Strength of Meterials

7/22/2019 Strength of Meterials

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STRENGTH OF MATERIALS

InstructorENGR. EFREN A. DELA CRUZ


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Engineering Mechanics(Mechanics of Materials)

Rigid Body

Mechanics

Deformable Body

Mechanics

Strength of Materials-

deals with the relation

between the externallyapplied loads and their

internal effects on

bodies assumed not

ideally rigid

Statics

Dynamics

Fluid

Mechanics

Strength of Materials- part of engineering

mechanics that deals with the relation of

externally applied loads and their internaleffects in bodies assumed not ideally rigid


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STRESSIntensity of forces distributed over a given section

COMPRESSIVE STRESS TENSILE STRESS

SHEAR STRESS

BEARING STRESS

larperpendicuA

Pcort

COMPRESSIVE STRESS TENSILE STRESS SHEAR STRESS

parallelA

P

s

bearingAPb


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STRESS

THERMAL STRESS

TORSIONAL SHEAR STRESS

FLEXURAL (BENDING) STRESS Pressure vessel

Spring

Beams


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Normal/Axial Stress

stressA

P A

P

A

P

2

2

A

P

ForcesPare applied normal to the memberBC.

Corresponding internal forces act in the section are calledAxial

force

The corresponding average axial stress is,


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Shearing Stress ForcesPand Pare applied transversely to the

memberAB.

A

Pave

The corresponding average shear stress is,

The resultant of the internal shear force

distribution is defined as theshearof the section

and is equal to the loadP.

Corresponding internal forces act in the plane

of section Cand are calledshearingforces.

Shear stress distribution varies from zero at themember surfaces to maximum values that may be

much larger than the average value.

parallelA

P

s


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Shearing Stress Examples

A

F

A

Pave

Single Shear

A

F

A

P

2ave

Double Shear


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Bearing Stress in Connections

Bolts, rivets, and pins create

stresses on the points of contactor bearing surfacesof the

members they connect.

dt

P

A

P

b

Corresponding average force

intensity is called the bearing

stress,

The resultant of the force

distribution on the surface isequal and opposite to the force

exerted on the pin.


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Sample Problem

The structure is designed tosupport a 30 kN load

Perform a static analysis to

determine the internal force in

each structural member and the

reaction forces at the supports

The structure consists of a steel

boom and rod joined by pins

(zero moment connections) at

the junctions and supports


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Structure Free-Body Diagram

Structure is detached from supports and

the loads and reaction forces are indicated

Ayand Cycan not be determined from

these equations

kN30

0kN300

kN40

0

kN40

m8.0kN30m6.00

yy

yyy

xx

xxx

x

xC

CA

CAF

AC

CAF

A

AM

Conditions for static equilibrium:


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Component Free-Body Diagram

In addition to the complete structure, each

component must satisfy the conditions forstatic equilibrium

Results:

kN30kN40kN40 yx CCA

Reaction forces are directed along

boom and rod

0

m8.00

y

yB

A

AM

Consider a free-body diagram for the boom:

kN30yC

substitute into the structure

equilibrium equation


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Method of Joints The boom and rod are 2-force members, i.e.,

the members are subjected to only two forceswhich are applied at member ends

kN50kN40

3

kN30

54

0

BCAB

BCAB

B

FF

FF

F

Joints must satisfy the conditions for static

equilibrium which may be expressed in the

form of a force triangle:

For equilibrium, the forces must be parallel to

to an axis between the force application points,

equal in magnitude, and in opposite directions


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Stress Analysis

Conclusion: the strength of memberBCisadequate

MPa165all

From the material properties for steel, theallowable stress is

Can the structure safely support the 30 kN

load?

MPa159m10314

N105026-

3

A

PBC

At any section through member BC, the

internal force is 50 kN with a force intensity

or stress of

dBC= 20 mm

From a statics analysis

FAB= 40 kN (compression)

FBC= 50 kN (tension)


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Design Design of new structures requires selection of

appropriate materials and component dimensions

to meet performance requirements

For reasons based on cost, weight, availability,

etc., the choice is made to construct the rod from

aluminum all= 100 MPa). What is an

appropriate choice for the rod diameter?

mm2.25m1052.2m10500444

m10500Pa10100

N1050

226

2

26

6

3

Ad

dA

PA

A

P

allall

An aluminum rod 26 mm or more in diameter is

adequate


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Problem The lap joint shown is fastened by 3-20mm

diameter joints. If a 50 kN load is applied asshown, determine:

a. Shearing stress in each rivet

b. Bearing stress in each plate

c. Maximum tensile stress in each plate.Assume the thickness of the plate is 25mm.

