice, shown in Figure 6.2.2.a, comprised of two (2) spliced bars, 1 & 2, and two

connection bars, 3 & 4, having cross-sections of 260mm x 18mm and 260mm x 8mm,

respectively. Ten bolts mm22=Φ , symmetrically located relatively to the splice in

two groups, are used.

Figure 6.2.2.a


A.1 – The system contains three structural elements: (a) the spliced bars 1 & 2, (b) the

connection bars 3 & 4, and (c) the bolted connection assembled from 10 bolts

mm22=Φ symmetrically located in two groups relatively to the splice. The

“transmission” diagram of the axial force between the structural components is shown

in Figure 6.2.2.b. Three values, sbcapP _ , cbcapP _ and boltP , corresponding to the

22

capable axial force pertinent to each of the structural components are calculated and

the minimum is retained as the capable axial force capP characterizing the entire

system capacity. The “transmission” diagram of the axial load between the connected

parts is shown in Figure 6.2.2.b.

Figure 6.2.2.b Axial Force Transmission Diagram


σall 210 MPa⋅:= - allowable stress of the material;

nbolt 5:= - number of bolts transmitting the force between the spliced bar 1 to the

connection plates and from the connection plates 3 & 4 to the spliced bar

2;

dbolt 22 mm⋅:= - bolt diameter;

ncon_bar 2:= - number of connection bars;

lcon 260 mm⋅:= - width of the connection bars 3 & 4;

tcon 8 mm⋅:= - thickness of the connection bars 3 & 4;

lplate 260 mm⋅:= - width of the spliced bars 1 & 2;

tplate 18 mm⋅:= - thickness of the spliced bars 1 & 2;

23

B. Calculations

B.1 –Calculation of the Capacity of the Spliced Bars 1 & 2 (see Figure 6.2.2.a)

The area of the spliced bars 1 & 2 critical cross-sections’ 1-1, 2-2 and 3-3,

respectively, are calculated as:

Asb1_1 lplate tplate⋅ 2 dbolt⋅ tplate⋅−:= →−⇒ ]8.1*2.2*28.1*26[

Asb1_1 38.88cm2=

Asb2_2 lplate tplate⋅ dbolt tplate⋅−:= →−⇒ ]8.1*2.28.1*26[ Asb2_2 42.84cm2=

Asb3_3 Asb1_1:= → Asb3_3 38.88cm2=

The axial force diagram, shown in Figure 6.2.2.b, indicates that only a fraction of the

total axial force capP

corresponds to each one of the cross-sections. Consequently, the

following equations are obtained:

Pcap_sb1_1 Asb1_1 σall⋅

35Pcap_sb2_2 Asb2_2 σall⋅

25Pcap_sb3_3 Asb3_3 σall⋅

The axial capable force in each cross-section is:

Pcap_sb1_1 Asb1_1 σall⋅:= →⇒ − 64 10*210*10*88.38

Pcap_sb1_1 8.165 105× N=

Pcap_sb2_253

Asb2_2 σall⋅:= →⇒ − 64 10*210*10*84.42*35

Pcap_sb2_2 1.499 106× N=

24

Pcap_sb3_352

Asb3_3 σall⋅:= →⇒ − 64 10*210*10*88.38*25

Pcap_sb3_3 2.041 106× N=

The capable axial force sbcapP _ carried by the spliced bars 1 & 2 is calculated as the

minimum of the values of the axial forces calculated above:

Pcap_sb min Pcap_sb1_1 Pcap_sb2_2, Pcap_sb3_3,( ):= → Pcap_sb 8.165 105× N=

B.2 – Calculation of the Capacity of the Connection Bars 3 & 4 (see Figure 6.2.2.a)

The area of the cross-sections of the each connection bar is obtained as:

Acb1_1 lcon tcon⋅ 2 dbolt⋅ tcon⋅−:= →−⇒ 8.0*2.2*28.0*26 Acb1_1 17.28cm2=

Acb2_2 lcon tcon⋅ dbolt tcon⋅−:= →−⇒ 8.0*2.28.0*26 Acb2_2 19.04cm2=

Acb3_3 lcon tcon⋅ 2 dbolt⋅ tcon⋅−:= →−⇒ 8.0*2.2*28.0*26 Acb3_3 17.28cm2=

From the axial force diagram, shown in Figure 6.2.2.b each of the cross-sections are

subjected to the following fraction of the total axial force as indicated by the

equations written below:

25Pcap_cb1_1 ncon_barAcb1_1 σall⋅

35Pcap_cb2_2 ncon_barAcb2_2 σall⋅

Pcap_cb3_3 ncon_barAcb3_3 σall⋅

where ncon_bar 2:= is the number of connection plates used.

Consequently, these forces are calculated as:

25

Pcap_cb1_152

ncon_bar Acb1_1⋅ σall⋅:= →⇒ − 64 10*210*10*28.17*2*25

Pcap_cb1_1 1.814 106× N=

Pcap_cb2_253

ncon_bar Acb2_2⋅ σall⋅:= →⇒ − 64 10*210*10*04.19*2*35

Pcap_cb2_2 1.333 106× N=

Pcap_cb3_3 ncon_bar Acb3_3⋅ σall⋅:= →⇒ − 64 10*210*10*28.17*2

Pcap_cb3_3 7.258 105× N=

The capable axial force cbcapP _

characterizing the connection plates is obtained as

follows:

Pcap_cb min Pcap_cb1_1 Pcap_cb2_2, Pcap_cb3_3,( ):= → Pcap_cb 7.258 105× N=

B.3 –Calculation of the Capacity of the Multiple-Bolt Connection

B.3.a – Calculation of the Individual Bolt Capacity

B.3.a.1 – Calculation of the Bearing Capacity

The allowable bearing stress used in the calculation is obtained:

σall_bearing 1.6 σall⋅:= →⇒ 610*210*6.1 σall_bearing 3.36 108× Pa=

The bearing capacity of the bolt bearboltP _

is calculated as:

Pbolt_bear dbolt min 2 tcon⋅ tplate,( )⋅ σall_bearing⋅:=

( ) →⇒ 810*36.3*018.0,008.0*2min*022.0 Pbolt_bear 1.183 105× N=

B.3.a.2 – Calculation of the Shear Capacity

The allowable shear stress used in the calculation is obtained:

τall_shear 0.6 σall⋅:= →⇒ 610*210*6.0 τall_shear 1.26 108× Pa=

26

The shear capacity of the bolt shearboltP _ is calculated considering two (2)

2=fn shearing sections:

nf 2:=

Pbolt_shear nfπ dbolt

2⋅

4⋅ τall_shear⋅:= →⇒ 8

2

10*26.1*4022.0**2 π

Pbolt_shear 9.579 104× N=

The capacity of the bolt boltP

is obtained:

Pbolt min Pbolt_bear Pbolt_shear,( ):= ( ) →⇒ 45 10*579.9,10*183.1min

Pbolt 9.579 104× N=

B.3.b – Multiple-Bolt Capacity

The transfer of the axial force from the spliced bar 1 to the connection bars 3 & 4 and

from the connection bars 3 & 4 to the spliced bar 2 is realized by a group of boltn

bolts. The total axial force

conboltP _ carried through the bolts group is:

nbolt 5:=

Pbolt_con nbolt Pbolt⋅:= →⇒ 410*579.9*5 Pbolt_con 4.79 105× N=

B.4 –Calculation of the Connection Capacity

The capacity of the entire connection capP

is calculated as the minimum value of the

capable axial forces calculated above for the splice bars, connection bars and

multiple-bolt connection, respectively.

Pcap min Pcap_sb Pcap_cb, Pbolt_con,( ):= → Pcap 4.79 105× N=

27

Problem 6.2.3 A detail of the hanger to ceiling connection is shown in Figure 6.2.3.a.

The hanger is fabricated from 2U20 structural steel shapes and is attached to a gusset

plate 260mm x 18mm by a bolted connection composed of eight (8) bolts mm16=Φ .

Assuming that the hanger is subjected to an axial tensile force kNP 400= , verify the

strength of the detail.


A.1 - The detail proposed for investigation is composed from three structural

elements: (a) the gusset plate, (b) the hanger and (c) the bolted connection. The first

step of the verification consists in the identification of the critical cross-sections

characterizing each structural component.

From the axial force “transmission” diagram, shown in Figure 6.2.3.b, the axial forces

pertinent to each critical cross-section is identified and, consequently, the normal

stresses are calculated. For the system to be safe the calculation must conclude that

the maximum axial stress induced by the axial force in all structural components must

be limited to the value of the allowable stress MPaall 200=σ .

Figure 6.2.3.a

.A.2 – Numerical Data

P 400 103⋅ N⋅:= - the force acting on the hanger;

28

σall 210 MPa⋅:= - the allowable normal stress of the material;

dbolt 16 mm⋅:= - diameter of the bolts;

nbolt_1_1 3:= - number of bolts in section 1-1 (see Figure 6.2.3);



nbolt nbolt_1_1 nbolt_2_2+ nbolt_3_3+:= → nbolt 8= - total number of bolts;

• U shape data

AU 32.2 cm2⋅:= - area;

tU 8.5 mm⋅:= - web thickness;

• Gusset Plate data

lplate 260 mm⋅:= - width;

tplate 18 mm⋅:= - thickness;

B. Calculations

The “transmission” diagram of the axial force trough the connection is shown in

Figure 6.2.3.b.

Figure 6.2.3.b Axial Force “Transmission” Diagram

29

B.1. Verification of the Gusset Plate

Note: The verification of the gusset plate must conclude that the maximum axial

stress induced by the axial force in the gusset plate is limited to the value of the

allowable stress MPaall 210=σ

The areas of the critical cross-sections characterizing the gusset plate are calculated

as:

Agp_1_1 lplate tplate⋅ 3 dbolt⋅ tplate⋅−:= →−⇒ 8.1*6.1*38.1*26

Agp_1_1 38.16cm2=

Agp_2_2 lplate tplate⋅ 3dbolt tplate⋅−:= →−⇒ 8.1*6.1*38.1*26

Agp_2_2 38.16cm2=

Agp_3_3 lplate tplate⋅ 2dbolt tplate⋅−:= →−⇒ 8.1*6.1*28.1*26

Agp_3_3 41.04cm2=

Accordingly to the axial force diagram shown in Figure 6.2.3.b, the axial forces

pertinent to the above cross-sections are:

Pgp_1_1 P:= → Pgp_1_1 4 105× N=

Pgp_2_2nbolt nbolt_1_1−

nboltP:= →

−⇒ 310*400*

838

Pgp_2_2 2.5 105× N=

Pgp_3_3nbolt nbolt_1_1− nbolt_2_2−

nboltP:= →

−−⇒ 310*400*

8338

Pgp_3_3 1 105× N=

The normal stress induced in the characteristic cross-sections is calculated as:

30

σgp_1_1Pgp_1_1Agp_1_1

:= →⇒ −−

624

5

10*10*16.38

10*4m

Nσgp_1_1 104.822MPa=

if σgp_1_1 σall≤ "ok", "no good",( ) "ok"=


:= →⇒ −−

624

5

10*10*16.3810*5.2

mN

σgp_2_2 65.514MPa=



:= →⇒ −−

624

5

10*10*04.41

10*1m

Nσgp_3_3 24.366MPa=


Conclusion: The maximum normal stress induced in the gusset plate is

MPa822.104

a value less than

MPaall 210=< σ .

B.2. Verification of the 2U shape member

The hanger is composed from two U20 ( nU_bar 2:= ) steel shapes. The area of the

critical cross-sections characterizing the hanger is calculated as:

AU_1_1 nU_bar AU nbolt_1_1 dbolt⋅ tU⋅−( )⋅:= ( ) →−⇒ 85.0*6.1*32.32*2

AU_1_1 56.24cm2=


AU_2_2 56.24cm2=


AU_3_3 58.96cm2=

Accordingly to the axial force diagram shown in Figure 6.2.3.b, the axial forces

pertinent to the above cross-sections are:

31

PU_1_1nbolt_1_1

nboltP:= →⇒ 310*400*

83

PU_1_1 1.5 105× N=

PU_2_2nbolt_1_1 nbolt_2_2+

nboltP:= →

+⇒ 310*400*

833

PU_2_2 3 105× N=

PU_3_3 P:= →⇒ 310*400 PU_3_3 4 105× N=

The normal stress induced in the characteristic cross-sections is calculated as:

σU_1_1PU_1_1AU_1_1

:= →⇒ −−

624

5

10*10*24.5610*5.1

mN

σU_1_1 26.671MPa=

if σU_1_1 σall≤ "ok", "no good",( ) "ok"=

σU_2_2PU_2_2AU_2_2

:= →⇒ −−

624

5

10*10*24.56

10*3m

NσU_2_2 53.343MPa=


σU_3_3PU_3_3AU_3_3

:= →⇒ −−

624

5

10*10*96.58

10*4m

NσU_3_3 67.843MPa=


Conclusion: The maximum normal stress induced in the gusset plate is MPa843.67 , a

value less than

MPaall 210=< σ

B.3. Verification of the Bolt connection

The bolted connection is realized with eight bolts mm16=Φ .

B.3.1 Capacity of Individual Bolt

B.3.1.a Bearing Capacity

32

The allowable bearing stress is calculated as:

σall_bearing 1.6 σall⋅:= →⇒ 610*210*6.1 σall_bearing 3.36 108× Pa=

and, consequently, the shearing capacity of the bolt is:

Pbolt_bear dbolt min 2 tU⋅ tplate,( )⋅ σall_bearing⋅:=

( ) →⇒ 810*36.3*018.0,0085.0*2min*016.0 Pbolt_bear 9.139 104× N=

B.3.1.b Shearing Capacity

The shear force is equally distributed on two ( nf 2:= ) shearing surfaces. The

allowable shear stress of the bolt material is calculated as:

τall_shear 0.6 σall⋅:= →⇒ 610*210*6.0 τall_shear 1.26 108× Pa=

and, consequently, the shear capacity of the bolt is:

Pbolt_shear nfπ dbolt

2⋅

4⋅ τall_shear⋅:= →⇒ 8

2

10*26.1*4016.0**2 π


The capacity of the bolt is obtained:

Pbolt min Pbolt_bear Pbolt_shear,( ):= ( ) →⇒ 44 10*067.5,10*139.9min

Pbolt 5.067 104× N=

3.2 Multiple Bolt Capacity

Pbolt_con nbolt Pbolt⋅:= →⇒ 410*067.5*8 Pbolt_con 4.053 105× N=

if Pbolt_con P≥ "ok", "no good",( ) "ok"=

33

Conclusion: The system composed of gusset plate, hanger (2U20 steel shapes) and

bolted connection is able to carry the axial force of kNP 400= . The individual

verification of the system components indicates that the bolted connection is the

weakest link of the system, characterized by the smallest capacity.

Problem 6.2.4 Determine the size and the length of the fillet weld necessary to realize

the lap-splice connection shown in Figures 6.2.4.a and 6.2.4.b assuming that an axial

force kNP 180= acts along the axis described by the centroid of the L shape.

Figure 6.2.4


A.1 – Numerical Data

P 180 kN⋅:= - the axial force;

σall 210 MPa⋅:= - allowable normal stress of the material;

τall_weld 0.7 σall⋅:= → τall_weld 147MPa= - allowable shear stress of the weld;

• L70x70x7 Shape Data (data is extracted from tables)

AL 9.4 cm2⋅:= - area;

34

lL 70 mm⋅:= - flange width;

tL 7 mm⋅:= - flange thickness;

eL 19.7 mm⋅:= - eccentricity;

• Gusset Plate Data

lgp lL 2 60⋅ mm⋅+:= →+⇒ 60*270 lgp 190mm= - minimum width of the gusset

tgp 10 mm⋅:= - thickness of the gusset

B. Calculations

B.1 - Gusset Plate Verification

Agp lgp tgp⋅:= →⇒ 1*19 Agp 19cm2=

σgpP

Agp:= →⇒ −

−6

24

3

10*10*19

10*180mN

σgp 94.737MPa=

if σgp σall≤ "ok", "no good",( ) "ok"=

B.2 – L shape Verification

σ ls

PAL

:= →⇒ −−

624

3

10*10*4.910*180

mN

σ ls 191.489MPa=

if σls σall≤ "ok", "no good",( ) "ok"=

B.3 – Welds’ Sizing

B.3.a - Two fillet welded connection (see Figure 6.2.4.c)

The free-body diagram of the L shape is pictured in Figure 6.2.4.c. The equations of

equilibrium are written as:

35

F1a 50.66kN=

F2a 129.3kN=

The length of the weld1 is obtained:

aweld1a 0.707 tL⋅:= →⇒ 7*707.0 aweld1a 4.949mm=

lweld1aF1a

aweld1a τall_weld⋅( ):= →⇒ −2

63

4

10*10*147*10*949.4

10*066.5Pam

N

lweld1a 6.964cm=

The length of the weld2 is calculated:

aweld2a aweld1a:= → aweld2a 4.949mm=

lweld2aF2a

aweld2a τall_weld⋅( ):= →⇒ −2

63

5

10*10*147*10*949.4

10*293.1Pam

N

lweld2a 17.773cm=

Conclusion: The following effective dimensions are chosen for weld1 and weld2,

respectively: mma aweld 51 = , mma aweld 52 = ,

mml aweld 701 = and

mml aweld 1802 = .

B.3.b - Three fillet welded connection (see Figure 6.2.4.d)

The capacity of the weld3, the weld normal to the longitudinal axis of the L shape, is

calculated as:

aweld3b 0.707 tL⋅:= →⇒ 7*707.0 aweld3b 4.949mm=

F3b aweld3b lL⋅ τall_weld⋅:= →⇒ −−− 3623 10*10*147*10*7*10*949.4 Pamm

F3b 50.925kN=

36

The free-body diagram of the L shape is pictured in Figure 6.2.4.d. The equations of

equilibrium are written as:

F1b 25.2kN=

F2b 103.84kN=

The length of the weld1 is obtained:

aweld1b aweld1a:= → aweld1b 4.949mm=

lweld1bF1b

aweld1b τall_weld⋅( ):= →⇒ −2

63

3

10*10*147*10*949.4

10*2.25Pam

N

lweld1b 3.464cm=

The length of the weld2 is calculated:

aweld2b aweld2a:= → aweld2b 4.949mm=

lweld2bF2b

aweld2b τall_weld⋅( ):= →⇒ −2

63

3

10*10*147*10*949.4

10*84.103Pam

N

lweld2b 14.273cm=

Conclusion: The following effective dimensions are chosen for weld1 and weld2,

respectively: mma bweld 51 = , mma bweld 52 = ,

mml bweld 401 = and

mml bweld 1502 = .

Problem 6.2.5 The rigid beam, shown in Figure 6.2.5.a, is subjected by vertical

concentrated force kNP 50= acting on the tip of the cantilever at point C. The rigid

37

beam is supported at point A by a hinge and at point B, located at distance mb 2=

from the cantilever’s tip, by a hanger. The Detail A, pictured in Figure 6.2.5.b,

representing the hinge detailing has 2 gusset plates of 100mm x 8mm. The hanger,

made of two rounded bars, shown in Figure 6.2.5.b, welded in Detail B to the gusset

1. The two gusset plates contained in Detail B, gusset 1 and gusset 2, have each cross-

sectional dimensions of 100mm x 10mm. Calculate: (a) the diameter of the hanger’s

rounded bars, (b) the diameter of the bolt used in Detail B and verify the gusset plates,

(c) size the length of the welds between the rounded bars and gusset 1 considering the

effective thickness of the weld mmt Bweld 5_ = , (d) the bolt diameter used, verify the

attachment gusset plates and the welds of Detail A considering an effective thickness

mmt Aweld 5_ = .

Figure 6.2.5.a

38

Figure 6.2.5.b


A.1 – Numerical Data

P 50 kN⋅:= - the acting force;

a 4 m⋅:= b 2 m⋅:= - beam lengths;

σall 210 MPa⋅:= - allowable stress of the material;

τall_shear 0.6 σall⋅:= → τall_shear 126MPa= - allowable shear of the material;

σall_bearing 1.6 σall⋅:= → σall_bearing 336MPa= - allowable stresses for the bolt;

• Detail A

ngusset_A 2:= - two similar gussets;

lgusset_A 100 mm⋅:= - length of the gusset plates;

tgusset_A 8 mm⋅:= - width of the gusset plates;

tweb 12 mm⋅:= - presumed thickness of the rigid beam web;

• Detail B

39

nhang 2:= - two rounded bars welded to the Gusset 1;

ngusset_B 2:= - two similar gussets (Gusset1 and Gusset2);

lgusset_B 100 mm⋅:= - length of the gusset plates;

tguss_B 10 mm⋅:= - width of the gusset plates;

tgusset1 tguss_B:= - width of the Gusset 1;

tgusset2 tguss_B:= - width of the Gusset 2;

• Welds

nwelds_A 2:= - number of welds in Detail A;

teff_weld_A 5 mm⋅:= - effective thickness of the welds in Detail A;

lweld_A_eff lgusset_A:= - effective length of the welds in Detail A;

nwelds_B 4:= - number of welds in Detail B;

teff_weld_B 5 mm⋅:= - effective thickness of the welds in Detail B;

τshear_weld 0.7 σall⋅:= -allowable stresses of the weld

B. Calculations

B.1 – Sizing of the Hanger (2 rounded bars nhang 2:= )

Solving the equilibrium equation (1), expressed below, the axial force in the hanger is

obtained:

(1)

Fhang 75000 N⋅:=

The necessary area of the rounded bar is then calculated:

Ahang_necFhang

nhang σall⋅:= →⇒ 4

6 10*10*210*2

75000Pa

NAhang_nec 1.786cm2=

P a b+( )⋅ Fhang a⋅− 0 solve Fhang, 75000 N⋅→

40

The resulting diameter of the rounded bar is:

dhang_nec4

πAhang_nec:= →⇒ 786.1*4

πdhang_nec 1.508cm=

The diameter of the rounded bar is chosen to be a rounded value of the previously

calculated value:

dhang_eff 1.6 cm⋅:=

Consequently, the effective area of each rounded bars comprising the hanger is

calculated as:

Ahang_effπ dhang_eff

2⋅

4:= →⇒

46.1* 2π

Ahang_eff 2.011cm2=

The normal stress existing in the rounded bars comprising the hanger must be

checked:

σhang_effFhang

nhang Ahang_eff⋅:= →⇒ − 2410*011.2*2

75000m

Nσhang_eff 1.865 108× Pa=

if σhang_eff σall< "ok", "no good",( ) "ok"=

B.2 – Sizing of the bolt located in Detail B

B.2.a – Failure in shearing

The bolt’s capacity in shear is equal to the axial force of the hanger:

Pcap_bolt_shear Fhang

The capacity of the bolt in shear is calculated considering only one shearing surface,

( nshear_sect_B 1:= ), located at the contact between the two gusset plates.

Pcap_bolt_shear nshear_sect_B Ashear_bolt_nec⋅ τall_shear⋅

From the above equation one can calculate the required area of the bolt:

41

Ashear_bolt_necFhang

nshear_sect_B τall_shear⋅( ):= →⇒ 46 10*

10*126*175000

PaN

Ashear_bolt_nec 5.952cm2=

The corresponding diameter is:

dshear_bolt_B_nec4π

Ashear_bolt_nec:= →⇒ 952.5*4π

dshear_bolt_B_nec 2.753cm=

B.2.b – Failure in bearing

The bolt’s capacity in bearing is equal to the axial force of the hanger:

Pcap_bolt_bear Fhang

The bearing capacity of the bolt is obtained as:

Pcap_bolt_bear dbear_bolt_B_nec min tguss1 tguss2,( )⋅ σall_bearing⋅

Consequently, the corresponding bolt diameter is:

dbear_bolt_B_necFhang

min tguss1 tguss2,( ) σall_bearing⋅( ):=

→⇒ −4

64 10*10*336*10*1

75000Pa

Ndbear_bolt_B_nec 2.232cm=

B.2.c - The final diameter of the bolt

The final diameter of the bolt is obtained by choosing the maximum value between

the two diameters calculated for the shearing and bearing failure:

dbolt_B_nec max dshear_bolt_B_nec dbear_bolt_B_nec,( ):= →

dbolt_B_nec 2.753cm=

42

An effective diameter of the bolt is chosen to be:

dbolt_B_eff 28 mm⋅:=

B.2.d - Verification of the bolt

The capacities of the bolt in shearing and bearing are:

Fshear_cap_bolt_B nshear_sect_Bπ dbolt_B_eff

2⋅

4⋅ τall_shear⋅:=

→⇒ − Pam 6242

10*126*10*4

8.2**1 πFshear_cap_bolt_B 7.758 104× N=

Fbear_cap_bolt_B dbolt_B_eff min tguss1 tguss2,( )⋅ σall_bearing⋅:=

→⇒ − Pam 624 10*336*10*1*8.2 Fbear_cap_bolt_B 9.408 104× N=

The bolt’s capacity is:

Fcap_bolt_B min Fshear_cap_bolt_B Fbear_cap_bolt_B,( ):= →

Fcap_bolt_B 7.758 104× N=

The capacity of the bolt is checked:

if Fhang Fcap_bolt_B< "ok", "no good",( ) "ok"=

B.3 - Verification of the gusset plates located in Detail B

The area of the gusset plate’s cross-section (both gussets are similar) is calculated:

Agusset_B lgusset tgusset⋅ dbolt_B_eff tgusset⋅−:= →−⇒ 1*8.21*10

Agusset_B 7.2cm2=

The normal stress in the gusset plate is obtained:

σgusset_BFhang

Agusset_B:= →⇒ − 2410*2.7

75000m

Nσgusset_B 1.042 108× Pa=

43

Note: The axial force is transmitted through one gusset at the time.

if σgusset_B σall< "ok", "no good",( ) "ok"=

B.4 – Calculation of the length of the welds connecting the hanger bars to the gusset

plates.

There are four (4) fillet welds, nwelds_B 4:= , connecting the hanger bars to the

gusset plates. The length of the weld is calculated as:

lweld_B_necFhang

nwelds_B teff_weld_B⋅ τshear_weld⋅:=

→⇒ − PamN

*10*147*10*5.0*475000

622 lweld_B_nec 2.551cm=

An effective length is chosen by rounding the above obtained value:

lweld_B_eff 2.6 cm⋅:=

Finally, the verification of the welds is conducted as:

Pcap_weld_B nwelds_B teff_weld_B⋅ lweld_B_eff⋅ τshear_weld⋅:=

→⇒ − Pam *10*147*10*6.2*5.0*4 624 Pcap_weld_B 7.644 104× N=

if Pcap_weld_B Fhang≥ "ok", "no good",( ) "ok"=

B.5 – Sizing the diameter of the bolt located in Detail A

From the equilibrium equation (2) the vertical reaction at point A is calculated:

P b⋅ VA a⋅− 0 solve VA, 25000 N⋅→ → VA 25000 N⋅:= (2)

B.5.a – Failure in shearing

The bolt’s capacity in shear is equal to the vertical reaction at point A:

Pcap_bolt_shear VA

44

The capacity of the bolt in shear is calculated considering two shearing surface

( nshear_sect_A 2:= ) located at the contact between the two gusset plates and the web

of the beam:

Pcap_bolt_shear nshear_sect_A Ashear_bolt_nec⋅ τall_shear⋅

From the above equation the required area of the bolt is calculated:

Ashear_bolt_necVA

nshear_sect_A τall_shear⋅( ):= →⇒ 46 10*

10*126*225000

PaN

Ashear_bolt_nec 0.992cm2=

The corresponding diameter is:

dshear_bolt_A_nec4π

Ashear_bolt_nec:= →⇒ 992.0*4π

dshear_bolt_A_nec 1.124cm=

B.5.b – Failure in bearing

The bolt’s capacity in bearing is equal to the axial force of the hanger:

Pcap_bolt_bear VA

The bearing capacity of the bolt is obtained as:

Pcap_bolt_bear dbear_bolt_A_nec min 2tgusset_A tweb,( )⋅ σall_bearing⋅

Considering that the gusset plates’ and web’s thicknesses are

tgusset_A 8 mm⋅:=

and tweb 12 mm⋅:= , respectively, the bolt’s necessary diameter is obtained:

45

dbear_bolt_A_necVA

min 2tgusset_A tweb,( ) σall_bearing⋅( ):=

→⇒ −4

64 10*10*336*10*2.1

25000Pa

N

dbear_bolt_A_nec 0.62cm=

B.5.c - The final diameter of the bolt

The final bolt diameter is obtained by choosing the maximum value between the

diameter calculated for the shearing and bearing failure:

dbolt_A_nec max dshear_bolt_A_nec dbear_bolt_A_nec,( ):= →

dbolt_A_nec 1.124cm=

Consequently, the effective diameter of the bolt is chosen to be:

dbolt_A_eff 12 mm⋅:=

B.5.d - Verification of the bolt

The capacities of the bolt in shearing and bearing are:

Fshear_cap_bolt_A nshear_sect_Aπ dbolt_A_eff

2⋅


→⇒ − Pam 6242

10*126*10*4

2.1**2 π Fshear_cap_bolt_A 2.85 104× N=

Fbear_cap_bolt_A dbolt_A_eff min 2tgusset_A tweb,( )⋅ σall_bearing⋅:=

→⇒ − Pam 624 10*336*10*2.1*2.1 Fbear_cap_bolt_A 4.838 104× N=

The bolt’s capacity is:

Fcap_bolt_A min Fshear_cap_bolt_A Fbear_cap_bolt_A,( ):= →

Fcap_bolt_A 2.85 104× N=

46

The verification is checked as:

if VA Fcap_bolt_A< "ok", "no good",( ) "ok"=

B.6 - Verification of the gusset plates located in Detail A

The two (2) gusset plates,

ngusset_A 2:= , are subjected to single-shear (the entire

cross-section is used) and double-shear (just half of the cross-section) failures by the

reaction force AV . The shear stress resulting from single-shear failure is calculated as:

Agusset_A_s_shear ngusset_A lgusset_A tgusset_A⋅ dbolt_A_eff tgusset_A⋅−( )⋅:=

→−⇒ )8.0*2.18.0*10(*2 Agusset_A_s_shear 14.08cm2=

τgusset_A_s_shearVA

Agusset_A_s_shear:= →⇒ − 2410*08.14

25000m

N

τgusset_A_s_shear 1.776 107× Pa=

The shear stress resulting from double-shear failure is obtained as:

Agusset_A_d_shear ngusset_A 2lgusset_A

2tgusset_A⋅

⋅:= →⇒ )8.0*

210*2(*2

Agusset_A_d_shear 16cm2=

τgusset_A_d_shearVA

Agusset_A_d_shear:= →⇒ − 2410*16

25000mN

τgusset_A_d_shear 1.563 107× Pa=

The maximum shear stress in the gusset plates is:

τgusset_A_shear max τgusset_A_s_shear τgusset_A_d_shear,( ):= →

τgusset_A_shear 1.776 107× Pa=

The stress verification is conducted:

if τgusset_A_shear τall_shear< "ok", "no good",( ) "ok"=

47

B.7 – Verification of the welds in Detail A

There are two welds,nwelds 2:= , connecting the gusset plates to the rigid element.

The welds’ capacity is calculated:

Pcap_weld_A nwelds_A teff_weld_A⋅ lweld_A_eff⋅ τshear_weld⋅:=

→⇒ − Pam 624 10*147*10*10*5.0*2 Pcap_weld_A 1.47 105× N=

Finally, the weld’s verification is conducted:

if Pcap_weld_A VA≥ "ok", "no good",( ) "ok"=

Problem 6.2.6 The system, shown in Figure 6.2.6.a, is comprised from two identical

hangers equally inclined with an angle α from the vertical axisOY . The system is

subjected to a vertical load kNP 270= . The hangers are fabricated from U8 steel

shapes and are lap-spliced as indicated in Details A and B, respectively. Detail A

represents a bolted connection realized in two alternatives from four bolts with a

diameter mm20=Φ . The Detail B is a welded connection with the leg thickness of

mmtweld 5= .

Figure 6.2.6.a

48

Assuming that the gusset plates have a cross-section of 120mm x 10mm and the

allowable stress of the steel is MPaall 210=σ calculate: (a) the angle α , (b) the

length of the welds contained in Detail B, Figure 6.2.6.b, and (c) plot the variation of

the capable load P function of the angle α variation.

Figure 6.2.6.b


A.1 – General Observations

The truss shown in Figure 6.2.6.a is comprised from two U8 steel shapes. The

equilibrium equations written around point O indicate that the axial force in the two

hangers is equal. Consequently, it can be determined calculating the capacity of the

Detail A. Once determined, the axial force in the hanger is used to size the length of

the welds of Detail B.

A.2– Numerical Data

P 270 kN⋅:= - acting force;

σall 210 MPa⋅:= - the allowable normal stress for steel ;

• U8 Shape

49

tweb 6 mm⋅:= - web thickness;

AU 11 cm2⋅:= - area;

• Gusset Plates

tguss 10 mm⋅:= - thickness;

lguss 120 mm⋅:= - width;

• Bolted Connection

nbolt 4:= - number of bolts;

db 20 mm⋅:= - bolt diameter;

σall_bearing 1.6 σall⋅:= → σall_bearing 336MPa= - bearing allowable stress;

τall_shear 0.6 σall⋅:= → τall_shear 126MPa= - shear allowable stress;

• Weld Connection

nweld 2:= - number of welds;

tweld 5 mm⋅:= - thickness of the weld;

τall_weld 0.8 σall⋅:= →τall_weld 168MPa= - shear allowable stress.

B. Calculations

B.1 – Equations of equilibrium

Two projection equations of equilibrium written for the isolated point O are:

Fhang_1− sin α( )⋅ Fhang_2 sin α( )⋅+ 0 - projection on Ox (1)

Fhang_1 cos α( )⋅ Fhang_2 cos α( )⋅+ P− 0- projection on Oy (2)

From equation (1) results:

Fhang_1 Fhang_2 (3)

50

Using the relation (3) into equation (1) the axial force in the hanger is obtained:

Fhang_1P

2 cos α( )⋅ (4)

B.2 – Calculation of the capacity of the multiple-bolts connection

The capacity of a single bolt in bearing is calculated as:

Pbolt_bear db min tweb tguss,( )⋅ σall_bearing⋅:= →⇒ − Pam 624 10*336*10*6.0*2

Pbolt_bear 4.032 104× N=

The shearing capacity of the bolt is determined considering the existence of only one

nshear 1:= shearing cross-section:

Pbolt_shear nshearπ db

2⋅

4⋅ τall_shear⋅:= →⇒ − Pam 624

2

10*126*10*42**1 π


The capacity of a single bolt is obtained as:

Pcap_bolt min Pbolt_shear Pbolt_bear,( ):= →Pcap_bolt 3.958 104× N=

The total capacity of the entire group of bolts, nbolt 4:= , transmitting the axial force

from the U shape into the connecting plate and back is:

Pcap_conn nbolt Pcap_bolt⋅:= →⇒ N410*958.3*4 Pcap_conn 1.583 105× N=

B.3 – Calculation of the U shape capacity

The critical section of the Detail A, marked aa − on the Figure 6.2.6, is the cross-

section with two bolt holes. This cross-section transfers the full axial force existing in

the hanger. The corresponding area is:

51

Amin_U AU 2 db⋅ tweb⋅−:= →−⇒ 6.0*2*211 Amin_U 8.6cm2=

The U shape’s capacity is obtained as:

Pcap_U Amin_U σall⋅:= →⇒ − Pam 624 10*210*10*6.8 Pcap_U 1.806 105× N=

B.4 – Calculation of the gusset plate’s capacity

The gusset plate has two possible critical cross-sections. The capable axial force

characterizing the cross-section aa − (2 bolt holes) is calculated as:

Aguss_a_a lguss tguss⋅ 2 db⋅ tguss⋅−:= →−⇒ 1*2*21*12 Aguss_a_a 8cm2=

Pcap_guss_a_a Aguss_a_a σall⋅:= →⇒ − Pam 624 10*210*10*8

Pcap_guss_a_a 1.68 105× N=

Considering the cross-section

bb − (one bolt hole) a similar calculation conducts to:

Aguss_b_b lguss tguss⋅ db tguss⋅−:= →−⇒ 1*21*12 Aguss_b_b 10cm2=

Pcap_guss_b_b Aguss_b_b σall⋅:= →⇒ − Pam 624 10*210*10*10

Pcap_guss_b_b 2.1 105× N=

The capacity of the gusset plate is obtained:

Pcap_guss min Pcap_guss_a_a Pcap_guss_b_b,( ):= →Pcap_guss 1.68 105× N=

B.5 - The axial force transmitted by the hanger

Fcap_hang min Pcap_U Pcap_conn, Pcap_guss,( ):= →Fcap_hang 1.583 105× N=

The angle α is obtained using formula (4):

52

α acosP

2 Fcap_hang⋅

:= →

⇒

NN

5

3

10*583.1*210*270arccos α 31.503deg=

B.5 – Sizing the welds of Detail B

The weld must be able to transfer the axial force existing in the hanger and,

consequently, the following equality is written:

Pcap_weld Fcap_hang where the weld capacity is expressed as:

Pcap_weld 2 lweld⋅ teff_weld⋅ τall_weld⋅

The length of the welds is calculated as:

teff_weld 0.707 tweld⋅:= →teff_weld 3.535mm=

lweldFcap_hang

nweld teff_weld⋅ τall_weld⋅:= →⇒ − Pam

N623

5

10*168*10*535.3*210*583.1

lweld 13.331cm=

The effective length of the welds is chosen:

lweld 14 cm⋅:=

B-6 – The variation of the angle α and force P

A series of twelve values, varying from 0 to 50 degrees, is assumed for the angleα

and the corresponding value of the force P is calculated, tabulated and plotted in the

sketch shown below.

i 0 11..:=

anglei 5 i⋅:= Pi 2 cos α i( )⋅ Fcap_hang⋅:= α i 5

π180

⋅ i⋅:=

53

Pcapi270 kN⋅:=

angle

001234567891011

0510152025303540455055

=

P

001234567891011

316.673315.468311.862305.882297.575287.003274.246259.403242.585223.921203.553181.636

kN=

Note: The sketch can be used to verify the numerically obtained result indicating that

the force kNP 270= corresponds to an angle deg31≈α .

0 8.33 16.67 25 33.33 41.67 501.5 .10 5

1.9 .10 5

2.3 .10 5

2.7 .10 5

3.1 .10 5

3.5 .10 5

P

P cap

angle


Problem 6.3.1 Determine the maximum allowable tensile load of the bolted

connection shown in Figure 6.3.1 assuming that all the components are made of steel,

54

the holes correspond to 22 mm diameter bolts and the U30 shape is fastened to a 12

mm thick gusset plate.

Figure 6.3.1

Problem 6.3.2 Two flat bars loaded in tension by forces P are spliced, as shown in

Figure 6.3.2, using two rectangular splice plates and two 13 mm diameter rivets. The

bars, except at the splice where they are wider, are 24 mm x 10 mm and are made of

steel with an ultimate stress in tension of 415 MPa. The ultimate stresses in shear and

bearing for the rivet steel are 207 MPa and 552 MPa, respectively. Calculate the

dimension of the wider section of the connection and then determine the connection’s

allowable axial load considering that a safety factor of 2.5 is desired with respect to

the ultimate load.

Figure 6.3.2

Problem 6.3.3 Figure 6.3.3 illustrates the bolted connection between a vertical column

and a diagonal brace. The compressive load P carried by the brace is 25 kN. The

connection is built from three 16mm bolts that join together two 8 mm plates and a

55

gusset plate welded to the column. An end plate is welded to the brace. Calculate: (a)

the average shear stress in each bolt and (b) the average bearing stress between the

bolt and the plates.

Figure 6.3.3

Problem 6.3.4 Two bars of rectangular cross-section and thickness t = 8 mm are

connected by a bolt in the manner shown in Figure 6.3.4. The allowable shear and

bearing stresses in the bolt are 83 MPa and 138 MPa, respectively. What is the

minimum required diameter of the bolt assuming a tensile load P = 8 kN? Size the

connecting parts of the connection considering for the material an allowable tensile

stress of 100 MPa.

Figure 6.3.4

56

Problem 6.3.5 A bar with rectangular cross-section is subjected to an axial load P as

shown in Figure 6.3.5. The bar has a width b = 60 mm and thickness t = 10 mm. A

hole of diameter d is drilled through the bar to provide a pin support. The allowable

tensile stress of the bar’s material is 140 MPa, while the allowable shear stress in the

pin and bar is 80 MPa. Considering a bearing stress of 220 MPa calculate the

necessary diameter of the pin for which the load P will be a maximum.

Figure 6.3.5

Problem 6.3.6 Two bolts are used to connect, as shown in Figure 6.3.6, three

rectangular bars in tension. The thickness of the middle bar is 60 mm. Calculate: (a)

the required diameter of the bolts if the average shear and bearing stresses for the

bolts are 140 MPa and 336 MPA, respectively, (b) size the exterior connecting bars

considering 210 MPa allowable tensile stress for the material and the bolt diameter

previously determined, and (c) verify the middle bar.