130mm


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Solutiona. Shearing stress in rivets

50x103N / 3

Ss= P/A = -------------------- = 53.05 N/mm2 = 53.05MPa

/4(20mm)2

b. Bearing stress in each plate50x103NSs= P/A = -------------------- = 33.33 N/mm2 = 33.33 MPa

25mmx20mmx3

c. Maximum tensile stress in each plate50x103N

Ss= P/A = -------------------- = 15.38 N/mm2 = 15.38 MPa

130mmx25mm

50x103N

Ss= P/A = -------------------------------- = 18.18 N/mm2 = 18.18 MPa

(130mm-20mm) x 25mm

Given:

P = 50kN w = 130mm

t = 25mm # of rivets = 3

= 20mm

El ti it


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Elasticity

All solid materials deform when they are stressed, and

as stress is increased, deformation also increases.The deformation per unit length of a material is called

strain .

If a material returns to its original size and shape on

removal of load causing deformation, it is said to beelastic.

If the stress is steadily increased, a point is reached

when, after the removal of load, not all the inducedstrain is removed.

This is called the elastic limit.


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Hookes Law

States that providing the limit of proportionality of amaterial is not exceeded, the stress is directlyproportional to the strain produced.

If a graph of stress and strain is plotted as load isgradually applied, the first portion of the graphwill be a straight line.

The slope of this line is the constant of

proportionality called modulus of Elasticity, E orYoungs Modulus.

It is a measure of the stiffness of a material.


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Hookes Law

Modulus of Elasticity, E = Direct stressDirect strain

Also: For Shear stress: Modulus of rigidity or shear modulus, G = Shear stressShear strain

Deformations Under A ial


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Deformations Under Axial

Loading

AE

P

EE

From Hookes Law:

From the definition of strain:

L

Equating and solving for the deformation,

AE

PL

With variations in loading, cross-section ormaterial properties,

i ii

ii

EA

LP


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Poisson's Ratio

W l l t th h i l th f d b t ki th


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We can calculate the change in length of a rod by taking the

definition of strain,

Or,

In the same way, we can calculate the change in diameter by

Substituting for , and diameter d for length L

d


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Problem

A 10mm x 6m steel rod is subjected to an axial

tension of 10 kN. If v= 0.30 and E = 200 GPa,find the change in the diameter of the rod.


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Solution

v= /

= d / 10mm

E = / = P/A/

200,000 N/mm2= {10,000 N / [/4 (10mm)2]} /

= 0.0006366

v= / 0.30 =d / 10mm / 0.0006366

d = 0.00191mm

Given:

= 10mm P = 10 kN L = 6mE = 200GPa = 0.30

An aluminum rod has a cross sectional area of 0 19635 in 2 An


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An aluminum rod has a cross-sectional area of 0.19635 in.2. An

axial tensile load of 6000 lb. causes the rod to stretch along its

length, and shrink across its diameter. What is the diameter before

and after loading when v= 0.33 and E = 30x106Psi? Report the

answer in inches.


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Factor of Safety

The load which any member of a machine/structurecarries is called working load, and stress produced by

this load is the working stress.

Obviously, the working stress must be less than the

yield stress, tensile strength or the ultimate stress.

This working stress is also called the permissible stress

or the allowable stress or the design stress.


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Stress-Strain Relations of Mild

Steel


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Factor of Safety Contd.

Some reasons for factor of safety include the

inexactness or inaccuracies in the estimation

of stresses and the non-uniformity of some

materials.Factor of safety =

Ultimate or yield stress

Design or working stress

Note: Ultimate stress is used for materials e.g.concrete which do not have a well-defined yield point,

or brittle materials which behave in a linear manner

up to failure. Yield stress is used for other materials

e.g. steel with well defined yield stress.


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Problem

A rigid bar is hinged at A

and supported by a steelrod at B. A strain gauge at

the rod indicates a strain of

0.0003. If the rod is 75mm2

in cross section, calculate

the applied load W.

Assume E = 200GPa.

600

450

W

3.5m

2.5m

A

B


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E = S /

= T/A

---------

200,000N/mm2 = T / 75mm2 / 0.0003

T = 4500N

From the Figure

= 180 - 6045 = 750

MA= 0

W (3.5sin 60)4500 (6 sin 75) = 0

w = 8604.17 N

SolutionGiven:

A = 75mm2 = 0.0003

E = 200GPa

600

450

W

3.5m

2.5m

A

B

Ty

Tx


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a. Required from stressP = AS

2000 N = /4 d2(140 N/mm2)

d = 4.26mm

b. Required from deformation

y = PL/AE

5mm = 2000 N ( 10000 mm )/4 d2(200 x103N/mm2)

d = 5.05mm

SolutionGiven:

P = 2000 N

S = 140 MPa

E = 200 GPa

L = 10m

y = 5mm

E l


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Example

Determine the deformation

of the steel rod shown

under the given loads.

in.618.0in.07.1psi1029

6

dDE

SOLUTION: Divide the rod into components at

the load application points.