Figure 6.3.6

Problem 6.3.7 Three steel plates with 16 mm equal thickness are joined together, as

shown in Figure 6.3.7, by two 13 mm diameter rivets. Calculate: (a) the largest

57

bearing stress acting on the rivets if the tensile load is P = 45 kN, (b) the force P

required to cause the shear failure of the rivets, considering that the ultimate shear

stress for the rivets is 220 MPa and the safety factor is 2.5, (c) the capable axial force

carried by the rivets connection, and (d) size the steel plates considering that the axial

force P is equal to the capable force carried by the rivets.

Figure 6.3.7

Problem 6.3.8 The connection shown in Figure 6.3.8 consists of five steel plates, each

5 mm thick, joined by a single 6 mm diameter bolt. The total load transferred between

the plates is 6000 N and is distributed among the plates as shown in the figure.

Calculate: (a) the largest shear stress in the bolt, disregarding the friction between the

plates, (b) the largest bearing stress acting against the bolt, and (c) size the steel plates

considering for steel an allowable tensile stress of 210 MPa.

Figure 6.3.8

Problem 6.3.9 A contractor constructs a long tension member from two wood boards

with a cross-section of 50x100 mm. The connection is realized, as shown in Figure

6.3.9, employing four 10 mm diameter bolts and two steel splice plates. Calculate: (a)

the axial force carried by the bolted connection if the only possible failure scenario is

58

the shearing failure of the bolts (the allowable shear stress for the bolts of 17.25

MPa), (b) the tensile load P carried by the wood boards if the allowable tensile stress

in the wood is 30 MPa, and (c) size the splice steel plates considering an allowable

tensile steel stress of 180 MPa.

Figure 6.3.9

Problem 6.3.10 Considering a tensile load of P=6.7 kN and the fact that the

contractor uses glue to attach the two steel splice plates to the wood boards instead of

using the bolted connection as described in Problem 6.3.9, what is the value of the

average direct shear stress in the glued joints if the slice plates are 200 mm long?

Problem 6.3.11 An angle bracket having a thickness t = 12 mm is attached to the

flange of a column by two 15 mm diameter bolts, as shown in Figure 6.3.11. A

uniformly distributed load p = 2.0 MPa acts on the top face of the bracket. The top

face of the bracket has a length and width of L = 150 mm and b=60 mm, respectively.

Determine: (a) the average bearing pressure between the angle bracket and the bolts

and (b) the average shear stress the bolts.

59

Figure 6.3.11

Problem 6.3.12 An L150x90x12 steel shape used as a tension member carrying a

tensile load of 300 kN is attached to a gusset plate as shown in Figure 6.3.12.

Calculate: (a) the bolt’s diameter necessary to carry the load if the shear and bearing

allowable stresses are 126 MPa and 336 MPa, respectively, and (b) verify the

connection members if the tensile allowable stress for steel is 210 MPA, the distance

s=60mm, the angle deg15=α , and the gusset thickness is 14 mm.

Figure 6.3.12

Problem 6.3.13 Two 13 mm nylon rods are spliced together, as shown in Figure

6.3.14, by gluing a 50 mm section of plastic pipe over the ends of the rods. If a tensile

60

force of P = 3 kN is applied to the spliced nylon rod, what is the average shear stress

occurring in the glue in the joint between the pipe and the rods? Considering that the

allowable tensile stress in the rods and plastic pipe is 20 MPa verify the rods and

calculate the required thickness of the plastic pipe.

Figure 6.3.13

Problem 6.3.14 Determine the size and the length of the necessary fillet weld to

realize the connection on the lap splice show in Figure 6.3.14, by considering first

only the welds A, and secondly, the welds A and B. The allowable shear stress in the

welds is 168 MPa. Verify the steel bars and the gusset plate if the allowable tensile

stress is 210 MPa.

Figure 6.3.14

Problem 6.3.15 The high-wire illustrated in Figure 6.3.15 is attached to a vertical

beam AC and is kept tensioned by the cable BD. At point C, the beam AC is attached

by a 10 mm diameter bolt to the bracket shown in view a-a. Determine the average

shear stress in the bolt at C if the tension in the high-wire is 5 kN, assuming that the

high-wire is horizontal and neglecting the weight of AC. Calculate the area of the

61

cables’ cross-sections if they are made of steel with an allowable tensile stress of 300

MPA.

Figure 6.3.15

Problem 6.3.16 A hollow box beam ABC, illustrated in Figure 6.3.16, has a length L

and is supported at end A by a 22 mm diameter pin, which passes through the beam

and its supporting pedestals. The roller support at B is located at distance 0.3L from

the end A. Determine: (a) the average shear stress in the pin due to a load P equal to

15 kN and (b) the average bearing stress between the pin and the box beam if the wall

thicknesses of the beam and supporting pedestals are equal to 14 mm.

62

Figure 6.3.16

Problem 6.3.17 A special-purpose hexagonal head bolt of diameter d = 12 mm, shown

in Figure 6.3.17, passes through a hole drilled in a steel plate and bears directly

against the steel plate. The hexagonal head has a radius of the circumscribed circle

equal to r = 10 mm (which means that each side of the hexagon has a length of 10

mm) and the thickness t = 6 mm. Considering that the tensile force P in the bolt is

9kN, determine: (a) the average bearing stress between the hexagonal head of the bolt

and the plate and (b) the average shear stress head of the bolt.

Figure 6.3.17

Problem 6.3.18 The 250x125x10 cm steel plate is hoisted by a cable as illustrated in

Figure 6.3.18. The pins holding the cables are 18 mm in diameter and are located 200

cm apart. Considering for the steel a specific weight of 77 kN/m3 calculate: (a) the

axial forces in the cables, (b) size the cables considering an allowable tension stress of

300 MPa, (d) the elongation of the cable and the vertical displacement of the steel

plate, (d) the average shear stress in the pins and (e) the average bearing stress

between the steel plate and the pins.

63

Figure 6.3.18

Problem 6.3.19 A reinforcing bar is casted in concrete, as shown in Figure 6.3.19,

with the intent to conduct a “pull-out test” capacity test. Assuming that the rebar has a

diameter d = 12 mm, an embedment length L = 30 cm and is subjected to a tensile

force P = 17.5 kN determine: (a) the normal stress in the reinforced bar, (b) the

average shear stress, the bond stress, developed between the concrete and the steel if

the uniform distribution is considered and (c) the maximum shear stress if the

following distribution of the shear bond is accepted as

)*6**9*4(**4

3233

max xxLLL

+−=ε

τ , where x is measured from the interior end of

the bar towards the concrete surface.

Figure 6.3.19

64

Problem 6.3.20 The system, shown in Figure 6.3.20, comprised of two hangers with

equal circular cross-sections, EC and AF, and a rigid beam AD, is subjected to a

vertical force kNP 50= .

Figure 6.3.20

Assuming that the length is ml 4= , all gusset plates ( mmt guss 8= ) and the hangers

are fabricated from steel characterized by an allowable normal stress MPaall 210=σ

size the following elements: (a) all hangers, (b) all bolts, (c) all welds, (d) all gusset

plates. Calculate the vertical displacements at points A, C and D ( GPAEsteel 200= ).

Typical details, AA, BB and CC, for the connections are pictured below.

65

Usage of the following allowable stresses is considered: MPabearingboltall 330__ =σ ,

MPashearboltall 140__ =τ and MPaweldall 100_ =τ .

Problem 6.3.21 The system, shown in Figure 6.3.21, comprised of two 30 mm

diameter hangers, AF and CG, and a rigid beam AD, is subjected to two vertical

forces kNP 25= acting at points B and E, respectively.

Figure 6.3.21

Assuming the following data: the length is ml 2= , the typical 100x8 mm gusset plates

used and the hangers are fabricated from steel with an allowable normal stress

MPaall 210=σ and MPaall 120=τ , the 28 mm bolts are characterized by

MPabearingboltall 330__ =σ and MPashearboltall 140__ =τ ., and the typical 3mm weld

has MPaweldall 100_ =τ , determine: (a) the force P carried by the system, (b) the

length of the welds and (c) the vertical displacements at points A, B, C and E

( GPAEsteel 200= ). Typical details, AA, BB and CC, for the connections are pictured

below.

66

Problem 6.3.22 A tied-down of a fiberglass sailboat is shown in Figure 6.3.22. The

bend bar with diameter 10mm is attached to a 12mm thick fiberglass board using a

washer with a diameter of 22mm.

Figure 6.3.22

Calculate the maximum force P allowed in the cable considering the following

material data: the allowable shear stresses in the fiberglass board is 4MPa, the bend

bar is made of steel with an allowable tensile stress of 180MPa, and the allowable

bearing stress between the washer and the fiberglass is 20 MPA.

67

CHAPTER 7 Bending of Plane Linear

Beams – Stress Distribution


The problems comprised in this chapter are exclusively concerned with the

investigation of the stress distribution on the cross-section of the plane linear beam

subjected to bending. By definition, bending is a deformation induced into a plane

linear beam by transversal acting loads.

A number of important assumptions are made:

(a) the beam has a constant cross-section along its entire length;

(b) the material is isotropic along the entire length of the beam;

(c) the cross-section is characterized by an axis of symmetry;

(d) the transversal loads are acting in the plane of symmetry;

(e) the cross-sections remain plane and perpendicular to the deflection curve after

the deformation (the Bernoulli-Euler hypothesis).

Note: These assumptions are restricting somehow the generality of the theoretical

frame, but the majority of the beams encountered in the structural engineering practice

are complying with these limitations. The above mentioned assumptions must be

always verified in order to validate the calculation results.

Consider a local coordinate system Oxyz attached to the left end point of the beam and

the axis Oy as the symmetry axis of the cross-section. The transversal loads, acting on

the vertical plan of symmetry Oxy induce onto the cross-section of the beam, located at

distance x from the origin of the coordinate system, only two cross-sectional resultants:

the vertical shear force Vy(x) and the bending moment Mz(x). The bending moment

Mz(x) and the shear force Vy(x) conduct to the existence on the cross-section of two

type of stresses: the normal stress σx and the shear stress τxy. In general both cross-

sectional resultants, Vy(x) and Mz(x), are simultaneous present on the cross-section. The

68

cross-sectional resultants are related to the stresses through the following integral

relations:

dAyxxVA

xyy *),()( ∫= τ (7.1)

dAyxxMA

xz *),()( ∫= σ (7.2)

They are not independent functions and are interrelated through the differential relation

expressed in Chapter 3 of the volume I. The cross-section is then subjected to non-

uniform bending. If the shear force Vy(x) is absent from the cross-section (Vy(x)=0),

then, the bending moment is constant (Mz(x)=constant) and the cross-section is

subjected to pure bending.

Due to the symmetry against the vertical plane Oxy imposed on the beam and the

existence of only transversally acting loads, the beam deflects only in the vertical plane

Oxy with some of its longitudinal fibers shortening and others lengthening.

Consequently, there are a number of fibers which conserve their length. These fibers

constitute the neutral plane and its intersection with the plane of symmetry Oxy defines

the deflection curve.

A complete theoretical discussion of the phenomena induced by bending is found in

the Chapter 7 of the textbook entitled Lectures in Mechanics of Materials, volume I. In

this chapter only the practical aspect of the application of the formulae related to the

normal stress σx and the shear stress τxy distributions on the cross-section of linear

planar beams is discussed.

7.1.1 Pure Bending

As discussed above, for the case of pure bending remains only one nonzero cross-

sectional resultant, the bending moment Mz(x). Consequently, the cross-section is

subjected only to a normal stress σx.

69

7.1.1.a Distribution of the Normal Stress (the Navier’s formula)

If the local coordinate system Oxyz is coincidental with the centroidal coordinate

system Cxyz, the normal stress σx in a point P(y,z) located on the cross-section x of a

linear planar beam is calculated using the most famous formula universally known as

the Navier’s formula:

yI

xMyxz

zx *)(),(

'

−=σ (7.3)

where Mz(x) is the bending moment pertinent to cross-section x;

zI is the moment of inertia against the centroidal axis Cz of the cross-section x;

y is the coordinate of the point of calculation measured on the centroidal axis

Cy of the cross-section x.

Note: Equation (7.3), the Navier’s formula, indicates that the normal stress σx is a

function of only two independent variables, x and y, varies linearly on variable y and is

independent of the variable z.

By definition the neutral axis of the cross-section is the axis where the normal stress σx

is zero.

Note: In the case of pure bending, the neutral axis passes through the centroid of the

cross-section and it is parallel to Cz axis.

The linearity of the normal stress σx function on variable y required two values for the

distribution on the cross-section to be identified. Those values are, for practical

reasons, the maximum normal stresses in tension and compression, maxsup__xσ and

maxinf__xσ , which are obtained at the extreme locations positioned from the centroid C.

The ratio obtained by dividing the moment of inertia zI by the distance y measured

from the centroid C of the cross-section to the extreme fiber is called section modulus.

The variation of the normal stress for positive and negative bending moments is

illustrated in Figure 7.1.1.

70

Figure 7.1.1 Normal Stress Distributions

• If the cross-section has only one axis of symmetry two section moduli can be

defined:

)()(

)(inf

inf

'

xyxI

xW z= (7.4)

supsup

)()(

'

yxI

xW z= (7.5)

For a bending moment assumed positive, 0)( ⟩xM z , as illustrated in Figure 7.1.1.a, the

maximum tensile and compressive stresses are obtained as:

0)()(

)(sup

maxsup__ <=xWxM

x zxσ compression (7.6)

0))()(

)(inf

maxinf__ >=xWxM

x zxσ tension (7.7)

For a bending moment assumed negative, 0)( ⟨xM z , the maximum stresses are as

shown in Figure 7.1.1.b:

0)(

)()(maxsup__ >=

xWxM

xbottom

zxσ tension (7.8)

0)()(

)(maxinf__ <=xWxM

xtop

zxσ compression (7.9)

• If the cross-section has two axes of symmetry the section moduli are equal and,

consequently, the magnitudes of the maximum normal stresses are equal. The

71

compressive and tensile areas corresponding to positive and negative bending

moments remain as discussed above and pictured in Figure 7.1.1.

7.1.2 Nonuniform Bending

As discussed above, for the case of nonuniform bending the cross-section is

characterized by the existence of both cross-sectional resultants, the bending moment

Mz(x) and the shear force Vy(x). Consequently, the cross-section is subjected to normal

and shear stresses, σx and τxy, respectively.

7.1.2.a Distribution of the Normal Stress (the Navier’s formula)

The finding that during nonuniform bending the cross-sections of the beam does not

remain plane after the deformation contradicts the basic kinematic assumption (e) of

pure bending. However, detailed investigation using advanced methods of the Theory

of Elasticity indicate that the warping of the cross-section due to the shear stress is

insignificant and does not affect the longitudinal strain when the ratio of the cross-

section’s height to the length of the beam is small. This is the case for most ordinary

structural members.

To extend the usage of the distribution of the normal stress ),( yxxσ derived for the

case of pure bending, formula 7.3, in the case of a beam subjected to nonuniform

bending, a supplementary assumption must be made: the distribution of the normal

stress ),( yxxσ in a given cross-section is independent of the shear stress induced

deformation. As a direct consequence of imposing the above stated assumption, the

pure bending strain-displacement relation remains valid and the influence of the

angular deformations on the longitudinal strain is neglected and the Navier’s formula

can be also used for the case of nonuniform bending.

7.1.2.b Distribution of the Shear Stress (the Jurawski’s formula)

The distribution of the shear stress is obtained from equilibrium considerations. The

procedure is applied first to a beam characterized by a rectangular cross-section and

72

later extended to the general case of beams with the cross-section composed from a

multitude of rectangular shapes.

7.1.2.b.1 Rectangular Cross-Section

Figure 7.1.2 illustrates the procedure for the case of the singular rectangular cross-

section with height h and width b.

Figure 7.1.2 Stress Resultants and Stresses

(a) Cross-section Resultants, (b) Stresses and (c) Free-body Diagram

The main steps are:

1. An infinitesimal volume of length dx is obtained by sectioning the beam at cross-

sections x and x+dx with vertical planes parallel to the centroidal plane Cyz. The

equilibrium condition of the infinitesimal obtained volume is pictured in Figure

7.1.2.a, where the cross-sectional resultants Mz(x) and Vy(x) and Mz(x+dx) and

Vy(x+dx) pertinent to x and x+dx cross-sections are indicated. The distribution of the

normal stress ),( yxxσ on both cross-sections is illustrated in Figure 7.1.2.b;

2. A second cut is made this time with a horizontal plane parallel to the centroidal

horizontal plane Cxz and passing through the point P(y,z) where the shear stress is

sought. Two infinitesimal volumes are obtained. The upper infinitesimal volume is

pictured in Figure 7.1.2.c together with the forces realizing the equilibrium. The two

horizontal forces xF and xxF ∆+ , generated by the existence of the normal stress

),( yxxσ on the remaining vertical areas 'A initially belonging to the cross-sections x

and x+∆x, must be in equilibrium with the longitudinal force H∆ acting on the newly

73

created horizontal section ab characterized by a length dx and width b.. The horizontal

force H∆ represents the resultant force of the longitudinal shear stresses ),,( zyxyxτ .

3. The equilibrium equation for the newly created body is:

xxx FFH −=∆ ∆+ (7.10)

After algebraic manipulations in-detail explained in the textbook one obtains:

)(*'

ySIMH A

zz

z∆=∆ (7.11)

where )('

yS Az and zI are the static moment of the section 'A and the moment of inertia

of the entire cross-section with respect to the neutral axis, respectively. The distance y

is measured from the centroid of the cross-section to the horizontal cut ab.

The increase of the bending moment from )(xM z to )( xxM z ∆+ is expressed as:

xxVxMxxMM yzzz ∆=−∆+=∆ *)()()( (7.12)

and can be positive or negative function of the shear force’s )(xVy sign.

The shear flow ),( yxq represents the shear force per unit length and is found by

taking the limit of the expression (7.11):

z

Azy

x IySxV

xHyxq

)(*)( lim),(

'

0 =∆∆

= →∆ (7.13)

In the absence of the shear stress distribution on the cut cross-section, the average

shear stress ),(_ yxavryxτ is used:

bIySxV

bxHyx

z

Azy

xavryx *)(*)(

*lim),(

'

0_ =∆∆

= →∆τ (7.14)

Note: It has to be emphasized that the real distribution of the shear stress is not known

and it is approximated by an average shear stress uniformly distributed across the

74

horizontal section ab. For the approximation to approach the real distribution, the

width of the rectangular cross-section b must be narrow. This finding is reinforced by

other theoretical investigations.

4. Due to the shear stress duality principle:

),,(),,( zyxzyx xyyx ττ = (7.14)

it results that the average shear stress in point P(y,z) is obtained:

tIySxV

yxyxz

Azy

avryxavrxy *)(*)(

),( ),('

__ ==ττ (7.15)

This formula is known in the technical literature as Jurawski’s formula. After algebraic

manipulations the distribution of the average shear stress ),(_ yxavrxyτ on the

rectangular cross-section is calculated:

]*41[**2

)(*3),( 2

2

_ hy

AxV

yx yavrxy −=τ (7.16)

where A is the area of the entire cross-section. The distribution of the shear stress on

the cross-section is illustrated in Figure 7.1.3.b.

Figure 7.1.3 Shear Stress Distribution for a Rectangular Cross-Section

75

Note: (a) On a particular cross-section (x=constant) the shear stress is mathematically

represented by a second order parabola depending only on the variable y. The extreme

values, zero and maximum, are attended on the contour and at the neutral axis,

respectively. The maximum shear stress is:

AxV

yx yavrxy

)(*5.1)0,(max

_ ==τ (7.17)

(b) The direction of the shear flow and average shear stress are coincidental to that of

the shear force Vy(x) direction pertinent to the cross-section investigated. The value of

the shear force and its direction are obtained from the shear force diagram.

7.1.2.b.2 Thin-Wall Cross-Sections

The majority of beams used in steel structures are classified as beams with “thin-wall”

cross-sections. The thin-wall cross-section is categorized in (a) open and (b) closed

cross-sections as illustrated in Figures 7.1.4.a and 7.1.4.b, respectively.

Figure 7.1.4 Thin-Wall Cross-Sections

The reduced thickness of the wall of the cross-section facilitates the introduction of

two important assumptions:

(a) the shear flow is always tangent to the local centerline of the cross-section;

(b) the shear stress is constant in the thickness of the cross-section.

These two assumptions mimic the situation described in the case of the rectangular

cross-section.

The shear flow ),( sxq and the shear stress ),( sxτ characteristic to a thin-wall cross-

section are shown in Figure 7.1.5

76

Figure 7.1.5 Shear Flow and Shear Stress in Thin-Wall Cross-Section (a) Shear Flow and (b) Shear Stress

They are related through the following formula:

),(),(),(

sxtsxqsx =τ (7.18)

7.1.2.b.2.a Open Thin-Wall Cross Section

To obtain the shear stress distribution for a cross-section composed from multiple

rectangular narrow shapes, the case in point for the majority of beams employed in

the structural analysis, the above explained procedure is entirely followed. The

complexity of the procedure is increased due to the fact that the second cut can be

either horizontally orientated parallel to the plane Cxz or vertically orientated parallel

to the plane Cxy, function of the positioning of the rectangular shape in the general

arrangement of the cross-section. It has to be emphasized that the shape of the cut is

always a rectangular shape.. This situation indicates that on the composed cross-

section concomitantly appear ),,( zyxxyτ and ),,( zyxxzτ shear stresses. Their values

are also calculated employing Jurawski’s formula (7.15) where the area 'A and the

width of the cut t, replacing the dimension b previously used in formula (7.14), vary

function of the employed cut’s location and direction (vertical or horizontal). The static

moment )('

yS Az is calculated:

'0

' *)('

yAyS Az = (7.19)

where '0y is the distance from the centroid of the area 'A to the neutral axis of the

cross-section.

77

It has to be emphasized that the area 'A is the area obtained from sectioning the entire

cross-section above or below the cut for the case of a horizontal cut or to the left or the

right of the cut in the case of a vertical cut. The maximum shear stress is obtained

when the static moment )('

yS Az reaches its maximum value. This is the case when the

horizontal cut coincides with the neutral axis Cz.

Due to the imposed restriction on the width t of the rectangular shape where the cut is

made to be narrow, the only significant cut direction is that parallel to the width of the

rectangular shape considered. Examples of horizontal and vertical cuts are shown in

Figure 7.1.6: a, a’, b, b’, f , f’, g and g’ are vertical cuts, while c, d and e are horizontal

cuts. A1, A2, A3 and A4 are the compounding shapes of the cross-section.

Figure 7.1.6 Example of Cuts

Taking a broad view, if the rectangular shape investigated is located with its length

vertically, parallel to Cy axis, the significant cut is horizontal, parallel to Cz axis (the

neutral axis). The shear flow ),( yxq and the corresponding shear stress ),,( zyxxyτ

are both orientated parallel to the Cy axis and are characterized by second-order

parabolic variations relatively to variable y. If the rectangular shape investigated is

located with its length horizontally, parallel to Cz axis, the significant cut is vertical,

parallel to Cy axis (the symmetry axis). The shear flow ),( yxq and the shear stress

),,( zyxxzτ are both orientated parallel to the Cz axis and are characterized by linear

variations relatively to variable y.

78

The value and orientation of the flow ),( yxq and shear stresses, ),,( zyxxyτ or

),,( zyxxzτ , are established and only their senses remain to be decided. The correct

sense of the shear stresses is judged following the mathematical rational explained

below function of the sign of the bending moment Mz(x) and shear force Vy(x) acting

on the particular cross-section x. Consequently, the cross-sectional resultants Mz(x) and

Vy(x) pertinent to the investigated cross-section x are extracted from the diagrams and

plotted on the cross-section with the positive normal x>0. The mathematical rational is

accompanied by graphical representations, the Figures 7.1.7 through 7.1.10. For

generality and clarity, in those figures representing the general case of a composed

cross-section assembled from rectangular shapes, only the last rectangular shape of the

area 'A , through where the cut is made, is illustrated. The following cases are

analyzed:

(1) if 0)( >xM z and 0)( >xVy (see Figure 7.1.7)

→ 0*)( >∆=∆ xxVM yz

→ 0)()( >∆+=∆+ zzz MxMxxM

→ )()( xMxxM zz >∆+

→ xxx FF >∆+

• for A’ located above the neutral axis (NA) and left to Cy axis (symmetry axis)

→ 0>xF (compression and parallel to x>0)

→ 0<∆+ xxF (compression and parallel to x<0)

It results that:

→ 0>∆H (parallel to x>0)

→ 0>q (parallel to x>0)

→ 0_ >avrzxτ → 0_ <avrxzτ (parallel to z<0) - vertical cut (Figure 7.1.7.a)

→ 0_ >avryxτ → 0_ <avrxyτ (parallel to y<0) - horizontal cut (Figure 7.1.7.b)

79

Figure 7.1.7 Mz(x)>0 and Vy(x)>0

• for A’ located below the neutral axis (NA) and left to Cy axis (symmetry axis)

→ 0<xF (tension and parallel to x<0)

→ 0>∆+ xxF (tension and parallel to x>0)

It results that:

→ 0<∆H (parallel to x<0)

→ 0<q (parallel to x<0)

→ 0_ <avrzxτ → 0_ >avrxzτ (parallel to z>0) - vertical cut (Figure 7.1.7c)

80

→ 0_ <avryxτ → 0_ <avrxyτ (parallel to y<0) - horizontal cut (Figure 7.1.7.d)

(2) if 0)( >xM z and 0)( <xV y (see Figure 7.1.8)

→ 0*)( <∆=∆ xxVM yz

→ zzz MxMxxM ∆+=∆+ )()(

→ )()( xMxxM zz <∆+

→ xxx FF <∆+




It results that:



→ 0_ <avrzxτ → 0_ >avrxzτ (parallel to z>0) - vertical cut (Figure 7.1.8.a)

→ 0_ <avryxτ → 0_ >avrxyτ (parallel to y>0) - horizontal cut (Figure 7.1.8.b)




It results that:



81

→ 0_ >avrzxτ → 0_ <avrxzτ (parallel to z<0) - vertical cut (Figure 7.1.8.c)

→ 0_ >avryxτ → 0_ >avrxyτ (parallel to y>0) - horizontal cut (Figure 7.1.8.d)

Figure 7.1.8 Mz(x)>0 and Vy(x)<0

(3) if 0)( <xM z and 0)( >xV y (see Figure 7.1.9)

→ 0*)( <∆=∆ xxVM yz

→ zzz MxMxxM ∆+=∆+ )()(

→ )()( xMxxM zz <∆+

→ xxx FF <∆+

82




It results that:



→ 0_ >avrzxτ → 0_ <avrxzτ (parallel to z<0) - vertical cut (Figure 7.1.9.a)

→ 0_ >avryxτ → 0_ <avrxyτ (parallel to y<0) - horizontal cut (Figure 7.1.9.b)

Figure 7.1.9 Mz(x)<0 and Vy(x)>0

83




It results that:



→ 0_ <avrzxτ → 0_ >avrxzτ (parallel to z>0) - vertical cut (Figure 7.1.9.c)

→ 0_ <avryxτ → 0_ <avrxyτ (parallel to y<0) - horizontal cut (Figure 7.1.9.d)

(4) if 0)( <xM z and 0)( <xV y (see Figure 7.1.10)

→ 0*)( <∆=∆ xxVM yz

→ 0)()( <∆+=∆+ zzz MxMxxM

→ )()( xMxxM zz >∆+

→ xxx FF >∆+




It results that:



→ 0_ <avrzxτ → 0_ >avrxzτ (parallel to z>0) - vertical cut (Figure 7.1.10.a)

→ 0_ <avryxτ → 0_ >avrxyτ (parallel to y>0) - horizontal cut (Figure 7.1.10.b)

84




It results that:



→ 0_ >avrzxτ → 0_ <avrxzτ (parallel to z<0) - vertical cut (Figure 7.1.10.c)

→ 0_ >avryxτ → 0_ >avrxyτ (parallel to y>0) - horizontal cut (Figure 7.1.10.d)

Figure 7.1.10 Mz(x)<0 and Vy(x)<0

85

7.1.2.b.2.b Closed Thin-Wall Cross-Section

The most commonly used closed thin-wall cross-sections are the box and tubular cross-

sections. The analysis of the distribution of shear stress for a closed thin-wall cross-

section is conducted in a similar manner with the theoretical development employed

for open cross-sections. A thin slice of length x∆ is isolated from the body of the

closed thin-wall beam. As illustrated in Figure 7.1.11 the beam has the longitudinal

vertical plane as a plane of symmetry. The shear flow shown in Figure 7.1.11 must be

directed in such a manner that the resultant is equal to the vertical shear force acting on

the cross-section.

Figure 7.1.11 Closed Thin-Wall Cross-Section

The main problem arising in the evaluation of shear stress of the closed thin-wall

members is how to position the second cutting plane. If the vertical cutting plane

passing through the axis of symmetry Cy is used, the beam is separated in two

symmetric halves. The shear flow q is tangent to the centerline of the cross-section.

Thus, for example, in Figure 7.1.11.c it is obviously that the shear flow is horizontal at

ends a andb . The shear flow must be zero in the plane of symmetry and as a

consequence of the duality principle the shear flow in the cross-section at these

locations is also zero. Since the shear flow is zero at the cross-section points

intersecting the axis of symmetryCy , the general distribution is obtained by using

cutting planes symmetrical to this axis and writing the horizontal equilibrium for the

beam segment contained between these planes.

86

7.1.3 Strength Condition

7.1.3.a Material with Different Tension and Compression Allowable Normal Stresses

• if the maximum absolute bending moment is positive 0M max

z >

then

→ 0maxsup__ <xσ (compression)

→ 0maxinf__ >xσ (tension)

and the strength conditions are:

compallx _maxsup__ σσ ≤ (7.20)

tensallx _maxinf__ σσ ≤ (7.21).

allxy ττ ≤max_ (7.22)

where compall _σ , tensall _σ and allτ are the allowable normal stress in compression, in

tension and shear, respectively.

• if the maximum absolute bending moment is negative 0M maxz <

then

→ 0maxsup__ >xσ (tension)

→ 0maxinf__ <xσ (compression)

and the strength conditions are:

tensallx _maxsup__ σσ ≤ (7.23)

compallx _maxinf__ σσ ≤ (7.24)


7.1.3.b Material with Equal Tension and Compression Allowable Normal Stresses

→ 0maxsup__ <xσ (compression)

87

→ 0maxinf__ >xσ (tension)

and the strength condition is:

allxx σσσ ≤),max( maxinf__maxsup__ (7.26)


where allσ is the allowable normal stress.

7.1.4 Built-Up Beam Connectors

The beams made from continuous material without any interruptions are called

homogeneous cross-sections. In general, the industrial steel buildings are subjected to

heavy loads and the most commonly used cross-sections, such as the wide-flange and

closed thin-wall, are fabricated from parts joined together by connectors. These types

of beams are called build-up members. Glue (adhesives) and nails are connectors

proper to be used in the wood built-up members, while welds, rivets and bolts are

commonly used for the steel built-up members. Examples of build-up members are


Figure 7.1.12 Built-Up Beams (a) Glued Wood Beam, (b) Box Wood Beam, (c) Welded Steel Beam and

(d) Reinforced Steel Beam The shear flow induced by nonuniform bending must be transferred in-between the

adjoining parts comprising the beam’s cross-section. This shear transfer is realized in

three ways: (a) by shear distributed over the interface areas, (b) by shear distributed

along a connector line and (c) by discrete shear connectors. The shear transfer type (a)

is common for the laminated wood beams composed of multiple pieces as shown in

88

Figure 7.1.12.a. The linear transfer, type (b), is proper to the welded beams as pictured

in Figure 7.1.12.c, while the discrete type shear transfer, shown in Figures 7.1.12.b and

7.1.12.d, are pertinent to bolt or nail connectors. There are special cases where welds

are discontinuously applied, referred to as stitch welds.

The shear flow determined using equation (7.13) is employed to calculate the required

shear flow recq , which represents the shear force per unit length that is transferred from

one part of the cross-section to the adjacent part under the given loading condition.

The required shear is calculated as:

z

y

z

Azy

rec IyAV

ISV

q'*** max_max_

== (7.28)

where 'A is the area of the connected part, max_yV is the maximum shear force acting

on the beam and y is the distance measured from the neutral axis to the centroid of the

connected part.

7.1.4.a Linear Shear Connectors

The required shear flow calculated with equation (7.28) determines the actual

necessary force per unit length which must be transmitted by the welded connection.

The shear capacity of the weld weldq must be larger than the required shear flow:

weldreq qq ≤ (7.29)

The weld capacity is calculated as:

weldallweldsweld efftnq _*.* τ= (7.30)

where weldsn , ffet. and weldall _τ are the number of the welds connecting the part, the

effective thickness of the weld and the allowable shear stress of the weld material,

respectively. The allowable shear stress of the weld material is determined from

laboratory test data.

89

7.1.4.b Discrete Shear Connectors

The discrete connectors have a shear capacity connectorQ , also established by laboratory

testing, and are assumed to be spaced at a equal distance s∆ . For the discrete

connector to transfer the shear the following relation must be maintained:

connectorreq Qsq ≤∆* (7.31)

Equation (7.31) is applied in different zones of the beams, considering in the

calculation the maximum value of the vertical shear force in the particular zone. This

way, the distance between the connectors may be increased or decreased within the

various zones of the beam in accordance with the total shear flow requirement for the

zone.


Problem 7.2.1 The cantilever beam, shown in Figure 7.2.1, is subjected to a

concentrated moment mkNM ⋅= 200 acting at the free end. The cross-section has a

square shape with an edge length d = 120 mm and length L = 5 m. Calculate: (a) the

cross-sectional resultants diagrams and (b) the strain and stress distribution.

Figure 7.2.1


90

A.1 The cantilever is subjected to pure bending conditions and, consequently, the

Navier’s Formula (7.3) is applied.

A.2 Numerical Application

M0 20 kN⋅ m⋅:= - the bending moment

L 5 m⋅:= - cantilever length

d 120 mm⋅:= - edge length

E 200 GPa⋅:= - modulus of elasticity

B. Calculations

B.1 Reactions Calculation (see Figure 7.2.1.a)

The reactions, AV and AM , are obtained using the following two equilibrium

equations:

∑ =A

M 0 →

MA M0− 0

∑ =B

M 0 →

VA L⋅ MA+ M0− 0

Figure 7.2.1.a

Solving the above equilibrium equations the reactions are obtained as:

MA M0:=

91

VA 0 kN⋅:=

B.2 Calculation of the Cross-Sectional Resultants (see Figure 7.2.1.a)

The cross-sectional resultants, the shear force )(xVy and bending moment )(xM z ,

acting on a particular cross-section located at distance x from end A, representing the

origin of the coordinate system, are:

Vy x( ) 0

Mz x( ) M0 → Mz 2 104× N m⋅=

The moment diagram, the only diagram different than zero, is illustrated in Figure

7.2.1.a

B.3 The Stress and Strain Distributions (see Figure 7.2.1.b)

The stress distribution is obtained using the Navier’s Formula (7.3). The distribution of

the normal stress identical on all cross-sections is linear and, consequently, only the

values at the extreme points, 2supdyy ==

or

2infdyy −== ,

are necessary to be

calculated. The maximum normal stresses

maxsup__xσ and

maxinf__xσ are obtained as:

ysupd2

:= → ysup 6cm=

yinfd2

−:= → yinf 6− cm=

σx_sup_maxM0Izc

− ysup⋅:= →⇒ −− m

mmN 2

48

3

10*6*10*1728

*10*20

σx_sup_max 6.944− 107× Pa=

σx_inf_maxM0Izc

− yinf⋅:= →−⇒ −− m

mmN )10*6(*

10*1728*10*20 2

48

3

σx_inf_max 6.944 107× Pa=

92

Figure 7.2.1.b

Note: (a) It is obvious that the maximum normal stresses,

maxsup__xσ and

maxinf__xσ ,

illustrated in Figure 7.2.1.b, are equal in absolute value due to the fact that they

are located at equal distance from the cross-section’s centroid. It can be

concluded that when the cross-section is also symmetrical against the Zc axis

only one value is necessary to be calculated;

(b) the area above and below the neutral axis CZc are in compression and

tension, respectively.

The linear strain distribution, shown in Figure 7.2.1.b, is calculated as:

ε x_sup_maxσx_sup_max

E:= →

−⇒

PaPa

9

7

10*20010*944.6

ε x_sup_max 3.472− 10 4−×=

ε x_inf_maxσx_inf_max

E:= →⇒

PaPa

9

7

10*20010*944.6

ε x_inf_max 3.472 10 4−×=

Problem 7.2.2 The aluminum part with a cross-section as illustrated in Figure 7.2.2 is

subjected to pure bending. Conduct the following tasks: (a) calculate the maximum

bending moment capM applied to the member if the allowable flexural stresses in

tension and compression are σall_tens = 200 MPa and σall_compr = 100 MPa, respectively,

and (b) draw the flexural stress and strain distributions on the cross-section

corresponding to the bending moment determined at point (a).

93

Figure 7.2.2


A.1 The cross-section is subjected to pure bending, but a special attention should be

given to the fact that the material has different normal allowable stresses for tension

and compression.


• material data

σall_tens 200 MPa⋅:= - allowable normal tensile stress

σall_compr 100 MPa⋅:= - allowable compressive stress


• cross-sectional dimensions

hw 80 mm⋅:= - web height

tw 20 mm⋅:= - web thickness

bf 160 mm⋅:= - flange width

tf 20 mm⋅:= - flange thickness

94

aw 80 mm⋅:= - interior distance between webs

B. Calculations

B.1 Calculation of Cross-section Geometrical Characteristics (see Figure 7.2.2.a)

Figure 7.2.2.a

Note: All dimensions shown on Figure 7.2.2.a are in cm.

The cross-section is considered composed from three individual shapes: two webs and

a flange. The areas of the webs, flange and the total area of the cross-section are

calculated as:

Aw tw hw⋅:= →⇒ 8*2 Aw 16cm2= - the web area

Af tf bf⋅:= →⇒ 16*2 Af 32cm2= - the flange area

Atotal 2 Aw⋅ Af+:= →+⇒ 3216*2 Atotal 64cm2=

The centroid C is located on the axis of symmetry and consequently, only the vertical

position of it is necessary to be calculated. The initial coordinate system OYZ is

considered located at the level of the centroid of the webs. The vertical position Cy of

the cross-sectional centroid is obtained:

95

yC2 Aw⋅ yC_w⋅ Af yC_f⋅+

Atotal:= →

−+⇒

64)5(*320*16*2

yC 2.5− cm=

where

yC_w 0 cm⋅:= - distance from O to the web centroid

yC_fhw2

tf2

+

−:= →

+−⇒

22

28

yC_f 5− cm= - distance from O to the flange

centroid

The moment of inertia of the entire cross-section against the CZc axis zCI is obtained:

Izc 2tw hw

3⋅

12Aw yC_w

2⋅+

⋅

bf tf3⋅

12Af yC_f

2⋅+

+:=

→

−++

+⇒ 2

32

3

)5.2(*3212

2*165.2*1612

8*2*2 Izc 581.333cm4=

where:

yC_w yC_w yC−:= →−−⇒ )5.2(0 yC_w 2.5cm= - distance from C to the web

centroid

yC_f yC_f yC−:=→−−−⇒ )5.2(5

yC_f 2.5− cm=- distance from C to the flange

centroid

The section modulus W is calculated as follows:

W min Wsup Winf,( ):=

where:

WsupIzc

ysup:= →⇒ 3

5.6333.581 cm Wsup 89.436cm3=

- for the superior fiber

ysup yC_whw2

+:= →+⇒285.2 ysup 6.5cm=

WinfIzcyinf

:= →⇒ 3

5.6333.581 cm Winf 166.095cm3=

- for the inferior fiber

96

yinf yC_ftf2

−:= →−−⇒225.2 yinf 3.5− cm=

W min Wsup Winf,( ):= →⇒ )095.166,436.89min( W 89.436cm3=

B.2 Calculation of the Capable Bending Moment

Due to the fact that the material considered is characterized by different allowable

normal stress in tension and compression the calculation has to identify the maximum

tensile and compressive normal stress existing on the cross-section. The bending

moment M, indicated in Figure 7.2.2, is negative and consequently, the area located

above the neutral axis (CZc axis) is in tension, while the area located below is in

compression.

The verification formulae are:

σx_supM0Izc

ysup⋅M0

Wsupσall_tens≤ -for normal tensile stress area

σx_infM0Izc

yinf⋅M0Winf

σall_compr≤

-for normal compressive stress area

The unknown values of the corresponding bending moments are at limite:

M0_sup Wsup σall_tens⋅ →⇒ − Pam 636 10*200*10*436.89

M0_sup 1.789 104× N m⋅=

M0_inf Winf σall_compr⋅:= →⇒ − Pam 636 10*100*10*095.166

M0_inf 1.661 104× N m⋅=

The capable bending moment capOM _ characterizing the entire cross-section is:

M0_cap min M0_sup M0_inf,( ):= → M0_cap 1.661 104× N m⋅=

B.3 Normal Stress and Strain Diagrams (see Figure 7.2.2.a)

97

The linear distribution of the normal stress on the cross-section corresponding to the

capable bending moment capOM _ is obtained:

σx_supM0_cap−

Izc− ysup⋅:= →

−−⇒ −

− mmN 28

4

10*5.6*10*333.581

*10*661.1

σx_sup 1.857 108× Pa=

σx_infM0_cap−

Izc− yinf⋅:= →−

−−⇒ −

− mmN 28

4

10*)5.3(*10*333.581

*10*661.1

σx_inf 1− 108× Pa=

Note: All true that the maximum normal stress is located at the superior extreme fiber,

the strength limit (the allowable normal compressive stress) is firstly reached at the

inferior fiber.

The corresponding linear distributed normal strains are calculated as:

ε x_supσx_sup

E:= →⇒

PaPa

9

8

10*20010*857.1

ε x_sup 9.286 10 4−×=

ε x_infσx_inf

E:= →

−⇒

PaPa

9

8

10*20010*000.1

ε x_inf 5− 10 4−×=

The final stress and strain distributions are plotted in Figure 7.2.2.a.

Problem 7.2.3 The timber beam made of four planks tied together with screws to form

a box section, as shown in Figure 7.2.3, is subjected to pure bending. If the flexural

stress at point B of the cross-section is a 6.2 MPa tensional stress, determine (a) the

bending stress and strain distribution on the cross-section, (b) the bending moment

carried by the cross-section, (c) the flexural stresses at points A and D of the cross-

section, (d) the axial force acting on the top and bottom planks and (e) the capable

bending moment of the cross-section considering that the wood has allowable flexural

stresses in tension and compression σall_tens =30 MPa and σall_compr =20 MPa.

98

Figure 7.2.3


A.1 The cross-section is subjected to pure bending, but a special attention should be

given to the fact that the material has different normal allowable stresses.


• material data

σall_tens 30 MPa⋅:= - allowable normal tensile stress

σall_compr 20 MPa⋅:= - allowable compressive stress



hw 20 cm⋅:= - web height

tw 4 cm⋅:= - web thickness

bf 18 cm⋅:= - flange width

tf_s 2 cm⋅:= - top flange thickness

tf_i 4 cm⋅:= - bottom flange thickness

99

• point B and D

dB 6 cm⋅:= - distance measured from the bottom of the box beam

σx_B 6.2 MPa⋅:= - effective normal stress at point B

dD 1 cm⋅:= - distance measured from the bottom of the box beam

B. Calculations

B.1 Calculation of the Cross- Section Geometrical Characteristics (see Figure 7.2.3.a)

Figure 7.2.3.a

The cross- section is composed from four rectangles: top flange, two webs and bottom

flange. The corresponding areas of the components and the total area of the cross-

section are calculated:

Note: All dimensions used in the calculations of the cross-section geometrical

characteristics are in cm.

Aw tw hw⋅:= →⇒ 20*4 Aw 80cm2= - area of the web

Af_s tf_s bf⋅:= →⇒ 18*2 Af_s 36cm2= - area of the top flange

Af_i tf_i bf⋅:= →⇒ 18*4 Af_i 72cm2= - area of the bottom flange

Atotal 2 Aw⋅ Af_s+ Af_i+:= →++⇒ 723680*2 Atotal 268cm2= - total area

100

The cross- section is symmetric against the vertical axis OYc and consequently, only

the vertical position of the cross- section centroid is required to be calculated. The

general coordinate system OYcZ is anchored in the centroid of the webs.

yC_w 0 cm⋅:= - distance from the point O to the web centroid.

yC_f_shw2

tf_s2

+:= →+⇒22

220 yC_f_s 11cm=

- distance from the point O to the

top flange centroid.

yC_f_ihw2

tf_i2

+

−:= →

+−⇒

24

220

yC_f_i 12− cm= - distance from the point O

to the botom flange

centroid

yC2 Aw⋅ yC_w⋅ Af_s yC_f_s⋅+ Af_i yC_f_i⋅+

Atotal:=

→−++

⇒268

)12(*7211*360*80*2yC 1.746− cm=

- the vertical position of the centroid

Note: The neutral axis, NA, is identical to the central horizontal axis of the coordinate

system CYcZc.

The new coordinate system CYcZc is moved in the centroid C of the cross- section.

The moment of inertia, Izc, of the entire cross-section calculated against the axis CZc is

obtained:

yC_w yC_w yC−:= →−−⇒ )746.1(0 yC_w 1.746cm= - distance from the point C

to the web centroid.

yC_f_s yC_f_s yC−:= →−−⇒ )746.1(11 yC_f_s 12.746cm=

- distance from the

point C to the top

flange centroid.

yC_f_i yC_f_i yC−:= →−−−⇒ )746.1(12 yC_f_i 10.254− cm=

- distance from the point C to the botom flange centroid

101

Izc 2tw hw

3⋅

12Aw yC_w

2⋅+

⋅

bf tf_s3⋅

12Af_s yC_f_s

2⋅+

+

bf tf_i3⋅

12Af_i yC_f_i

2⋅+

+

...:=

→

−++

++

+⇒ 2

32

32

3

)254.10(*7212

4*18746.12*3612

2*18746.1*801220*4*2

Izc 1.935 104× cm4= - cross- section moment of inertia

The section modulus W is obtained:

ysup yC_f_stf_s2

+:= →+⇒22746.12 ysup 13.746cm=

- distance from NA to the

upper edge of the top

flange

yinf yC_f_itf_i2

−:= →−−⇒24254.10 yinf 12.254− cm=


lower edge of the bottom

flange

WsupIzc

ysup:= →⇒

746.1310*935.1 4

Wsup 1.408 103× cm3=

WinfIzcyinf

:= →⇒254.12

10*935.1 4

Winf 1.579 103× cm3=

W min Wsup Winf,( ):= →⇒ )10*579.1,10*408.1min( 33 W 1.408 103× cm3=

- sectional modulus

B.2 Normal Stress and Strain Diagrams

According to Navier’s Formula the normal stress distribution on the cross- section is

linear. To draw the diagram, two point are necessary: the first point is the centroid and

the second is the given normal stress σx_B 6.5 MPa⋅ at point B. The resulting

diagram of the normal stress is shown in Figure 7.2.3.a. The maximum values of the

102

normal stresses, tx max__σ and cx max__σ , corresponding to the upper edge of the top

flange and the lower edge of the bottom flange, respectively, are calculated using

simple geometrical proportions.

σx_max_t σx_ByinfyB

⋅:= →−−

⇒−

−

mmPa 2

26

10*254.610*254.12*10*5.6

σx_max_t 1.215 107× Pa=

σx_max_c σx_max_tysupyinf

⋅:= →−

⇒−

−

mmPa 2

27

10*254.1210*746.13*10*215.1

σx_max_c 1.363− 107× Pa=

where By is the distance measured from the neutral axis, NA, to point B.

yBhw2

tf_i+ dB−

− yC−:= →−−

−+−⇒ )746.1(64

220

yB 6.254− cm=

The corresponding maximum strains, tx max__ε and cx max__ε calculated at the upper

edge of the top flange and the lower edge of the bottom flange, respectively, are

obtained as:

ε x_supσx_max_t

E:= →⇒

PaPa

9

7

10*3010*215.1

ε x_sup 4.049 10 4−×=

ε x_infσx_max_c

E:= →

−⇒

PaPa

9

7

10*3010*363.1

ε x_inf 4.543− 10 4−×=

Note: The values of the obtained maximum normal stresses and corresponding strains

indicate that the area above the neutral axis is in compression, while the area below

the neutral axis is in tension.

B.3 Calculation of the Bending Moment

The bending moment, Mo, acting on the CZc axis and corresponding to the normal

stress distribution calculated in section B.2 is obtained:

103

M0IzcyB

− σx_B⋅:= →−

⇒−

−

Pamm 6

2

484

10*5.6*10*254.610*10*935.1

M0 1.918 104× N m⋅=

Note: Due to the rounding to the third decimal of the numbers used in the manual

calculation an insignificant numerical error appears in comparison with similar

Mathcad results.

To verify the maximum values of the normal stresses diagram, calculated in section

B.2, these values are recalculated using the moment Mo and Navier’s Formula:

σx_max_tM0Izc

− yinf⋅:= →−−⇒ −−

)10*254.12(*10*10*935.1

*10*918.1 2484

4

mmmN

σx_max_t 1.215 107× Pa=

σx_max_cM0Izc

− ysup⋅:= →−⇒ −−

)10*746.13(*10*10*935.1

*10*918.1 2484

4

mmmN

σx_max_c 1.363− 107× Pa=

B.4 Calculation of Normal Stress at points A and D

The normal stresses at points A and D can be calculated using two methods: (a)

Navier’s Formula or (b) geometrical proportions of the linear diagram shown in Figure

7.2.3.a.

(a) using Navier’s Formula

σx_AM0Izc

− yA⋅:= →−⇒ −−

)10*746.13(*10*10*935.1

*10*918.1 2484

4

mmmN

σx_A 1.363− 107× Pa=

σx_DM0Izc

− yD⋅:= →−−⇒ −−

)10*254.11(*10*10*935.1

*10*918.1 2484

4

mmmN

σx_D 1.116 107× Pa=

104

where: yA yC_f_stf_s2

+:= →+⇒22746.12 yA 13.746cm=

yD yC_f_i dD−:= →−−⇒ 1254.10 yD 11.254− cm=

(b) using the “Geometrical Method”

σx_A σx_max_c:= → σx_A 1.363− 107× Pa=

σx_D σx_ByDyB

⋅:= →−−

⇒−

−

mmPa 2

26

10*254.610*254.11*10*5.6 σx_D 1.116 107× Pa=

B.5 Calculation of the Axial Forces of the Flanges

The axial forces in the flanges are obtained using the normal stress diagram shown in

Figure 7.2.3.b.

(a) The axial force on the top flange

σx_f_s_top σx_max_c:= → σx_f_s_top 1.363− 107× Pa=

yf_s_bot yA tf_s−:= →−⇒ 2746.13 yf_s_bot 11.746cm=

σx_f_s_botM0Izc

− yf_s_bot⋅:= →−⇒ −−

)10*746.11(*10*10*935.1

*10*918.1 2484

4

mmmN

σx_f_s_bot 1.165− 107× Pa=

Ff_s12

tf_s σx_f_s_top σx_f_s_bot+( )⋅ bf⋅:= ⇒

( ) →−−⇒ −− mPaPam 2772 10*18*10*165.110*363.110*2*21

Ff_s 4.549− 104× N=

(b) The axial force on the bottom flange

yf_i_top yinf tf_i+:= →+−⇒ 4254.12 yf_i_top 8.254− cm=

σx_f_i_topM0Izc

− yf_i_top⋅:= →−−⇒ −−

)10*254.8(*10*10*935.1

*10*918.1 2484

4

mmmN

105

σx_f_i_top 8.183 106× Pa=

σ x_f_i_bot σ x_max_t:= → σx_f_i_bot 1.215 107× Pa=

Ff_i12

σx_f_i_top σx_f_i_bot+( )⋅ bf⋅ tf_i⋅:=

( ) →+⇒ −− mPaPam 2762 10*18*10*215.110*183.810*4*21

Ff_i 7.319 104× N=

Figure 7.2.3.b.

B.6 Calculation of the Capable Bending Moment

In the previous sections the sign of the bending moment Mo is positive, as shown in

Figure 7.2.3.b. The sign of the bending moment is established by the sign of the

normal stress Bx _σ imposed at point B. The calculation of the capable bending

moment pertinent to the cross-section, due to the asymmetry of the cross-section

against the centroidal axis CZc and material allowable stresses for compression and

tension, requires the examination of both conditions: the positive and negative

directions of the centroidal acting bending moment.

(a) If M0_cap>0

The verification formulas corresponding to the extreme fibers are:

106

σx_max_cM0_cap

Izc− ysup⋅

M0_capWsup

− σall_compr≤

σx_max_tM0_cap

Izc− yinf⋅

M0_capWinf

σall_tens≤

The capable bending moment calculated under the assumption that the bending

moment is positive, a vector parallel with the positive CZc axis, is:

M0_cap_poz min σall_compr Wsup⋅ σall_tens Winf⋅,( ):=

→⇒ −− )10*10*579.1*10*30,10*10*408.1*10*20min( 36363636 mPamPa

M0_cap_poz 2.815 104× N m⋅=

(b) If M0_cap<0

The verification formulae corresponding to the extreme fibers are:

σx_max_cM0_cap−

Izc− ysup⋅

M0_cap−

Wsup− σall_tens≤

σx_max_tM0_cap−

Izc− yinf⋅

M0_cap−

Winfσall_compr≤

The capable bending moment calculated under the assumption that the bending

moment is negative, a vector parallel with the negative CZc axis, is:

M0_cap_neg min σall_tens Wsup⋅ σall_compr Winf⋅,( )−:=

→−⇒ −− )10*10*579.1*10*20,10*10*408.1*10*30min( 36363636 mPamPa

M0_cap_neg 3.158− 104× N m⋅=

Finally, the capable bending moment pertinent to the cross-section is obtained:

M0_cap if M0_cap_poz M0_cap_neg> M0_cap_neg, M0_cap_poz,( ):= ⇔

M0_cap min Winf Wsup,( ) min σall_compr σall_tens,( )⋅:= ⇔

M0_cap W min σall_compr σall_tens,( )⋅:= → M0_cap 2.815 104× N m⋅=

107

Problem 7.2.4 The cross- section shown in Figure 7.2.4 is considered subjected to non-

uniform bending characterized by a bending moment Mo=55 kNm and shear force

Vo=10 kN,. Conduct the following tasks: (a) calculate the normal stress distribution on

the cross- section, (b) calculate the shear stress distribution on the cross- section and

(c) verify the cross- section, considering MPaall 210=σ and MPaall 120=τ .

Figure 7.2.4


A.1 The cross-section is subjected to non-uniform bending. Both Navier’s and

Jourawski’s Formulas are applicable. The beam is made of steel and consequently, the

allowable bending stress has an equal value for both tensile and compressive normal

stress.


• cross- sectional resultants

M0 55kN m⋅:= - bending moment

V0 10 kN⋅:= - shear force

• material data

σall 210MPa:= - allowable normal tensile and compressive stress

τall 120MPa:= - allowable shear stress

108

E 200GPa:= - modulus of elasticity


t 1 cm⋅:=

ν1 1.5:=

ν2 1.2:=

α 20:=

β 1 15:=

β 2 10:=

hsect α t⋅:= → hsect 20cm= - section height

tweb t:= → tweb 1cm= - web thickness

bf_s β 1 t⋅:= → bf_s 15cm= - top flange width

bf_i β 2 t⋅:= → bf_i 10cm= - bottom flange width

tf_s ν1 t⋅:= → tf_s 1.5cm= - top flange thickness

tf_i ν2 t⋅:= → tf_i 1.2cm= - bottom flange thickness

hweb hsect tf_s− tf_i−:=

→−−⇒ 2.15.120 h web 17.3cm= - web height

B. Calculations

B.1 Calculation of the Cross- Section Geometrical Characteristics (see Figure 7.2.4.a)

The cross- section consists on three individual rectangular areas: upper and lower

flanges and the web. The corresponding areas are calculated:

Af_s bf_s tf_s⋅:= →⇒ 5.1*15 Af_s 22.5cm2= - area of the top flange

Af_i bf_i tf_i⋅:= →⇒ 2.1*10 Af_i 12cm2= - area of the bottom flange

Aweb hweb tweb⋅:= →⇒ 1*3.17 Aweb 17.3cm2= - area of the web

109

Atotal Aweb Af_s+ Af_i+:= →++⇒ 125.223.17 Atotal 51.8cm2= - total area

Figure 7.2.4.a

The initial coordinate system OYcZ passes through the centroid of the web. The

distances from point O to the lower and upper flange centroids are obtained:

yC_web 0 cm⋅:=

yC_f_shweb

2

tf_s2

+:= →+⇒25.1

23.17

yC_f_s 9.4cm= distance from the point O to

the upper flange centroid.

yC_f_ihweb

2

tf_i2

+

−:= →

+−⇒

22.1

23.17

yC_f_i 9.25− cm= distance from the

point O to the lower flange

centroid.

yCAweb yC_web⋅ Af_s yC_f_s⋅+ Af_i yC_f_i⋅+

Atotal:=

→−++

⇒8.51

)25.9(*124.9*5.220*3.17 yC 1.94cm=


system CYcZc.

110

The new coordinate system CYcZc is moved in the centroid C of the cross-section. The

moment of inertia, Izc, of the entire cross-section calculated against the axis CZc is

obtained:

yC_web yC−:= → yC_web 1.94− cm=

yC_f_shweb

2

tf_s2

+

yC−:= →−

+⇒ 94.1

25.1

23.17

yC_f_s 7.46cm=

yC_f_ihweb

2

tf_i2

+

− yC−:= →−

+−⇒ 94.1

22.1

23.17

yC_f_i 11.19− cm=

Iz_Cbf_s tf_s

3⋅

12Af_s yC_f_s

2⋅+

tweb hweb3⋅

12Aweb yC_web

2⋅+

+

bf_i tf_i3⋅

12Af_i yC_f_i

2⋅+

+

...:=

→

−++

−++

+⇒ 2

32

32

3

)19.11(*1212

2.1*10)94.1(*3.1712

3.17*146.7*5.2212

5.1*15

Iz_C 3.257 103× cm4=

The sectional modulus of the cross- section is obtained as:

ysup yC_f_stf_s

2+:= →+⇒

25.146.7 ysup 8.21cm=

- distance from NA to the upper

edge of the top flange

yinf yC_f_itf_i2

−:= →−−⇒22.119.11 yinf 11.79− cm=


lower edge of the bottom

flange

WsupIz_Cysup

:= →⇒21.8

10*257.3 3

Wsup 396.719cm3=

WinfIz_Cyinf

:= →−

⇒79.1110*257.3 3

Winf 276.247cm3=


-sectional modulus

111

B.2 Calculation of the Normal Stress Distribution ( xσ Diagram)

The stress distribution is obtained using the following formula (Navier’s Formula).

The maximum values of the normal stresses, tx max__σ and cx max__σ , corresponding to

the upper edge of the top flange and the lower edge of the bottom flange, respectively,

are calculated:

σx_max_tM0Iz_C

− yinf⋅:= →−−⇒ −−

)*10*79.11(**10*10*257.3

**10*55 2483

3

mm

mN

σx_max_t 1.991 108× Pa=

σx_max_cM0Iz_C

− ysup⋅:= →−⇒ −−

mm

mN *10*21.8**10*10*257.3

**10*55 2483

3

σx_max_c 1.386− 108× Pa=

The corresponding strains are obtained:

ε x_max_tσx_max_t

E:= →⇒

PaPa

9

8

10*20010*991.1

ε x_max_t 9.955 10 4−×=

ε x_max_cσx_max_c

E:= →−⇒

PaPa

9

8

10*20010*386.1

ε x_max_c 6.932− 10 4−×=

The stress and strain distributions are shown in Figure 7.2.4.b.

Figure 7.2.4.b

112

B.3 Calculation of Shear Stress Distribution (τ Diagram)

The shear flow and stress diagram is calculated employing the Jurawski’s Formula.

The cross-section is divided, as shown in Figure 7.2.4.c, into four distinct rectangular

areas: A1 through A4. Each area is delineated by two cuts (A1:a-b or a’-b’, A2:c-d or

c’-d’, A3:e-f and A4:g-f or g’-f’). The variation of the corresponding shear stress is

calculated on each of the four areas.

Figure 7.2.4.c

B.3.a Upper Flange ( xzτ shear stress)

The horizontal cut 1-1 is made, as shown in Figure 7.2.4.d, at distance s1 measured

from a and increasing towards b.

Figure 7.2.4.d

113

Theoretical rational:

0_ >xzM → 0>xF (area located above the NA is in compression).

0_ >xyV → 0*__ >∆=∆ xVM Syxz → 0___ >∆+=∆+ xzxzxxz MMM → 0<∆+ xxF

xzxxz MM __ >∆+ → xxx FF >∆+ → 0>∆H

0>∆H → 0_ >avrzxq → 0_ >avrzxτ → 0_ <avrxzτ

Numerical calculations:

tab tf_s:= → tab 1.5cm= - thickness of the cut at s1

SZc_ab s1( ) s1 tf_s⋅ ysuptf_s2

−

⋅:= - static moment about the neutral axis

qzx_ab s1( )V0 SZc_ab s1( )⋅

Iz_C:= - shear flow on the cut area

τzx_ab_avg s1( )qzx_ab s1( )

tab:= - shear stress on the cut area

The above functions are particularized for the two limiting cuts of the segment ab.

Only two values are sufficient because the variation of the shear stress )1(szxτ is

linear.

• cut a

s1 0 cm⋅:=

SZc_ab s1( ) 0cm3=

qzx_ab s1( ) 0kNm

=

τzx_ab_avg s1( ) 0MPa=

τxz_ab_avg s1( ) τzx_ab_avg s1( )−:= → τxz_ab_avg s1( ) 0MPa=

- shear stress at the face of the flange

114

• cut b

s1bf_s

2

tweb2

−:= →−⇒21

215

s1 7cm=

SZc_ab s1( ) 78.328cm3=

qzx_ab s1( ) 24.049kNm

=

τzx_ab_avg s1( ) 1.603MPa=

τxz_ab_avg s1( ) τzx_ab_avg s1( )−:= → τxz_ab_avg s1( ) 1.603− MPa=

- shear stress at the face of the web

The above calculations refer only to the left area of the upper flange. Due to the

symmetry exhibited by the upper flange against the CYc axis, the shear stress )1(sxzτ

calculated for the left side of the upper flange is symmetric with the shear stress on the

right side.

B.3.b Lower Flange ( xzτ shear stress)

The calculation of the shear stress )3(sxzτ corresponding to the lower flange is

conducted in a similar manner with the calculation developed in section B.3.a. The

horizontal cut 3-3 is made, as shown in Figure 7.2.4.e, at distance s3 measured from g

and increasing towards f.


0_ >xzM → 0<xF (area below the NA is in tension)

0_ >xyV → 0*__ >∆=∆ xVM xySz → 0___ >∆+=∆+ xzxzxxz MMM → 0>∆+ xxF

xzxxz MM __ >∆+ → xxx FF >∆+ → 0<∆H

0<∆H → 0_ <avrzxq → 0_ <avrzxτ → 0_ >avrxzτ


tgf tf_i:= → tgf 1.2cm= - thickness of the cut at s3

115

SZc_gf s3( ) s3 tf_i⋅ yinftf_i2

+


qzx_gf s3( )V0 SZc_gf s3( )⋅


τzx_gf_avgs3( )qzx_gf s3( )

tgf:= - shear stress on the cut area

Figure 7.2.4.e

The above functions are particularized for the two limiting cuts of the segment gf. Only

two values are sufficient because the variation of the shear stress )3(szxτ is linear.

• cut g

s3 0 cm⋅:=

SZc_gf s3( ) 0cm3=

qzx_gf s3( ) 0kNm

=

τzx_gf_avgs3( ) 0MPa=

τxz_gf_avgs3( ) τzx_gf_avgs3( )−:= → τxz_gf_avgs3( ) 0MPa=

- shear stress at the face of the flange

116

• cut f

s3bf_i2

tweb2

−:= →−⇒21

210

s3 4.5cm=

SZc_gf s3( ) 60.427− cm3=

qzx_gf s3( ) 18.553−kNm

=

τzx_gf_avgs3( ) 1.546− MPa=

τxz_gf_avgs3( ) τzx_gf_avgs3( )−:= → τxz_gf_avgs3( ) 1.546MPa=

- shear stress at the face of the web

The above calculations refer only to the left area of the upper flange. Due to the

symmetry exhibited by the upper flange against the CYc axis, the shear stress )3(sxzτ

calculated for the left side of the lower flange is symmetric with the shear stress on the

right side.

B.3.c Web ( xyτ shear stress)

The calculation of the shear stress )2(sxyτ on the area of the web located above the

neutral axis, area A2, is conducted by considering the equilibrium of the infinitesimal

three-dimensional body as shown in Figure 7.2.4.f. The horizontal cut 2-2 is made at

distance s2 measured from c and increasing towards d.


0_ >xzM → 0>xF (area above the NA is in compression).

0_ >xyV → 0*__ >∆=∆ xVM xyxz → 0___ >∆+=∆+ xzxzxxz MMM → 0<∆+ xxF

xzxxz MM __ >∆+ → xxx FF >∆+ → 0>∆H

0>∆H → 0_ >avryxq → 0_ >avryxτ → 0_ <avrxyτ


tcd tweb:= → tcd 1cm= - thickness of the cut at s3

117

SZc_cd s2( ) bf_s tf_s⋅ ysuptf_s2

−

⋅ s2 tweb⋅ ysup tf_s−

s22

−

⋅+:=

- static moment about the neutral axis

qyx_cd s2( )V0 SZc_cd s2( )⋅


τyx_cd_avg s2( )qyx_cd s2( )

tcd:= - shear stress on the cut area

Figure 7.2.4.f

The variation of the shear stress is a second- order parabola. Only two values of the

shear stress are necessary and they are calculated at the location of the junction

between the upper flange and the web and at the neutral axis location.

• cut c

s2 0 cm⋅:=

SZc_cd s2( ) 167.847cm3=

qyx_cd s2( ) 51.534kNm

=

τyx_cd_avg s2( ) 5.153MPa=

τxy_cd_avg s2( ) τyx_cd_avg s2( )−:= → τxy_cd_avg s2( ) 5.153− MPa=

118

- shear stress at the junction between upper flange and web

• cut d

s2 ysup tf_s−:= → s2 6.71cm=

SZc_cd s2( ) 190.358cm3=

qyx_cd s2( ) 58.446kNm

=

τyx_cd_avg s2( ) 5.845MPa=

τxy_cd_avg s2( ) τyx_cd_avg s2( )−:= → τxy_cd_avg s2( ) 5.845− MPa=

- shear stress at the neutral axis

The variation of the shear stress )4(sxyτ on the area below the neutral axis, shown in

Figure 7.2.4.g, is parametrically calculated using the following rational:


0_ >xzM → 0<xF (area below the NA is in tension).

0_ >xyV → 0*__ >∆=∆ xVM xyxz → 0___ >∆+=∆+ xzxzxxz MMM → 0>∆+ xxF

xzxxz MM __ >∆+ → xxx FF >∆+ → 0<∆H

0<∆H → 0_ <avryxq → 0_ <avryxτ → 0_ <avrxyτ

Figure 7.2.4.g

119


ted tweb:= → ted 1cm= - thickness of the cut at s3

SZc_ed s4( ) bf_i tf_i⋅ yinftf_i2

+

⋅ s4 tweb⋅ yinf tf_i+

s42

+

⋅+:=


qyx_ed s4( )V0 SZc_ed s4( )⋅


τyx_ed_avg s4( )qyx_ed s4( )

ted:= - shear stress on the cut area

The variation of the shear stress is a second-order parabola. Only two values of the

shear stress are necessary to be calculated and they are expressed at the location of the

junction between the lower flange and the web and at the neutral axis location.

• cut e

s4 0 cm⋅:=

SZc_ed s4( ) 134.282− cm3=

qyx_ed s4( ) 41.229−kNm

=

τyx_ed_avg s4( ) 4.123− MPa=

τxy_ed_avg s4( ) τyx_ed_avg s4( ):= → τxy_ed_avg s4( ) 4.123− MPa=

- shear stress at the junction between lower flange and web

• cut d

s4 yinf tf_i+:= → s4 10.59cm=

SZc_ed s4( ) 190.358− cm3=

qyx_ed s4( ) 58.446−kNm

=

τyx_ed_avg s4( ) 5.845− MPa=

120

τxy_ed_avg s4( ) τyx_ed_avg s4( ):= → τxy_ed_avg s4( ) 5.845− MPa=

- shear stress at the neutral axis

The shear stress distribution on the entire cross-section, calculated above, is shown in

Figure 7.2.4.h.

Figure 7.2.4.h

B.4 Verification of the Cross-Section

B.4.a Verification of the Normal Stress

The verification of the cross-section is conducted considering the following formula:

σx_max max σx_max_c σx_max_t,( ) σall≤:=

)10*991.1,10*386.1max( 88 PaPa−⇒ MPaMPa 21010*991.1 2 <⇒

if max σx_max_c σx_max_t,( ) σall≤ "OK", "NG",( ) "OK"=

B.4.b Verification of the Maximum Shear Stress

The verification of the cross-section is conducted considering the following formula:

τxy_max τall≤

where:

τxy_max τxy_ed_avg s4( ):= → τxy_max 5.845 106× Pa= MPa120<

- maximum shear stress on the web

121

if τxy_max τall≤ "OK", "NG",( ) "OK"=

B.5 The Web Shear Force

The shear force carried by the web is obtained by integrating on the web thickness the

xyτ shear stress diagram pictured in Figure 7.2.4.h:

Vweb τxy_min_u ysup tf_s−( )⋅ 23

τxy_max τxy_min_u−( )⋅ ysup tf_s−( )

+

...

τxy_min_l yinf tf_i−( )⋅ +

...

23

τxy_max τxy_min_l−( )⋅ yinf tf_i−( )

+

...

tweb⋅:=

( )[ ]( ) ( )

( )[ ]( ) ( )

→

−−−+

−−+

−−+

−

⇒ −

−−

−−

−−

−−

m

mPa

mPa

mPa

mPa

2

2266

226

2266

226

10*1*

10*2.110*79.11*10*123.410*845.532

10*2.110*79.11*10*123.4

10*5.110*21.8*10*153.510*845.532

10*5.110*21.8*10*153.5

Vweb 9.349 103× N= - the shear force in the web of the cross- section

where:

τxy_min_u τxy_cd_avg 0 cm⋅( ):= → τxy_min_u 5.153− MPa=

τxy_min_l τxy_ed_avg 0 cm⋅( ):= → τxy_min_l 4.123MPa=

The ratio between the web shear force and the total shear force acting on the cross-

section is:

→= 4

3

0

_

1010*349.9

VV lweb

VwebV0

0.935=

The ratio indicates that the web transfers 93% of the total shear force acting on the

cross-section. For this reason the codes employed for the calculations of steel

structures are imposing that the entire shear force is transferred only to the web.

122

Then, the average shear stress in the web is:

τav_webV0

hsect tweb⋅:= →=

−− mmN

22

4

10*1*10*2010

τav_web 5 106× Pa=

Problem 7.2.5 The cross-section shown in Figure 7.2.5 (all dimensions in mm) is

subjected to non-uniform bending condition: bending moment Mo>0 and shear force

Vo=50 kN. Calculate: (a) the bending capacity, (b) the normal stress distribution on the

cross- section, (c) the maximum shear stress and verify the cross-section, and (d) the

shear flow in the welds considering MPaall 210=σ and MPaall 120=τ .

Figure 7.2.5



Jourawski’s Formulae are applicable. The beam is made of steel and

consequently, the allowable bending stress has an equal value for both tensile and

compressive normal stress.

The shear stress induced by the existence of the shear force Vo is calculated using

the observation that for a thin- wall closed cross- section the shear flow is null in

the axis of symmetry of the cross- section.

123


• cross- sectional resultants

V0 50 kN⋅:= - shear force

• material data

σall 210MPa:= - allowable normal tensile and compressive stress

τall 120MPa:= - allowable shear stress

E 200GPa:= - modulus of elasticity


brs 200 mm⋅:= - tubular cross- section width

hrs 300 mm⋅:= - tubular cross- section height

trs 10 mm⋅:= - tubular cross- section thickness

fLs 80 mm⋅:= - equal legs L shape width

tLs 10 mm⋅:= - L shape thickness

dLs 140 mm⋅:= - position of the L shape (see Figure 7.2.5)

B. Calculations

B.1 Calculation of the Cross- Section Geometrical Characteristics

The cross-section consists of three individual areas: a tubular cross-section and two

equal L shaped legs. Their corresponding geometrical characteristics are calculated as

shown below.

B.1.1 Calculation of the L Shape Geometrical Characteristics (see fig. 7.2.5.a)

AL_1 fLs tLs−( ) tLs⋅:= ( ) →−⇒ 1*18 AL_1 7cm2=

AL_2 fLs tLs⋅:= →⇒ 1*8 AL_2 8cm2=

124

ALs AL_1 AL_2+:= →+⇒ 87 ALs 15cm2= - area of the L shape

Figure 7.2.5.a

yC_l_lfLs2

tLs2

−:= →−⇒21

28 yC_l_l 3.5cm= - distance from the point O to the

upper flange centroid.

yC_l_2 0 cm⋅:= - distance from the point O to the lateral flange centroid.

yC_lAL_1 yC_l_l⋅ AL_2 yC_l_2⋅+

ALs:= →

+⇒

150*85.3*7

yC_l 1.633cm=

yC_l_l yC_l_l yC_l−:= →−⇒ 633.15.3 yC_l_l 1.867cm= - distance from the point

C to the upper flange centroid.

yC_l_2 yC_l_2 yC_l−:= →−⇒ 633.10 yC_l_2 1.633− cm= - distance from the point

C to the upper flange centroid.

Iz_LtLs

3 fLs tLs−( )⋅

12AL_1 yC_l_l

2⋅+

tLs fLs3⋅

12AL_2 yC_l_2

2⋅+

+:=

( ) ( ) →

−+−

+

−⇒ 2

32

3

633.1*812

8*1867.1*712

18*1Iz_L 88.983cm4=

-

moment of inertia of the L shape computed with respect to its local centroid

e1 yC_l_ltLs2

+:= →+⇒21867.1 e1 2.367cm= - eccentricity of the L shape

e2 fLs e1−:= →−⇒ 367.28 e2 5.633cm=

125

B.1.2 Calculation of the Tubular Cross- Section Geometrical Characteristics (see fig.

7.2.5.b)

Figure 7.2.5.b

Aext hrs brs⋅:= →⇒ 20*30 Aext 600cm2= - area of the exterior rectangle

Aint hrs 2trs−( ) brs 2trs−( )⋅:= ( )( ) →−−⇒ 1*2201*230 Aint 504cm2= - area of the

exterior rectangle

Ars Aext Aint−:= →−⇒ 504600 Ars 96cm2= - area of the tubular cross- section

Iz_rshrs

3 brs⋅

12

hrs 2trs−( )3 brs 2trs−( )⋅

12−:=

( ) ( )→

−−−⇒

121*220*1*230

1220*30 33

Iz_rs 12072cm4=

- moment of inertia of the tubular cross- section

B.1.3 Calculation of the Cross- Section Geometrical Characteristics (see fig. 7.2.5.c)

Atot 2 ALs⋅ Ars+:= →+⇒ 9615*2 Atot 126cm2= - total area

The initial coordinate system OYcZ passes through the centroid of the web. The

distances from point O to the lower and upper flange centroids are obtained:

yO_rs 0cm:=

126

yO_L dLsfLs2

− yC_l+:= →+−⇒ 633.12814 yO_L 11.633cm= - distance from the

point O to the center of the L

shape

yC2 ALs⋅ yO_L⋅ Ars yO_rs⋅+

Atot:= →

+⇒

1260*96633.11*15*2 yC 2.77cm= -

centroid of the cross- section


system CYcZc.

Figure 7.2.5.c

The new coordinate system CYcZc is moved in the centroid C of the cross- section. The

moment of inertia, Izc, of the entire cross-section calculated against the axis CZc is

obtained:

yC_rs yO_rs yC−:= →−⇒ 77.20 yC_rs 2.77− cm= - distance from the point C to

the centroid of the tubular cross- section

yC_L yO_L yC−:= →−⇒ 77.2633.11 yC_L 8.863cm= - distance from the point C

to the centroid of the L shape

Iz_C 2 Iz_L ALs yC_L2⋅+( )⋅ Iz_rs Ars yC_rs

2⋅+( )+:=

( ) ( ) →−+++⇒ 22 )77.2(*9612072863.8*15983.882 Iz_C 15343.325cm4=

127

The sectional modulus of the cross-section is obtained as:

ysuphrs2

yC−:= →−⇒ 77.22

30ysup 12.23cm=

- distance from NA to the upper

edge of cross- section

yinfhrs2

yC+

−:= →

+−⇒ 77.2

230 yinf 17.77− cm=


lower edge of the cross- section

WsupIz_Cysup

:= →⇒23.12325.15343

Wsup 1254.548cm3=

WinfIz_Cyinf

:= →−

⇒77.17325.15343

Winf 863.448cm3=


-sectional modulus

B.2 Calculation of Bending Capacity and Normal Stress Diagram (see fig. 7.2.5.d)

The capable moment is obtained:

Mz_cap W σall⋅:= →⇒ −

mNm *10*210**10*448.863 66 Mz_cap 181.324kN m⋅=

The stress distribution is obtained using the Navier’s Formula:

Figure 7.2.5.d

128

The maximum values of the normal stresses, supmax__xσ and infmax__xσ , corresponding

to the upper edge of the top flange and the lower edge of the bottom flange,

respectively, are calculated:

σx_max_supMz_cap−

Iz_Cysup⋅:= →−⇒ −

− mmmN *10*23.12*

*10*325.15343**10*324.181 2

48

3

σx_max_sup 144.533− MPa=

σx_max_infMz_cap−

Iz_Cyinf⋅:= ( ) →−−⇒ −

− mmmN *10*77.17*

*10*325.15343**10*324.181 2

48

3

σx_max_inf 210MPa=

B.3 Calculation of Maximum Shear Stress and Verification

The maximum shear stress on the cross-section is calculated employing the Jurawski’s

Formula. Due to the vertical symmetry of the cross-section, symmetry against the CYc

axis, the shear flow at the symmetry axis CYc locations is zero. Consequently, the

calculation of the shear flow is conducted only on the left half of the cross-section. The

maximum static moment used in the Jurawski’s Formula represents the static moment

of the half-area located above the neutral axis (NA) and is calculated as:

Sz_C_max ALs yC_L⋅brs2

trs⋅ ysuptrs2

−

⋅+ ysup trs−( ) trs⋅ysup trs−( )

2⋅+:=

Sz_C_max 313.312cm3=

The maximum shear stress located at the neutral axis position is obtained:

τxy_maxV0 Sz_C_max⋅

Iz_C trs⋅:= →⇒ −

−

mmmN*01.0**10*325.15343

10*312.313**10*5048

363

τxy_max 10.21MPa=

The strength verification:

if τxy_max τall≤ "OK", "NoGood",( ) "OK"=

129

B.4 Shear Flows in the Welds(see Figure 7.2.5.e )

The total shear in the L shape is calculated as:

qL_maxV0 Sz_C_L⋅

Iz_C:= →−⇒ −

−

48

363

*10*325.15343*10*952.132**10*50

mmN

qL_max 43.326

kNm

=

where Sz_C_L ALs yC_L⋅:= →⇒ 863.8*15 Sz_C_L 132.952cm3=

Figure 7.2.5.e

To calculate the shear flow in the welds two equilibrium equations are used:

x

F∑ 0 - projections on x axis

CL

Mz∑ 0 - moment about z axis passing through the L shape centroid

Explicitly the two equations, containing the shear flow in the welds as unknown

quantities are written as:

qweld_1 qweld_2+ qL_max

qweld_1 e1⋅ qweld_2 e2⋅− 0

130

Find qweld_1 qweld_2,( )589810001933259

Nmm

⋅

34900027229

Nmm

⋅

→ float 3,

30.5N

mm⋅

12.8N

mm⋅

→

The solutions, representing the shear flow in each weld, are:

qweld_1 30.5kNm

:=

qweld_2 12.8kNm

:=

Problem 7.2.6 The longitudinal view and the cross-section of the structural steel bridge

beam are shown in Figure 7.2.6. The upper and lower bolts are of 10 mm and 12 mm

diameter, respectively, while the steel is characterized by the following allowable

stresses: MPaall 210=σ and MPaall 126=τ . Calculate: (a) the geometrical

characteristics of the cross-section with and without the bolts existence, (b) the capable

bending moment and shear force of the cross-section with and without considering the

bolt existence, (c) the necessary spacing between the bolts ( MPabearall 336_ =σ and

MPashearall 126_ =τ ) and (d) if the bolted connection is replaced by a welded

connection, the effective thickness of the welds ( MPaweldall 147_ =τ ). For the

calculations required at point (d) consider a shear force equal to half of the capable

shear force of the cross-section.

Figure 7.2.6

131



Jourawski’s Formulas are applicable. The beam is made of steel and

consequently, the allowable bending stress has an equal value for both tensile

and compressive normal stress. Usually the existence of the bolts, which are

reducing the effective area of the cross-section, is neglected when the cross-

sectional geometrical properties are calculated. The error introduced by

neglecting the bolt existence is negligible and is covered by the safety factor

used in determining the allowable stresses of the material.

The calculation required at (a) evaluates the geometrical characteristics for both

cases: a full cross-section and a reduced cross-section. A comparison is made.

In engineering practice this dilemma is solved by placing the full-section in the

cross-section of maximum moment. As it is concluded in a previous problem

the shear force is primarily carried by the web, area little affected by the

reduction.