Apply a free-body analysis on each

component to determine theinternal force

Evaluate the total of the component

deflections.


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SOLUTION:

Divide the rod into three

components:

221

21

in9.0

in.12

AA

LL

23

3

in3.0

in.16

A

L

Apply free-body analysis to each

component to determine internal forces,

lb1030

lb1015

lb1060

33

32

31

P

P

P

Evaluate total deflection,

in.109.75

3.0

161030

9.0

121015

9.0

121060

1029

1

1

3

333

6

3

33

2

22

1

11

A

LP

A

LP

A

LP

EEA

LP

i ii

ii

in.109.75 3


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Sample Problem

The rigid bar BDEis supported by twolinksABand CD.

LinkABis made of aluminum (E= 70

GPa) and has a cross-sectional area

of 500 mm2. Link CDis made of steel

(E = 200 GPa) and has a cross-sectional area of (600 mm2).

For the 30-kN force shown, determine

the deflection a) of B, b) of D, and c)

of E.

SOLUTION:

Apply a free-body analysis to the bar

BDEto find the forces exerted by

linksABandDC.

Evaluate the deformation of linksAB

andDC or the displacements ofB

andD.

Work out the geometry to find the

deflection at E given the deflections

at B and D.


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Displacement of B:

m10514

Pa1070m10500

m3.0N1060

6

926-

3

AE

PLB

mm514.0B

Displacement of D:

m10300

Pa10200m10600m4.0N1090

6

926-

3

AE

PLD

mm300.0D

Free body: Bar BDE

ncompressioF

F

tensionF

F

M

AB

AB

CD

CD

B

kN60

m2.0m4.0kN300

0M

kN90

m2.0m6.0kN300

0

D

SOLUTION:

Sample Problem


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Displacement of D:

mm7.73

mm200

mm0.300

mm514.0

x

x

x

HD

BH

DD

BB

mm928.1E

mm928.1mm7.73

mm7.73400

mm300.0

E

E

HD

HE

DD

EE

Sample Problem


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Static Indeterminacy Structures for which internal forces and reactions

cannot be determined from statics alone are said

to bestatically indeterminate.

0 RL

Deformations due to actual loads and redundant

reactions are determined separately and then added

orsuperposed.

Redundant reactions are replaced with

unknown loads which along with the other

loads must produce compatible deformations.

A structure will be statically indeterminate

whenever it is held by more supports than

what are required to maintain its equilibrium.

Example


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Example

Determine the reactions atAandBfor the steel

bar and loading shown, assuming a close fit at

both supports before the loads are applied.

Solve for the reaction atAdue to applied loads

and the reaction found atB.

Require that the displacements due to the loadsand due to the redundant reaction be compatible,

i.e., require that their sum be zero.

Solve for the displacement atBdue to the

redundant reaction atB.

SOLUTION:

Consider the reaction atBas redundant, release

the bar from that support, and solve for the

displacement atBdue to the applied loads.


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SOLUTION:

Solve for the displacement atBdue to the applied

loads with the redundant constraint released,

EEA

LP

LLLL

AAAA

PPPP

i ii

ii 9L

4321

2643

2621

34

3321

10125.1

m150.0

m10250m10400

N10900N106000

Solve for the displacement atBdue to the redundant

constraint,

i

B

ii

iiR

B

E

R

EA

LP

LL

AA

RPP

3

21

262

261

21

1095.1

m300.0

m10250m10400

Example


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Require that the displacements due to the loads and due to

the redundant reaction be compatible,

kN577N10577

01095.110125.1

0

3

39

B

B

RL

R

E

R

E

Find the reaction atAdue to the loads and the reaction atB

kN323

kN577kN600kN3000

A

Ay

R

RF

kN577

kN323

B

A

R

R

Example


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Thermal Stresses A temperature change results a change in length or

thermal strain. There is no stress associated with thethermal strain unless the elongation is restrained by

the supports.

coef.expansionthermal

AEPLLT PT

Treat the additional support as redundant and apply

the principle of superposition.

0

0

AE

PLLT

PT

The thermal deformation and the deformation from

the redundant support must be compatible.