• material data

σall 210 MPa⋅:= - allowable normal tensile and compressive stress

τall 126MPa= - allowable shear stress


tf_s_1 1.2 cm⋅:= - thickness of the first part of the upper flange

bf_s_1 16 cm⋅:= -width of the first part of the upper flange

tf_s_2 1.5 cm⋅:= - thickness of the second part of the upper flange

bf_s_2 20 cm⋅:= -width of the second part of the upper flange

tweb 1 cm⋅:= - web thickness

132

hweb 65 cm⋅:= - web height

tf_i 2 cm⋅:= - thickness of the first part of the lower flange

bf_i 15 cm⋅:= - web of the first part of the lower flange

bangle 6 cm⋅:= - length of the angle

tangle 6 mm⋅:= - thickness of the angle

eccangle 1.69 cm⋅:= - eccentricity of the angle

Aangle 6.91 cm2⋅:= - area of the angle

Iz_angle 22.8 cm4⋅:= - moment of inertia of the angle

• bolts

τall_shear 126MPa= - allowable shear stress

σall_bear 336MPa= - allowable bearing stress

nbolt_sup 2:= - number of bolts used to connect the parts of the upper flange

φbolt_sup 10 mm⋅:= - diameter of the upper bolts

hbolt_sup tf_s_1 tf_s_2+:= → hbolt_sup 2.7cm= - length of the upper bolts

nbolt_inf 1:= - number of bolts used to connect the parts of the lower flange

φbolt_inf 12 mm⋅:= - diameter of the lower bolts

hbolt_inf 2 tangle⋅ tweb+:= → hbolt_inf 2.2cm= - length of the upper bolts

posbolt_inf 30 mm⋅:= - position of the lower bolts

• weld

τall_weld 147MPa= - allowable shear stress

133

B. Calculations

B.1 Calculation of the Cross- Section Geometrical Characteristics

Note: the calculations are conducted using cm as unit.

B.1.a Full Cross-section (see Figure 7.2.6.a)

The area of the component parts and the total area of the cross-section, calculated

under the assumption that the bolt holes are neglected, are:

Af_s_1 bf_s_1 tf_s_1⋅:= →⇒ 2.1*16 Af_s_1 19.2cm2=

- area of the first part of the upper flange

Af_s_2 bf_s_2 tf_s_2⋅:= →⇒ 5.1*20 Af_s_2 30cm2=

- area of the second part of the upper flange

Aweb hweb tweb⋅:= →⇒ 1*65 Aweb 65cm2= - area of the web

Aangle 6.91cm2= - area of the angle

Af_i bf_i tf_i⋅:= →⇒ 2*15 Af_i 30cm2= - area of the first part of the lower flange

Atotal Af_s_1 Af_s_2+ Aweb+ 2 Aangle⋅+ Af_i+:=

→++++⇒ 3091.6*265302.19 Atotal 158.02cm2= - total area

The cross-section is symmetric against the vertical axis OYc and, consequently, only

the vertical position of the centroid is necessary to be calculated. The origin of the

coordinate system OYcZ coincides with the centroid of the web. The distances from the

origin O to the centriods of the parts are:

yC_f_s_1hweb

2tf_s_2+

tf_s_12

+:= →++⇒22.15.1

265

yC_f_s_1 34.6cm=

yC_f_s_2hweb

2

tf_s_22

+:= →+⇒25.1

265

yC_f_s_2 33.25cm=

yC_anglehweb

2eccangle−

−:= →−−⇒ )69.1

265( yC_angle 30.81− cm=

134

yC_f_ihweb

2

tf_i2

+

−:= →+−⇒ )

22

265( yC_f_i 33.5− cm=

Calculation of the static moment of inertia about the axis CZ:

Sz Af_s_1 yC_f_s_1⋅ Af_s_2 yC_f_s_2⋅+ Aweb yC_web⋅+

Af_i yC_f_i⋅ 2 Aangle⋅ yC_angle⋅++

...:=

→−+−+++⇒ )5.33(*30)81.30(*91.6*20*6525.33*306.34*2.19

Sz 231.026cm3= - static moment

yCSz

Atotal:= →⇒

02.158026.231 yC 1.462cm=

- centroid position

The coordinate system is moved in the centroid C (the new coordinate system CYcZc)

and the distances to the centroid of the parts are accordingly adjusted:

yC_web yC−:= → yC_web 1.462− cm=

yC_f_s_1 yC_f_s_1 yC−:= →−⇒ 462.16.34 yC_f_s_1 33.138cm=

yC_f_s_2 yC_f_s_2 yC−:= →−⇒ 462.125.33 yC_f_s_2 31.788cm=

yC_angle yC_angle yC−:= →−−⇒ 462.181.30 yC_angle 32.272− cm=

yC_f_i yC_f_i yC−:= →−−⇒ 462.15.33 yC_f_i 34.962− cm=

The centroidal moment of inertia about the axis CZc is obtained:

Iz_Cbf_s_1 tf_s_1

3⋅

12Af_s_1 yC_f_s_1

2⋅+

bf_s_2 tf_s_2

3⋅

12Af_s_2 yC_f_s_2

2⋅+

+

...

tweb hweb3⋅

12Aweb yC_web

2⋅+

+

...

2 Iz_angle Aangle yC_angle2⋅+( )⋅+

...

bf_i tf_i3⋅

12Af_i yC_f_i

2⋅+

+

...

:=

135

→−++−++

++++++⇒

))962.34(*3012

2*15())272.32(*91.68.22(*2

)462.1*651265*1()788.33*30

125.1*20()138.33*2.19

122.1*16(

23

2

23

23

23

Iz_C 1.255 105× cm4=

The section modulus considering the distances measured from the centroid to the

extreme fibers of the cross-section:

ysup yC_f_s_1tf_s_1

2+:= →+⇒

22.1138.33 ysup 33.738cm=

yinf yC_f_itf_i2

−:= →−−⇒22962.34 yinf 35.962− cm=

WsupIz_Cysup

:= →⇒738.33

10*255.1 5

Wsup 3.721 103× cm3=

WinfIz_Cyinf

:= →⇒738.33

10*255.1 5

Winf 3.491 103× cm3=

W min Wsup Winf,( ):= → W 3.491 103× cm3=

B.1.b Effective Cross-section (see Figure 7.2.6.b)

The calculation is repeated considering the missing areas of the holes of the bolts. The

position of the holes is measured in the coordinate system OYcZ:

yC_bolt_suphweb

2

hbolt_sup2

+:= →+⇒27.2

265 yC_bolt_sup 33.85cm=

yC_bolt_infhweb

2posbolt_inf−

−:= →−−⇒ )3

265( yC_bolt_inf 29.5− cm=

The corresponding hole areas are calculated as:

Abolt_sup φbolt_sup hbolt_sup⋅:= →⇒ 7.2*1 Abolt_sup 2.7cm2= - upper bolt

Abolt_inf φbolt_inf hbolt_inf⋅:= →⇒ 2.2*2.1 Abolt_inf 2.64cm2= - lower bolt

136

Atotal_nett Atotal nbolt_sup Abolt_sup⋅− nbolt_inf Abolt_inf⋅−:=

→−−⇒ 64.27.2*202.158 Atotal_nett 149.98cm2=

Note: A reduction area by 5% is calculated due the bolt existence.

The centroid of the effective area is obtained:

Sz_nett Sz nbolt_sup Abolt_sup−( )⋅ yC_bolt_sup⋅+ nbolt_inf Abolt_inf−( )⋅ yC_bolt_inf⋅+:=

→−−−⇒ )5.29(*64.2)85.33(*7.2*2026.231 Sz_nett 126.116cm3=

yC_nettSz_nett

Atotal_nett:= →⇒

98.14918.126 yC_nett 0.841cm=

- position of the centroid of the effective cross-section

The new coordinate system CYcZ is located in the centroid of the effective cross-

section. After correcting the centroidal distances measured from the centroid of the

effective cross-section to the centroid of the component areas, the centroidal moment

of inertia of the effective cross-section against the axis CZc is obtained as:

yC_web_n yC_nett−:= → yC_web_n 0.841− cm=

yC_f_s_1_n yC_f_s_1_0 yC_nett−:= →−⇒ 841.06.34 yC_f_s_1_n 33.759cm=

yC_f_s_2_n yC_f_s_2 yC_nett−:= →−⇒ 841.025.33 yC_f_s_2_n 32.409cm=

yC_angle_n yC_angle yC_nett−:= →−−⇒ 841.081.30 yC_angle_n 31.651− cm=

yC_f_i_n yC_f_i yC_nett−:= →−−⇒ 841.05.33 yC_f_i_n 34.341− cm=

yC_bolt_sup yC_bolt_sup_0 yC_nett−:= →−⇒ 841.06.34 yC_bolt_sup 33.009cm=

yC_bolt_inf yC_bolt_inf_0 yC_nett−:= →−⇒ 841.06.34 yC_bolt_inf 30.341− cm=

137

Iz_C_nettbf_s_1 tf_s_1

3⋅

12Af_s_1 yC_f_s_1_n

2⋅+

bf_s_2 tf_s_2

3⋅

12Af_s_2 yC_f_s_2_n

2⋅+

+

...

tweb hweb3⋅

12Aweb yC_web_n

2⋅+

+

...

2 Iz_angle Aangle yC_angle_n2⋅+( )⋅+

...

bf_i tf_i3⋅

12Af_i yC_f_i_n

2⋅+

+

...

2φbolt_sup hbolt_sup

3⋅

12− Abolt_inf yC_bolt_sup

2⋅−

⋅+

...

φbolt_inf3

hbolt_inf⋅

12− Abolt_inf yC_bolt_inf

2⋅−

+

...

:=

→−−−+

+−−+−++−++

+−+++++⇒

))341.30(*7.212

2.1*2.2(*2

)009.33*7.212

7.2*1(*2))341.34(*3012

2*15())651.31(*91.68.22(*2

))841.0(*651265*1()409.32*30

125.1*20()759.33*2.19

122.1*16(

23

23

23

2

23

23

23

Iz_C_nett 1.174 105× cm4=

Note: The ratio between the moments of inertia calculated is indicating a reduction

with 6.5% when the bolts are considered. In engineering practice the bolts are not all

staggered in the same cross-section.

Iz_C_nett

Iz_C0.935=

The effective section modulus is calculated as:

ysup_n yC_f_s_1_ntf_s_1

2+:= →+⇒

22.1759.33 ysup_n 34.359cm=

yinf_n yC_f_i_ntf_i2

−:= →−−⇒22341.34 yinf_n 35.341− cm=

138

Wsup_nettIz_C_nettysup_n

:= →⇒359.34

10*174.1 5

Wsup_nett 3.418 103× cm3=

Winf_nettIz_C_nett

yinf:= →⇒

341.3510*174.1 5

Winf_nett 3.265 103× cm3=

Wnett min Wsup_nett Winf_nett,( ):= → Wnett 3.265 103× cm3=

B.2 Calculation of Bending Capacity of the Cross-Section

B.2.a Full Cross-section

The capable bending moment of the full cross-section is obtained as:

Mcap σall W⋅:= →⇒ −− 33636 10*10*10*491.3*10*210 mPa

Mcap 733.147kN m⋅=

B.2.b Effective Cross-section

The capable bending moment of the effective cross-section is calculated in a similar

manner:

Mcap_nett σall Wnett⋅:= →⇒ −− 33636 10*10*10*265.3*10*210 mPa

Mcap_nett 685.695kN m⋅=

The ratio of the capable bending moments is:

Mcap_nettMcap

0.935=

Note: As expected, a reduction in the bending capacity of the effective cross-section is

emphasized. This reduction represents a small percentage of the bending capacity

when the bolts existence is neglected. For this reason and some other consideration

139

related to the safety factor used in the calculation of the allowable normal stress in the

engineering practice the bolts are not considered and the moment of inertia of the full

cross-section is usually used.

B.3 Calculation of Shear Capacity of the Cross-Section

B.3.a Full Cross-section

Using the Jourawski’s Formula the capable shear force is calculated:

Vcaptweb Iz_C⋅

Sz_C_maxτall⋅:= →⇒ −

−

−−36

363

4852

10*10*126**10*10*072.2

10*10*255.1*10*1 Pam

mm

Vcap 763.637kN=

where the maximum static moment about the neutral axis is obtained as:

Sz_C_max Af_s_1 yC_f_s_1⋅ Af_s_2 yC_f_s_2⋅+

ysup tf_s_1− tf_s_2−( ) tweb⋅ysup tf_s_1− tf_s_2−( )

2⋅+

...:=

→−−

−−++⇒2

)5.12.1738.33(*1*)5.12.1738.33(788.31*30138.33*2.19

Sz_C_max 2.072 103× cm3=

B.3.b Effective Cross-section

The calculation of the capable shear force of the cross-section is conducted:

Vcap_netttweb Iz_C_nett⋅

Sz_C_max_nettτall⋅:=

→⇒ −−

−−36

363

4852

10*10*126**10*10*943.1

10*10*174.1*10*1 Pam

mmVcap_nett 759.786kN=

Sz_C_max_nett Af_s_1 yC_f_s_1_n⋅ Af_s_2 yC_f_s_2_n⋅+

ysup_n tf_s_1− tf_s_2−( ) tweb⋅ysup_n tf_s_1− tf_s_2−( )

2⋅+

...

2 Abolt_inf− yC_bolt_sup⋅( )⋅+

...

:=

140

→−+

+−−

−−++⇒

)009.33*7.2(*22

)5.12.1356.34(*1*)5.12.1356.34(409.32*30759.33*2.19

Sz_C_max_nett 1.943 103× cm3=

The ratio of the capable shear forces calculated considering the full section and the

section with bolts is:

Vcap_nettVcap

0.997=

Note: The reduction induced by the bolts existence is insignificant and shows that the

bolts can be excluded from the calculation involving the shear force.

B.4 Calculation of the Bolts’ Spacing

The shear force crV required to conduct the calculation is:

Vcr 0.5 Vcap⋅:= →⇒ capV*5.0 Vcr 381.819kN=

B.4.a Upper Bolts

The shear flow corresponding to the upper bolts is calculated as:

qrec_bolt_supVcr Sz_C_f_s⋅

Iz_C:= →⇒ −

−

−3

485

363

10**10*10*255.1

10*25.636*10*819.381m

mN

qrec_bolt_sup 193.495kNm

=

where the static moment is:

Sz_C_f_s Af_s_1 yC_f_s_1⋅:= →⇒ 138.33*2.19 Sz_C_f_s 636.25cm3=

The upper 10 mm bolt capacity is obtained calculating the minimum between the bolt

capacities in bearing and shearing:

141

nf 1:= - number of shearing sections of the bolt

Pbolt_shear nfπ φbolt_sup

2⋅


→⇒ −−

36242

10*10*126*410*1**1 Pamπ

Pbolt_shear 9.896kN=

Pbolt_bear φbolt_sup min tf_s_1 tf_s_2,( )⋅ σall_bear⋅:=

→⇒ −− 3624 10*10*336*10*)5.1,2.1min(*1 Pam

Pbolt_bear 40.32kN=

Pbolt_sup min Pbolt_shear Pbolt_bear,( ):= →⇒ kN)32.40,896.9min(

Pbolt_sup 9.896kN=

The spacing of the upper bolts is obtained:

ssupnbolt_sup Pbolt_sup⋅

qrec_bolt_sup:= →⇒ 210*

/495.193896.9*2

mkNkN ssup 10.229cm=

B.4.b Lower Bolts

The shear flow corresponding to the upper bolts is calculated as:

qrec_bolt_infVcr Sz_C_f_i⋅

Iz_C:= →⇒ −

−

−3

485

3633

10**10*10*255.1

10*10*495.1*10*819.381m

mN

qrec_bolt_inf 454.613

kNm

=

where the static moment is:

Sz_C_f_i Af_i yC_f_i⋅ 2 Aangle⋅ yC_angle⋅+:=

→+⇒ )272.32(*91.6*2)962.34(*30 Sz_C_f_i 1.495 103× cm3=

142

The capacity of the 12 mm bolt is obtained calculating the minimum between the bolt

bearing and shearing capacities.

:

nf 2:= - number of shearing sections of the bolt

Pbolt_shear nfπ φbolt_inf

2⋅


→⇒ −−

36242

10*10*126*4

10*2.1**2 Pamπ Pbolt_shear 28.501kN=

Pbolt_bear φbolt_inf min tweb 2tangle,( )⋅ σall_bear⋅:=

→⇒ −− 3624 10*10*336*10*)6.0*2,1min(*2.1 Pam Pbolt_bear 40.32kN=

Pbolt_inf min Pbolt_shear Pbolt_bear,( ):= →⇒ kN)32.40,501.28min(

Pbolt_inf 28.501kN=

The spacing of the lower bolts is obtained:

sinfnbolt_inf Pbolt_inf⋅

qrec_bolt_inf:= →⇒ 210*

/613.454501.28*1

mkNkN

sinf 6.269cm=

The lower and upper bolts are both placed in the same cross-section as required by the

problem’s text. Consequently, the spacing is calculated as:

s min ssup sinf,( ):= →⇒ )269.6,229.10min( sinf 6.269cm=

A rounded value is chosen to match the spacing:

s 5 cm⋅:=

The verification of the bolts is conducted for the new spacing for exercising purposes

only:

if qrec_bolt_sup s⋅ nbolt_sup Pbolt_sup⋅< "OK", "NO GOOD",( ) "OK"=

143

if qrec_bolt_inf s⋅ nbolt_inf Pbolt_inf⋅< "OK", "NO GOOD",( ) "OK"=

B.5 Sizing of the Welds

B.5.a Upper Welds

The bolted connection is replaced by a welded connection. The shear flow in the upper

welds is calculated:

qrec_welds_supVcr Sz_C_f_s⋅

Iz_C:= →⇒ −

−

−3

485

3633

10**10*10*255.1

10*10*59.1*10*819.381m

mN

qrec_welds_sup 483.513

kNm

=

where the corresponding static moment is obtained:

Sz_C_f_s Af_s_1 yC_f_s_1⋅ Af_s_2 yC_f_s_2⋅+:= →+⇒ 788.31*30138.33*2.19

S z_C_f_s 1589.889cm3=

The upper welds capacity is calculated:

nfillet_welds 2:=

Pcap_weld_sup nfillet_weldstweld_sup⋅ τall_weld⋅

The effective thickness of the upper welds is obtained from the equation:

qrec_welds_sup Pcap_weld_sup≤

and consequently,

tweld_supqrec_welds_sup

nfillet_weldsτall_weld⋅:= →⇒ 3

6

3

10*10*147*210*513.483

PaN

tweld_sup 1.645mm=

B.5.b Lower Welds

The shear flow in the lower welds is calculated:

144

qrec_welds_infVcr Sz_C_f_i⋅

Iz_C:= →⇒ −

−

−3

485

3633

10**10*10*255.1

10*10*049.1*10*819.381m

mN

qrec_welds_inf 318.977

kNm

=

where the corresponding static moment is obtained:

Sz_C_f_i Af_i yC_f_i⋅:= →⇒ 962.34*30 Sz_C_f_i 1.049 103× cm3=

The lower welds capacity is calculated:

nfillet_welds 2:=

Pcap_weld_inf nfillet_weldstweld_inf⋅ τall_weld⋅

The effective thickness of the lower welds is obtained from the equation:

qrec_welds_inf Pcap_weld_inf≤

tweld_infqrec_welds_inf

nfillet_weldsτall_weld⋅:= →⇒ 3

6

3

10*10*147*210*977.318

PaN

tweld_inf 1.085mm=

Problem 7.2.7 Considering the steel beam shown in Figure 7.2.7 conduct the following

tasks: (a) calculate and draw the pertinent cross-sectional resultant diagrams, (b)

calculate the cross-section geometrical characteristics, (c) verify the beam for strength

employing the following allowable normal MPaall 210=σ and shear allall στ *6.0=

stresses, (d) size the welds (the effective throat) using the allowable shear stress

MPaweldall 147_ =τ , and (e) calculate and draw the normal and shear stresses in cross-

section S located at 1m right of end point of the uniform distributed load.

Figure 7.2.7

145


A.1 The loads are acting in the vertical plane and, consequently, the steel beam is

subjected to non-uniform bending. Both Navier’s and Jourawski’s Formulae

are applicable. The beam is made of steel and consequently, the allowable

normal stress has an equal value for both tensile and compressive normal

stress.


• loading and dimensions

P 115 kN⋅:=

q 40kNm

⋅:=

a 3 m⋅:=

b 2 m⋅:=

c 1 m⋅:=

• material data

σall 210 MPa⋅:= - allowable normal tensile and compressive stress

τall 0.6 σall⋅:= →⇒ MPa210*6.0 τall 126MPa= - allowable shear stress


- Shape 1: Rectangular Web (300x10 mm)

hpl 30 cm⋅:= - height

tpl 1 cm⋅:= - thickness

- Shape 2: U24

hU24 24 cm⋅:= - height of the shape

tw_U24 0.95 cm⋅:= - thickness of the web

bf_U24 8.5 cm⋅:= - length of the flange

146

tf_U24 1.262 cm⋅:= - thickness of the flange

eU24 2.23 cm⋅:= - eccentricity

AU24 42.3 cm2⋅:=

- area

Iz_U24 248 cm4⋅:= - moment of inertia about an axis parallel to z axis

- Shape 3: U12

hU12 12 cm⋅:= - height of the shape

tw_U12 0.70 cm⋅:= - thickness of the web

bf_U12 5.5 cm⋅:= - length of the flange

tf_U12 0.872 cm⋅:= - thickness of the flange

eU12 1.60 cm⋅:= - eccentricity

AU12 17 cm2⋅:= - area

Iz_U12 43.2 cm4⋅:= - moment of inertia about an axis parallel to z axis

• welds

nwelds_sup 2:= - number of superior welds

nwelds_inf 2:= - number of inferior welds

τall_weld 0.7 σall⋅:= →⇒ MPa210*7.0 τall_weld 147MPa=

- allowable shear stress

B. Calculations

B.1 Cross-Sectional Internal Resultants (see Figure 7.2.7.a)

Note: The cross-sectional internal resultants are calculated and graphically represented

as detailed explained in Chapter 3 of volume I.

147

B.1.a Calculation of the Reaction Forces

The support restrictions are replaced by the corresponding reaction forces. The reaction

forces together with the loads represent the free-body diagram of the beam. Due to the

fact that the loads and reaction forces are located in the vertical plane OXY only three

(3) global equilibrium equations are written for the entire free-body diagram of the

beam. They are:

x

Fx∑ 0 - projection of the forces on the horizontal direction OX

1

Mz∑ 0 - sum of moments about direction OZ in point 1

3

Mz∑ 0 - sum of moments about direction OZ in point 3

Explicitly, the above equations are written as:

H1 0 kN⋅

V1 a b+( )⋅ q a⋅a2

b+

⋅− P c⋅+ 0 solve V1, 61000 N⋅→

q a⋅a2

⋅ V3 a b+( )⋅− P a b+ c+( )⋅+ 0 solve V3, 174000 N⋅→

To verify the calculations conducted above the equilibrium on OY axis is employed:

V1 V3+ q a⋅− P− 0kN= →=−−+⇒ 01153*4017461 the calculations are OK

B.1.a Calculation of the Shear Force Vy

The value of the shear force at the locations indicated on Figure 7.2.7.a are obtained as:

V1_left 0 kN⋅:=

V1_right V1_left V1+:= →+⇒ 610 V1_right 61kN=

V2_left V1_right q a⋅−:= →−⇒ 3*4061 V2_left 59− kN=

148

V2_right V2_left:= → V2_right 59− kN=

V3_left V2_right:= → V3_left 59− kN=

V3_right V3_left V3+:= →+−⇒ 17459 V3_right 115kN=

V4_left V3_right:= → V4_left 115kN=

V4_right V4_left P−:=

V4_right 0kN=

The shear force on the interval 12 changes the sign from positive to negative, and

consequently, a value of zero appears in a cross-section located at distance x from point

1. The expression of the shear force on the interval 12 is:

V1_right q x⋅− 0

The distance x is calculated:

xV1_right

q:= →⇒

4061 x 1.525m=

B.1.b Calculation of the Bending Moment Mz

The value of the bending moment at the locations indicated on Figure 7.2.7.a are

obtained as:

M1 0 kN⋅ m⋅:=

M12_max M112

V1_right x⋅+:= →+⇒ 525.1*61*210 M12_max 46.513kN m⋅=

M2 M112

V1_right x⋅+12

V2_left a x−( )⋅+:=

→−++⇒ )525.13(*61*21525.1*61*

210 M2 3kN m⋅=

M3 M2 V2_right b⋅+:= →−⇒ 2*593 M3 115− kN m⋅=

M4 M3 V3_right c⋅+:= →−−⇒ 1*115115 M4 0kN m⋅=

149

Figure 7.2.7.a

B.2 Calculation of the Cross- Section Geometrical Characteristics (see Figure 7.2.7.b)

Note: The calculations are conducted using cm as unit.

The cross-section is composed from three shapes: a rectangular shape 300x10mm, U24

and U12. The area of the cross-section is calculated:

150

Atotal hpl tpl⋅( ) AU24+ AU12+:= →++⇒ 0.173.421*30 Atotal 89.3cm2=

The original coordinate system is attached to point C1, the centroid of the rectangular

shape. The distances from the origin of the coordinate system to the centroid of each

individual shape are:

yC1 0 cm⋅:=

yC2hpl2

tw_U24+ eU24−:= →−+⇒ 23.295.02

30yC2 13.72cm=

yC3hpl2

eU12+

−:= →+−⇒ )60.1

230( yC3 16.6− cm=

The position of the centroid is obtained:

yChpl tpl⋅( ) yC1⋅ AU24 yC2⋅+ AU12 yC3⋅+

Atotal:=

→−++

⇒3.89

)6.16(*0.1772.13*3.420*)1*30( yC 3.339cm=

zC 0 cm⋅:= (due to the cross-section symmetry)

Figure 7.2.7.b

A new coordinate system, attached to the cross-section centroid C, is now employed.

The new positions of the centroids of each individual shape are established as:

yC1 yC1 yC−:= →−⇒ 339.30 yC1 3.339− cm=

151

yC2 yC2 yC−:= →−⇒ 339.372.13 yC2 10.381cm=

yC3 yC3 yC−:= →−−⇒ 339.36.16 yC3 19.939− cm=

The only moment of inertia necessary to be calculated is:

IZchpl

3 tpl⋅

12hpl tpl⋅( ) yC1

2⋅+

Iz_U24 AU24 yC22⋅+( )+ Iz_U12 AU12 yC3

2⋅+( )+:=

( ) [ ]→−++++

−+⇒ 222

3

)939.19(*172.43381.10*3.42248)339.3(*)1*30(12

1*30

IZc 14192.717cm4=

The distances to the extreme fibers of the cross-section are:

ysup yC2 eU24+:= →+⇒ 23.2381.10 ysup 12.611cm=

yinf yC3 eU12+ bf_U12−:= →−+−⇒ 50.560.1939.19 yinf 23.839− cm=

The sectional modulus is calculated:

WsupIZcysup

:= →⇒611.12

717.14192Wsup 1125.407cm3=

WinfIZcyinf

:= →−

⇒839.23717.14192

Winf 595.362cm3=

W min Wsup Winf,( ):= → W 595.362cm3=

B.3 Verification of the Cross-Section for Strength

B.3.a Normal Stress Verification (σx Verification)

The verification of the normal stress is conducted in the cross-section where the

maximum absolute bending moment is reached. This cross-section is identified by

surveying the bending moment diagram shown in Figure 7.2.7.a.

152

σx_max max σx_sup σx_inf,( ) σall≤

where:

Mz_max M3:= → Mz_max 115− kN m⋅=

σx_supMz_max

IZc− ysup⋅:= →

−−⇒ −

− mm

N 248

3

10*611.12*10*717.14192

10*115

σx_sup 102.185MPa=

σx_infMz_max

IZc− yinf⋅:= →−

−−⇒ −

− mm

N 248

3

10*)839.23(*10*717.14192

10*115

σx_inf 193.16− MPa=

σx_max max σx_sup σx_inf,( ):= → σx_max 193.16MPa=

The normal stress diagram in the cross-section corresponding to the maximum bending

moment is shown in Figure 7.2.7c.

Figure 7.2.7c

An alternative approach, using the sectional modulus, is the following:

σx_maxMz_max

IZc− ymax⋅ σall≤

σx_maxMz_max

Wσall≤

153

where:

ymax max ysup yinf,( ):= → ymax 23.839cm=

σx_maxMz_max

W:= →

−⇒ − 36

3

10*362.59510*115

mN

σx_max 193.16MPa=

if σx_max σall≤ "OK", "NG",( ) "OK"=

B.3.b Shear Stress Verification (τxy Verification)

The shear stress verification is conducted in the cross-section where the shear force

reaches the absolute maximum. This cross-section is identified by surveying the shear

force diagram shown in Figure 7.2.7.a. The formula used in the shear stress verification

is:

τxy_maxVy_max SZc_NA⋅

IZc tpl⋅τall≤

where:

Vy_max V3_right:= → Vy_max 115kN=

SZc_NA AU24 yC2⋅ tplhpl2

yC−

⋅

12

⋅hpl2

yC−

⋅+:=

→

−

−+⇒ 339.3

230*

21*339.3

230*1381.10*3.42 SZc_NA 507.116cm3=

- the static moment of the area located

above the neutral axis (NA)

τxy_maxVy_max SZc_NA⋅

IZc tpl⋅:= →⇒ −−

−

mmmN248

363

10*1*10*717.1419210*116.507*10*115

τxy_max 41.09MPa=

if τxy_max τall≤ "OK", "NG",( ) "OK"=

154

B.4 Sizing of the Welds

B.4.a The Upper Welds

The sizing of the upper welds is performed using the following formula:

qweld_sup 1 m⋅( )⋅ Pcap_welds_sup≤

The shear capacity of the upper welds is calculated as:

Pcap_welds_sup nwelds_sup teff_sup⋅ 1 m⋅( )⋅ τall_weld⋅

The shear flow in the upper welds is calculated:

qweld_supVy_max Szc_U24⋅

IZc:= →⇒ −

−

48

363

10*717.1419210*124.439*10*115m

mN

qweld_sup 355.811kNm

=

where:

Szc_U24 AU24 yC2⋅:= →⇒ 381.10*3.42 Szc_U24 439.124cm3=

- the static moment of the U24 about the neutral axis (NA)

The effective throat of the upper weld is the only unknown quantity and it is calculated

as:

teff_supqweld_sup

nwelds_sup τall_weld⋅:= →⇒

PamN

6

3

10*147*2/10*811.355

teff_sup 1.21mmm=

B.4.bThe Lower Welds

The sizing of the lower welds is performed using the following formula:

qweld_inf 1 m⋅( )⋅ Pcap_welds_inf≤

The shear capacity of the lower welds is calculated as:

Pcap_welds_inf nwelds_inf teff_inf⋅ 1 m⋅( )⋅ τall_weld⋅

155

The shear flow in the lower welds is calculated:

qweld_infVy_max Szc_U12⋅

IZc:= →⇒ −

−

48

363

10*717.1419210*96.338*10*115m

mN

qweld_inf 274.651kNm

=

where:

Szc_U12 AU12 yC3⋅:= →−⇒ )939.19(*0.17 Szc_U12 338.96cm3=

The effective throat of the lower weld is the only unknown quantity and is calculated

as:

teff_infqweld_inf

nwelds_inf τall_weld⋅:= →⇒

PamN

6

3

10*147*2/10*651.274

teff_inf 0.934mm=

B.5 Stress Diagrams on Section S.

The bending moment and the shear force in the cross-section S are: s 1 m⋅:=

Mz_S M2 V2_right s⋅+:= →−⇒ 1*593 Mz_S 56− kN m⋅=

Vy_S V2_right:= → Vy_S 59− kN=

B.5.a Normal Stress Diagram (σx Diagram – see Figure 7.2.7.d)

The normal stress (σx) diagram in the cross-section S is constructed in a similar manner

with the σx diagram used in the calculation of the σx_max. The variation of the normal

stress on the cross-section is linear in accordance to the Navier’s Formula and only

the values at the extreme fibers are necessary to be calculated.

σx_sup_SMz_SIZc

− ysup⋅:= →−

−⇒ −− mm

N 248

3

10*611.12*10*717.14192

10*56

σx_sup_S 49.76MPa=

156

σx_inf_SMz_SIZc

− yinf⋅:= →−−

−⇒ −− mm

N 248

3

10*)839.23(*10*717.14192

10*56

σx_inf_S 94.06− MPa=

Figure 7.2.7.d

B.5.b Shear Stress Diagram (τ Diagram)

The shear flow and stress diagrams are calculated following the explanations regarding

the application of the Jurawski’s Formula detailed in the theoretical paragraph. An

infinitesimal element of the beam starting from the cross-section S and having a length

∆x is isolated from the entire beam.

The cross-section is divided into six distinct rectangular areas, A1 through A6, as

shown in Figure 7.2.7.e, where Jurawski’s Formula is applicable.

Each area is delineated by two cuts (A1:a-b or a’-b’, A2:c-d or c’-d’, A3:e-f, A4:g-f,

A5:k-j or k’-j’, A6:i-g or i’-h’). The variation of the corresponding shear stress is

calculated on each of the six areas.

For each cut the corresponding free body diagram is drawn. The resulting forces xF

and xxF ∆+ on each face of the resulting free body are indicated function of the signs of

the bending moment and shear force acting in the cross-section S.

157

Figure 7.2.7.e

B.5.b.1 Calculation of the Shear Stress on Area A1 (see Figure 7.2.7.f)

The horizontal cut 1-1 is made at distance s1 measured from a and increasing towards

b.

Figure 7.2.7.f


0_ <SzM → 0<xF (area above the NA is in tension).

0_ <SyV → 0*__ =<∆=∆ xVM SySz → 0___ <∆+=∆+ SzSzxSz MMM → 0>∆+ xxF

SzxSz MM __ >∆+ → xxx FF >∆+ → 0<∆H


158


tab tf_U24:= → tab 1.262cm= - thickness of the cut at s1

SZc_ab s1( ) s1 tf_U24⋅ ysup bf_U24−s12

+

⋅:=


qyx_ab s1( )Vy_S SZc_ab s1( )⋅

IZc:=

- shear flow on the cut area

τyx_ab_avg s1( )qyx_ab s1( )

tab:= - shear stress on the cut area

The above functions are particularized for the two limiting cuts of the segment ab.

• cut a

s1 0 cm⋅:=

SZc_ab s1( ) 0cm3=

qyx_ab s1( ) 0kNm

=

τyx_ab_avg s1( ) 0MPa=

τxy_ab_avg s1( ) τyx_ab_avg s1( )−:= → τxy_ab_avg s1( ) 0MPa=

• cut b

s1 bf_U24 tw_U24−:=

→ s1 7.55cm=

SZc_ab s1( ) 75.14cm3=

qyx_ab s1( ) 31.236−kNm

=

τyx_ab_avg s1( ) 2.475− MPa=

τxy_ab_avg s1( ) τyx_ab_avg s1( ):= → τxy_ab_avg s1( ) 2.475− MPa=

159

The calculation for the segment a’b’ is conducted in a similar manner as for the

segment ab and the shear stress is symmetric.

B.5.b.2 Calculation of the Shear Stress on Area A2 (see Figure 7.2.7.g)

The vertical cut 2-2 is made at distance s2 measured from c and increasing towards d.



0_ <SyV → 0*__ =<∆=∆ xVM SySz → 0___ <∆+=∆+ SzSzxSz MMM → 0>∆+ xxF

SzxSz MM __ >∆+ → xxx FF >∆+ → 0<∆H

0<∆H → 0_ <avrzxq → 0_ <avrzxτ → 0_ >avrxzτ

Figure 7.2.7.g


tcd tw_U24:= → tcd 0.95cm= - thickness of the cut at s2

SZc_cd s2( ) SZc_ab bf_U24( ) s2 tw_U24⋅ ysuptw_U24

2−

⋅+:=


qzx_cd s2( )Vy_S SZc_cd s2( )⋅

IZc:= - shear flow on the cut area

160

τzx_cd_avg s2( )qzx_cd s2( )

tcd:= - shear stress on the cut area

The above functions are particularized for the two limiting cuts of the segment cd:

• cut c

s2 0 cm⋅:=

SZc_cd s2( ) 89.69cm3=

qzx_cd s2( ) 37.285−kNm

=

τzx_cd_avg s2( ) 3.925− MPa=

τxz_cd_avg s2( ) τzx_cd_avg s2( )−:= → τxz_cd_avg s2( ) 3.925MPa=

• cut d

s2hU24

2tf_U24−

tpl2

−

:= → s2 10.238cm=

SZc_cd s2( ) 207.728cm3=

qzx_cd s2( ) 86.354−kNm

=

τzx_cd_avg s2( ) 9.09− MPa=


The calculation for the segment c’d’ is conducted in a similar manner as for the

segment cd and the shear stress is symmetric.

B.5.b.3 Calculation of the Shear Stress on Area A3 (see Figure 7.2.7.h)

The horizontal cut 3-3 is made at distance s3 measured from e and increasing towards

f.

161

Figure 7.2.7.h



0_ <SyV → 0*__ =<∆=∆ xVM SySz →

0___ <∆+=∆+ SzSzxSz MMM → 0>∆+ xxF

SzxSz MM __ >∆+ → xxx FF >∆+ → 0<∆H



tef tpl:= → tef 1cm= - thickness of the cut at s3

SZc_ef s3( ) AU24 yC2⋅ s3 tpl⋅ ysup tw_U24−s32

−

⋅+:=


qyx_ef s3( )Vy_S SZc_ef s3( )⋅


τyx_ef_avgs3( )qyx_ef s3( )

tef:= - shear stress on the cut area

The above functions are particularized for the two limiting cuts of the segment ef:.

162

• cut e

s3 0 cm⋅:=

SZc_ef s3( ) 439.124cm3=

qyx_ef s3( ) 182.547−kNm

=

τyx_ef_avgs3( ) 18.255− MPa=

τxy_ef_avgs3( ) τyx_ef_avgs3( )−:= → τxy_ef_avgs3( ) 18.255MPa=

• cut f

s3 ysup tw_U24−:= → s3 11.661cm=

SZc_ef s3( ) 507.116cm3=

qyx_ef s3( ) 210.811−kNm

=

τyx_ef_avgs3( ) 21.081− MPa=


B.5.b.4 Calculation of the Shear Stress on Area A5 (see Figure 7.2.7.i)

The horizontal cut 5-5 is made at distance s5 measured from k and increasing towards

j.


0_ <SzM → 0>xF (area below the NA is in compression).

0_ <SyV → 0*__ =<∆=∆ xVM SySz → 0___ <∆+=∆+ SzSzxSz MMM → 0<∆+ xxF

SzxSz MM __ >∆+ → xxx FF >∆+ → 0>∆H

0>∆H → 0_ >avryxq → 0_ >avryxτ → 0_ >avrxyτ

163


tkj tf_U12:= → tkj 0.872cm= - thickness of the cut at s5

SZc_kj s5( ) s5 tf_U12⋅ yinfs52

+


qyx_kj s5( )Vy_S SZc_kj s5( )⋅


τyx_kj_avg s5( )qyx_kj s5( )

tkj:= - shear stress on the cut area

Figure 7.2.7.i

The above functions are particularized for the two limiting cuts of the segment kj:.

• cut k

s5 0 cm⋅:=

SZc_kj s5( ) 0cm3=

qyx_kj s5( ) 0kNm

=

τyx_kj_avg s5( ) 0 MPa=

τxy_kj_avg s5( ) τyx_kj_avg s5( ):= → τxy_kj_avg s5( ) 0 MPa=

164

• cut j

s6 bf_U12 tw_U12−:= → s6 4.8cm= - thickness of the cut at s5

SZc_kj s5( ) 89.734− cm3=

qyx_kj s5( ) 37.303kNm

=

τyx_kj_avg s5( ) 4.278 MPa=

τxy_kj_avg s5( ) τyx_kj_avg s5( ):= → τxy_kj_avg s5( ) 4.278 MPa=

The calculation for the segment k’j’ is conducted in a similar manner as for the

segment kj and the shear stress is symmetric.

B.5.b.5 Calculation of the Shear Stress on Area A6 (see Figure 7.2.7.j)

The vertical cut 6-6 is made at distance s6 measured from i and increasing towards h.

Figure 7.2.7.j




165

SzxSz MM __ >∆+ → xxx FF >∆+ → 0>∆H

0>∆H → 0_ >avrzxq → 0_ >avrzxτ → 0_ <avrxzτ


tih tw_U12:= → tih 0.7cm= - thickness of the cut at s6

SZc_ih s6( ) SZc_kj bf_U12( ) s6 tw_U12⋅ yinf bf_U12+tw_U12

2−

⋅+:=


qzx_ih s6( )Vy_S SZc_ih s6( )⋅


τzx_ih_avgs6( )qzx_ih s6( )

tih:= - shear stress on the cut area

The above functions are particularized for the two limiting cuts of the segment kj:.

• cut i

s6 0 cm⋅:=

SZc_ih s6( ) 101.142− cm3=

qzx_ih s6( ) 42.045kNm

=

τzx_ih_avgs6( ) 6.006 MPa=

τxz_cd_avg s6( ) τzx_cd_avg s6( )−:= → τxz_cd_avg s6( ) 3.925 MPa=

• cut j

s6hU12

2tf_U12−

tpl2

−

:= → s6 4.628cm=

SZc_ih s6( ) 161.686− cm3=

qzx_ih s6( ) 67.214kNm

=

166

τzx_ih_avgs6( ) 9.602 MPa=


The calculation for the segment i’j’ is conducted in a similar manner as for the segment

ij and the shear stress is symmetric.

B.5.b.6 Calculation of the Shear Stress on Area A4 (see Figure 7.2.7.k)

The horizontal cut 4-4 is made at distance s3 measured from g and increasing towards

f.




SzxSz MM __ >∆+ → xxx FF >∆+ → 0>∆H

0>∆H → 0_ >avryxq → 0_ >avryxτ → 0_ >avrxyτ

Figure 7.2.7.k


tgf tpl:= → tef 1cm= - thickness of the cut at s4

167

SZc_gf s4( ) AU12 yC3⋅ s4 tpl⋅ yinf bf_U12+s42

+

⋅+:=


qyx_gf s4( )Vy_S SZc_gf s4( )⋅


τyx_gf_avgs4( )qyx_gf s4( )

tgf:= - shear stress on the cut area

• cut g

s4 0 cm⋅:=

SZc_gf s4( ) 338.96− cm3=

qyx_gf s4( ) 140.908kNm

=

τyx_gf_avgs4( ) 14.091MPa=

τxy_gf_avgs4( ) τyx_gf_avgs4( )−:= → τxy_gf_avgs4( ) 14.091− MPa=

• cut f

s4 yinf bf_U12+:= → s4 18.339cm=

SZc_gf s4( ) 507.116− cm3=

qyx_gf s4( ) 210.811kNm

=

τyx_gf_avgs4( ) 21.081MPa=

τxy_gf_avgs4( ) τyx_gf_avgs4( ):= → τxy_gf_avgs4( ) 21.081MPa=


Problem 7.3.1 The cross-section of the wide-flange beam I40 made of structural steel,

shown in Figure 7.3.1, is subjected to pure bending. Alternatively considering the

168

situations pictured in Figures 7.3.1.a and 7.3.1.b and assuming that the allowable

bending normal stress is σallow = 200 MPa, calculate: (a) the flexural stress and strain

distribution on the cross-section and (b) the maximum bending moment which can be

applied.

Figure 7.3.1

Problems 7.3.2 The cross-section shown in Figure 7.3.2 is subjected to pure bending.

Conduct the following tasks: (a) Draw the flexural stress and strain distributions

considering the following dimensions: bf = 20 cm, t f = 5 cm, hw =15 cm, and tw = 2

cm, (b) Calculate the capable bending moment of the cross-section if the allowable

tensile and compressive normal stress is σallow = 220 MPa, and (c) Determine the

stresses at points A and B if a bending moment mkNM ⋅= 5.13 is acting on the cross-

section.

Figure 7.3.2

169

Problems 7.3.3 A wood beam made from three boards glued together to form a single

beam as shown in Figure 7.3.3 is subjected to pure bending. Obtain an analytical

expression for: (a) the tensile and compressive maximum stresses and draw the normal

stress distribution on the cross-section, (b) the total force acting on the top board and

(c) for a=10 cm calculate the capable bending moment if the allowable tensile and

compressive stresses are σall_tens =35 MPa and σall_compr =25 MPa, respectively.

Figure 7.3.3

Problems 7.3.4 An I shaped beam with unequal flanges, with dimensions as shown in

Figure 7.3.4, is subjected to a bending moment of magnitude of M=40 kN. Draw the

normal stress and strain distributions and calculate the maximum tensile and flexural

stresses on this cross-section.

Figure 7.3.4

170

Problem 7.3.5 If the allowable shear stress in the timber beam cross-section shown in

Figure 7.3.5 is τall = 20 MPa, draw the shear stress diagram and calculate the

maximum transversal shear force V which can be sustained by the wood beam. For

calculation consider b=10 cm and h=15 cm, respectively.

Figure 7.3.5

Problem 7.3.6 Using Jurawski’s Formula, determine an expression for the shear stress

at the neutral-axis of the circular and tubular cross-sections shown in Figure 7.3.6.

Conduct a numerical application for the following data: d=d0=10 cm and di=5 cm.

Figure 7.3.6

Problem 7.3.7 The tee-shaped cross-section shown in Figure 7.3.7 is subjected to a

vertical shear force V. Calculate: (a) and draw the shear stress distribution and (b) the

vertical shear force carried by the web of the cross-section. Conduct a numerical

application for t=1 cm.

171

Figure 7.3.7

Problem 7.3.8 A vertical shear force V = 200 kN is applied to the cross-section shown

in Figure 7.3.8. Considering the following dimensions: h = 320 mm, hw = 300 mm,

b=160mm and tw = 8mm, calculate: (a) the distribution of the shear stress on the cross-

section and (b) the shear force carried by the web.

Figure 7.3.8

Problem 7.3.9 Two wide-flange beams I20 and I40, respectively, are subjected to the

same magnitude vertical shear force, V. Determine: (a) the distribution of the shear

stresses for each cross-section, (b) the maximum shear force allowed by each wide-

flange beam if the allowable shear stress is MPaall 140=τ , (c) the shear force carried

by the webs of each beam and (d) the ration of the maximum obtained shear stresses.

172

Figure 7.3.9

Problem 7.3.10 The channel cross-section shown in Figure 7.3.10 (dimensions in mm)

is subjected to a vertical shear force V = 10 kN. Calculate: (a) the shear stress

distribution on the cross-section, (b) the shear stress at points A and B and (c) the

vertical shear forces carried by the vertical positioned flanges.

Figure 7.3.10

Problem 7.3.11 Considering that the thin-wall closed cross-section shown in Figure

7.3.11 (dimensions in mm) is subjected to a vertical shear force V calculate: (a) the

shear stress distribution on the cross-section and (b) the vertical shear force carried by

the webs if the material has an allowable shear stress MPaall 140=τ .

173

Figure 7.3.11

Problem 7.3.12 The tubular thin-wall closed cross-sections shown in Figure 7.3.12 are

subjected to a vertical shear force V. Calculate the following: (a) the shear stress

distribution on the cross-sections, (b) the ratio of the maximum shear stresses and (c)

the ratio of the shear forces in the webs. Consider a numerical application for t=10

mm.

Figure 7.3.12

Problem 7.3.13 Repeat the calculations required in problem 7.3.9 for the cross-sections

shown in Figure 7.3.13. For numerical application consider a thickness t=1.5 cm.

174

Figure 7.3.13

Problem 7.3.14 Repeat the requirements of problem 7.3.13 considering the cross-

section shown in Figure 7.3.14.

Figure 7.3.14

Problem 7.3.15 The steel plate girder pictured in Figure 7.3.15 (all dimensions in mm)

is fabricated by welding together two horizontal plates to a vertical steel plate. Using

the dimensions in mm shown in Figure 7.3.15 and considering for the material an

allowable shear stress, MPaall 140=τ , (a) calculate the allowable shear force V

characterizing the cross-section and (b) size the welds considering an allowable shear

stress, MPaweldall 150_ =τ .

175

Figure 7.3.15

Problem 7.3.16 The tee-shaped steel beam shown in Figure 7.3.16, fabricated by

welding together two steel plates, is subjected to a vertical shear force V. Calculate: (a)

the distribution of the shear stress on the cross-section, (b) the allowable shear force

carried by the cross-section and (c) verify the welds (effective thickness mmteff 7=

and MPaweldall 160_ =τ ).

Figure 7.3.16

Problem 7.3.17 The I36 steel wide-flange beam, shown in Figure 7.3.17, is

strengthened by welding on its flanges two additional 175 mm x 10 mm cover steel

plates. Considering that a vertical shear force V is acting on the cross-section centroid

calculate: (a) the shear stress distribution in the original and strengthened cross-

sections, (b) the allowable vertical shear forces if the material is characterized by an

176

allowable shear stress MPaall 140=τ , and (c) size the weds considering that the weld

shear stress is MPaweldall 150_ =τ .

Figure 7.3.17

Problem 7.3.18 An I26 steel wide-flange beam subjected to a vertical shear force V is

strengthened by bolting two additional U14 steel channels. The strengthened cross-

section is shown in Figure 7.3.18.

Figure 7.3.18

Calculate: (a) the shear stress distribution of the web of the original and strengthened

cross-sections and (b) the maximum allowable vertical shear force of the strengthened

cross-section if the bolts have a diameter mmbolt 12=Φ , are longitudinally spaced at

mms 100= and are made of a material with allowable bearing and shear stresses of

MPabearall 400_ =σ and MPashearall 130_ =τ , respectively.

177

Problem 7.3.19 A steel beam is built up from a I40 wide-flange beam and two 155mm

x 15 mm cover plates as shown in Figure 7.3.19. Calculate the allowable shear force

and bending moment carried by the original and strengthened cross-sections

considering a material characterized by MPaall 400=σ , and MPaall 160=τ allowable

stresses. If the bolts are mmbolt 15=Φ in diameter and are made of a material with

allowable bearing and shear stresses of MPabearall 400_ =σ and MPashearall 130_ =τ

find the required spacing between the bolts.

Figure 7.3.19

Problem 7.3.20 Two wood box beams, shown in Figure 7.3.20, have identical cross-

sections.

Figure 7.3.20

178

Considering a vertical shear force V = 5 kN and a bending moment M=50 KN*m

acting at the centroid of the cross-sections calculate: (a) the normal and shear stress

distributions on the cross-sections, (b) the maximum spacing between the nails for each

nail arrangement, considering that they are equally spaced at 50 mm and an individual

nail can carry an allowable shear force of 250 kN.

Problems 7.3.21 and 7.3.22 For the wood beams shown in Figures 7.3.21 and 7.3.22

calculate: (a) the internal resultants diagrams and draw their plots, (b) the distributions

of the normal tensile and compressive stresses corresponding to the cross-section

where the maximum stresses appear, (c) calculate the distribution of the shear stress in

the cross-section where the maximum shear stress appears, (d) verify the beam using

the following allowable stresses MPAtensall 60_ =σ , MPAcompsall 40_ =σ , and

MPAall 25=τ .

Figure 7.3.21

Figure 7.3.22

179

Problem 7.3.23 The timber beam shown in Figure 7.3.23, with 200 mm x 300 mm

rectangular cross-section, is subjected to the following vertical concentrated loads: PB

= 15 kN and PC = 25 kN. Calculate: (a) the distribution of the normal bending stress in

the cross-section where the maximum bending moment appears, (b) the distribution of

the shear stress on the cross where the maximum appears and (c) verify the beam

considering the following allowable stresses: MPAtensall 50_ =σ , MPAcompsall 35_ =σ ,

and MPAall 25=τ .

.

Figure 7.3.23

Problem 7.3.24 The boxed type beam shown in Figure 7.3.25 is made of timber with

the following allowable stresses: MPAtensall 50_ =σ , MPAcompsall 35_ =σ , and

MPAall 25=τ .

180

Figure 7.3.24

Calculate: (a) the distribution of the flexural stress in the cross-section where the

maximum appears, (b) the distribution of the shear stress on the cross where the

maximum appears and (c) verify the beam and (d) find the necessary spacing between

the nails if nails mm2=Φ and MPAall 120=τ are employed to realize the connection.

Problems 7.3.25 through 7.3.30 For the steel beams shown in Figures 7.3.25 and 7.3.30

calculate: (a) the internal resultants diagrams and draw their plots, (b) the distributions

of the normal tensile and compressive stresses corresponding to the cross-section

where maximum stresses appear, (c) calculate the distribution of the shear stress in the

cross-section where the maximum shear stress appears, (d) verify the beam using the

following allowable stresses MPAall 210=σ and MPAall 160=τ , and (d) assuming an

welded connection between the web and flanges with MPAweldall 180_ =τ size the

welds .

Figure 7.3.25

181

Figure 7.3.26

Figure 7.3.27

Figure 7.3.28

Figure 7.3.29

182

Figure 7.3.30

Problem 7.3.31 and 7.3.32 The beams shown in Figures 7.3.31 and 7.3.32 are made of

I32 structural steel. Calculate: (a) the internal resultants diagrams and plot the

diagrams, (b) the flexural stress and shear stress distributions on the cross-section right

of B, (c) identify the cross-sections where the maximum normal and shear stresses are

and draw the corresponding distributions (d) verify the beam considering the following

allowable stresses MPAall 210=σ and MPAall 160=τ , (d) the vertical shear force in

the web.

Figure 7.3.31

Figure 7.3.32

183

Problems 7.3.33 through 6.3.35 The beams shown in Figures 7.3.33 through 6.3.35

have the cross-section made of I shape type laminated steel. Determine the profile

required if it is assumed that the allowable tensile and compressive stresses are both

equal to MPAall 210=σ and MPAall 160=τ . Draw the normal and shear stresses

diagrams in cross-section C.

Figure 7.3.33

Figure 7.3.34

Figure 7.3.35

184

Problem 7.3.36 The beam illustrated in Figure 7.3.36 is made of wood which allows

equal normal tensile and compressive stresses MPAall 60=σ and shear stress

MPAall 40=τ , respectively. Calculate the required dimensions for the cross-section,

assuming the ratio height to width is 1.5. Draw the corresponding normal stress and

shear stress diagrams for the cross-section located at the right of point C.

Figure 7.3.36

Problem 7.3.37 The laminated beam shown in Figure 7.3.37 is made by gluing together

n planks of width b and thickness t. Derive a formula for a cross-section with height

h=n*t and width b considering that the wood is characterized by the following

allowable stresses: tension MPAtensall 60_ =σ , compression MPAcompall 50_ =σ and

shear MPAall 40=τ , respectively. Assume that the thickness of the glue layer is

negligible and that the obtained laminated beam behaves like a compact ordinary wood

beam. Calculate the number of required planks to create the lightest timber beam. if the

fallowing data is considered: b = 200 mm, L= 5 m, P = 9 kN and t = 50 mm.

Figure 7.3.37

185

Problem 7.3.38 Locate the roller support B of the beam shown in Figure 7.3.38 such

that the beam AD has an equal magnitude of the maximum positive and negative

moments. Assuming PA = 30 kN, PC = 40 kN, L = 1.5 m and equal flexural tensile and

compressive stresses MPAall 60=σ for the steel, determine the required I shape.

Figure 7.3.38

Problem 7.3.39 An electrical switch is fabricated as shown in Figure 7.3.39. In the

“on” position the button D moves down and applies a vertical force P=50kN on the tip

C of the copper plate AC. Assuming that the plate AC has a rectangular cross-section

and the copper is characterized by the following allowable stresses MPAall 100=σ

and MPAall 60=τ , calculate the required dimensions of the cross-section.

Figure 7.3.39

186

Problem 7.3.40 and 7.3.41 The beams illustrated in Figures 7.3.40 and 7.3.41 are made

of rectangular wood which allows equal flexural tensile and compressive stresses

MPAall 60=σ and shear stress MPAall 40=τ . Calculate the required dimensions for

the rectangular cross-section, assuming the ratio height to width is 1.5.

Figure 7.3.40

Figure 7.3.41

Problem 7.3.42 through 7.3.44 The steel beams shown in Figures 7.3.42 through 7.3.44

are assembled from same type structural shapes. Determine the size of the necessary

structural shapes, neglecting the bolt existence and considering the following allowable

stresses for the steel: normal stress MPAall 210=σ and shear stress MPAall 160=τ .

Calculate the spacing between the bolts assuming a mm12=Φ bolt diameter and the

following allowable stresses for the bolts: MPAbearlal 360_ =σ and

MPAshearall 140_ =τ .

187

Figure 7.3.42

Figure 7.3.43

Figure 7.3.44

Problem 7.3.45 through 7.3.49 The beams shown in Figures 7.3.45 through 7.3.49 are

made of structural steel characterized by following allowable stresses:

MPAall 210=σ and MPAall 160=τ . Calculate: (a) the internal resultants diagrams,

(b) the geometrical characteristics of the cross-section, (c) the flexural and shear stress

distribution in the cross-sections where the maximum values are obtained, (d) the

maximum distributed force w carried by the beam, (e) the vertical shear force in the

web and (f) the effective thickness of the weld considering MPAweldall 180_ =τ .

188

Figure 7.3.45

Figure 7.3.46

Figure 7.3.47

Figure 7.3.48

189

Figure 7.3.49

Problem 7.3.50 Two wood boards are attached together by screws, spaced at regular

intervals of s=10 cm along the length of the beam, to form the inverted T-shaped beam

shown in Figure 7.3.50. Each screw has an allowable shear capacity of 2 k N. Calculate

the maximum allowable load P carried by the composed beam.

Figure 7.3.50

Problem 7.3.51 An inverted structural steel tee beam with dimensions shown in Figure

73.51 is subjected at end C to a concentrated force P.

Figure 7.3.51

190

What is the maximum load P assuming a material with allowable stresses of

MPAall 210=σ and MPAall 160=τ is used? Repeat the calculation for the case when

the allowable flexural tensile and compressive stresses are MPAtensall 180_ =σ and

MPAcompall 90_ =σ , respectively. Draw the stress distributions corresponding to the

cross-section located at the left of point B. Size the welds if MPAweldall 180_ =τ .

Problem 7.3.52 Repeat the calculations required at problem 7.3.51 considering the

beam shown in Figure 7.3.52.

Figure 7.3.52

Problem 7.3.53 A timber beam, with 15cm x 25cm nominal dimensions, is supported

and loaded as shown in Figure 7.3.53. If the allowable flexural and shear stresses of the

wood are: MPAtensall 60_ =σ , MPAcompall 40_ =σ and MPAall 25=τ , calculate the

maximum load P possible to be applied to this beam.

Figure 7.3.53

191

Problem 7.3.54 The simply supported beam shown in Figure 7.3.54 has a rectangular

cross-section. Consider for the wood the following allowable stresses:

MPAtensall 60_ =σ , MPAcompall 40_ =σ and MPAall 25=τ . Calculate: (a) Pσ, based

on the allowable flexural stress, (b) Pτ, based on the allowable shear stress and (c) the

ratio Pσ / Pτ of the previously calculated loads. Discuss the implications of the beam

dimensions and calculate the length L for which the shear-stress or the flexural stress

criterion governs.

Figure 7.3.54

Problem 7.3.55 As pictured in Figure 7.3.55, a set of wheels on the one-axle railway

car passes directly over one of the rectangular cross ties with 18cm height and 20 cm

width.

Figure 7.3.55

If the following calculation scheme, as pictured in Figure 7.3.55a, is assumed, then the

load exercised by each wheel is represented by a uniformly distributed load pw = 500

kN/m applied over the 0.3m width of the tie plate and while the cross tie distributes this

load as an uniformly distributed pressure pb to the ballast on which the cross tie rests.

192

Figure 7.3.55.a

Calculate the following: (a) the allowable distributed load pb, (b) the cross-section

internal resultants diagrams, (c) the maximum normal bending stress and (d) maximum

shear stress in the wood cross tie.

193

CHAPTER 8 Bending of Plane Linear Beams – Deflection

The application of the second-order and forth-order differential equations of the

deflection of linear beams is presented.


The equation relating the bending moment )(xM z with the radius of curvature

)(xρ is:

)(*)(*)(

1)( xIxEx

xM zz ρ= (8.1)

where )(xE - the modulus of elasticity;

)(xI z - the moment of inertia about the horizontal centroidal axis of the cross-

section;

)(xM z - the cross-section’s bending moment;

)(xρ - the radius of curvature of the deflection curve.

Note: If the material is a homogeneous, linear elastic one and the beam is of uniform

cross-section, the product zIE * is called the flexural rigidity and is

constant. Then,

zz IEx

xM **)(

1)(ρ

= (8.2)

The sign of the deflection v(x), expressed by equation (8.1) or (8.2) is determined

using the sign convention established during the theoretical analysis induced by the

bending deformation , Lecture 7, and re-plotted below in Figure 8.1.

Note: The moment diagram is always plotted at the beam side where the fibers are

subjected to tension, while the curvature center is placed on the opposite side. When

194

the bending moment is positive the deflection curvature is concave, while for the

negative bending moment the deflection curvature is convex.

Figure 8.1 Sign Convention for Beams in Bending

The relation between the radius of curvature )(xρ and the deflection )(xv is

established in calculus as:

23

2

2

2

]))((1[

)(

)(1

dxxdv

dxxvd

x+

=ρ

(8.3)

where:

dxxdvx )()(tan =θ (8.4)

Under the small displacement assumption the first derivative is a very small value:

1)(<<

dxxdv (8.5)

and consequently, the products of the deflection derivative can be neglected

0)( 2

≅

dxxdv (8.6)

and the angle approximated by its tangent:

dxxdvxx )()(tan)( =≈ θθ (8.7)

195

Note: Relation (8.7) is valid if the influence of the shear force on the deflection )(xv

is neglected.

Substituting (8.6) into equation (8.3) a more simplified relation between the radius of

curvature )(xρ and the deflection )(xv is obtained:

2

2 )()(

1dx

xvdx

≅ρ

(8.8)

The moment-curvature equation, a second order differential equation, is obtained

substituting equation (8.8) into (8.1):

)(*)(*)()( " xvxIxExM z= (8.9)

where the notation 2

2" )()(

dxxvdxv = is employed.

Consider the differential relation, previously obtained from the equilibrium of the

beam infinitesimal volume element, between the transverse loading )(xpn and the

cross-section stress resultants, the shear force )(xVy and bending moment )(xM z :

)()(

xpdx

xdVn

y = (8.10)

)()(

xVdx

xdMy

z = (8.11)

Note: Distributed normal load )(xpn acting on the longitudinal axis Ox is orientated

in the positive direction of the vertical axis Oy, 0)( >xpn .

Differentiating the moment-curvature equation (8.9) and using equations (8.10) and

(8.11) two new differential equations, the shear-deflection and load-deflection

equations, are obtained:

)]'("*)(*)([)( xvxIxExV zy = (8.12)

)]"("*)(*)([)( xvxIxExp zn = (8.13)

The load-deflection equation (8.13) is a forth-order differential equation.

196

If the beam is made from a homogeneous linear elastic material and has a constant

cross-section in the continuity interval of the bending moment or transversal loading

equations (8.9) and (8.13) become:

)(**)( " xvIExM z= (8.14)

)(**)( xvIExp IVzn = (8.15)

where constant* =zIE .

The continuity intervals pertinent to the functions involved in the differential

equations (8.14) and (8.15), the bending moment )(xM z and transverse load )(xpn ,

must be recognized in order for the integration process to be properly conducted. In

the general case, when the flexural rigidity zIE * is not constant, its functional

continuity must be also considered in the integration process.

8.1.1 Integration of the Moment - Curvature Differential Equation (8.9)

Integration of the second-order differential equation (8.9) on an interval of continuity

for the bending moment )(xM z yields the following relations:

1' *

)(*)()(

)()( CdxxIxE

xMxvx

z

z∫ +==θ (8.16)

21 ***)(*)(

)()( CxCdxdx

xIxExM

xvz

z ++= ∫ ∫ (8.17)

For the case of a beam with constant flexural rigidity constIE z =)*( , the integrals

expressed in equations (8.16) and (8.17) simplify as follows:

1' *)(

*1)()( CdxxM

IExvx z

z∫ +==θ (8.18)

21 ***)(*1)( CxCdxdxxM

IExv z

z

++= ∫ ∫ (8.19)

The integration constants 1C and 2C are calculated by imposing the boundary

conditions of the specific continuity interval. The boundary conditions are

197

represented by known values of the rotation and deflection at the end locations of the

bending moment )(xM z continuity interval.

For practical purposes the above formulae (8.16) through (8.19) are re-written

considering that the values of the rotation 0)0( θθ ==x and deflection 0)0( vxv == ,

respectively, at the beginning of the continuity interval, at 0=x , are known values.

Consequently, the expressions (8.16), (8.17), (8.18) and (8.19), respectively, become:

0' *

)(*)()(

)()( θθ ∫ +== dxxIxE

xMxvx

z

z (8.20)

00 ***)(*)(

)()( vxdxdx

xIxExM

xvz

z ++= ∫ ∫ θ (8.21)

and

0' *)(

*1)()( θθ ∫ +== dxxM

IExvx z

z

(8.22)

00 ***)(*1)( vxdxdxxM

IExv z

z

++= ∫ ∫ θ (8.23)

8.1.2 Integration of the Load-Deflection Differential Equation (8.13)

Successive integration of the forth-order differential equation (8.13) in a continuity

interval of the transverse loading function yields the following relations:

1'" *)()](*)(*)([)( CdxxpxvxIxExV nzy +== ∫ (8.24)

21" ***)()(*)(*)()( CxCdxdxxpxvxIxExM nzz ++== ∫ ∫ (8.25)

321

3'

*)(*)(

1**)(*)(

*

*]**)([)(*)(

1

*)(*)(

)()()(

CdxxIxE

CdxxIxE

xC

dxdxdxxpxIxE

CdxxIxE

xMxvx

nz

z

z

+++

+=

=+==

∫∫

∫∫∫

∫θ

(8.26)

43

21

*

**)(*)(

1***)(*)(

*

****)()(*)(

1)(

CxC

dxdxxIxE

CdxdxxIxE

xC

dxdxdxdxxpxIxE

xv nz

++

+++

+=

∫ ∫∫ ∫

∫∫∫∫

(8.27)

198

The integration constants 1C , 2C , 3C and 4C are calculated by imposing the boundary

conditions applicable to the specific problem.

For the case of a beam with constant flexural rigidity constIE z =)*( , the integrals

expressed in equations (8.24) through (8.27) may be considerably simplified as

follows:

1''' *)()(**)( CdxxpxvIExV nzy +== ∫ (8.28)

21" ***)()(**)( CxCdxdxxpxvIExM nzz ++== ∫∫ (8.29)

3

2

21

3'

*

*2

***)(*1

*)(*1)()(

CxC

xCdxdxdxxpIE

CdxxMIE

xvx

n

z

zz

+

+

++=

=+==

∫ ∫ ∫

∫θ

(8.30)

43

2231

*

*2

*6

****)(*1)(

CxC

xCxCdxdxdxdxxpIE

xv nz

++

+

++= ∫ ∫ ∫ ∫ (8.31)

The integration constants 1C , 2C , 3C and 4C are calculated by imposing the

boundary conditions of the specific continuity interval. The boundary conditions are

represented by known values of the shear force, bending moment, rotation and

deflection at the end locations of the transverse load )(xpn continuity interval.

For practical purposes the above formulae (8.24) through (8.31) are re-written

considering that the values of the shear force 0)0( VxVy == , bending moment

0)0( MxM z == , the rotation 0)0( θθ ==x and deflection 0)0( vxv == ,

respectively, at the beginning of the continuity interval, at 0=x , are known values.

Consequently, the expressions (8.24) and (8.27) and (8.28) and (8.31), respectively,

become:

0*)()( VdxxpxV ny += ∫ (8.32)

00 ***)()( MxVdxdxxpxM nz ++= ∫ ∫ (8.33)

199

0

00 *)(*)(

1**)(*)(

*

*]**)([)(*)(

1)(

θ

θ

+

+++

+=

∫∫

∫∫∫

dxxIxE

MdxxIxE

xV

dxdxdxxpxIxE

x nz

(8.34)

00

00

*

**)(*)(

1***)(*)(

*

****)()(*)(

1)(

vx

dxdxxIxE

MdxdxxIxE

xV

dxdxdxdxxpxIxE

xv nz

++

+++

+=

∫ ∫∫ ∫

∫∫∫∫

θ

(8.35)

and

0*)()( VdxxpxV ny += ∫ (8.36)

00 ***)()( MxVdxdxxpxM nz ++= ∫∫ (8.37)

0020 **

2***)(

*1)( θθ +

++= ∫ ∫ ∫ xMx

Vdxdxdxxp

IEx n

z

(8.38)

00

2030

*

*2

*6

****)(*1)(

vx

xM

xV

dxdxdxdxxpIE

xv nz

++

+

++= ∫ ∫ ∫ ∫

θ(8.39)

General Conclusion: The moment-curvature equation (8.9) can be used only if the

bending moment )(xM z variation is known, in the case of statically determinate

structures. The load-deflection curve equation (8.13) requires the variation of

transverse load )(xpn to be known and, consequently, can be used for either

statically determinate or indeterminate beams.

8.1.3 Boundary and Continuity Integration Conditions

As previously stated, the functions (8.16) though (8.27) obtained above, represent the

general solutions. Only after the boundary conditions are imposed and the integration

constants are determined, the solutions become representative for a specific variation

interval. The boundary conditions commonly encountered in the application of the

equations (8.20) and (8.23) or equations (8.32) through (8.39) are presented in Table

8.1.

200

Table 8.1 Boundary Conditions

Table 8.2 Continuity Conditions

201

In general, the transverse load )(xpn and the bending moment )(xM z functions for a

particular loaded beam are described by a number of continuity intervals which are

priory determined to the beginning of the integration. Therefore, the continuity

conditions at the common ends of the intervals must be described in order for the

constants to be calculated. Each continuity interval is treated as an independent

interval and then, the boundary and continuity conditions are applied. The most

common continuity conditions are summarized in Table 8.2.

8.1.4 Superposition Method

In the technical literature, a large number of deflection )(xv and rotation )(xθ curves

corresponding to simple cases of restrained beams subjected to simple types of

loadings are calculated and tabulated. Tables T1 and T2 of the Appendix A contain

frequently encountered cases of loadings pertinent to the cases of cantilever, simply

supported and fixed-fixed beams. Due to the linearity of the differential equations

describing the deflection curve, a complicated case can be decomposed in simple

cases with deflection curves tabulated. By summation of these deflection functions

pertinent to each simple loading case, the final expression of the deflection curve is

obtained. This method is extensively used in structural engineering practice.


Problem 8.2.1 Considering the cantilever beam illustrated in Figure 7.2.1 calculate:

(a) the radius of curvature, (b) the deflection function and (c) the maximum

displacement.

.


A.1 The cantilever is subjected to pure bending conditions. Consequently the Navier’s

formula is applied.


M0 20 kN⋅ m⋅:= - the bending moment

202

L 5 m⋅:= - cantilever length

d 120 mm⋅:= - edge length


B. Calculations

B.1 Calculation of the Radius of Curvature

The radius of curvature )(xρ is calculated using the following formula:

ρ x( )E

Mz x( )Izc⋅

where: zcI is the moment of inertia of the entire cross-section about the central axis z.

The central moment of inertia is calculated:

Izcd4

12:= →⇒ 4

4

1212 cm Izc 1.728 103× cm4=

The radius of curvature

)(xρ is then obtained as:

ρE

M0Izc⋅:= →⇒ − 48

3

9

10*1728**10*20

10*200 mmN

Paρ 172.8m=

Note: the radius of curvature )(xρ is constant. It can be concluded that when the

beam is in a pure bending condition the radius of curvature is always constant and

the defection curve is a circle.

The curvature )(xk is:

k x( )1

ρ x( )→⇒ −1

8.1721 m k 5.787 10 3−× m-1=

B.2 Calculation of Maximum Displacement

Because the bending moment is known and constant the second-order differential

equation of the deflection curve is used:

203

2xv x( )d

d

2 M0E Izc⋅

By integration the slope

)(')()( xvxvdxdx ==θ and the vertical displacement

)(xv are

obtained as:

xv x( )d

d

M0E Izc⋅

x⋅ C0+

v x( )M0

2E Izc⋅x2⋅ C0 x⋅+ C1+

where

0C and

1C are integration constants, which are calculated from the boundary

conditions imposed at 0=x :

v' 0( ) 0 → C0 0

v 0( ) 0 → C1 0

Substituting the constants , the expression of the vertical displacement is:

v x( )M0

2E Izc⋅x2⋅

The maximum vertical displacement

Bvxv =)(

is obtained at the cantilever tip

mLx 5==

→⇒ −

22489

3

5*10*1728*10*200*2

*10*20 mmPa

mN vB 0.072m=

To verify the vertical displacement formula obtained above, the function )(xv is also

calculated employing Mathcad differential equation package, and then plotted:

l0L

1 m⋅:=

Given

204

2xv x( )d

d

2 M0E Izc⋅

− 0

v 0( ) 0

v' 0( ) 0

v Odesolve x l0,( ):=

0 1 2 3 4 50

0.018

0.036

0.054

0.072

v x( )

x

B.1 Maximum Displacement

The maximum value Bvxv =)( obtained from the graph is:

vB v 5( ) m⋅:= → vB 0.072m=

Problem 8.2.2 The deflection v(x) of the simply supported beam loaded as illustrated

in Figure 8.2.2 is:

−

+

−=

Lx

Lx

Lx

IELMxv

2320 *3*2*

**3*

)(

205

Considering that the flexural rigidity of the beam constant * =zIE along its entire

length, obtain the following: (a) the expressions of the rotation )(xθ , shear force V(x)

and bending moment M(x), (c) the expressions of the applied moment M1 and the

reactions V1 and V2 and (d) the position and maximum absolute rotation and vertical

deflection, maxϑ and maxv , respectively. Plot all the graphs using the Mathcad graphic

capabilities.

Figure 8.2.2


A.1. The beam illustrated in Figure 8.2.2 is investigated using the moment - curvature

second-order differential equation (8.9) since it is a statically determinate system.

Consequently, the equation (8.14) is employed.

A.2 The boundaries are:

- at point 1 v1 0 ;

- at point 2 v3 0 .

A.3 In the absence of any transversal distributed force, the beam is characterized by

only one continuity interval, the interval 1-2.

A.4 For simplicity of the calculations the following notation is used:

zIEK

*1

=

206

B. Calculations

B.1 Calculation of the Rotation, Bending Moment and Shear Force

The expression of the vertical deflection v(x) characterizing the beam shown in Figure

8.2.1 is assumed known on the continuity interval 1-2:

v x( )M0 l2⋅

3 E⋅ I⋅2−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅

where x represents the position of a particular cross-section located between the end

points 1 and 2 and is measured starting from point 1. Replacing zIE

K*1

= into the

expression of the vertical deflection v(x) a new expression is obtained:

v x( )M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅

Using the formula (8.7) the rotation function )(xθ is calculated:

θ x( )x

v x( )dd →

The bending moment function M(x) is obtained employing the equation (8.9) as

follows:

K M x( )⋅2xv x( )d

d

2

xθ x( )d

d→

→ M x( ) 2M0

l⋅ 2− x⋅ l+( )⋅

207

The shear force V(x) and the distributed transversal load p(x) are calculated by

successive differentiation from the bending moment function M(x):

V x( )x

M x( )dd

→ V x( )x

2M0

l⋅ 2− x⋅ l+( )⋅

dd

V x( ) 4−M0

l⋅→

p x( )x

V x( )dd

→ p x( )x

4−M0

l⋅

dd

p x( ) 0→

Note: The absence of the transversal distributed force p(x) is verified by the above

obtained results.

B.2 Calculation of the Reaction Forces 1V and 2V

The reaction forces are calculated particularizing the position x of the cross-section

into the shear force function V(x):

- at point 1

V1 V x 0( ) → V1 4−M0

l⋅ substitute x 0, V1 4−

M0l

⋅→

- at point 2

V2 V x l( ) → V2 4−M0

l⋅ substitute x l, V2 4−

M0l

⋅→

B.3 Calculation of the Concentrated Moment 1M

The bending moment acting at point 1 is calculated:

M1 M x 0( ) → M1 2M0

l⋅ 2− x⋅ l+( )⋅ substitute x 0, M1 2 M0⋅→

To verify the validity of the above obtained moment formula, the value of the

concentrated moment acting at point 2 is calculated:

208

M2 M x l( ) → M2 2M0

l⋅ 2− x⋅ l+( )⋅ substitute x l, M2 2− M0⋅→

B.4 Important Values of the Rotation )(xθ and Vertical Deflection )(xv Functions

The value of the rotations of the beam at point 1 and 2 are:

θ11−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅ substitute x 0, θ11−

3M0⋅ l⋅ K⋅→

θ21−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅ substitute x l, θ21−

3M0⋅ l⋅ K⋅→

The rotation function

)(xθ is described by a second-order polynomial and

consequently, a local maximum value, located between point 1 and 2, is expected.

The location is obtained:

xθ x( )d

d0 →

x1−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅

dd

0 solve x,12

l⋅→

Substituting the above calculated location into the rotation expression the maximum

value is obtained:

θmax1−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅ substitute x 12

l⋅, θmax16

M0⋅ l⋅ K⋅→

The values of the vertical displacement at points 1 and 2 are known and the

calculation is conducted just for verification purposes:

v1M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅ substitute x 0, v1 0→

v2M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅ substitute x l, v2 0→

The vertical deflection function

)(xv is described by a third-order polynomial and

consequently, local extremum values, located between point 1 and 2, are expected.

209

The locations are obtained as:

xv x( )d

d0 →

x

M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅

dd

0 solve x,

12

16

3⋅+

l⋅

12

16

3⋅−

l⋅

→ →

The value of the vertical deflection v(x) at those locations is calculated:

B.5 Numeric Application

The numeric application is conducted for the following data:

K K 1

N m2⋅⋅:=

l 6 m⋅:= M0 1000 N⋅ m⋅:=

The shear force, bending moment, rotation and vertical deflection functions are

particularized for the data set indicated above:

210

V x( ) 4−M0

l⋅

substitute M0 1000 N⋅ m⋅,

substitute l 6 m⋅,

float 4,

V x( ) 666.7− N⋅→

M x( ) 2M0

l⋅ 2− x⋅ l+( )⋅



substitute x x m⋅,

simplify

collect m,

float 4,

M x m⋅( ) 666.7− N⋅ m⋅ x 3.−( )⋅→

Numerical values of the above functions calculated in the important points of their

definition are:

V1 4−M0

l⋅



float 4,

V1 666.7− N⋅→

211

V2 4−M0

l⋅



float 4,

V2 666.7− N⋅→

M1 2M0

l⋅ 2− x⋅ l+( )⋅



substitute x 6 m⋅,

float 4,

M1 2000.− N⋅ m⋅→

θ11−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅




substitute K K 1

N m2⋅⋅,

simplify

float 4,

θ1 2000.− K⋅→

θ21−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅




substitute K K 1

N m2⋅⋅,

simplify

float 4,

θ2 2000.− K⋅→

xθ12

l⋅ substitute l 6 m⋅, xθ 3 m⋅→

θmax1−

3

M0l

⋅ 6 x2⋅ 6 x⋅ l⋅− l2+( )⋅ K⋅




substitute K K 1

N m2⋅⋅,

simplify

float 4,

θmax 1000. K⋅→

212

v1M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅




substitute K K1

N m2⋅⋅,

simplify

float 4,

v1 0→

v2M0 l2⋅

32−

xl

3

⋅ 3xl

2

⋅+xl

−

⋅ K⋅




substitute K K1

N m2⋅⋅,

simplify

float 4,

v2 0→

xv_112

16

3⋅−

l⋅


simplify

float 4,

xv_1 1.268 m⋅→

xv_212

16

3⋅+

l⋅


simplify

float 4,

xv_2 4.732 m⋅→

213

B.6 Plotted Diagrams

The shear force, bending moment, rotation and vertical displacement functions

obtained in the previous section are plotted using the Mathcad graphical capabilities.

l 6 m⋅:= M0 1000 N⋅ m⋅:=

nint 14:= nint1 3:= nint2 nint 2 nint1⋅−:= → nint2 8=

xv_112

16

3⋅−

l⋅:= → xv_1 1.268m= xv_2

12

16

3⋅+

l⋅:= → xv_2 4.732m=

lint2 l xv_1 l+ xv_2−( )−:= → lint2 3.464m=

dlint1xv_1nint1

:= → dlint1 0.423m= dlint2lint2nint2

:= → dlint2 0.433m=

pos0 0:= position of the origin point

Initial values:

shear02000−3

:= → shear0 666.667−= mom0 2000:=

214

rot0 2000−:= disp0 0:=

i 1 nint..:=

dlinti if i 3≤ dlint1, if i 12≥ dlint1, dlint2,( ),( ) 1

m⋅:=

posi posi 1− dlinti+:=

sheari shear0:=

momi2000−3

posi 3−( )⋅:=

roti1000−3

posi( )26 posi⋅− 6+ ⋅:=

dispi1000−9

posi⋅ posi( )2 9 posi⋅− 18+ ⋅:=

Position (m) Shear Force (kN) Bending Moment (kN*m)

pos

001234567891011121314

00.4230.8451.2681.7012.1342.567

33.4333.8664.2994.7325.1555.577

6

=

shear

001234567891011121314

-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667-666.667

=

mom

001234567891011121314

2·10 3

1.718·10 3

1.436·10 3

1.155·10 3

866.025577.35

288.6752.961·10 -13

-288.675-577.35

-866.025-1.155·10 3

-1.436·10 3

-1.718·10 3

-2·10 3

=

215

0 1 2 3 4 5 62000

1000

0

1000

2000

Shear Force and Bending Moment Diagrams

Position (m)

Shea

r For

ce (k

N),

Ben

ding

Mom

ent (

kN*m

)

shear

mom

pos

Position (m) Rotation (rad) Vert Displacement (m)

pos

001234567891011121314

00.4230.8451.2681.7012.1342.567

33.4333.8664.2994.7325.1555.577

6

=

rot

001234567891011121314

-2·10 3

-1.214·10 3

-547.5780

437.5750

937.51·10 3

937.5750

437.51.184·10 -12

-547.578-1.214·10 3

-2·10 3

=

disp

001234567891011121314

0-675.055

-1.043·10 3

-1.155·10 3

-1.055·10 3

-793.857-423.992

0423.992793.857

1.055·10 3

1.155·10 3

1.043·10 3

675.0554.737·10 -12

=

216

0 1 2 3 4 5 62000

1000

0

1000

2000

Rotation and Vert Displacement Diagrams

Position (m)

Rot

atio

n/K

(rad

), D

ispl

acem

ent/K

(m)

rot

disp

pos

Problem 8.2.3 The beam illustrated in Figure 8.2.3 has a constant cross-section along

its entire length. Using the second-order differential equation conduct the following

tasks: (a) derive mathematical expressions for rotation )(xθ and vertical deflection

)(xv functions, (b) develop expressions for rotation and vertical deflection

characterizing all important points of the beam and (c) conduct a numeric application

considering m

kNp *8= and ml *6= .