TEA

P

TAEPPT

0


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Problem

Steel railroad rails 10m long are laid with a

clearance of 3mm at a temperature of 150C.

At what temperature will the rails just touch?

What stress will be induced in the rails at thattemperature if there where no initial

clearance? Assume = 11.7 x 10-6m/ (m0C)

and E = 200 GPa.


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YT= LT

3mm = 11.7 x 10-6m /(m0C) 10,000mm (T-15)

T = 40.640C

Y = SL/E

3mm = S (10,000 mm) / 200 x 103N/mm2

S = 60 MPa

SolutionGiven:

L = 10m

= 11.7 x 10-6m /(m0C)

E = 200 MPa

Yt = 3mm


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Problem 13

A steel rod is stretched between two

rigid walls and carries a tensile load of

5000N at 200C. If the allowable stress

is not to exceed 130 MPa at - 200C,what is the minimum diameter of the

rod? Assume = 11.7 x 10-6m/ (m0C)

and E = 200 GPa.


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Y = Yt+ Y1

SL/E = LT + P1L /AE

130 N/mm2 5000N

------------------------ = 11.7 x 10-6(20 + 20) + ----------------------------

200 x 103N/mm2 A (200 x 103N/mm2)

A = 137.4 mm2

137.4 mm2= /4 d2

d = 13.22mm

5000N 5000N

Yt Y1

Y

Solution Given:P = 5000N

= 11.7 x 10-6m /(m0C)

E = 200 MPa

S = 130 MPa

Thin-Walled Pressure Vessels


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Thin-Walled Pressure Vessels

A pressure vessel is a container that holds

a fluid (liquid or gas) under pressure.

Examples include carbonated beveragebottles, propane tanks, and water supply

pipes. Drain pipes are not pressure vessels

because they are open to the atmosphere.

If the thickness of the wall is less than 10%


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If the thickness of the wall is less than 10%

of the internal radius of the pipe or tank, then

the pressure vessel is described as a thin-

walled pressure vessel.We can assume that the stress in the wall is

the same on the inside and outside walls.

(Thick-walled pressure vessels have a higher

stress on the inner wall than on the outer wall,

so cracks form from the inside out.)

Imagine cutting a thin-walled pipe lengthwise


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Imagine cutting a thin walled pipe lengthwise

through the pressurized fluid and the pipe wall:

the force exerted by the fluid must equal the

force exerted by the pipe walls (sum of the forces

equals zero). The force exerted by the fluid ispA= p diL

where diis the inside diameter of the pipe, and L

is the length of the pipe.

The stress in the walls of the pipe is equal to the


51/142

The stress in the walls of the pipe is equal to the

fluid force divided by the cross-sectional area of

the pipe wall. This cross-section of one wall is

the thickness of the pipe, t, times its length L.Since there are two walls, the total cross-

sectional area of the wall is 2tL. The stress is

around the circumference or the hoopdirection,so

hoop=pdiL / 2tL.

Notice that the length cancels: hoop stress is

independent of the length of the pipe, so

hoop=pdi/ 2t.

Example #1


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Example #1A pipe with a 14 inch inside diameter carries pressurized

water at 110 psi. What is the hoop stress if the wall

thickness is 0.5 inches?SolutionFirst, check if the pipe is thin-walled. The ratio

of the pipe wall thickness to the internal radius is

What if the pipe has a cap on the end? If the cap


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What if the pipe has a cap on the end? If the cap

were loose, pressure would push the cap off the

end. If the cap is firmly attached to the pipe, then

a stress develops along the length of the pipe toresist pressure on the cap. Imagine cutting the

pipe and pressurized fluid transversely. The force

exerted by the fluid equals the force along thelength of the pipe walls. Pressure acts on a

circular area of fluid, so the force exerted by the

fluid is

Ffluid=pA=

The cross-sectional area of the pipe wall


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e c oss sec o a a ea o e p pe a

We can estimate the cross-sectional area of athin-walled pipe pretty closely by multiplying the

wall thickness by the circumference, so

The stress along the length of the pipe is,

Compare the hoop stress and longitudinal stress


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p p g

equations: in a thin-walled pipe, hoop stress is

twice as large as longitudinal stress. If the

pressure in a pipe exceeds the strength of thematerial, then the pipe will split along its length

(perpendicular to the hoop direction).

In conclusion, if you have a pipe or a tubular tank,use

If you have a spherical tank, use


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Problem

A cylindrical pressure vessel is fabricated

from steel plates which have a thickness

of 20mm. The diameter of the vessel is

500mm and its length is 3m. Determinethe maximum internal pressure which can

be applied if the stress in the steel is

limited to 140MPa.