Figure 8.2.3

217


A.1. The beam illustrated in Figure 8.2.3 is investigated using the moment- curvature

second-order differential equation (8.9). The differential equation (8.9) can be used

only if the investigated system is a statically determinate system. The beam proposed

for investigation is characterized by a constant cross-section along the entire length

and represents a statically determinate system. Consequently, the equations (8.22) and

(8.23) are employed.


- at point 1 v1 0 ;

- at point 3 v3 0 .

A.3 The acting load is characterized by two continuity intervals, 1-2 and 2-3. In the

absence of concentrated bending moments, it is expected that the bending moment

function is also characterized by the same continuity intervals. This fact will be

confirmed by the calculations. Consequently, two continuity equations representing

the equality of the rotations and deflections at the left and right locations of the

common end, point 2, can be written:

- v2_12 v2_23;

- θ2_12 θ2_23.

A.4 For simplicity of the calculations the following notation is used in the formulae

(8.22) and (8.23):

zIEK

*1

=

B. Calculations

B.1. Calculation of the reactions forces

It is obvious that in the absence of the horizontal load the horizontal component of the

reaction force existing at point 1 of the beam is null. The magnitudes of the two

vertical reactions forces 1V and 3V are calculated as:

218

03

=∑ zM → V1 l⋅ p l2

⋅34

⋅ l⋅− 0 solve V1,38

p⋅ l⋅→

01

=∑ zM → V3− l⋅ p l2

⋅14

⋅ l⋅+ 0 solve V3,18

p⋅ l⋅→

The expressions of the obtained vertical reaction forces are:

V138

p⋅ l⋅

V318

p⋅ l⋅

B.2 Calculation of the shear forces and bending moments

B.2.a Continuity Interval 1-2 ]2

,0[ lx ∈

The origin of the continuity interval 1-2 is point 1. The dummy variable ]2

,0[ lx ∈ is

specific to the investigated interval and changes from interval to interval. The

expressions of the shear force )(12 xV and bending moment )(12 xM on the continuity

interval 1-2 are:

V12 x( ) V1 xp−⌠⌡

d+ substitute V138

p⋅ l⋅, V12 x( )38

p⋅ l⋅ p x⋅−→

and

M12 x( ) M1 xV12 x( )⌠⌡

d+ →

M12 x( ) M1 x38

p⋅ l⋅ p x⋅−

⌠⌡

d+ substitute M1 0, M12 x( ) 38

p⋅ l⋅ x⋅12

p⋅ x2⋅−→

where

lpxVV **83)0(121 === is the shear force at point 1

219

0)0(121 === xMM is the bending moment at point 1

The values of the shear force and bending moment at the end of the continuity

interval, the point 2 )2

( lx = are calculated:

V2 V12 x( )

substitute V12 x( )38

l⋅ x−

p⋅,

substitute x l2

,V2

1−8

p⋅ l⋅→

M2 M12 x( )

substitute M12 x( )38

p⋅ l⋅ x⋅12

p⋅ x2⋅−,

substitute x l2

,M2

116

p⋅ l2⋅→

The location maxx and the value of the maximum bending moment max_12M are

obtained:

V12 x( ) 0 →38

p⋅ l⋅ p x⋅− 0 solve x,38

l⋅→ → xmax38

l⋅

M12_max38

p⋅ l⋅ x⋅12

p⋅ x2⋅− substitute x38

l⋅, M12_max9

128p⋅ l2⋅→

B.2.b Continuity Interval 2-3 ]2

,0[ lx ∈

The origin of the continuity interval 2-3 is point 2. The dummy variable ]2

,0[ lx ∈ is

specific to the investigated interval and changes from interval to interval. The

expressions of the shear force )(23 xV and bending moment )(23 xM on the continuity

interval 2-3 are:

V23 x( ) V2 x0−⌠⌡

d+ substitute V21−

8p⋅ l⋅, V23 x( )

1−8

p⋅ l⋅→

and

M23 x( ) M2 xV23 x( )⌠⌡

d+ →

220

where

lplxVV **81)

2(122 −=== the shear force at point 2 (previously calculated)

2122 **

161)

2( lplxMM === the bending moment at point 2

The values of the shear force and bending moment at the end of the continuity

interval, the point 3 )2

( lx = are calculated:

V3 V23 x( )

substitute V23 x( ) 1−8

p⋅ l⋅,

substitute x l2

,V3

1−8

p⋅ l⋅→

M3 M23 x( )

substitute M23 x( ) 116

p⋅ l2⋅18

p⋅ l⋅ x⋅−,

substitute x l2

,M3 0→

Note: The bending moment function is characterized by two continuity interval. The

expression of the bending moments is known and consequently, the integration of the

equations (8.22) and (8.23) can proceed.

B.3 Calculation of the rotation and deflection curves


,0[ lx ∈

The initial conditions for the interval 1-2 are:

θ1 real unknown value and must be calculated

221

v1 0

The expressions of the rotation )(12 xθ and deflection )(12 xv curves are:

θ12 x( ) θ1 K xM12 x( )⌠⌡

d⋅+ →

θ 12 x( ) θ 1 K x38

p⋅ l⋅ x⋅12

p⋅ x2⋅−

⌠⌡

d⋅+ →

θ 12 x( ) θ 1 K

316

p⋅ l⋅ x2⋅16

p⋅ x3⋅−

⋅+

and

v12 x( ) v1 xθ12 x( )⌠⌡

d+ →

The rotation and deflection at end 2 are obtained:

222


,0[ lx∈

The initial conditions for the interval 2-3 are calculated using the continuity of the

deformation in the common point 2 identified in section A.3:

θ2 θ15

192K⋅ p⋅ l3⋅+

v212

θ1⋅ l⋅1

192K⋅ p⋅ l4⋅+

The expressions of the rotation )(23 xθ and deflection )(23 xv curves are:

θ23 x( ) θ2 K xM23 x( )⌠⌡

d⋅+ →

and

v23 x( ) v2 xθ23 x( )⌠⌡

d+ →

223

The rotation and deflection at end 3 are obtained:

Note: It can be remarked that the expressions of the rotations and deflections pertinent

to both continuity intervals, 1-2 and 2-3, are dependent on the unknown rotation 1θ .

B.4. Solution

The only unknown value, the rotation 1θ , is obtained by imposing the second boundary

condition represented by the nullity of the displacement in point 3. Consequently, the

unknown rotation 1θ is found:

224

v3 0 →3

128K⋅ p⋅ l4⋅ θ1 l⋅+ 0 solve θ1,

3−128

K⋅ p⋅ l3⋅→

θ13−

128K⋅ p⋅ l3⋅

B.5. Final Expressions

The expression of the rotation 1θ is replaced in the previously determined functions

of the rotations and vertical displacements pertinent to the continuity intervals 1-2 and

2-3.


,0[ lx∈

The final expressions of the rotation

)(12 xθ and )(12 xv are:


,0[ lx∈

The final expressions of the rotation

)(23 xθ and deflection )(23 xv curves are:

225

Note: It has to be remarked that the position of the cross-section is described in each

interval by different measurements; the dummy variable x stars at point 1 and 2 for the

intervals 1-2 and 2-3, respectively.

B.5.c Rotations and Vertical Displacements at Points 1, 2 and 3

• at point 1

θ13−

128K⋅ p⋅ l3⋅

calculated above

v1 0 boundary condition

• at point 2

θ2 θ15

192K⋅ p⋅ l3⋅+ substitute θ1

3−128

K⋅ p⋅ l3⋅, θ21

384K⋅ p⋅ l3⋅→

226

• at point 3

θ3124

K⋅ p⋅ l3⋅ θ1+ substitute θ13−

128K⋅ p⋅ l3⋅, θ3

7384

K⋅ p⋅ l3⋅→

v3 0 boundary condition

Note: It has to be remarked that all four functions, V , M , ϑ and v , are

proportional with the load intensity p , while the functions representing the

deformation only, ϑ and v , are additionally proportional with the inverse of the

flexural rigidity K . This proportionality is a characteristic of the elastic behavior.

B.6. Usage of Unique Origin ],0[ lx∈

A unique origin, point 1, can be used by changing the variable x to x x l2

− in the

interval 2-3. No change is operated in interval 1-2, where the origin is point 1. The

representative functions, the shear force, the bending moment, the rotation and the

vertical displacement are expressed as:

V23 x( ) 1−8

p⋅ l⋅ substitute x x l2

−, V23 x 12

l⋅−

1−8

p⋅ l⋅→

227

B.7. Numeric Application

The following data is used to obtain numerical values for the functions calculated

above:

l 6 m⋅ p 8 kNm

⋅

K K 1

kN m2⋅⋅

B.7. a Reactions

V138

p⋅ l⋅substitute l 6 m⋅,

substitute p 8 kNm

⋅,V1 18 kN⋅→

V318


substitute p 8 kNm

⋅,V3 6 kN⋅→

B.7. b Interval 1-2 ]2

,0[ lx∈

V12 x( )38

p⋅ l⋅ p x⋅−


substitute p 8 kNm

⋅,


collect kN,

V12 x m⋅( ) 18 8 x⋅−( ) kN⋅→

228

B.7. c Interval 2-3 ],2

[ llx∈

V23 x( )1−

8p⋅ l⋅


substitute p 8 kNm

⋅,V23 x( ) 6− kN⋅→

229

B.7.d Shear Forces, Bending Moments, Rotations and Vertical Displacements at

Points 1, 2 and 3

• at point 1

V138


substitute p 8 kNm

⋅,V1 18 kN⋅→

230

mkNM ∗∗= 01

θ13−

128K⋅ p⋅ l3⋅


substitute p 8 kNm

⋅,

substitute K K1

kN m2⋅⋅,

float 3,

θ1 40.5− K⋅→

mv ∗= 01

• at point 2

V2 18 8 x⋅−( ) kN⋅ substitute x 3, V2 6− kN⋅→

M2 18 x⋅ 4 x2⋅−( ) kN⋅ m⋅ substitute x 3, M2 18 kN⋅ m⋅→

xmax38

l⋅ substitute l 6 m⋅, xmax94

m⋅→

M12_max9

128p⋅ l2⋅


substitute p 8 kNm

⋅,

float 4,

M12_max 20.25 kN⋅ m⋅→

θ21−

6243 54 x2⋅− 8 x3⋅+( )⋅ K⋅

substitute x 3,

float 3,θ2 4.50 K⋅→

v21−

6m⋅ x⋅ 243 18 x2⋅− 2 x3⋅+( )⋅ K⋅

substitute x 3,

float 4,v2 67.50− m⋅ K⋅→

• at point 3

V3 6− kN⋅ substitute x 6, V3 6− kN⋅→ M3 36 6 x⋅−( ) kN m⋅⋅ substitute x 6, M3 0→

θ33−

251 24 x⋅− 2 x2⋅+( ) K⋅⋅

substitute x 6,

float 3,θ3 31.5 K⋅→

v31−

2m⋅ 54− 153 x⋅ 36 x2⋅−+ 2 x3⋅+( )⋅ K⋅

substitute x 6,

float 3,v3 0→

231



obtained in the previous section are plotted using the Mathcad graphic capabilities.

l 6 m⋅:= xmax38

l⋅:=

→ xmax 2.25m= lintxmax

6:=

nint llint

:= → nint 16= lspan lm

:=

i 0 nint..:=

posilspannint

i⋅:=

shear12i 18 8 posi( )⋅−:= shear23i 6−:=

mom12i 18 posi( )⋅ 4 posi( )2⋅−:=

mom23i 36 6 posi( )⋅−:=

rot12i 9 posi( )2⋅43

posi( )3⋅− 40.5−:=

rot23i 76.5− 36 posi( )⋅+ 3 posi( )2⋅−:=

disp12i 40.5− posi( )⋅ 3 posi( )3⋅+13

posi( )4⋅−:=

disp23i 27 76.5 posi( )⋅− 18 posi( )2⋅+ posi( )3−:=

sheari if posilspan

2≤ shear12i, shear23i,

:=

momi if posilspan

2≤ mom12−( )i, mom23−( )i,

:=

roti if posilspan

2≤ rot12i, rot23i,

:=

dispi if posilspan

2≤ disp12i, disp23i,

:=

232


pos

0012345678910111213141516

00.375

0.751.125

1.51.875

2.252.625

33.375

3.754.125

4.54.875

5.255.625

6

=

shear

0012345678910111213141516

1815129630

-3-6-6-6-6-6-6-6-6-6

= mom

0012345678910111213141516

0-6.188-11.25

-15.188-18

-19.688-20.25

-19.688-18

-15.75-13.5

-11.25-9

-6.75-4.5

-2.250

=

0 1 2 3 4 5 630

20

10

0

10

20


Position (m)

Shea

r For

ce (k

N),

Bend

ing

Mom

ent (

kN*m

)

shear

mom

pos

233


pos

0012345678910111213141516

00.375

0.751.125

1.51.875

2.252.625

33.375

3.754.125

4.54.875

5.255.625

6

=

rot

0012345678910111213141516

-40.5-39.305

-36-31.008-24.75

-17.648-10.125-2.602

4.510.82816.31320.95324.75

27.70329.81331.078

31.5

=

disp

0012345678910111213141516

0-15.036-29.215-41.825-52.313-60.282-65.496-67.876

-67.5-64.6

-59.484-52.471-43.875-34.014-23.203-11.76

0

=

0 1 2 3 4 5 680

60

40

20

0

20

40

Rotation and Vert Displacement DiagramsPosition (m)

Rot

atio

n/K

(rad)

, Dis

plac

emen

t/K (m

)

rot

disp

pos

234

Problem 8.2.4 Repeat the calculations required by the problem 8.2.3 using the load-

deflection fourth-order differential equation.


A.1. The beam illustrated in Figure 8.2.3 is investigated using the load-deflection

fourth-order differential equation (8.13). The differential equation (8.13) can be used

to investigate the deformation of any system. The beam proposed for investigation is

characterized by a constant cross-section along the entire length and represents a

statically determinate system. Consequently, the equations (8.36) and (8.39) are

employed.


- at point 1 M1 0 , v1 0 ;

- at point 3 M3 0 , v3 0 .

A.3 The acting load is characterized by two continuity intervals, 1-2 and 2-3. In the



conformed by the calculations. Consequently, four continuity equations representing

the equality of shear forces, bending moments, rotations and deflections at the left and

at the right locations of the common point 2, can be written:

- V2_12 V2_23 ;

- M2_12 M2_23 ;

- θ2_12 θ2_23 ;

- v2_12 v2_23 .

A.4 For simplicity of the calculations the following notation is used when applying

the formulae (8.22) and (8.23):

zIEK

*1

=

235

B. Calculations

B.1 Calculation of the Shear Forces, Bending moments, Rotations and Vertical

Displacements


,0[ lx∈

The distributed loading existing on this continuity interval is a uniformly distributed

load:

pn x( ) p−

To apply the formulae (8.36) to (8.39) the values of the shear force, bending moment,

rotation and vertical displacement pertinent to the beginning of the interval, at

0=x ,

has to be identified. They are:

V0 V1 unknown value

M0 0 boundary condition in point 1

θ0 θ1:= θ1 unknown value

v0 0

boundary condition in point 1

The following functions are calculated using the expressions (8.36) through (8.39):

V12 x( ) xp−( )⌠⌡

d V1+ V12 x( ) p− x⋅ V1+→

236

The values of the shear force, bending moment, rotation and vertical displacement at

the end of the interval 1-2, at point 2, are obtained substituting 2lx = into the above

formulae:

V2_12 V12 x( )substitute V12 x( ) p− x⋅ V1+,

substitute x l2

,V2_12

1−2

p⋅ l⋅ V1+→

M2_12 M12 x( )

substitute M12 x( ) 1−2

p⋅ x2⋅ V1 x⋅+,

substitute x l2

,M2_12

1−8

p⋅ l2⋅12

V1⋅ l⋅+→

237


,0[ lx∈

No distributed load is acting on the continuity interval 2-3 and consequently:

pn x( ) 0



0=x ,

has to be identified. They are obtained employing the continuity conditions existing at

point 2:

V0 V2_23 V2_121−

2p⋅ l⋅ V1+

continuity condition in point 2

M0 M2_23 M2_121−

8p⋅ l2⋅

12

V1⋅ l⋅+


θ0 θ2_23 θ2_121−

48l3⋅ p⋅

18

V1⋅ l2⋅+

K⋅ θ1+:= θ2_23


v0 v2_23 v2_12 K 1−384

p⋅ l4⋅148

V1⋅ l3⋅+

⋅

12

θ1⋅ l⋅+

continuity condition in point 2.


V23 x( ) x0( )⌠⌡

d V2_23+substitute V2_23

1−2

p⋅ l⋅ V1+,

collect p,V23 x( ) 1−

2p⋅ l⋅ V1+→

238

239

The values at the end of the continuity interval 2-3 are obtained substituting 2lx =

into the above formulae:

V3 V23 x( )


p⋅ l⋅ V1+,

substitute x l2

,V3

1−2

p⋅ l⋅ V1+→

240


to both continuity intervals, 1-2 and 2-3, are dependent on the unknown value of the

shear force 1V and rotation 1θ .

B.2. Solution

The unknown values of the shear force 1V and rotation 1θ are obtained by imposing

the boundary conditions at point 3. They are:

M3 0 →

V1 l⋅38

p⋅ l2⋅− 0 solve V1,38

p⋅ l⋅→

v3 0 →

5−128

l4⋅ p⋅16

V1⋅ l3⋅+

K⋅ θ1 l⋅+ 0

substitute V138

p⋅ l⋅,

solve θ1,3−

128p⋅ l3⋅ K⋅→

θ13−

128K⋅ p⋅ l3⋅


241

The final expressions of the shear force 1V and of the rotation 1θ , respectively, are

replaced in the previously determined functions of the shear forces, bending moments,

rotations and vertical displacements pertinent to continuity intervals 1-2 and 2-3.


,0[ lx∈

The final expressions are:

V12 x( ) V1 p x⋅−substitute V1

38

p⋅ l⋅,

collect p,V12 x( ) x−

38

l⋅+

p⋅→

M12 x( ) 1−2

p⋅ x2⋅ V1 x⋅+substitute V1

38

p⋅ l⋅,

collect p,M12 x( ) 1−

2x2⋅ 3

8l⋅ x⋅+

p⋅→

The values calculated at the end of the continuity interval are:

V2_121−

2p⋅ l⋅ V1+ substitute V1

38

p⋅ l⋅, V2_121−

8p⋅ l⋅→

M2_121−

8p⋅ l2⋅ 1

2V1⋅ l⋅+ substitute V1

38

p⋅ l⋅, M2_12116

p⋅ l2⋅→

242

θ2_12 θ15

192K⋅ p⋅ l3⋅+ substitute θ1

3−128

K⋅ p⋅ l3⋅, θ2_121

384p⋅ l3⋅ K⋅→

The position and value of maximum bending moment are calculated as:

V12 x( ) 0 → 38

l⋅ x−

p⋅ 0 solve x, 3

8l⋅→ → xmax

38

l⋅

M12_max38

l⋅ x⋅12

x2⋅−

p⋅ substitute x 38

l⋅, M12_max9

128p⋅ l2⋅→


,0[ lx∈

The final expressions are calculated:

V23 x( ) V112

p⋅ l⋅− substitute V138

p⋅ l⋅, V23 x( ) 1−8

p⋅ l⋅→

243

Note: It has to be remarked that all four functions V , M , ϑ and v are proportional

with the load intensity p , while the functions representing the deformation only, ϑ

and v , are additionally proportional with the inverse of the flexural rigidity K . This

proportionality is a characteristic of the elastic behavior.

B.4 Usage of Unique Origin ],0[ lx∈


− in the




V23 x( ) 1−8

p⋅ l⋅ substitute x x l2

−, V23 x 12

l⋅−

1−8

p⋅ l⋅→

M23 x( ) 116

l⋅ 2− x⋅ l+( )⋅ p⋅substitute x x l

2−,

simplifyM23 x 1

2l⋅−

18

l⋅ x− l+( )⋅ p⋅→

244

V3 V23 x( )substitute V23 x( ) 1−

8p⋅ l⋅,

substitute x l,V3

1−8

p⋅ l⋅→

M3 M23 x( )substitute M23 x( ) 1

8l⋅ x− l+( )⋅ p⋅,

substitute x l,M3 0→

verification

verification



l 6 m⋅ p 8kNm

⋅

K K 1

kN m2⋅⋅

B.5. a Interval 1-2 ]2

,0[ lx∈

V12 x( ) 38

l⋅ x−

p⋅

substitute p 8kNm

⋅,



simplify

V12 x m⋅( ) 2− 4 x⋅ 9−( )⋅ kN⋅→

245

M12 x( ) 38

l⋅ x⋅ 12

x2⋅−

p⋅

substitute p 8kNm

⋅,



simplify

M12 x m⋅( ) 2− x⋅ m⋅ 2 x⋅ 9−( )⋅ kN⋅→

B.5. b Interval 2-3 ],2

[ llx∈

V23 x( ) 1−8

p⋅ l⋅

substitute p 8kNm

⋅,


simplify

V23 x( ) 6− kN⋅→

M23 x( ) 18

l⋅ x− l+( )⋅ p⋅

substitute p 8kNm

⋅,



simplify

M23 x m⋅( ) 6− m⋅ x 6−( )⋅ kN⋅→

246

B.5. c Values at Important Points

V138

p⋅ l⋅substitute p 8

kNm

⋅,

substitute l 6 m⋅,V1 18 kN⋅→

V21−

8p⋅ l⋅

substitute p 8kNm

⋅,

substitute l 6 m⋅,V2 6− kN⋅→

M2116

p⋅ l2⋅substitute p 8

kNm

⋅,

substitute l 6 m⋅,M2 18 kN⋅ m⋅→

247

M12_max9

128p⋅ l2⋅

substitute p 8kNm

⋅,


float 5,

M12_max 20.250 kN⋅ m⋅→

θ13−

128K⋅ p⋅ l3⋅

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 3,

θ1 40.5− K⋅→

θ21

384K⋅ p⋅ l3⋅

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 3,

θ2 4.50 K⋅→

v25−

768K⋅ p⋅ l4⋅

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 4,

v2 67.50− K⋅ m⋅→

V31−

8p⋅ l⋅

substitute p 8kNm

⋅,

substitute l 6 m⋅,V3 6− kN⋅→

M3 0

substitute p 8kNm

⋅,

substitute l 6 m⋅,M3 0→

θ37

384K⋅ p⋅ l3⋅

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 4,

θ3 31.50 K⋅→

248

v3 0

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 4,

v3 0→




l 6 m⋅:= xmax38

l⋅:=

→ xmax 2.25m= lintxmax

6:=

nint llint

:= → nint 16= lspan lm

:=

i 0 nint..:=

posilspannint

i⋅:=

shear12i 2− 4 posi⋅ 9−( )⋅:=

shear23i 6−:=

mom12i 2− posi( )⋅ 2 posi⋅ 9−( )⋅:=

mom23i 6− posi( ) 6− ⋅:=

sheari if posilspan


:=

momi if posilspan

2≤ mom12−( )i, mom23−( )i,

:=

249


pos

0012345678910111213141516

00.375

0.751.125

1.51.875

2.252.625

33.375

3.754.125

4.54.875

5.255.625

6

=

shear

0012345678910111213141516

1815129630

-3-6-6-6-6-6-6-6-6-6

=

mom

0012345678910111213141516

0-6.188-11.25

-15.188-18

-19.688-20.25

-19.688-18

-15.75-13.5

-11.25-9

-6.75-4.5

-2.250

=

0 1 2 3 4 5 630

20

10

0

10

20


Position (m)

Shea

r For

ce (k

N),

Bend

ing

Mom

ent (

kN*m

)

shear

mom

pos

250

rot12i1−

68 posi( )3⋅ 54 posi( )2⋅− 243+ ⋅:=

rot23i3−

22 posi( )2⋅ 24 posi( )⋅− 51+ ⋅:=

disp12i1−

6posi( )⋅ 2 posi( )3⋅ 18 posi( )2⋅− 243+ ⋅:=

disp23i1−

22 posi( )3⋅ 36 posi( )2⋅− 153 posi( )⋅ 54−+ ⋅:=

roti if posilspan


:=

dispi if posilspan


:=

Position (m) Rotation/K (rad) Vertical Displacement/K (kN*m)

pos

0012345678910111213141516

00.3750.75

1.1251.5

1.8752.25

2.6253

3.3753.75

4.1254.5

4.8755.25

5.6256

=

rot

0012345678910111213141516

-40.5-39.305

-36-31.008-24.75

-17.648-10.125-2.602

4.510.82816.31320.95324.75

27.70329.81331.078

31.5

=

disp

0012345678910111213141516

0-15.036-29.215-41.825-52.313-60.282-65.496-67.876

-67.5-64.6

-59.484-52.471-43.875-34.014-23.203-11.76

0

=

251

0 1 2 3 4 5 680

60

40

20

0

20

40

Rotation and Vert Displacement Diagrams

Position (m)

Rot

atio

n/K

(rad)

, Ver

t Dis

pl/K

(m)

31.5

67.876−

rot

disp

60 pos

Problem 8.2.5 The beam pictured in Figure 8.2.5 has a constant cross-section along

its entire length. Using the fourth-order differential equation conduct the following

tasks: (a) determine the mathematical expressions of the rotation )(xθ and vertical

deflection )(xv curves, (b) determine the expressions of the rotation and deflection in

all important points of the beam and (c) conduct a numeric application considering

mkNp *8= , ml *6= .

Figure 8.2.5

252


A.1. The beam illustrated in Figure 8.2.5 is investigated using the load-deflection

fourth-order differential equation (8.13). The differential equation (8.13) can be used

to investigate the deformation of any system. The beam proposed for investigation is

characterized by a constant cross-section along the entire length and represents a

statically indeterminate system. Consequently, equations (8.36) and (8.39) are

employed.

A.2. The boundaries are:

- at point 1 θ1 0, v1 0 ;

- at point 3 M3 0 , v3 0 .

A.3. The acting load is characterized by two continuity intervals, 1-2 and 2-3. In the



conformed by the calculations. Consequently, four continuity equations representing

the equality of shear forces, bending moments, rotations and deflections at the left and

right locations of the common point 2, can be written:

- V2_12 V2_23 ;

- M2_12 M2_23 ;

- θ2_12 θ2_23 ;

- v2_12 v2_23 .

A.4. For simplicity of the calculations the following notation is used when applying

the formulae (8.22) and (8.23):

zIEK

*1

=

B. Calculations

B.1 Calculation of the Shear Forces, Bending moments, Rotations and Vertical

Displacements

253


,0[ lx∈

The distributed loading existing on this continuity interval is an uniform distributed

load:

pn x( ) p−

To apply the formulae (8.36) to (8.39), the values of the shear force, bending moment,


0=x ,

have to be identified. They are:

V0 V1 unknown value

θ0 0 boundary condition in point 1

M0 M1:= M1 unknown value

v0 0 boundary condition in point 1


V12 x( ) xp−( )⌠⌡

d V1+ V12 x( ) p− x⋅ V1+→

M12 x( ) xxp−⌠⌡

d⌠⌡

d V1 x⋅+ M1+

M12 x( ) xxp−⌠⌡

d⌠⌡

d V1 x⋅+ M1+ collect p, M12 x( ) 1−2

p x2⋅⋅ V1 x⋅+ M1+→

254

The values of the shear force, bending moment, rotation and vertical displacement at

the end of the interval 1-2, at point 2, are obtained substituting 2lx = into the above

formulae:

V2_12 V12 x( )substitute V12 x( ) p− x⋅ V1+,

substitute x l2

,V2_12

1−2

p⋅ l⋅ V1+→

M2_12 M12 x( )

substitute M12 x( ) 1−2

p⋅ x2⋅ V1 x⋅+ M1+,

substitute xl2

,M2_12

1−8

p l2⋅⋅12

V1 l⋅⋅+ M1+→

255


,0[ lx∈

No distributed load is acting on the continuity interval 2-3 and consequently:

pn x( ) 0



0=x ,

have to be identified. They are obtained employing the continuity conditions existing

at point 2:

V0 V2_23 V2_121−

2p⋅ l⋅ V1+


M0 M2_23 M2_121−

8p⋅ l2⋅

12

V1⋅ l⋅+ M1+ continuity condition in point 2

θ0 θ2_23 θ2_121−

48l3⋅ p⋅ 1

8V1⋅ l2⋅+ 1

2M1⋅ l⋅+

K⋅:= θ2_23


v0 v2_23 v2_12 K1−

384p⋅ l4⋅

148

V1⋅ l3⋅+18

M1⋅ l2⋅+

⋅

continuity condition in point 2.


V23 x( ) x0( )⌠⌡

d V2_23+substitute V2_23

1−2

p⋅ l⋅ V1+,

collect p,V23 x( ) 1−

2p⋅ l⋅ V1+→

256

+

257

The values at the end of the continuity interval 2-3 are obtained substituting 2lx =

into the above formulae:

V3 V23 x( )


p⋅ l⋅ V1+,

substitute x l2

,V3

1−2

p⋅ l⋅ V1+→

+

+

+

258


to both continuity intervals, 1-2 and 2-3, are dependent on the unknown values of the

shear force 1V and moment 1M .

B.2. Solution

The unknown values of the shear force 1V and moment 1M are obtained by imposing

the boundary conditions at point 3. They are:

M3 0 →

v3 0 →

V1 l⋅ 38

p⋅ l2⋅− M1+ 0 solve V1, 18

3 p l2⋅⋅ 8 M1⋅−

l⋅→

5−128

l4⋅ p⋅ 16

V1⋅ l3⋅+ 12

M1⋅ l2⋅+

K⋅ 0

substitute V118

3 p⋅ l2⋅ 8 M1⋅−

l⋅,

solve M1,9−

128p l2⋅⋅→

V118

3 p⋅ l2⋅ 8 M1⋅−

l⋅ substitute M1

9−128

p⋅ l2⋅, V157128

p l⋅⋅→


The final expressions of shear force 1V and moment 1M , respectively, are replaced in

the previously determined functions of the shear forces, bending moments, rotations

and vertical displacements pertinent to continuity intervals 1-2 and 2-3.


,0[ lx∈

The final expressions are:

V12 x( ) V1 p x⋅−substitute V1

57128

p⋅ l⋅,

collect p,V12 x( ) x−( ) 57

128l⋅+

p⋅→

259

The values calculated at the end of the continuity interval are:

260

The position and value of maximum bending moment are calculated as:

V12 x( ) 0 → x− 57128

l⋅+

0 solve x, 57

128l⋅→

xmax57128

l⋅

The position of zero moment is computed as:


,0[ lx∈

The final expressions are calculated:

V23 x( )1−

2p⋅ l⋅ V1+ substitute V1

57128

p l⋅, V23 x( )7−

128p l⋅⋅→

+

261

(verification)

(verification)

Verify the continuity conditions:

262

Note: It has to be remarked that all four functions V , M , ϑ and v are proportional

with the load intensity p , while the functions representing the deformation only, ϑ

and v , are additionally proportional with the inverse of the flexural rigidity EI. This

proportionality is a characteristic of the elastic behavior.

B.4 Usage of Unique Origin ],0[ lx∈


− in the




V23 x( ) 7−128

p⋅ l⋅ substitute x xl2

−, V23 x 12

l⋅−

7−128

p l⋅⋅→

V3 V23 x( )substitute V23 x( ) 7−

128p⋅ l⋅,

substitute x l,V3

7−128

p l⋅⋅→

263

M3 M23 x( )substitute M23 x( ) 7−

128p⋅ l⋅ x⋅

7128

p⋅ l2⋅+,

substitute x l,M3 0→

(verification)

(verification)



l 6 m⋅ p 8kNm

⋅

K K 1

kN m2⋅⋅

B.5. a Interval 1-2 ]2

,0[ lx∈

V12 x( ) x− 57128

l⋅+

p⋅

substitutep 8kNm

⋅,

substitutel 6 m⋅,

substitutex x m⋅,

simplify

V12 x m⋅( ) 1−8

64 x⋅ 171−( ) kN⋅⋅→

264

B.5. b Interval 2-3 ],2

[ llx∈

V23 x( )7−

128p⋅ l⋅

substitute p 8kNm

⋅,


simplify

V23 x( )21−8

kN⋅→

265

B.5. c Values at Important Points

The maximum moment:

266

M12_max945

32768p⋅ l2⋅

substitute p 8kNm

⋅,


float 5,

M12_max 8.3057 kN m⋅⋅→

The position of zero moment:

M12_max945

32768p⋅ l2⋅

substitute p 8kNm

⋅,


float 5,

M12_max 8.3057 kN m⋅⋅→

267

M3 0

substitute p 8kNm

⋅,

substitute l 6 m⋅,M3 0→

(verification)

v3 0

substitute p 8kNm

⋅,


substitute K K 1

kN m2⋅⋅,

float 4,

v3 0→

(verification)




l 6 m⋅:= nint 16:= nint1 6:=

nint2nint2

nint1−:= nint2 2= nint3 8:=

xmax57128

l⋅:= xmax 2.672m= dlint1xmaxnint1

:=

dlint1 0.445m= dlint2

l2

xmax−

nint2:= dlint2 0.164m=

dlint3

l2

nint3:= dlint3 0.375m=

268

lspanlm

:= pos0 0:= shear01718

:=

mom0162−8

−:= rot0 0:= disp0 0:=

i 1 nint..:=

dlinti if i nint1≤ dlint1, if inint2

≤ dlint2, dlint3,

,

:=

posi posi 1−dlintim

+:=

shear12i1−

864 posi⋅ 171−( )⋅:=

shear23i21−8

:=

mom12i1−

832 posi( )2⋅ 171 posi⋅− 162+ ⋅:=

mom23i21−8

posi 6−( )⋅:=

sheari if posilspan


:=

momi if posilspan

2≤ mom12−( )i, mom23−( )i,

:=


pos

0012345678910111213141516

00.4450.8911.3361.7812.2272.6722.836

33.375

3.754.125

4.54.875

5.255.625

6

= shear

0012345678910111213141516

21.37517.813

14.2510.688

7.1253.563

0-1.313-2.625-2.625-2.625-2.625-2.625-2.625-2.625-2.625-2.625

= mom

0012345678910111213141516

20.2511.525

4.386-1.167-5.133-7.512-8.306-8.198-7.875-6.891-5.906-4.922-3.938-2.953-1.969-0.984

0

=

269

0 1 2 3 4 5 610

5

0

5

10

15

20

25


Position (m)

Shea

r For

ce (k

N),

Ben

ding

Mom

ent (

kN*m

)

shear

mom

pos

rot12i1−

48posi⋅ 64 posi( )2⋅ 513 posi⋅− 972+ ⋅:=

rot23i3−

167 posi( )2⋅ 84 posi⋅− 192+ ⋅:=

disp12i1−

48posi( )2⋅ 16 posi( )2⋅ 171 posi⋅− 486+ ⋅:=

disp23i1−

16126− posi( )2⋅ 576 posi⋅ 432−+ 7 posi( )3⋅+ ⋅:=

roti if posilspan


:=

dispi if posilspan


:=

270

Position (m) Rotation/K (rad) Vertical Displacement/K (kN*m)

pos

0012345678910111213141516

00.4450.8911.3361.7812.2272.6722.836

33.375

3.754.125

4.54.875

5.255.625

6

= rot

0012345678910111213141516

0-7.016

-10.5-11.157-9.696-6.822-3.241-1.884-0.5632.2064.6056.6368.2979.589

10.51211.06511.25

= disp

0012345678910111213141516

0-1.706-5.724

-10.638-15.347-19.064-21.317-21.738-21.938-21.618-20.329-18.21

-15.398-12.033-8.253-4.196

0

=

Problem 8.2.6 The beam illustrated in Figure 8.2.6 has a constant cross-section along

its entire length. Using the Superposition Method conduct the following tasks: (a)

derive mathematical expressions for rotation )(xθ and vertical deflection )(xv

functions, (b) develop expressions for rotation and vertical deflection characterizing

271

all important points of the beam, (c) conduct a numeric application considering

mkNp *80 = and ml *6= and (d) draw the deflection curve and rotation for various

values of the parameter α .

Figure 8.2.6


A.1. The loading of the cantilever beam illustrated in Figure 8.2.4 is composed from

the addition of two individual cases: (a) Load Case I - a uniform distributed force 0p

applied on the left side of the beam between points 1 and 2 and (b) Load Case II - a

concentrated force 0** plP α= acting on the tip of the cantilever, in point 3.

The loading presents two continuity intervals, 1-2 and 2-3.

A.2 The expressions of the deflection curve )(xv and rotation )(xθ for Load Case I

are obtain from Table T1, case 6, of the Appendix A by replacing 2la = into the

formulae:

• for the continuity interval 1-2 2

0 lx ≤≤

)**2*23(**

*24)( 2220 xxllx

EIpxv +−=

)**23*

43(**

*6)( 220 xxllx

EIpx +−=θ

• for the continuity interval 2-3 lxl≤≤

2

)*81(*

*48*

)(3

0 lxEIlpxv −=

272

EIlpx

*48*

)(3

0=θ

The above obtained formulae are changed, for future graphical representation, into a

more manageable format by replacing the variable lx *ς= :

• for the continuity interval 1-2 210 ≤≤ ς

+−= 22

40 *2

23**

*24*)( ςςςEIlpxv

)*23

43(**

*6*

)( 23

0 ςςςθ +−=EIlpx


≤≤ ς

)81(*

*48*

)(4

0 −= ςEIlpxv

EIlpx

*48*

)(3

0=θ

A.3 The expressions of the deflection curve )(xv and rotation )(xθ for Load Case II

are obtain from Table T1, case 3, of the Appendix A by replacing 0** plP α−= into

the formulae. The formulae are identical for both continuity intervals 1-2 and 2-3:

)*3(***6

**)( 20 xlx

EIplxv −−=

α

)*2(***2

**)( 0 xlx

EIplx −−=

αθ

After replacing lx *ς= into the above expressions the deflection curve )(xv and

rotation )(xθ are:

)3(***6

**)( 2

40 ςς

ας −−=

EIlpv

)2(***2

**)(

30 ςς

αςθ −−=

EIlp

where the parameter ς is described by the following set: 10 ≤≤ ς

273

B. Calculations

B.1. Expressions of the deflection curve )(xv and rotation )(xθ

The expressions of the deflection curve )(xv and rotation )(xθ are obtained for each

of the two continuity intervals, 1-2 and 2-3, by summations of the formulae obtained

above. After algebraic manipulations, they are:


+−−

−= 22

40 )*21(**2*12

23**

*24*

)( ςαςαςςEIlpv

+

−−

−= 2

30

21**3*6

43**

*6*

)( ςαςαςςθEIlp


≤≤ ς

+−+−= )**8**24

81*

*48*

)( 324

0 ςαςαςςEIlpv

[ ])2(***241*48*

)(3

0 ςςαςθ −−=EIlp

B.2. Deflection curve )(xv and rotation )(xθ in Points 1, 2 and 3

These expressions are calculated by replacing the corresponding ς coordinate

pertinent to each point in the above obtained formulae:

9 for Point 1 (fixed point) 00 =⇒= ςx

00 =⇒= ςx is substituted into the formulae pertinent to continuity interval 1-2:

0)0(1 === ςvv

0)0(1 === ςθθ

• for Point 2 21

2=⇒= ς

lx

274

21

2=⇒= ς

lx is substituted into the formulae pertinent to continuity interval 1-2 or

continuity interval 2-3. Using the formulae pertinent to the continuity interval 1-2, the

following values are calculated:

)*403(**384*

21)*21(*

21*2*12

23*

21*

*24*

)21(

40

2240

2

α

αας

−=

=

+−

−

−

===

EIlp

EIlpvv

)*181(**48*

21

21*

21*3*6

43*

21*

*6*

)21(

30

230

2

α

ααςθθ

−=

=

+

−

−

−

===

EIlp

EIlp

Using the formulae pertinent to the continuity interval 2-3, the following values are

calculated:

)*403(**384*

)21**8

21**24

21

81*

*48*

)21(

40

3240

2

α

αας

−=

=

+

−

+−===

EIlp

EIlpvv

)*181(**48*

)212(*

21**241

*48*

)21(

30

30

2

α

αςθθ

−=

=

−

−===

EIlp

EIlp

• for Point 3 1=⇒= ςlx

1=⇒= ςlx is substituted into the formulae pertinent to continuity interval 2-3:

( ) ( ) ( )

)*1287(**384*

)1**81**24181*

*48*

)1(

40

324

02

α

αας

−=

=

+−+−===

EIlp

EIlp

vv

( ) ( )[ ]

)*241(**48*

)12(*1**241*48*

)1(

30

30

2

α

αςθθ

−=

=−−===

EIlp

EIlp

275

B.3. Deflection Curve and Rotation Graphical Representation

In preparation for plotting, the deflection curve )(xv and rotation )(xθ formulae

pertinent to the continuity intervals 1-2 and 2-3 obtained above are modified to have

the same common factor EIlp

*48* 4

0 and EIlp

*48* 3

0 , respectively:


),(**48*

)*21(**2*1223**2*

*48*

)(

12

40

224

012

αζ

ςαςας

cvEIlp

EIlp

xv

=

=

+−−

−=

),(**48*

21**3*6

43**8*

*48*

)(

12

30

23

012

αζθ

ςαςαςθ

cEIlp

EIlp

x

=

=

+

−−

−=


≤≤ ς

),(**48*

)**8**2481*

*48*

)(

23

40

324

023

αζ

ςαςας

cvEIlp

EIlp

xv

=

=

+−+−=

[ ]

),(**48*

)2(***241*48*

)(

23

30

30

23

αζθ

ςςαθ

cEIlp

EIlp

x

=

=−−=

The coefficients ),(12 αζcv , ),(12 αζcv , ),(12 αζθc and ),(12 αζθc are used to draw

the variation of the normalized deflection curve EIlp

xv*48*

/)(4

0 and rotation

EIlp

x*48*

/)(3

0θ . The plot is conducted for a value α 0.5:= and a number of cross-

sections nint 11:= .

276

i 0 nint 1−..:= ; ζi 0.1 i⋅:=

cv12i 2 ζi( )2 32

12 α⋅−

2 ζi⋅ 1 2 α⋅−( )⋅− ζi( )2+

⋅:=

cθ12i 8 ζi⋅34

6 α⋅−

3 ζi⋅

12

α−

⋅− ζi( )2+

⋅:=

cv23i1−8

ζi+ 24 α⋅ ζi( )2⋅− 8 α⋅ ζi( )3⋅+:=

cθ23i 1 24 α⋅ ζi⋅ 2 ζi−( )⋅−:=

cvi if ζi12

≤ cv12i, cv23i,

:=

cθi if ζi

12

≤ cθ12i, cθ23i,

:=

ζ

0012345678910

00.10.20.30.40.50.60.70.80.91

=

cv

0012345678910

0-0.09-0.357-0.794-1.389-2.125-2.981-3.933-4.957-6.029-7.125

=

cθ

0012345678910

0-1.792-3.536-5.184-6.688

-8-9.08-9.92-10.52-10.88-11

=

277

8.3. Proposed Problems

Problem 8.3.1 The deflection v(x) of the simply supported beam illustrated in Figure

8.3.1 is:

+

−

=

Lx

Lx

Lx

IELq

xv *6*7***840

*)(

3740

Considering that the flexural rigidity of the beam E*I = constant along its entire

length, determine: (a) the maximum absolute deflection maxv , (b) the expression of the

rotation )(xθ , (c) the expressions of the shear force V(x), bending moment M(x) and

distributed load p(x)and (c) the expressions of the reactions RA and RB. Plot all the

graphs using the Mathcad graphic capabilities.

Figure 8.3.1

Problem 8.3.2 The deflection v(x) of the cantilevered beam illustrated in Figure 8.3.2

is:

−

+

+

−= 4*5*2*

**48*

)(344

0

Lx

Lx

Lx

IELp

xv

.

Figure 8.3.2

Considering the flexural rigidity of the beam E*I = constant along its entire length

calculate: (a) the maximum absolute deflection maxv , (b) the expressions of the shear

278

force V(x) and bending moment M(x), (c) the expressions of the concentrated force P,

reactions R2 and M2 and (d) the expression of the rotation )(xθ . Plot all the graphs

using the Mathcad graphic capabilities

Problems 8.3.3 – 8.3.20 Integrating the moment-curvature second-order differential

equation and considering that the flexural rigidity E*I = constant along the entire

length of the beam obtain the expressions of: (a) the rotation )(xϑ (b) deflection v(x)

and (c) maximum rotation maxϑ and deflection maxv for the beams illustrated in

Figures 8.3.3 through 8.3.20. Plot all the pertinent graphs using Mathcad graphic

capabilities.

Figure 8.3.3

Figure 8.3.4

Figure 8.3.5

279

Figure 8.3.6

Figure 8.3.7

Figure 8.3.8

Figure 8.3.9

280

Figure 8.3.10

Figure 8.3.11

Figure 8.3.12

Figure 8.3.13

281

Figure 8.3.14

Figure 8.3.15

Figure 8.3.17

Figure 8.3.18

282

Figure 8.3.19

Figure 8.3.20

Problems 8.3.21 – 8.3.33 Integrating the load-deflection fourth-order differential

equation and considering that the beams illustrated in Figures 8.3.21 through 8.3.33

are characterized by a flexural rigidity E*I = constant along the entire length of the

beam, determine the expressions of: (a) the shear force V(x), (b) bending moment

M(x), (c) rotation )(xϑ , (d) deflection v(x), (e) maximum rotation maxϑ and (f)

deflection maxv . Plot all the pertinent graphs using Mathcad graphic capabilities.

Figure 8.3.21

283

Figure 8.3.22

Figure 8.3.23

Figure 8.3.24

Figure 8.3.25

284

Figure 8.3.26

Figure 8.3.27

Figure 8.3.28

Figure 8.3.29

285

Figure 8.3.30

Figure 8.3.31

Figure 8.3.32

Figure 8.3.33

286

Problem 8.3.34 For the cantilever beam illustrated in Figure 8.3.34, obtain the

expressions of the rotation )(xϑ and vertical deflection v(x) by integrating the

moment-curvature second-order differential equation. Find the maximum rotation

maxϑ and vertical displacement maxv of the beam.

Figure 8.3.34

Problem 8.3.35 The uniform cylindrical beam shown in Figure 8.3.35 is subjected,

through the pulley located at point 2, to a downward concentrated load P. Neglecting

the width of the pulley and of the bearings and assuming that the bearings provide

only vertical support for the beam, calculate the expressions of: (a) the deflection

curve v(x), (b) rotation )(xϑ and (c) maximum rotation maxϑ and deflection maxv .

Conduct the calculation by integrating the moment-curvature second-order

differential equation. Plot all the graphs using the Mathcad graphic capabilities.

Repeat the calculations for the case when an additional weight W=2P acts in point D.

Figure 8.3.35

Problem 8.3.36 A number of cement bags are stacked on the over-hanging segment,

B-C, of the beam shown in Figure 8.3.36, producing a uniform distributed load of

magnitude p0 per unit length. Using the moment-curvature second-order differential

equation determine the following expressions: (a) the deflection curve v(x), (b)

rotation )(xϑ and (c) maximum rotation maxϑ and deflection maxv . Repeat the

287

calculations using the load-deflection fourth-order differential equation. Plot all the

pertinent graphs using the Mathcad graphic capabilities.

Figure 8.3.36

Problem 8.3.37 The exercising bar of diameter d = 25 mm, shown in Figure 8.3.37,

rests on a rigid stand.

Figure 8.3.37

Assimilating the exercising bar as a simply supported beam with a span of L = 1 m

and assuming that each pair of weights exerts a concentrated force W = 100 N at a

distance a = 200 mm beyond the support points, calculate the following expressions:

(a) the deflection curve v(x), (b) rotation )(xϑ and (c) maximum rotation maxϑ and

deflection maxv . Assume that the bar is made of steel with a modulus of elasticity E =

210 GPa.

Problem 8.3.38 The wide-flange beam AB, shown in Figure 8.3.38, is supported at

both ends by two rods AC and BD. The rods are fabricated from circular bars with

equal diameters. Considering that a uniform distributed force p0 acts on the entire

length of the beam and that the beam and rods are fabricated from the same type of

288

steel, calculate the total elongation of the rods and the maximum deflection in the

beam by integrating the moment-curvature second-order differential equation.

Figure 8.3.38

Problems 8.3.39 – 8.3.40 Determine the expressions of the deflection curve v(x) and

rotation )(xϑ for the beams illustrated in Figures 8.3.38 and 8.3.40 by employing the

Superposition Method and pertinent expressions obtained from Table T1 contained in

Appendix A. Particularize the obtained expressions for the locations of points A, B, C

and D and draw the deflection curve v(x) and rotation )(xϑ using Mathcad graphical

capabilities.

Figure 8.3.39

Figure 8.3.40

289

Problems 8.3.41 Determine the expressions of the deflection curve v(x) and rotation

)(xϑ for the system illustrated in Figure 8.3.41 by employing the Superposition

Method and pertinent expressions obtained from Table T1 contained in Appendix A.

Particularize the obtained expressions for the locations of points A, B, C and D and

draw the deflection curve v(x) and rotation )(xϑ using Mathcad graphical

capabilities.

Figure 8.3.41

290

APPENDIX 8.1 Deflection Formulae

Table T1 - Cantilever Beam

Notation: v(x) = deflection curve θ(x) = rotation vB = v(L) = deflection at end B θB = θ(L) = rotation at end B

291

292

293

294

Table T2 - Simple Supported Beam Notation: v(x) = deflection θ(x) = rotation θA = θ(0) = rotation at end A θB = - θ(L) = rotation at end B xm = location of the point of maximum deflection

vC = vL2

= deflection at the center of the beam

vmax = max v x( ) = maximum deflection

295

296

297

298

299

Table T3 Fixed-Fixed Beam Notation: v(x) = deflection θ(x) = rotation xm = location of the point of maximum deflection vmax = max v x( ) = maximum deflection MC M x a( ) l = left cross-section

r = right cross-section

300

301

302

303

304

305

CHAPTER 9 Torsion 9.1. Theoretical Background

A plane linear member subjected to a torsional moment )(xT undergoes a torsional

deformation if after the deformation:

(a) the axis of the member remains straight and without longitudinal

extension;

(b) the cross-sections remain plane and perpendicular to the longitudinal

axis of the beam;

(c) radial lines remain straight and radial as the cross-section rotates about.

It is necessary to establish a consistent sign convention which will be used throughout

the theoretical development. The torque )(xT and rotation angle )(xφ are positive on

the cross-section if they rotate in the right-hand rule sense of the outer normal to the

cross-section. The sign convention established for the torque )(xT and angle of

rotation )(xφ is shown in Figure 9.1.1.

Figure 9.1.1 Torque and Rotation Angle Sign Convention

9.1.1 Member with Circular Cross-Section

The strain-displacement equation for torsional deformation of a circular cross-

section is represented by the relation between the shear strain ),( ργ x and the rotation

angle )(xφ pertinent to cross-section x , is expressed as:

dxdx φ

ρργ *),( = (9.1)

306

where x is the position of the cross-section and ρ varies between zero and the radius

of the cross-section r.

Note: Accordingly to relation (9.1), in a particular cross-section, the shear

strain ),( ργ constx = varies linearly with ρ and reaches its maximum value on the

periphery of the cross-section. The shear strain distribution pertinent to circular and

tubular circular cross-sections is depicted in Figure 9.1.2.

Figure 9.1.2 Shear Strain Distribution for Circular Cross-Sections

(a) Solid Cross-Section and (b) Tubular Cross-Sections

If the homogeneous linear elastic material behavior is considered, the relationship

between shear stress and strain is written according to Hook’s Law as:

),(*),( ργρτ xGx = (9.2)

whereG is the shear modulus and is calculated as:

)1(*2 ν+=

EG (9.3)

where E and ν are the modulus of elasticity and Poisson’s ratio, respectively.

Note: Equation (9.2) indicates that the shear stress follows the linear distribution of

the shear strain when the material is linear elastic and homogeneous. The distribution

of the shear stress on solid and tubular cross-sections is shown in Figure 9.1.3.

Substituting equation (9.1) into equation (9.2), the shear stress ),( ρτ x is expressed as

a function of the angle of rotation )(xφ :

307

dxdGx φ

ρρτ **),( = (9.4)

The torque )(xT is related with the shear stress ),( ρτ x by the following integral

equation:

∫=A

dAxxT *),(*)( ρτρ (9.5)

Figure 9.1.3 Shear Stress Distribution for Circular Cross-Sections

(b) Solid Cross-Section and (b) Tubular Cross-Section

Substituting equation (9.4) into equation (9.5) it results:

pIdxdGxT **)( φ

= (9.6)

where AdIA

p ∫= *2ρ represents the polar moment of inertia of the cross-section.

Rewriting the equation (9.6) the torque-twist equation is obtained:

pIG

xTdxd

*)(

=φ (9.7)

The torque )(xT is a continuous real function on a given interval of continuity [0,L],

and consequently, equation (9.7) can be integrated. The angle of rotation )(xφ at a

particular cross-section x is calculated:

∫ =+= )0(**

)()( xdxIGxTxp

φφ (9.8)

308

Consequently, the total angle of twist between the ends of the continuity interval is

obtained:

∫==−==L

ptotal dx

IGxTxLx

0

**

)()0()( φφφ (9.9)

where L is the length of the continuity interval characterizing the torque )(xT .

Combining the equations (9.4) and (9.7), a new expression for the shear stress is

obtained:

ρρτ *)(),(pIxTx = (9.10)

The maximum value of the shear stress )(max xτ is calculated:

rIxTrxxp

*)(),()(max === ρττ (9.11)

9.1.2 Member with Solid Rectangular Cross-Section

This type of torsional problem is exactly solved using methods of the Theory of

Elasticity. A member characterized by a solid rectangular cross-section when

subjected to torque application suffers not only relative rotation of adjacent cross-

sections, but also warping of the cross-sections. Only some practical formulae are

presented herein.

The distribution of shear stress for a rectangular cross-section, illustrated in Figure

9.1.4, is expressed as:

2max ** tdT

ατ = (9.12)

where α is calculated as function of the ratio of the dimensions d and t .

309

Note: The shear stress attains its absolute maximum in the middle of the longer edge,

while becoming zero at the corners.

The rotation angle φ between the ends of the member is calculated:

JGLT

**

=φ (9.13)

where the torsional moment of inertia J is obtained as: 3** tdJ β= (9.14)

Figure 9.1.4 Shear Stress on Rectangular Cross-Section

The constants α and β are listed in Table 9.1.

Note: For ratios 10≥td both coefficients α and β tends to 333.0 .

Table 9.1 Torsional Constants for Rectangular Cross-Section

310

9.1.3 Member with Solid Elliptical Cross-Section

The shear stress distribution on a solid elliptical cross-section is illustrated in Figure

9.1.5 The maximum value of the shear stress occurs on the periphery of the cross-

section at the two ends of the ellipse’s minor axis and is calculated using the

following formula:

2max ***2

baT

πτ = (9.15)

Figure 9.1.5 Shear Stress in Elliptical Cross-Section

The rotation angle φ measured between the ends of the member with elliptical cross-

section is calculated as:

JGLT

**

=φ (9.16)

where the torsional moment of inertia J is obtained as:

22

33 **ba

baJ+

=π (9.17)

9.1.4 Member with Closed Thin-Wall Cross-Section

The torsional deformation of thin-wall members can be treated using the method

employed in Mechanics of Materials only if some simplifying assumptions are

imposed. They are:

(a) the member is cylindrical and the cross-section is constant its entire

length;

311

(b) the cross-section is closed and is comprised of a single-cell having a

small thickness relative to the general dimensions of the cross-section;

(c) the shear stress is constant through the thickness and parallel to the

median curve defining the cross-section;

(d) the member is subjected to end torque only;

(e) the warping of the cross-section is unrestrained at both ends;

A single-cell thin-wall member is illustrated in Figure 9.1.6.

Figure 9.1.6 Single Cell Thin-Wall Cross-Section

(a) Geometry and Notation, (b) Shear Flow Distribution and

(c) Shear Stress Distribution

Note: An important aspect of this discussion is that the cross-section is not required

to be circular.

The shear flow q is defined as:

tq *τ= (9.18)

where τ is the shear stress.

The equilibrium of the shear forces in the longitudinal direction, in the absence of any

forces acting on the exterior surface planes, implies that the shear flow q is constant

on the entire cross-section and, consequently, the shear stress is expressed as:

consttq==τ (9.19)

Note: Equation (9.19) reflects the underlying assumption (c) that the shear stress is

constant across the thickness of the thin-wall.

312

Following the notation shown in Figure 9.1.7 the integral relation between the torque

T and the shear flow q is written as:

∫∫∫ ===mmm CCsC

dsqdsqFdT ***** ρρρ (9.20)

where dsdA **21

ρ= (9.21)

Substituting equation (9.21) into the (9.20) the torque T is expressed as:

mAqT **2= (9.22)

where mA is the area enclosed by the median curve depicted in Figure 9.1.7.c.

Figure 9.1.7 Thin-Wall Cross-Section Notations

The expression of the shear flow is obtained as:

mA

Tq*2

= (9.23)

and consequently, the shear stress τ is calculated:

tAT

m **2=τ (9.24)

The rotation angle φ between the ends of the member is calculated:

JGLT

**

=φ (9.25)

313

where the torsional moment of inertia J is calculated as:

m

m

PtAJ **4 2

= (9.26)

where mP is the perimeter of the area enclosed by the median curve depicted in

Figure 9.1.7.c.

9.1.5 Member with Open Thin-Wall Cross-Section

If the cross-section of the member subjected to torque is assembled from n thin walled

rectangular shapes and all component shapes equally rotate, the formulae developed

for the rectangular cross-section in section 9.1.2 can be used in the evaluation of the

shear stress distribution of the composed open thin-wall cross-section. The rotation

angle iφ pertinent to component shape “i” is obtained as:

φφφ ****

LJGT

JGLT i

ii

ii =⇒== (9.27)

where JGLT

**

=φ is the rotation angle and J is the torsional moment of inertia of the

entire cross-section.

From equilibrium and substituting equation (9.27) the torque T applied on the cross-

section is:

JLGJ

LGTT

n

ii

n

ii ****

11

φφ ∑∑==

=== (9.28)

where the torsional moment of inertia J is calculated:

∑∑==

==n

iii

n

ii dtJJ

1

3

1

**31 (9.29)

Note: The 1/3 factor used in equation (9.29) reflects the assumption b. In this case the

ratio of the rectangular component dimensions are larger that 10.

314

Then, from equation (9.27), the torque iT corresponding to each shape is obtained:

TJJ

T ii *= (9.30)

On each of the rectangular component shapes the stress distribution is identical to that

depicted in Figure 9.1.4. Consequently the maximum shear stress is obtained in the

middle of the longer dimension id of the shape and is calculated:

i

ii

i

ii

ii t

JT

Jt

TJJ

td

T*

*1

*

**31 2

max_ ===τ (9.31)

Note: The maximum shear stress for the entire cross-section is obtained in the middle

of the longer dimension of the thicker shape comprised on the cross-section.


Problem 9.2.1 A 4 m long circular member with a diameter D = 30 cm, illustrated in

Figure 9.2.1, is loaded by a constant torque of magnitude T = 100 kNm. Considering

that the member is made of steel characterized by a shear modulus G = 80 MPA, an

allowable shear stress MPatorqueall 60_ =τ and an allowable rotation

angle mtorqueall deg/1_ =φ , verify its strength and deformability. Repeat the strength

and deformability calculations considering the following types of cross-sections: (a)

tubular cross-section (De = 30 cm and Di = 25 cm), (b) rectangular cross-section ( d =

30 cm and t = 13 cm), closed thin-wall rectangular cross-section (b=30cm, h=25cm ,

t1=t2=2cm), “imperfect” closed thin-wall rectangular cross-section(b=30cm, h=25cm,

t1=1.5cm and t2=2cm), splinted” thin-wall rectangular cross-section and I-type thin-

wall cross-section (b1=20cm, b2=26cm, b3=20cm, t1=2m, t2=1cm and t3=2cm).

Figure 9.2.1

315


A.1- The member illustrated in Figure 9.2.1 is successively verified for strength and

deformability considering that its pertinent cross-section is:

(a) circular with the diameter D=30 cm;

(b) tubular with the exterior and interior diameters De = 30 cm and Di = 25

cm;

(c) rectangular with the length d = 30 cm and the height t = 13 cm;

(d) closed thin-wall rectangular cross-section characterized by length b=30cm,

width h=25cm and wall thicknesses t1=t2=2cm;

(e) “imperfect” closed thin-wall rectangular cross-section characterized by

length b=30cm, width h=25cm and wall thicknesses t1=1.5cm and t2=2cm;

(f) “splinted” thin-wall rectangular cross-section;

(g) I-type thin-wall cross-section with the following dimensions: b1=20cm,

b2=26cm, b3=20cm, t1=2m, t2=1cm and t3=2cm.

The geometrical significance of the dimensions listed, pertinent to each type of cross-

section, are graphically represented in figures located in the particular calculation

area.

A.2 – The material is characterized by the following constants: (a) the shear modulus

G = 80 MPA, (b) the allowable shear stress MPatorqueall 60_ =τ and (c) the allowable

rotation angle mmtorqueall deg/1__ =φ .

A.3 – For each type of cross-section the strength and deformability verifications are

conducted considering the following requirements:

τmax τall_torque≤

and

φmax φall_torque≤ φall_torque_m L⋅

where

φall_torque 0.07 rad=

316

B. Calculations

B.1 Circular Cross-Section (Figure 9.2.1.a)

Figure 9.2.1.a – Circular Cross-Section

The circular cross-section is characterized by a diameter D=30 cm. The polar moment

of inertia pI is calculated:

IPπ D4⋅

32:=

3230* 4

→⇒π IP 79521.564cm4=

The maximum shear stress is obtained using formula (9.11) for rmax 15cm= :

τmaxTIP

rmax⋅:= *10*15**10*564.79521

**10*100 248

3

→⇒ −− mmmN

τmax 18.863MPa=

Consequently,

τmax 18.863MPa= < τall_torque 60MPa:= ok

The maximum rotation angle is calculated using formula (9.9):

φmaxT

G IP⋅L⋅:= *4*

*10*564.79521**10*80

**10*10048

29

3

→⇒−

mm

mN

mN

φmax 6.288 10 3−× rad= = φmax 0.36deg=

Consequently,

φmax 6.288 10 3−× rad= < φall_torque 0.07 rad= ok

317

B.2: Tubular Circular Cross-Section (Figure 9.2.1.b)

Figure 9.2.1.b – Tubular Circular Cross-Section

The tubular circular cross-section has an interior and exterior diameters, Di 25cm:=

and De 30cm:= , respectively. The polar moment of inertia pI is calculated:

IPπ De

4⋅

321

DiDe

4

−

⋅:=

30251*

3230* 44

→

−⇒

π IP 41172.044cm4=

The maximum shear stress is obtained using formula (9.11) for rmax 15cm= :

τmaxTIP

rmax⋅:= *10*15**10*044.41172

**10*100 248

3

→⇒ −− mmmN τmax 36.432MPa=

Consequently,



φmaxT

G IP⋅L⋅:= *4*

*10*044.41172**10*80

**10*10048

29

3

→⇒−

mm

mN

mN

φmax 0.012 rad=

Consequently,

φmax 0.012 rad= < φall_torque 0.07 rad= ok

318

B.3 Rectangular Cross-Section (Figure 9.2.1.c)

Figure 9.2.1.c – Rectangular Cross-Section

The dimensions of the rectangular cross-section ared 30cm:= and t 13cm:= , it

resulting into a ratio dt

2.308= . From Table 9.1, by interpolation, the coefficients

α and β are obtained: α 0.253= and β 0.241= . The torsional moment of inertia

is then calculated:

J β d⋅ t3⋅:= 13*30*241.0 3 →⇒ J 15904.59cm4=

The maximum shear stress is calculated accordingly to formula (9.12):

τmaxT

α d⋅ t2⋅:=

*10*13*30*253.0**10*100

362

3

→⇒ − mmN

τmax 77.842MPa=

and is located in the middle of the largest edge.

Consequently,

τmax 77.842MPa= < τall_torque 60MPa:= not ok???

The cross-section needs to be resized. The resizing is conducted in the following

manner:

Wt_necT

τall_torque:=

*10*60**10*100

6

3

→⇒PamN Wt_nec 1666.667cm3=

tnec

3 Wt_nec

αdt

⋅:=

308.2*253.0667.1666

3 →⇒ tnec 14.178cm=

319

dnecdt

tnec⋅:= 178.14*308.2 →⇒ dnec 32.72cm=

The following new dimensions are obtained by rounding up the exact above

calculated values to: d 33cm:= and t 14.5cm:= . The new dimensional ratio

calculated is dt

2.276= and therefore, the new calculated coefficients, after three

digits rounding, are: α 0.253= and β 0.24= . Consequently, the new value of the

torsional moment of inertia is:

J β d⋅ t3⋅:= 5.14*33*24.0 3 →⇒ J 24148.579cm4=

The maximum shear stress is recalculated using the new cross-sectional dimensions:

τmaxT

α d⋅ t2⋅:=

*10*5.14*33*253.0**10*100

362

3

→⇒ − mmN

τmax 5.705 107× Pa=



φmaxT

G J⋅L⋅:= *4*

*10*579.24148**10*80

**10*10048

29

3

→⇒−

mm

mN

mN

φmax 0.021 rad= = φmax 1.186deg=

Consequently,

φmax 0.021 rad= < φall_torque 0.07 rad= ok

B.4: Rectangular Closed Thin-Wall Cross-Section (Figure 9.2.1.d)

Figure 9.2.1.d - Rectangular Closed Thin-Wall Cross-Section

320

The thin-wall rectangular cross-section is characterized by the following dimensions:

b 30cm:= h 25cm:= t1 2cm:=

t2 2cm:=

Computation of the average area :

bm b t2−:= → bm 28cm=

hm h t1−:= → hm 23cm=

Am bm hm⋅:= 23*28 →⇒ Am 644cm2=

The shear flow q and maximum shear stress τmax are obtained using the formulae

(9.23) and (9.24), respectively:

tmin min t1 t2,( ):=

t tmin:= → t 2 cm=

qT

2 Am⋅:=

*10*644*2**10*100

24

3

→⇒ − mmN q 7.764 105×

Nm

=

τmaxq

tmin:=

*10*2

*10*764.72

5

→⇒ − mmN

τmax 3.882 107× Pa=



The torsional moment of inertia is calculated using formula (9.26):

J4 Am

2⋅ t⋅

Pm:=

*10*102*10*2**)10*644(*4

2

2424

→⇒ −

−−

mmm J 3.253 104× cm4=

where: Pm 2 bm hm+( )⋅:= )2328(*2 →+⇒ Pm 102cm=

Computation of the rotation angle is conducted accordingly to (9.25):

φmaxT

G J⋅L⋅:= *4*

*10*253.3**10*80

**10*10044

29

3

→⇒−

mm

mN

mN


Consequently, φmax 0.015 rad= < φall_torque 0.07 rad= ok

321

B.5. Imperfect Rectangular Closed Thin-Wall Cross-Section (Figure 9.2.1.e)

Figure 9.2.1.e – Imperfect Rectangular Closed Thin-Wall Cross-Section

The thin-wall rectangular cross-section considered above is modified to create an

“imperfect” cross-section characterized by the following dimensions:

b 30cm:= h 25cm:= t1 1.5cm:=

t2 2cm:=

Computation of the average area :

bm b t2−:= → bm 28cm=

hm h t1−:= → hm 23.5cm=

Am bm hm⋅:= 5.23*28 →⇒ Am 658cm2=

The shear flow q and maximum shear stress τmax are obtained using the formulae

(9.23) and (9.24), respectively:

tmin min t1 t2,( ):=

t tmin:= → tmin 1.5cm=

qT

2 Am⋅:=

*10*658*2**10*100

24

3

→⇒ − mmN

q 7.599 105×Nm

=

τmaxq

tmin:=

*10*5.1

*10*599.72

5

→⇒ − mmN

τmax 5.066 107× Pa=



The torsional moment of inertia is calculated using the following formula (9.26):

J4 Am

2⋅ tmin⋅

Pm:=

*10*103*10*5.1**)10*658(*4

2

2424

→⇒ −

−−

mmm J 2.522 104× cm4=

322

where: Pm 2 bm hm+( )⋅:= )5.2328(*2 →+⇒ Pm 103cm=

Computation of the rotation angle is conducted accordingly to (9.25):

φmaxT

G J⋅L⋅:= *4*

*10*522.2**10*80

**10*10044

29

3

→⇒−

mm

mN

mN


Consequently, φmax 0.02 rad= < φall_torque 0.07 rad= ok

B.6. Rectangular Closed Thin-Wall Cross-Section with a Split (Figure 9.2.1.f)

The closed thin-wall cross-section analyzed in section B.4 is considered being splinted

in the middle of its height h as illustrated in Figure 9.2.1.f. The existence of the split

changes the thin-wall closed cross-section into an open cross-section, and,

consequently, the analysis is conducted as indicated in the theoretical section 9.1.5.

Figure 9.2.1.f – Rectangular Thin-Wall Cross-Section with a Split

The cross-sectional dimensions are:

b 30 cm⋅:= h 25 cm⋅:= t1 2cm:=

t2 2cm:=

The dimensions of the average perimeter are:

bm b t2−:= → bm 28cm=

hm h t1−:= → hm 23cm=

323

The shape of the cross-section is decomposed in five individual rectangular shapes.

The individual torsional moments of inertia of the vertical uncut and splinted flanges

are calculated:

J113

hm⋅ t23⋅:= 2*23*

31 3 →⇒ J1 61.333cm4=

J213

hm2

⋅ t23⋅:= 2*

223*

31 3 →⇒ J2 30.667cm4=

J313

hm2

⋅ t23⋅:= 2*

223*

31 3 →⇒ J3 30.667cm4=

The individual torsional moments of inertia of the horizontal flanges, lower and

upper, are calculated as:

J413

bm⋅ t13⋅:= 2*28*

31 3 →⇒ J4 74.667cm4=

J5 J4:= → J5 74.667cm4=

The total torsional moment of inertia pertinent to the entire cross-section is obtained

by summing the individual torsional moment of inertia of the component shapes:

J J1 J2+ J3+ J4+ J5+:= → J 272cm4=

The maximum shear stress on the cross-section is:

τmaxTJ

tmax⋅:= *10*2**10*272

**10*100 248

3

→⇒ −− mmmN

τmax 7.353 108× Pa=

where tmax max t1 t2,( ):= → tmax 2cm=


τmax 735.294MPa= > τall_torque 60MPa:= NG!!!!

The maximum angular rotation is calculated:

φmaxT

G J⋅L⋅:= *4*

*10*272**10*80

**10*10048

29

3

→⇒−

mm

mN

mN

324


Consequently,

φmax 1.838 rad= > φall_torque 0.07 rad= NG????

Note: The shaft with an open cross-section has a bigger than allowable deformation

and should be redesigned.

If the split is welded, as illustrated in Figure 9.2.1.g, the calculations are conducted

as if the cross-section is the closed thin-wall cross-section analyzed in section B.4.

The shear stress calculated for the closed thin-wall cross-section, constant on the

entire cross-section, was previously calculated as:

τmax 38.82MPa:=

The shear flow in the weld is obtained:

qweld τmax t2⋅:= *10*2*10*82.38 26 →⇒ − mPa qweld 7.764 105×Nm

=

Figure 9.2.1.g – Rectangular Thin-Wall Cross-Section with a Welded Split

If the weld is made of a material characterized by an allowable shear

stressτall_weld 160MPa:=

, then the effective thickness of the weld is calculated

employing the following formula:

q Rweld≤ tweld τall_weld⋅

Consequently,

325

tweldq

τall_weld:=

*10*160

*10*764.76

5

→⇒PamN

tweld 4.749mm=

A rounded value for the weld thickness is chosen:

tweld 5mm:=

B.7 I-type Thin-Wall Cross-section (Figure 9.2.1.h)

All true in section B.6 an open thin-wall cross-section was already analyzed, in this

section the case of I-type cross-section, a shape often encountered in structural

engineering, is investigated using the same approach. The I-type cross-section

illustrated in Figure 9.2.1.g has the following dimensions:

b1 20cm:= b2 26cm:= b3 20cm:=

t1 2cm:= t2 1cm:= t3 2cm:=

The torsional moment of inertia pertinent to the entire cross-section is calculated:

J13

1

3

i

bi ti( )3⋅∑=

⋅:= )]2*20()1*26()2*20[(*31 333 →++⇒ J 115.333cm4=

Figure 9.2.1.h – Open Contour Thin-Wall Cross-Section

The maximum shear stress on each individual part, as illustrated in Figure 9.2.1.h, is

located in the middle of the larger length of the rectangular part and is obtained as:

τ1TJ

t1⋅:= *10*2**10*333.115**10*100 2

48

3

→⇒ −− mmmN

τ1 1.734 109× Pa=

326

τ2TJ

t2⋅:= *10*1**10*333.115**10*100 2

48

3

→⇒ −− mmmN

τ2 8.671 108× Pa=

τ3TJ

t3⋅:= *10*2**10*333.115**10*100 2

48

3

→⇒ −− mmmN

τ3 1.734 109× Pa=

The maximum shear stress pertinent to the entire cross-section is realized in the

flanges, at points 1 and 3:

τmaxTJ

tmax⋅:= *10*2**10*333.115**10*100 2

48

3

→⇒ −− mmmN

τmax 1734.104MPa=


τmax 1734.104MPa= > τall_torque 60MPa:= NG!!!

The maximum angular rotation is:

φmaxT

G J⋅L⋅:= *4*

*10*272**10*80

**10*10048

29

3

→⇒−

mm

mN

mNφmax 4.335 rad=


Consequently,

φmax 4.335 rad= > φall_torque 0.07 rad= NG????

Note: The cross-section analyzed in sections B6 and B7 indicate a violation of the strength or deformation verification criteria and, under normal circumstances, should be resized. This task remains to be conducted by the student. 9.3 Proposed Problems

Problem 9.3.1 The solid shaft, illustrated in Figure 9.3.1, with a diameter d = 40 mm

and a length L = 800 mm is made of an aluminum alloy. The material is characterized

by an allowable shear stress τall = 70 MPa and a shear modulus of elasticity G = 26

GPa. If the allowable angle of twist over the entire length of the shaft is Фall = 0.10

rad. Draw the variation of the torque magnitude along the entire length of the shaft.

Calculate: (a) the maximum allowable torque Tall sustained by the shaft and draw the

327

corresponding shear stress distribution on the cross-section, and (b) the angle of twist

induced by a torque T = 0.9 Tall.

Figure 9.3.1

Problem 9.3.2 The 2 m long solid shaft shown in Figure 9.3.1 is made of brass

characterized by an allowable shear stress τall = 120 MPa and a shear modulus of

elasticity G = 26 GPa. The shaft is subjected at both ends by a torque T = 25 kNm.

Calculate the required diameter of the shaft considering the strength and deformation

of the material. The allowable angle of twist over the entire length of the shaft is Фall

= 0.10 rad/m.

Problem 9.3.3 The shaft, shown in Figure 9.3.3, with an outside diameter d0 = 60 mm

is made of a steel alloy characterized by an allowable shear stress of τall = 80 MPa and

a shear modulus of elasticity G = 200 GPa. The shaft is to be subjected at both ends

to a torque T = 2 kNm. Draw the variation of the torque magnitude along the entire

length of the shaft. Calculate: (a) the required interior diameter di, (b) the required

exterior diameter de if the shaft has a full cross-section, (c) the ratio of the weight of

the shaft obtained in the previous two conditions, and (d) the angle of twist between

the two ends if the shaft has a length L=1m.

Figure 9.3.3

Problem 9.3.4 The solid circular shaft, illustrated in Figure 9.3.4.a, reaches the

allowable shear stress τall under the action of a torque Ta. Later, the solid shaft is

328

replaced by a tubular shaft, shown in Figure 9.3.4.b, with a ratio of outside diameter

to inside diameter of 5.0=i

o

dd

but weighing the same as the solid shaft.

Figure 9.3.4

Draw the variation of the torque magnitude along the entire length of the shaft and

calculate: (a) the magnitude of the torque Tb necessary to produce in the tubular cross-

section the same allowable shear stress and draw the distribution of the shear stress on

the cross-section, and (b) the twisting angles Фa and Фb corresponding to torques Ta

and Tb, respectively.

Problem 9.3.5 The circular stepped shaft AC, illustrated in Figure 9.3.5, is made of

steel and is subjected to two torques located at sections B and C. The diameters of the

shaft segments are: d1 = 40 mm and d2 = 25 mm. Draw the variation of the torque

magnitude along the entire length of the shaft and calculate: (a) the magnitude of the

torques TB and TC necessary to induce in both segments a maximum shear stress of

200 MPa, (b) considering the magnitude of the torques previously calculated and a

shear modulus of elasticity G = 200 GPa, the angular rotations at B and C, and (c) the

ratio of the torques for the angular rotation in C to be zero.

Figure 9.3.5

329

Problem 9.3.6 The stepped shaft AC, shown in Figure 9.3.6, consisting of two solid

circular segments is subjected to torques TB = 5 kNm and TC = 1.7 kNm. The lengths

and the diameters of the segments are: d1 = 65 mm, d2 = 50 mm, L1 =650 mm and L2

= 450 mm. The shaft is made of steel with an allowable shear stress τall = 70 MPa and

a shear modulus G = 80 GPa.

Figure 9.3.6

Draw the variation of the torque magnitude along the entire length of the shaft and (a)

verify the shaft cross-sections and draw the corresponding shear stress distributions on

the cross-section, and (b) calculate the angle of twist at locations B and C.

Problem 9.3.7 The allowable shearing stress in the circular stepped steel shaft shown

in Figure 9.3.7 is τall = 200 MPa. Consider that the diameters of the shaft segments

are d1 = 50 mm, d2 = 100 mm and d3 = 150 mm and that the shaft is made of steel

with a shear modulus G = 200 GPa. Draw the variation of the torque magnitude along

the entire length of the shaft and calculate: (a) the magnitudes of the torques TA, TB

and TC necessary to induce in all segments a shear stress equal to τall, (b) the angular

rotation at locations A, B and C induced by the torques TA, TB and TC previously

calculated.

Figure 9.3.7

330

Problem 9.3.8 The stepped shaft ABCD, illustrated in Figure 9.3.8, consisting of

three solid circular segments, is subjected to torques of magnitudes TB=12 kNm, TC=

6 kNm and TD =2 kNm, respectively.

Figure 9.3.8

The shaft is made of steel characterized by an allowable shear stress τall = 200 MPa

and a shear modulus of elasticity G = 200 GPa. Considering that the segments have

equal lengths of 0.5 m and that the diameters of the segments are 80 mm, 60 mm, and

40 mm, respectively, conduct the following tasks: (a) draw the variation of the torque

magnitude along the entire length of the shaft, (b) verify the shear stress in the shaft,

(c) calculate the shaft angle of twist at locations B, C and D, and (d) indicate an

allowable angle of twist per unit length.

Problem 9.3.9 A circular tube of outer diameter d3 = 70 mm and inner diameter d2 =

60 mm is welded at the right end to a fixed plate and at the left end to a rigid end plate

as illustrated in Figure 9.3.9. The second bar, a solid circular bar with a diameter d1 =

40 mm, located inside and concentric with the tube, passes through a hole in the fixed

plate and is welded to the left end plate. The lengths of the tube and solid circular bar

are L and 0.5L, respectively. A torque of magnitude T acts at the tip A of the circular

bar. Both members are made of steel characterized by an allowable shear stress τall

and shear modulus of elasticity G. Conduct the following tasks: (a) draw the variation

of the torque magnitude along the entire lengths of both members, (b) derive a

formula for the maximum shear stresses in both the bar and tube, (c) derive a formula

for the angle twist at tip A of the bar, (d) considering the length of the solid bar L= 1

m, the allowable shear stress τall = 200 MPa calculate the allowable torque of the

system, and (e) for shear modulus of elasticity G = 200 GPa and the magnitude of the

torque previously calculated determine the angle of twist at the left plate location and

at tip A.

331

Figure 9.3.9

Problem 9.3.10 Three identical circular rigid disks A, B, and C are welded to the ends

of three identical solid circular bars as shown in Figure 9.3.10. The bars are welded at

their intersection D to form a rigid connection. The bars have a diameter d2 = 10 mm

and a length of L = 1 m, while the rigid disks have a diameter d1 = 75 mm.

Figure 9.3.10

Considering that the forces P1, P2, and P3 are applied on the disks A, B, and C,

respectively, conduct the following tasks: (a) draw the variation of the torque

magnitude along each bar of the system, (b) derive a relation between the forces P1,

P2, and P3, (c) verify the bars for strength if P3 = 200 N and τall =80 MPa, and (d)

calculate the angle of twist at A, B and C locations considering P1 = 140 N and the

shear modulus of elasticity G = 70 GPa.

Problem 9.3.11 The vertical pole of solid circular cross-section and height H,

illustrated in Figure 9.3.11, is twisted by two horizontal forces P acting at the ends of

a horizontal arm AB. Considering that the vertical pole is made of a material with an

allowable shear stress τall and a shear modulus of elasticity G conduct the following

tasks: (a) find the horizontal force P inducing in the pole a shear stress equal to τall, (b)

332

find the expression of the maximum angle of twist of the vertical pole, (c) considering

the following values: τall = 90 MPa, c = 0.25 m and P = 2 kN and d=50 mm verify

the pole for strength, and (d) considering G = 70 GPa and Фall=0.1 rad/m calculate

the required height of the pole.

Figure 9.3.11

Problem 9.3.12 The tapered solid shaft, shown in Figure 9.3.12, has a variable

diameter linearly changing from d0 at tip A (x = 0) to 2*d0 at point B (x =L). The shaft

is attached in point B, at x = L, to a rigid wall and is subjected in point A, at x = 0, to

a torque of magnitude T0. Draw the variation of the torque magnitude along the entire

length of the shaft and (a) derive an expression for the maximum cross-sectional shear

stress in the tapered shaft as a function of the distance x, (b) derive an expression for

the angle of twist of the shaft at the tip A considering the shear modulus of elasticity G

of the material known, (c) repeat the tasks (a) and (b) if the shaft has a circular tubular

cross-section of a constant thickness t, and (d) repeat the tasks (a) and (b) if a hole of

constant diameter di = 0.5* d0 is drilled out along the entire length of the shaft

(resulting in a tubular shaft of linearly varying wall thickness).