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ST= PD / 2t

140 N/mm2= P (500mm) / 2 (20mm)

P = 11.2 N/mm2

P = 11.2 MPa.

SL= PD / 4t

140 N/mm2= P (500mm) / 4 (20mm)

P = 22.4 N/mm2P = 22.4 MPa

Given:

t = 20mm

= 500mm

L = 3m

S = 140MPa

Solution


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Problem

A water tank is 8m in diameter and 12m

high. If the tank is to be completely filled

with water, determine the minimum

thickness of the tank plates if the stress islimited to 40MPa.


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S = PD / 2t

From Thermo and Fluid Mechanics

Pmax@ the bottom of tank:

P = h = 9810N/m3

(12m)= 117720 N/m2= 0.117N/mm2

S = PD / 2t

40 N/mm2

= 0.117 N/mm2

(8000mm) / 2tt = 11.7mm

Given:

= 8m

H = 12m

S = 40MPa

Solution

8m

12m


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Problem

A pipe carrying steam at 3.5 MPa has an outside

diameter of 450mm and wall thickness of 10mm.

A gasket is inserted between the flange at one

end of the pipe and a flat plate was used to holdthe cap end. How many 40mm bolts must be

used to hold the cap on if the allowable stress in

the bolts is 80MPa, of which 55 MPa is the initial

stress? What circumferential stress is developedin the pipe?

S l ti


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S = PD/2tS = 3.5N/mm2(430mm)/2(10mm)

S = 72.25MPa (Circumferential stress)

Force due to pressure

F = AS

F = (/4) (450mm - 2*10mm)2(3.5 N/mm2)

F = 508,270.42N

Force that can be carried by each bolt

T = AS

= (/4) (40mm)2(80-55N/mm2)

T = 31,415.93 NNo. of Bolts

nT = F

n = F/T = 508,270.42N / 31,415.93N

n = 16.2 say 17 bolts.

Given:

out= 450mm

t = 10mm

Sbolt= 80MPa

Sini= 55MPa

bolt= 40mm

P = 3.5 MPa

Solution

Torsional Loads on Circular Shafts


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Torsional Loads on Circular Shafts

Many machine parts are loaded in

torsion, either to transmit power (likea driveshaft or an axle shaft in a

vehicle) or to support a dynamic load

(like a coil spring or a torsion bar).

Power transmission parts are

typically circular solid shafts orcircular hollow shafts because these

shapes are easy to manufacture and

balance, and because the outermost

material carries most of the stress.For a given maximum size, more

material is available along the entire

surface of a circle than at the four

corners of a square.

Shear Stress on Circular Shafts


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Shear Stress on Circular ShaftsApply a torque T to a round shaft, and

the shaft will twist through an angle .

Twisting means the material isdeforming, so we have strain in the

material. The greatest strain is at the

surface, while strain is zero at the

center of the shaft. The strain varies

linearly from the center to the surface of

the shaft. We learned that materials

follows Hooke's law: the ratio of

stress/strain is Young's modulus, a

constant. Therefore, the stress in theshaft also varies linearly from the center

to the surface of the shaft, and that the

shearing stress, =0 at the center, and

= max at the surface.



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Shear Stress on Circular ShaftsConsider a small area a at a distance r from the

center of the circle. If we define c = the distance

from the centroid to the surface of the circle,then the shear stress at r is:

Since shear stress is force divided by area, the

shear force acting on area a is:

The torque on area a is the force times the distance from the

centroid:

The total torque on the entire circular area about the centroid is

the sum of the torques on all the small areas that comprise the

circle, so



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Shear Stress on Circular ShaftsThe last part of this equation is the polar

moment of inertia of a circle is:

For design purposes, only the maximum stress matters, so we

usually drop the subscript from the shear stress, understandingthat we mean the stress at the surface, so we write:

In many problems, we know the applied torque and dimensions;

we need the stress. Rewriting the equation to solve for stress:



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ksi

Where:

torsional shear stress

Tapplied torque

cradial distance from neutral axis tooutermost fiber

JPolar moment of inertia

For solid circular section

J = D4/32

For hollow circular section

J = (D4d4)/32



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S ea S ess o C cu a S a s

Angle of Twist in Circular Shafts


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g

Normal strain is defined as the change in length of tensile member

divided by its original length: =/L. We can define shear strain on

a torsion member as the change in location of a point on the

surface of the shaft divided by the length of the shaft: =shear/L.

The angle of twist, ,is measured in radians, so we can substitute

shear= c . Now the shear strain is =c/L.



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g

The ratio of normal stress to normal strain is Young's modulus,

E=/. We have a similar ratio for shear stress and shear strain:

the shear modulus G=/. The eq. is the same; just the symbols

have changed. Shear modulus is a materials property, just like

Young's modulus. Substitute the expression for shear strain, and

we have Or,

Since shear stress

Then,



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gWhere:

Angular displacement in radian

LShaft length

GModulus of elasticity in torsion ormodulus of rigidity

RELATION OF TORQUE SPEED AND POWER

ENGLISH UNIT

Hp = (2TN)/33000 Where Torque is in Ft-lb and speed in rpmHp = (TN)/63000 Where Torque is in in-lb and speed in rpm

SI UNIT

kW = (2TN)/60 Where Torque is in kN-m and speed in rpm

kW = (TN)/9550 Where Torque is in N-m and speed in rpm



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g



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g

Sample Problem


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Shaft BC is hollow with inner and outer

diameters of 90 mm and 120 mm,

respectively. Shafts AB and CD are

solid of diameter d. For the loading

shown, determine (a) the minimum and

maximum shearing stress in shaft BC,

(b) the required diameter dof shaftsAB

and CDif the allowable shearing stress

in these shafts is 65 MPa.

Sample Problem

SOLUTION:

Cut sections through shaftsABandBCand perform static

equilibrium analysis to find

torque loadings

Given allowable shearing stress

and applied torque, invert the

elastic torsion formula to find the

required diameter

Apply elastic torsion formulas to

find minimum and maximumstress on shaftBC

SOLUTION

Sample Problem


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SOLUTION:

Cut sections through shaftsABand

BCand perform static equilibrium

analysis to find torque loadings

CDAB

ABx

TT

TM

mkN6

mkN60

mkN20

mkN14mkN60

BC

BCx

T

TM

Sample Problem

Sample Problem


75/142

Apply elastic torsion formulas to

find minimum and maximum

stress on shaftBC

46

4441

42

m1092.13

045.0060.022

ccJ

MPa2.86

m1092.13

m060.0mkN20

46

22max

J

cTBC

MPa7.64

mm60

mm45

MPa2.86

min

min

2

1

max

min

c

c

MPa7.64

MPa2.86

min

max

Given allowable shearing stress and

applied torque, invert the elastic torsion

formula to find the required diameter

m109.38

mkN665

3

3

2

4

2

max

c

cMPa

c

Tc

J

Tc

mm8.772

cd

Sample Problem

Statically Indeterminate Shafts


76/142

Given the shaft dimensions and the applied

torque, we would like to find the torque reactions

atAandB.

Statically Indeterminate Shafts

From a free-body analysis of the shaft,

which is not sufficient to find the end torques.

The problem is statically indeterminate.

ftlb90 BA TT

ftlb9012

21 AA TJL

JLT

Substitute into the original equilibrium equation,

ABBA T

JL

JLT

GJ

LT

GJ

LT

12

21

2

2

1

121 0

Divide the shaft into two components which

must have compatible deformations,

Sample Problem


77/142

p

Two solid steel shafts are connected

by gears. Knowing that for each

shaft G= 11.2 x 106psi and that the

allowable shearing stress is 8 ksi,

determine (a) the largest torque T0that may be applied to the end of

shaft AB, (b) the corresponding

angle through which end A of shaft

ABrotates.

SOLUTION:

Apply a static equilibrium analysis onthe two shafts to find a relationship

between TCDand T0

Find the corresponding angle of twist

for each shaft and the net angular

rotation of endA

Find the maximum allowable torque

on each shaftchoose the smallest

Apply a kinematic analysis to relate

the angular rotations of the gears

Sample Problem


78/142

SOLUTION:

Apply a static equilibrium analysis on

the two shafts to find a relationship

between TCDand T0

0

0

8.2

in.45.20

in.875.00

TT

TFM

TFM

CD

CDC

B

Apply a kinematic analysis to relate

the angular rotations of the gears

CB

CCB

CB

CCBB

r

r

rr

8.2

in.875.0

in.45.2

Sample Problem

Sample Problem


79/142

Find the T0for the maximum

allowable torque on each shaft

choose the smallest

in.lb561

in.5.0

in.5.08.28000

in.lb663

in.375.0

in.375.08000

0

4

2

0max

0

4

2

0max

T

Tpsi

J

cT

T

Tpsi

J

cT

CD

CD

AB

AB

inlb5610 T

Find the corresponding angle of twist for each

shaft and the net angular rotation of endA

oo

/

oo

o

64

2

/

o

642

/

2.2226.8

26.895.28.28.2

95.2rad514.0

psi102.11in.5.0

.24in.lb5618.2

2.22rad387.0

psi102.11in.375.0

.24in.lb561

BABA

CB

CD

CDDC

ABABBA

in

GJ

LT

in

GJ

LT

o48.10A

Sample Problem

P bl


80/142

A hollow steel shaft (G = 12 x 106psi)must transmit a torque of 300,000 in-lb.