Figure 9.3.12

Problem 9.3.13 The solid circular shaft attached to a rigid wall as illustrated in Figure

9.3.13 is subjected to a distributed external torque which varies linearly from intensity

of t0 per unit of length at point A (x = 0) to zero value at point B (x = L).

333

Figure 9.3.13

Draw the variation of the torque magnitude along the entire length of the shaft.

Considering that the shaft diameter and the material’s elastic shear modulus are d and

G, respectively, determine: (a) the expression of the cross-sectional shear stress in the

shaft as a function of the distance x, (b) the expression of the total angle of twist at the

free end of the shaft, (c) repeat the tasks (a) and (b) if the stated externally-applied

torque distribution is replaced by the cubically-varying torque ])(1[*)( 30 L

xtxt −= ,

and (d) repeat the tasks (a) and (b) if the stated externally-applied torque distribution

is replaced by the varying torque )*2*cos(*)( 0 L

xtxt π= .

Problem 9.3.14 – 9.3.23 A thin-wall tubular member with a cross-section illustrated

in Figures 9.3.14 through 9.3.23 is subjected at both ends by opposite torques of

equal magnitudes T. Conduct the following tasks: (a) derive and draw the expression

of the shear stress on the cross-section function of the indicated parameters

(T,t,a,b,etc.), (b) considering that the maximum shear stress reaches the allowable

shear stress τall of the material calculate the allowable torque magnitude Tall, and (c)

considering that the material shear modulus of elasticity G is known calculate the

angle of twist between the ends of the member. Conduct numerical evaluations for the

following set of geometrical data: a =100 mm, b = 80 mm, t = 2 mm, t1 =t and t2 =2*t

334

and a material characterized by τall = 90 MPa and G=200 GPa. Hint: for problem

9.3.19 the variation of the wall thickness is )2

sin1(*)( 0θ

θ += tt .

Figure 9.3.14 Figure 9.3.15



Figure 9.3.20

335


Figure 9.3.23

Problem 9.3.24 Consider that the bars with the thin-wall cross-sections illustrated in

Figures 9.3.14 through 9.3.19 are 1m long and are made of the same material

(densityγ , allowable shear stress τall and allowable angle of twist Фall). All bars are

subjected to the same magnitude torque T. Calculate the material weight

corresponding to each bar and conclude about the efficiency of the shape.

Problem 9.3.25-9.3.28 A thin-wall member with a cross-section illustrated in Figures

9.3.25 through 9.3.28 is subjected at both ends to opposite torques of equal

magnitudes T. Conduct the following tasks: (a) derive and draw the expression of the

shear stress on the cross-section function of the indicated parameters (T, t, a, b), (b)

considering that the maximum shear stress reaches the allowable shear stress τall of the

material calculate the allowable torque magnitude Tall, and (c) considering the

material shear modulus of elasticity G known calculate the angle of twist between the

ends of the member. Conduct numerical evaluations for the following set of

geometrical data: a =100 mm, b = 80 mm, t = 2 mm and a material characterized by

τall = 90 MPa and G=200 GPa

336



Problem 9.3.29 Calculate the allowable torque Tall of the thin-wall cross-sections

illustrated in Figure 9.3.29 considering that all are made of the same material and the

welds are broken. Considering that the welds are in-place calculate the shear forces in

the welds.

Figure 9.3.29

Problem 9.3.30 A solid circular bar with diameter d is to be replaced by a rectangular

tubular cross section with dimensions shown in Figure 9.3.30. Determine the

thickness t of the tube so that the maximum shear stress in the tube and in the circular

bar to be equal.

337

Figure 9.3.30

Problem 9.3.31 The statically indeterminate stepped shaft ACB, illustrated in Figure

9.3.31, is fixed at both ends, A and B, and is subjected to a torque To at point C. The

two segments of the shaft characterized by lengths LA and LB and polar moments of

inertia IpA and IpB, respectively, are made of the same material. Calculate the ratio

between the diameters of the two segments in order for the maximum shear stress to

reach the allowable limit in both segments.

Figure 9.3.31

Problem 9.3.32 As illustrated in Figure 9.3.32, a solid steel bar with a 30 mm diameter

is enclosed by a steel tube of 45 mm outer diameter and 36 mm inner diameter. Both

the bar and the tube are joined by a rigid plate at end B and fixed into a rigid wall at

end A. Considering that the rigid end plate is subjected to a torque T = 500 Nm and

the length of both members is 500 mm determine: (a) the maximum shear stresses in

the bar and tube, (b) the angle of rotation (in degrees) of the end plate, assuming that

the shear modulus of the steel is G = 200 GPa and (c) if the tube and the bar are

338

replaced by a single bar with solid circular cross-section, find its diameter d for the

angle of twisting to remain unchanged.

Problem 9.3.32

Problem 9.3.33 As illustrated in Figure 9.3.33 the hollow circular tube A fits over the

end of a solid circular bar B without being attached to each other. The far ends of both

bars are fixed into rigid walls. Then, bar B is twisted with an angle β and two holes

are made through both members and a rigid pin is placed through the holes. Later, the

bar B is released and the system returns to equilibrium. Considering that the members

have the same lengths and are characterized by polar moments of inertia IpA and IpB,

respectively, calculate: (a) the reaction torques acting at the fixed ends and (b) the

maximum shear stresses in the members.,

Problem 9.3.33

Problem 9.3.34 The shaft AB of diameter d1 attached to two circular disks of diameter

d2, as illustrated in Figure 9.3.34.a, is subjected at both ends by a torques To. While

these torques are acting, a tube of the same material as the shaft and with an outer

diameter d3 and an inner diameter d2, is fitted over the disks and welded securely to

them, as pictured in Figure 9.3.34.b, and then, the torques To are removed.

339

Problem 9.3.34

Assuming that the flanges are rigid, derive a formula for the maximum shear stress in

the shaft before and after the fitting of the tube.

340

CHAPTER 10 Plane Stress Transformation 10.1. Theoretical Background

The three-dimensional state of stress is illustrated in Figure 10.1.1 where, for clarity,

only the stresses drawn on the faces with positive normal are represented.

Figure 10.1.1 Three-Dimensional State of Stress

Mathematically the three-dimensional state of stress is represented by a generalized

stress tensor ),,( zyxTσ defined by nine distinct stress components:

zzyzx

yzyyx

xzxyx

zyxTστττστττσ

σ =),,( (10.1)

Applying the shear stress duality principle ( yxxy ττ = , zxxz ττ = and zyyz ττ = ) the

number of independent stress components is reduced from nine to six.

There are instances when some of the stress tensor components vanish and the general

three-dimensional stress tensor ),,( zyxTσ degenerates into a tensor characterized by

only three independent components all independent of the variable z. This condition

is called plane state of stress. For the present theoretical development, it is assumed

that all stress components pertinent to the planes having normal vector parallel to the

z axis are zero:

0=zσ (10.2)

0== xzzx ττ (10.3)

341

0== yzzy ττ (10.4)

and

),(),,( yxTzyxT σσ = (10.5)

where:

yyx

xyxyxTσττσ

σ =),( (10.6)

Consequently, the plane stress tensor ),( yxTσ , defined by three non-zero components,

can be represented in plane oxy as depicted in Figure 10.1.2.

Figure 10.1.2 Plane State of Stress

The beam deformation models encountered in previous chapters, including the study

of the axial and torsional deformation and pure and non-uniform bending, are

characterized by a simpler state of plane stress where the normal stress 0=yσ has a

null magnitude. In this case the stress tensor ),( yxTσ is characterized by only two

non-zero stress components:

0),(

yx

xyxyxTτ

τσσ = (10.7)

10.1.1 Plane Stress Transformation Equations

The transformation relations between normal and shear stresses nσ and ntτ pertinent

to a rotated plane and the stresses xσ , yσ and xyτ are obtained by writing the

342

equilibrium equations for the infinitesimal triangular element depicted in Figure

10.1.3. The inclined plane is defined by its positive normal n which is rotated

counterclockwise with an angleθ from the horizontal direction x .

Figure 10.1.3 Equilibrium of the Infinitesimal Triangular Element

The normal nσ and shear ntτ stresses are:

θθτθσθσσ cos*sin**2sin*cos* 22xyyxn ++= (10.8)

)sin(cos*cos*sin*)( 22 θθτθθσστ −+−−= xyyxnt (10.9)

Equations (10.8) and (10.9), called the plane stress transformation equations, can be

re-written using the trigonometric relations between the angleθ and the double

angle )*2( θ :

)*2sin(*)*2cos(*22

θτθσσσσ

σ xyyxyx

n +−

++

= (10.10)

)*2cos(*)*2sin(*2

θτθσσ

τ xyyx

nt +−

−= (10.11)

Figure 10.1.4 Stresses on Orthogonal Rotated Faces

343

Two faces are needed to express the plane stress tensor around the particular

point ),( yxP . Consequently, the formulae (10.10) and (10.11) are applied twice, as

illustrated in Figure 10.1.4. First, considering the rotation angle 'xxθθ = and,

secondly, for the complementary angle 'xyθθ = . The stresses on two orthogonal

rotated faces are expressed as:

)*2sin(*

)*2cos(*22

)(

'

'''

xxxy

xxyxyx

xxnx

θτ

θσσσσ

θθσσ

+

+−

++

=== (10.12)

)*2cos(*

)*2sin(*2

)(

'

''''

xxxy

xxyx

xxntyx

θτ

θσσ

θθττ

+

+−

−=== (10.13)

)*2sin(*

)*2cos(*22

)90(

'

'''

xxxy

xxyxyx

xxny

θτ

θσσσσ

θθσσ

−

−−

−+

=+== •

(10.14)

)*2cos(*

)*2sin(*2

)90(

'

''''

xxxy

xxyx

xxntxy

θτ

θσσ

θθττ

−

−−

=+== •

(10.15)

If equations (10.12) and (10.14) are summed, the invariance of the summation of

normal stresses is established:

yxyxσσσσ +=+ '' (10.16)

10.1.2 Principal Stresses and Directions

The maximum and minimum normal stresses are called principal stresses and

mathematically represent the extreme values of the normal stress function )(θσ n . The

extreme values are obtained by imposing the condition that the first derivative of the

normal stress )(θσ n relative to the rotation angle θ is zero:

0)(

== ntn

dd

τθ

θσ (10.17)

Note: Equation (10.17) indicates that the principal normal stresses are obtained for a rotated plane where the shear stress is zero.

344

The principal stresses 1σ and 2σ , schematically depicted in Figure 10.1.5, 1σ are

calculated as:

Ravxyyxyx

pn +=+−

++

=== στσσσσ

θθσσ 22

11 4)(

2)( (10.18)

Ravxyyxyx

pn −=+−

−+

=== στσσσσ

θθσσ 22

22 4)(

2)( (10.19)

where

2yx

av

σσσ

+= (10.20)

22

4)(

xyyxR τ

σσ+

−= (10.21)

Figure 10.1.5 Principal Stresses and Directions

From the stationarity condition (10.17), after algebraic manipulations, the angle pθ ,

representing the angle for which the normal stress )(θσ n reaches its extreme value is

obtained:

2

)*2tan(yx

xyp σσ

τθ

−= (10.22)

Solution of the trigonometric equation (10.22) yields two solutions, )*2( 1pθ

and )*2( 2pθ , where the two angles are related as:

πθθ += 12 *2*2 pp (10.23)

From (10.23) the orthogonality of the two principal directions is established:

345

212π

θθ += pp (10.24)

To identify which of the two angles, 1pθ or 2pθ , corresponds to the maximum

principal stress 1σ , two methods can be employed:

(a) by successively assigning to the angle θ in equation (10.10) the values 1pθ

and 2pθ and observing which angle produces the maximum principal stress;

(b) by employing the inequality (10.25) relating the of the principal maximum

direction angle of rotation to the shear stress xyτ :

0tan

>xy

p

τ

θ (10.25)

For the inequality (10.24) to hold true, the signs of the tangent of the angle pθ

and shear stress xyτ must be identical.

10.1.3 Maximum Shear Stresses and Directions

The maximum shear stresses are determined in a similar manner as the principal stresses. The extreme condition for the shear stress function )(θτ nt contained in equation (10.17) is written as:

0)(

)(=

θθτ

dd nt (10.26)

From the stationarity condition (10.26) the angle sθ representing the angle for which the shear stress )(θτ nt reaches its extreme value is obtained:

)*2tan(12)*2tan(

pxy

yx

s θτ

σσ

θ −=

−−

= (10.27)

Solving the trigonometric equation (10.27), two solutions )*2( 1sθ and )*2( 2sθ , two l solutions are obtained. The angles orthogonal and are related as:

212π

θθ += ss (10.28)

The relationship between the angles of the maxim principal normal stress and shear stress directions, sθ and pθ , is expressed as:

346

411π

θθ −= ps (10.29)

The maximum and minimum shear stresses are calculated as:

max2

2

11 4)(

)( ττσσ

θθττ ==+−

=== Rxyyx

snts (10.30)

min2

2

22 4)(

)( ττσσ

θθττ =−=+−

−=== Rxyyx

snts (10.31)

The normal stresses corresponding to the maximum and minimum shear stresses are

non-null values and are obtained as:

2)( 11

yxsns

σσθθσσ

+=== (10.32)

2)( 22

yxsns

σσθθσσ

+=== (10.33)

The maximum and minimum shear stresses and the corresponding normal stresses are


Figure 10.1.6 Principal Planes and Maximum Shear Stress Planes

Note: From Figure 10.1.6 it can be concluded that in contrast to the principal planes

which are free of shear stress, the planes on which the shear stress achieves

extreme values are not necessarily free of normal stresses.

347

10.1.4 Mohr’s Circle for Plane Stresses

Mohr’s Circle is a graphical construction reflecting the variation of the plane state of

stress around a particular point, including information pertinent to the principal and

maximum shear stresses.

The Mohr’s Circle for plane stress condition is drawn relative to a Cartesian system

with the abscissa and the ordinate axis representing the normal stresses σ and the

shear stressτ , respectively. The following sign convention is employed as illustrated

in Figure 10.1.7:

(a) the positive shear stressτ axis is downward. This convention is elected in

order to be able to enforce the positive measurement of the angle (sign

convention (b));

(b) the positive angle is measured counterclockwise;

(c) the shear stress on a face plots as positive shear if tends to rotate the face

counterclockwise.

Morh’s Circle for plane stress is constructed in the following steps:

(a) The coordinate system is drawn as shown in Figure 10.1.7. The horizontal axis

represents the normal stress σ , while the vertical axis represents the shear

stress τ . To gain full advantage of the graphical benefits of the method it is

necessary that the drawing to be made on scale. The representation illustrated

considers the case yx σσ > and 0>xyτ ;

(b) Using the calculated values of the normal stresses xσ and yσ and the shear

stress xyτ two points noted as ),( xyxX τσ and ),( xyyY τσ − are placed on the

drawing. The line XY intersects the horizontal axis at point C which

represents the center of the Mohr’s Circle;

(c) The distance CX represents the radius of the circle. Using the radius CX and

the position of the center )0,( avgC σ the Mohr’s Circle is constructed. The

intersection points, 1P and 2P , between the circle and the horizontal axis

348

represent the maximum and the minimum principal stresses, 1σ and 2σ ,

respectively;

(d) The value of the angles of rotation pertinent to the principal stresses 1pθ and

2pθ can also be calculated from the graph. The third point ),( xyyZ τσ , called

the pole of normals , is constructed. The lines 1ZP and 2ZP represent the

principal direction 1 (associated with the maximum principal stress) and 2

(associated with the minimum principal stress), respectively. The rotation

angles of the principal directions, 1pθ or 2pθ , are measured from the line XZ to

1ZP and 2ZP , respectively;

Figure 10.1.7 Mohr’s Circle Notation

(e) Constructing two points, S1 and S2, at the intersection of the vertical diameter

of the Mohr’s circle with its circumference, the maximum shear stresses and

their corresponding normal stresses are obtained. The points S1 and S2 have as

coordinates the two pairs ( 1sσ , 1sτ ) and ( 2sτ , 2sσ ), respectively. The

directions of the maximum shear stresses are obtained by drawing the lines

349

ZS1 and ZS2. The angles 1sθ and 2sθ are measured from the reference line ZX

to the lines ZS1 and ZS2, respectively.

(f) The normal and shear stresses, )(θσ n and )(θτ nt , corresponding to an

inclined plane NN, characterized by the angle θ , are graphically obtained

constructing a line inclined with the angle θ measured from the line ZX. The

coordinates of the intersection point N between the inclined line and the circle

circumference represent the of the corresponding normal and shear stresses,

)(θσ n and )(θτ nt ;


Problem 10.2.1 The stress components of the plane stress tensor Tσ at a particular

point are shown in the Figures 10.2.1. Conduct the following tasks: (a) calculate the

principal stresses and directions, (b) calculate the maximum shear stresses and

directions, (c) calculate the stresses on the indicated inclined plane NN ( θ 30deg:= ),

(d) calculate the stress components of the stress tensor if the coordinate system is

rotated with 45 deg and (e) plot the corresponding Mohr’s Circle.

Figure 10.2.1


The plane stress tensor corresponding to the stress condition illustrated in Figure

10.2.1 is obtained as:

350

MpayxTyyx

xyx

32161648

),(−

==σττσ

σ

where the stress components are identified accordingly to the sign convention

pictured in Figure 10.1.3:

σx 48 MPa⋅:=

σy 32− MPa:=

τxy 16MPa:=

B. Calculations

B.1 Principal Stresses and Directions

The two main values, the average normal stress avσ and the radius R are calculated

accordingly to formulae (10.20) and (10.21):

σavg12

σx σy+( )⋅:= →−⇒ )3248(*21

σavg 8MPa=

R14

σx σy−( )2⋅ τxy

2+:= →++⇒ 22 16)3248(*

41

R 43.081MPa=

The principal stresses 1σ and 2σ are calculated using the formulae (10.18) and

(10.19):

σ1 σavg R+:= →+⇒ 081.438 σ1 51.081MPa=

σ2 σavg R−:= →−⇒ 081.438 σ2 35.081− MPa=

The principal directions are calculated solving the trigonometric equation (10.22)

tan θ2p( )τxy

σx σy−

2

:= →−−

⇒

2)32(48

16 τxyσx σy−

2

0.4=

Then, it results that the double angle is:

θ2p atanτxy

σx σy−

2

:= →−−

⇒ − )

2)32(48

16(tan 1 θ2p 21.801deg=

Two trigonometric solutions are possible:

351

θp0_1θ2p2

:= →⇒2801.21

θp0_1 10.901deg=

θp0_2π2

θp0_1+:= →+⇒ 901.102

180θp0_2 100.901deg=

To identify the direction angle of the maximum normal stress two methods can be used: (a) First method

The normal stresses corresponding to the two calculated angles are obtained using the formula (10.9):

σn1σx σy+

2

σx σy−

2cos 2 θp0_1⋅( )⋅+ τxy sin 2 θp0_1⋅( )⋅+:=

→++

+−

⇒ )901.10*2sin(*16)901.10*2cos(*2

32482

3248

σn1 51.081MPa=

σn2σx σy+

2

σx σy−

2cos 2 θp0_2⋅( )⋅+ τxy sin 2 θp0_2⋅( )⋅+:=

→++

+−

⇒ )901.100*2sin(*16)901.100*2cos(*2

32482

3248

σn2 35.081− MPa=

The principal direction angles are calculated:

θp1 θp0_1 σ1 σn1if

θp0_2 σ1 σn2if

:= →

angle of the maximum normal stress direction θp1 10.901deg=

θp2 θp0_1 θp1 θp0_2if

θp0_2 θp1 θp0_1if

:= →

angle of the minimum normal stress direction θp2 100.901deg=

(b) Second method The angle of the maximum normal stress direction is obtained using the formula (10.25):

352

θp1 θp0_1tan θp0_1( )

τxy0≥if

θp0_2tan θp0_1( )

τxy0<if

:= →


The complementary angle of the minimum normal stress direction is calculated: θp2 θp0_1 θp1 θp0_2if


:= →


The graphical representation of the principal stresses and directions is illustrated

below.

B.2 Maximum Shear Stresses and Directions

The magnitude of the maximum shear stresses 1τ and 2τ are calculated accordingly to

formulae (10.30) and (10.31): τ1 R:= → τ1 43.081MPa=

τ2 R−:= → τ2 43.081− MPa=

The corresponding normal stress is:

σs σavg:= → σs 8MPa=

The direction of maximum shear stress is calculated employing the formula (10.27):

353

tan θ2s( )σx σy−

2

τxy−:= →

−−

−⇒162

)32(48

-2.5

Then, it results:

: θ2s atan

σx σy−

2

τxy−

:= →

−−

−⇒ −

162

)32(48

tan 1 θ2s 68.199− deg=


θs0_1θ2s2

:= →−⇒2199.68

θs0_1 34.099− deg=

θs0_2π2

θs0_1+:= →−⇒ 099.342

180θs0_2 55.901deg=

Two methods can be used to identify the direction angle of the maximum shear stress:

(c) First method

The shear normal stresses corresponding to the two calculated angles are obtained using the formula (10.11):

τs1σx σy−

2− sin 2 θs0_1⋅( )⋅ τxy cos 2 θs0_1⋅( )⋅+:= → τs1 43.081MPa=

τs2σx σy−

2− sin 2 θs0_2⋅( )⋅ τxy cos 2 θs0_2⋅( )⋅+:= → τs2 43.081− MPa=

The direction angles are calculated: θs1 θs0_1 τ1 τs1if

θs0_2 τ1 τs2if

:= →

angle of the maximum shear stress direction θs1 34.099− deg=

θs2 θs0_1 θs1 θs0_2if

θs0_2 θs1 θs0_1if

:= →

angle of the minimum shear stress direction θs2 55.901deg=

354

(d) Second method

The angle of the maximum shear stress direction is obtained using the formula (10.29):

θs1 θp1π4

−:= →−⇒4

180901.10 θs1 34.099− deg=

The complementary angle corresponding to the minimum shear stress direction is calculated as:

θs2 θs1π2

+:= →+−⇒2

90099.34 θs2 55.901deg=

B.3 Stresses on an Inclined Plane (θ=30deg)

The rotation angle of the inclined plane is θ 30deg:= . Employing the formulae

(10.10) and (10.11) the normal and shear stresses, nσ and ntτ , are calculated:

σnσx σy+

2

σx σy−

2cos 2 θ⋅( )⋅+ τxy sin 2 θ⋅( )⋅+:=

→++

+−

⇒ )30*2sin(*16)30*2cos(*2

32482

3248σn 41.856MPa=

τntσx σy−

2− sin 2 θ⋅( )⋅ τxy cos 2 θ⋅( )⋅+:=

→++

−⇒ )30*2cos(*16)30*2sin(*2

3248τnt 26.641− MPa=

355

B.4 Modification of the Plane Stress Tensor with Rotated Coordinate System

The stresses on the infinitesimal square rotated with an angle θr 45deg:=

are

calculated accordingly to formulae (10.12) through (10.15). For calculation simplicity those formulae can be re-cast in a matrix notation as:

σr R σ0⋅

The extended notation of the above equation is:

σrx

σry

τrxy

12

1 cos 2 θr⋅( )+( )⋅

12

1 cos 2 θr⋅( )−( )⋅

12

sin 2 θr⋅( )⋅

12

1 cos 2 θr⋅( )−( )⋅

12

1 cos 2 θr⋅( )+( )⋅

12

− sin 2 θr⋅( )⋅

sin 2 θr⋅( )

sin 2 θr⋅( )−

cos 2 θr⋅( )−

σx

σy

τxy

⋅

For the case in point, after the algebraic manipulations and matricial multiplication, it

results:

σ0

48

32−

16

MPa⋅:=

R

0.5

0.5

0.5−

0.5

0.5

0.5

1

1−

0

= and

σr

24

8−

40−

MPa=

Consequently, the rotated stress components

σrx 24MPa= ,

σry 8− MPa=

and

τxy 40− MPa= are pictured below.

356

B.5 Mohr’s Circle Construction

The Morh’s Circle pertinent to this stress condition is constructed following the steps

described in the theoretical section. The results obtained in sections B3 and B4 are

illustrated by the coordinates of the points N( ntn τσ , ), P( rxrx τσ , ) and P’( rxry τσ −, ).

357

Problem 10.2.2 Considering the beam shown in Figure 7.2.7 and a point P located on

the cross-section S at 5 cm above the neutral axis calculate the followings: (a) the

pertinent stresses for point P, (b) the principal stresses and directions, (c) the

maximum shear stresses and directions, and (d) plot the corresponding Mohr’s Circle.


A1. The cross-section internal resultants pertinent to cross-section S are obtained from

problem 7.2.7 calculations, section B.5:

Mz_S 56− kN⋅ m⋅:=

Vy_S 59− kN⋅:=

A2. The following data necessary for the calculation of the normal stress in point P of

the cross-section S is retrieved from problem 7.2.7 calculations, section B.5.a and

Figure 7.2.7.d:

yP 5 cm⋅:= - the position of the point P measured from the neutral axis (NA).

IZc 14192.717 cm4⋅:=

A2. The following data necessary for the calculation of the shear stress in point P of

the cross-section S is retrieved from problem 7.2.7 calculations, section B.5.b.3 and

Figure 7.2.7.h:

AU24 42.3 cm2⋅:=

yC2 10.381 cm⋅:=

ysup 12.611 cm⋅:=

tw_U24 0.95 cm⋅:=

tpl 1 cm⋅:=

B. Calculations

B.1 Stresses in Point P of the Cross-section S

The normal stress in point P is calculated accordingly to Navier’s Formula:

358

σx_P_SMz_S−

IZcyP⋅:= →

−−⇒ −

− mm

mN *10*5*10*717.14192

*10*56 248

3

σx_P_S 19.728MPa=

The shear stress in point P is obtained employing the Jurawski’s Formula:

s3 ysup tw_U24− yP−:= →−−⇒ 595.0611.12 s3 6.661cm= - the height of the web above the cut

SZc_ef s3( ) AU24 yC2⋅ s3 tpl⋅ ysup tw_U24−s32

−

⋅+:=

→−−+⇒ )2661.695.0611.12(*1*661.6381.10*3.42 SZc_ef s3( ) 494.606cm3=

- the static moment of the area above the cut

qyx_ef s3( )Vy_S SZc_ef s3( )⋅

IZc:=

→

−⇒ −

−

48

363

10*717.14192*10*606.494*10*59

mmN

qyx_ef s3( ) 205.611−kNm

= - the flow

τyx_ef_avgs3( )qyx_ef s3( )

tpl:= →

−⇒ − m

mkN

210*1

*611.205τyx_ef_avgs3( ) 20.561− MPa=


In order to use the notation employed in the theoretical section the above calculated stresses are recast as: σx σx_P_S:= → σx 19.728MPa=

σy 0 MPa⋅:= → σy 0MPa=

τxy τxy_ef_avgs3( ):= → τxy 20.561MPa=

Note: For the case of the plane linear beams subjected to non-uniform bending the

normal stress yσ is always zero.

The plane stress tensor corresponding to the stress condition pertinent to point P,

illustrated in Figure 10.2.2, is:

MpayxTyx

xyx

0561.20561.20728.19

0),( ==

ττσ

σ

where the stress components are identified accordingly to the sign convention

pictured in Figure 10.1.3:

359

Figure 10.2.2

B.2 Principal Stresses and Directions

The two main values, the average normal stress avσ and the radius R are calculated

accordingly to formulae (10.20) and (10.21):

σavg12

σx σy+( )⋅:= →+⇒ )0728.19(*21

σavg 9.864MPa=

R14

σx σy−( )2⋅ τxy2+:= →++⇒ 22 561.20)0728.19(*

41 R 22.805MPa=

The principal stresses 1σ and 2σ are calculated using the formulae (10.18) and

(10.19):

σ1 σavg R+:= →+⇒ 805.22864.9 σ1 32.669MPa=

σ2 σavg R−:= →−⇒ 805.22864.9 σ2 12.941− MPa=

The principal directions are calculated solving the trigonometric equation (10.22)

tan θ2p( )τxy

σx σy−

2

:= →⇒

2728.19561.20 τxy

σx σy−

2

2.084=

Then, it results that the double angle is:

θ2p atanτxy

σx σy−

2

:= →⇒ − )

2728.19561.20(tan 1 θ2p 64.371deg=


θp0_1θ2p2

:= →⇒2371.64

θp0_1 32.185deg=

360

θp0_2π2

θp0_1+:= →+⇒ 185.322

180θp0_2 122.185deg=

To identify the direction angle of the maximum normal stress two methods can be used: (e) First method The normal stresses corresponding to the two calculated angles are obtained using the formula (10.9):

σn1σx σy+

2

σx σy−

2cos 2 θp0_1⋅( )⋅+ τxy sin 2 θp0_1⋅( )⋅+:=

→++⇒ )185.32*2sin(*561.20)185.32*2cos(*2728.19

2728.19

σn1 32.669MPa=

σn2σx σy+

2

σx σy−

2cos 2 θp0_2⋅( )⋅+ τxy sin 2 θp0_2⋅( )⋅+:=

→++⇒ )185.122*2sin(*561.20)185.122*2cos(*2728.19

2728.19

σn2 12.941− MPa=

The principal direction angles are calculated:

θp1 θp0_1 σ1 σn1if

θp0_2 σ1 σn2if

:= →


θp2 θp0_1 θp1 θp0_2if


:= →


(f) Second method The angle of the maximum normal stress direction is obtained using the formula (10.24):

θp1 θp0_1tan θp0_1( )

τxy0≥if

θp0_2tan θp0_1( )

τxy0<if

:= →

361


The complementary angle of the minimum normal stress direction is calculated: θp2 θp0_1 θp1 θp0_2if


:= →


The graphical representation of the principal stresses and directions is illustrated

below.

B.3 Maximum Shear Stresses and Directions

The magnitude of the maximum shear stresses 1τ and 2τ are calculated accordingly to

formulae (10.30) and (10.31): τ1 R:= → τ1 22.805MPa=

τ2 R−:= → τ2 22.805− MPa=

The corresponding normal stress is:

σs σavg:= → σs 9.864MPa=

The direction of maximum shear stress is calculated employing the formula (10.27):

tan θ2s( )σx σy−

2

τxy−:= →−⇒

561.202728.19

-0.480

Then, it results:

362

: θ2s atan

σx σy−

2

τxy−

:= →

−⇒ −

561.202728.19

tan 1 θ2s 25.629− deg=


θs0_1θ2s2

:= →−⇒2629.25

θs0_1 12.815− deg=

θs0_2π2

θs0_1+:= →−⇒ 815.122

180θs0_2 77.185deg=

Two methods can be used to identify the direction angle of the maximum shear stress: (g) First method The shear normal stresses corresponding to the two calculated angles are obtained using the formula (10.11):

τs1σx σy−

2− sin 2 θs0_1⋅( )⋅ τxy cos 2 θs0_1⋅( )⋅+:= → τs1 22.805MPa=

τs2σx σy−

2− sin 2 θs0_2⋅( )⋅ τxy cos 2 θs0_2⋅( )⋅+:= → τs2 22.805− MPa=

The direction angles are calculated: θs1 θs0_1 τ1 τs1if

θs0_2 τ1 τs2if

:= →

angle of the maximum shear stress direction θs1 12.815− deg=

θs2 θs0_1 θs1 θs0_2if

θs0_2 θs1 θs0_1if

:= →

angle of the minimum shear stress direction θs2 77.185deg=

(h) Second method

The angle of the maximum shear stress direction is obtained using the formula (10.29):

θs1 θp1π4

−:= →−⇒4

180185.32 θs1 12.815− deg=

363

The complementary angle corresponding to the minimum shear stress direction is calculated as:

θs2 θs1π2

+:= →+−⇒2

180815.12 θs2 77.185deg=

B.5 Mohr’s Circle Construction

The Morh’s Circle pertinent to the calculated stress condition is constructed following

the steps described in the theoretical section.

364


Problem 10.3.1- 10.3.3 The stress components of the plane stress tensor Tσ at a

particular point are shown in the Figures 19.3.1 through 10.3.3. Calculate the stresses

on the indicated inclined plane NN for each individual case and construct the

corresponding Morh’s circle.


Figure 10.3.3

Problem 10.3.4 The components of the stress tensor Tσ at point A of a bracket are

shown in Figure 10.3.4. Calculate and draw the stresses around the point A if the

coordinate system is rotated anti-clock-wise with an angle θ = 20º.

365

Figure 10.3.4

Problem 10.3.5 The hinged support of the steel plate girder illustrated in Figure 10.3.5

depicts the region of a plate girder where it transfers its vertical load through a roller

support to a concrete pier. The components of the stress tensor Tσ at point A are also

shown in the picture. Calculate and draw the components of the plane stress tensor if

the coordinate system is rotated anti-clock-wise with an angle θ = 30º. Recalculate for

the caseθ =-30º.

Figure 10.3.5

Problem 10.3.6 - 10.3.7 The components of the plane stress tensor Tσ around a given

point are shown in the corresponding figures. Calculate and draw the components of

the stress tensor if the coordinate system is rotated anti-clock-wise with an angle θ =

366

30º and then with an additional angle θ = 80º. Recalculate for a rotation angle θ = -

30º. Comment the results.


Problem 10.3.8 – 10.3.9 The components of the plane stress tensor Tσ at a given point

A and corresponding to a rotated coordinate system Ox’y’ are shown in Figures 10.3.8

and 10.3.9. Calculate and draw the components of the stress tensor corresponding to

the coordinate system Oxy.


Problem 10.3.10 – 10.3.11 The components of the plane stress tensor Tσ at point A

considering two coordinate systems, Oxy and Ox’y’, rotated from each other with an

angle θ, are shown in the following figures. Calculate the angle θ and the unknown

367

indicated stress components. Repeat the calculations graphically employing the

corresponding Mohr’s circle.

Figure 10.3.10

Figure 10.3.11

Problem 10.3.12 Considering the following generalized plane stress tensors

Tσ calculate and draw: (a) the stress tensor components, (b) the principal stresses and

principal directions, (c) the maximum shear stresses and directions and (d) the


(T1) σx = 20.68 MPa σy = -34.47 MPa τxy = -20.68 MPa;

(T2) σx = 48 MPa σy = -32 MPa τxy = 16 MPa;

(T3) σx = 12 MPa σy = -8 MPa τxy = -4 MPa;

368

(T4) σx = 55.16 MPa σy = -8.27 MPa τxy = -27.58 MPa;

(T5) σx =-27.58 MPa σy = 82.7 MPa τxy = 41.37 MPa;

(T6) σx = 16.54 MPa σy = 99.29 MPa τxy = 99.28 MPa.

Problem 10.3.13 – 10.3.17 Considering the stress tensors calculated at point A during

the life service of a suspension car component as indicated in Figures 10.3.13 through

10.3.17 calculate and draw the followings: (a) the generalized stress plane tensor, (b)

(b) the principal stresses and principal directions, (c) the maximum shear stresses and

directions and (d) the corresponding Mohr’s circle.

Figure 10.3.13


369


Problem 10.3.18 At a certain point A, shown in Figure 10.3.18, the normal stresses on

two mutually perpendicular x and y positive faces are 7 MPa (compression) and 3

MPa (tension). The maximum in-plane shear stress at this point is 10 MPa. Calculate:

(a) the magnitude of the shear stress acting on the x>0 and y>0 faces and (b) the

principal stresses and directions. Conduct the calculations graphically by constructing

the corresponding Mohr’s circle.

Figure 10.3.18

Problem 10.3.19 The components of generalized plane stress tensor at a certain point

A located on the surface of an airplane wing are illustrated in Figure 10.3.19. The

maximum in-plane shear stress at this point is 89.63 MPa. Calculate: (a) the

magnitude of the shear stress on the x and y faces, (b) the principal stresses and

370

directions at this point. Conduct the calculations graphically by constructing the


Figure 10.3.19

Problem 10.3.20 The state of plane stress at a point A is characterized by the stress

components shown in Figure 10.3.20. The maximum in-plane shear stress is 78 MPa

and one of the two in-plane principal stresses is 22 MPa (tension). Calculate: (a) the

values of the two unknown components of the generalized plane tensor, (b) the other

principal stress, (c) the principal directions and (d) the maximum shear directions.

Conduct the calculations graphically by constructing the corresponding Mohr’s circle.

Figure 10.3.20

Problem 10.3.21 For the plane generalized stress tensor, representing the uni-axial

state of stress illustrated in Figure 10.3.21, conduct the following tasks: (a) construct

Mohr's circle, (b) use the Mohr's circle to derive the equations for the normal and

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shear stresses on the n-face and (c) use the Mohr's circle to determine the planes on

which maximum shear stress acts. Sketch a properly oriented maximum shear stress

element and indicate the normal and shear stresses acting on its faces.

Figure 10.3.21

Problem 10.3.22 For the plane generalized stress tensor, representing the pure shear

state of stress illustrated in Figure 10.3.22, conduct the following tasks: (a) construct

Mohr's circle, (b) use the Mohr's circle to derive the equations for the normal and

shear stresses on the n-face and (c) use the Mohr's circle to determine the planes on

which maximum shear stress acts. Sketch a properly oriented maximum shear stress

element and indicate the normal and shear stresses acting on its faces.

Figure 10.3.22

Problem 10.3.23 The cantilever beam shown in Figure 10.3.23 is subjected to a

vertical uniform distributed load. The beam cross-section is of rectangular shape and

constant along its entire length. Calculate and draw: (a) the generalized plane stress

tensor, (b) the principal stresses and directions, (c) the maximum shear stresses and

372

directions and (d) the corresponding Mohr’s circle for all three points, A (x = 0, yA =

h/2), B (x = 0, yB = h/4) and C (x = 0, yC = 0) , as indicated in the figure.

Figure 10.3.23

Problem 10.3.24 The simply supported beam shown in Figure 10.3.24 is subjected to

a concentrated vertical load acting at the half-span. The beam cross-section is made of

a W8x40 and constant along its entire length. Calculate and draw the followings: (a)

the generalized plane stress tensor, (b) the principal stresses and directions, (c) the

maximum shear stresses and directions and (d) the corresponding Mohr’s circle for

three points: A (x = 0.6 m, yA = h/2), B (x = 0.6 m, yB = hw/2) and C (x = 0.6 m, yC =

hw/4) , as indicated in the figure, where h and hw are the W shape and the web heights,

respectively.

Figure 10.3.24

Problem 10.3.25 The simply supported beam loaded shown in Figure 10.3.25 is made

of I30 steel shape. At the cross-section located at 20 in from the right-hand support,

calculate: (a) the generalized plane stress tensor, (b) the principal stresses and

directions, (c) the maximum shear stresses and directions and (d) the corresponding

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Mohr’s circle at each of the following locations: (1) top of the beam, (2) top of the

web and (3) the neutral axis.

Figure 10.3.25

Problem 10.3.26 The overhanging beam illustrated in Figure 10.3.26 is loaded with

concentrated force P acting at the tip of the cantilever.

Figure 10.3.26

Considering that the cross-section is a rectangular shape h x b and constant along its

entire length, find the expressions of the plane stress tensor components, the principal

stresses and direction and the maximum shear stress around point D. Assuming the

following numerical data: h = 150 mm, b = 25 mm and L = 1.5 m, calculate the

magnitude of the concentrated load P, if the maximum principal stress (tension) at

point D is 49 MPa and construct the corresponding Mohr’s circle.

Problem 10.3.27 The simply supported beam shown in Figure 10.3.27 loaded with a 8

concentrated load P has a constant cross-section along its entire length. For the mid-

span cross-section calculate: (a) the generalized plane stress tensor, (b) the principal

374

stresses and directions, (c) the maximum shear stresses and directions and (d) the

corresponding Mohr’s circle at each of the points A, B and C.

Figure 10.3.27

Problem 10.3.28 A rectangular wooden simply supported beam shown in Figure

10.3.28 is loaded with 36 kN concentrated force acting at 1.22 m distance from the

left-hand support. Considering that the grain of the wood makes an angle of 20º with

the longitudinal axis of the beam conduct the following tasks: (a) find the stress tensor

components at points A and B along the grain fiber, (b) calculate the principal stresses

and directions at points A and B, and (c) calculate the maximum shear stress at points

A and B and (d) construct the corresponding Mohr’s circles.

Figure 10.3.28

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REFERENCES

1. Craig, Roy R., Mechanics of Materials, Second Edition, John Willey & Sons, New York, 2000.

2. Gere, James M., Mechanics of Materials, Fifth Edition, Brooks/Cole,

Pacific Grove, CA, 2001.

3. Higdon, A., Ohlsen, E.H., and Stiles, W.B., Mechanics of Materials, Third Edition, John Willey & Sons, New York, 1962.

4. Popov, E.P., Introduction to Mechanics of Solids, Prentice-Hall Inc.,

Englewood Cliffs, New Jersy, 1968.

5. Mazilu P., Posea N. and Iordachescu E., Probleme de Rezistenta Materialelor, Vol I, Editura Tehnica Bucuresti, 1969.

6. Hartog, J.P.D., Strength of Materials, Dover Publications Inc., New York,

1961.

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