The total angle of twist must not exceed 30

per 100ft. The maximum shearing stressmust not exceed 16,000 psi. Find the

inside diameter d if the outside diameter D

is 12.

Problem

Given:Solution


81/142

Ss = Tr/J16,000 psi= 300,000 in-lb (6)/ J

J = 112.5 in4

= TL/JG

3 /180 = 300,000 in-lb (100 x 12/1)J (12 x 106psi)

J = 572.96 in4

For hollow shaft *For Solid Shaft

J = /32 (D4d4) J = /32 D4527.96 in4= /32 (124d4)

d = 11.05

Given:

D = 12

G = 12x106psi

T = 300,000 in-lb

= 30

Ss = 16,000 psi

Solution

P bl


82/142

A steel marine propeller is to transmit4.5MW at 3r/s without exceeding a

shearing stress of 50MPa or twisting

through more than 10in a length of 25diameters. Compute the proper diameter if

G = 83 GPa.

Problem

Solution


83/142

P = T 2f N-m/sT = 4.5x106watts / 2(3 rev/s)

T = 238,732 N-m

From Shearing stress:

Ss = 16T / d3

50 N/mm2 = 16(238,732N-m) / d3

d = 290mm

From deformation:

= TL/JG

1(/180) = 238,732(1000) N-mm(25d)

/32 d4(83x103)N/mm2

d = 347.5mm say 348 mm

Therefore use d = 348mm

Given:

P = 4.5MW

f = 3 rev/sec

= 10

G = 83 GPa

Ss = 50 MPa

Solution

Reactions, Shear and Moment in Beam


84/142

-If loading is perpendicular (transverse) to the axis so that the

member (machine or structural) tends to bends then the it is called

abeam.

TYPES OF BEAMS


85/142

I.STATICALLY DETERMINATE

-beams in which the reactions of the support maybe

determined by the use of equations of static equilibriuma. Cantilever

A beam that is supported only at one end and in such a

manner that the axis of the beam cannot rotate at that

points.

b. Simple Beam

A beam that is freely supported at both ends.

c. Over hanging Beams

A beam freely supported at two points and having one or

both ends extending from these supports.

II.STATICALLY INDETERMINATE

-If the number of reactions exerted upon the bean exceeds

the number of equations of statically equilibrium

Types of Beam


86/142

Classification of Beam Supports

Reactions in Beams (Statically Determinate)

W l l t th ti f i b i th


87/142

-We can calculate the reaction forces in beams using the

equation of static equilibrium.

INTERNAL FORCES AND MOMENTS IN BEAMS

When beams are loaded with forces and couples internal


88/142

When beams are loaded with forces and couples, internal

stress arise, i.e., both normal and shearing stress, to verify such

effect in any portion of the beam, it is necessary to know the

resultant force and the moment acting in that section.

x

A B C D A B

R1x

b

a

V

RESISTING MOMENTThis couple M is called

the resisting moment

M=R1(x)A(x-a)B(x-b)

RESISTING SHEARThe vertical force V called

the resisting forces or shear

V= R1AB

Shear Force and Moment Diagram


89/142

When we calculate reaction forces and torques on

tension members, we are calculating external forcesand torques. Unless the material has no strength at all,

the material resists these external loads by developing

internal loads. Beams in bending also develop internalforces to resist external forces. Since the external forces

on beams are transverse (perpendicular to the axis of

the beam), the internal resisting forces are also

transverse forces (shear Force or Resisting Shear).

Imagine a simply-supported beam with a point load at the mid-span. Cut the beam to the left of the point load, and draw a free-


90/142

p p

body diagram of the beam segment. In a free body diagram,

forces must balance. Therefore, a downward force at the cut edge

balances the support reaction RA. We call this shear force V. It is ashear force because the force acts parallel to a surface (the cut

edge of the beam).

The forces RA and V are in balance (equal in value; opposite insign), but our segment tend to spin clockwise about point A. To


91/142

g ) g p p

counteract this tendency to spin, a moment M develops within the

beam to prevent this rotation. The moment equals the shear force

times its distance from point A.

Cut the beam to the right of the point load, and draw the free-body diagram. Since P is larger than RA, force V points upwards.


92/142

A

Shear Diagram-We can sketch V as a function of location along the beam


93/142

g

using a Shear Diagram. Draw vertical construction lines below the

load diagram wherever the applied loads and reactions occur.

Draw a horizontal construction line, indicating zero shear load.Next, draw the value of V along the length of the beam, as follows:


94/142

Moment Diagrams-The moment about a point along a beam is defined as the


95/142

distance from that point to a force acting perpendicular to thebeam, so the units are force distance: lb.ft. (or ft.lb.the order

does not matter), lb.in., kipft., Nm, or kNm. We can graph thevalue of the bending moment along a beam by drawing a moment

diagram.


96/142


97/142


98/142


99/142


100/142


101/142


102/142


103/142


104/142


105/142


106/142

From shear diagram, we can identify the location and size ofthe largest shear load in a beam. Therefore, we know the location

f h l h d l l l h l f


107/142

of the largest shear stress, and we can also calculate the value of

this stress. Once we know the actual stress in the material, we

can compare this values with the shear strength of the material,and we can determine whether the beam will fail in shear.

Shear diagrams are necessary for drawing bending moment

diagrams (momentdiagrams),which we can use to identify the

location and magnitude of bending stresses that develop within

beams. We can compare the actual bending stresses with the

yield strength of the material, and we can determine whether the

beam will fail in bending.


108/142


109/142


110/142


111/142


112/142


113/142


114/142


115/142


116/142


117/142


118/142


119/142


120/142

Flexural Stress/Pure Bending


121/142

Pure Bending: members are subjected to

equal and opposite couples acting in the

same longitudinal plane

Bending DeformationsB i h l f i


122/142

Beam with a plane of symmetry in pure

bending:

member remains symmetric

bends uniformly to form a circular arc

cross-sectional plane passes through arc center

and remains planar

length of top decreases and length of bottom

increases

a neutral surfacemust exist that is parallel to the

upper and lower surfaces and for which the length

does not change

stresses and strains are negative (compressive)

above the neutral plane and positive (tension)

below it

Beam Stress/Section PropertiesTh i l t d t b di


123/142

The maximum normal stress due to bending,

modulussection

inertiaofmomentsection

c

IS

I

S

M

I

Mcm

A beam section with a larger section

modulus will have a lower maximum stress Consider a rectangular beam cross section,

Ahbhh

bh

c

IS

6

12

6

1

3

12

1

2

Between two beams with the same crosssectional area, the beam with the greater

depth will be more effective in resisting

bending. Structural steel beams are designed to have a

large section modulus.

Properties of American Standard Shapes


124/142

Deformations in a Transverse CrossSection

D f ti d t b di t M i tifi d


125/142

Deformation due to bending momentM is quantified

by the curvature () of the neutral surface

EI

M

I

Mc

EcEcc

mm

11

Sample ProblemSOLUTION:


126/142

A cast-iron machine part is acted upon

by a 3 kN-m couple. Knowing E= 165GPa and neglecting the effects of fillets,

determine (a) the maximum tensile and

compressive stresses, (b) the radius of

curvature.

SOLUTION:

Based on the cross section geometry,

calculate the location of the sectioncentroid and moment of inertia.

2dAIIA

AyY x

Apply the elastic flexural formula to

find the maximum tensile and

compressive stresses.

I

Mcm

Calculate the curvature

EI

M

1



127/142

SOLUTION:

Based on the cross section geometry,

calculate the location of the sectioncentroid and moment of inertia.

mm383000

10114 3

A

AyY

3

3

3

32

101143000

104220120030402

109050180090201

mm,mm,mmArea,

AyA

Ayy

49-3

2312123

121

231212

m10868mm10868

18120040301218002090

I

dAbhdAIIx


128/142

Apply the elastic flexural formula to find the

maximum tensile and compressive stresses.

49

49

mm10868

m038.0mkN3

mm10868

m022.0mkN3

I

cM

I

cM

IMc

BB

AA

m

MPa0.76A

MPa3.131B

Calculate the curvature

49- m10868GPa165mkN3

1

EI

M

m7.47

m1095.201 1-3



129/142

For the timber beam and loading

shown, draw the shear and bend-

moment diagrams and determine

the maximum normal stress due to

bending.

SOLUTION:

Treating the entire beam as a rigid

body, determine the reaction forces

Identify the maximum shear and

bending-moment from plots of their

distributions.

Apply the elastic flexure formulas to

determine the corresponding

maximum normal stress.

Section the beam at points near

supports and load application points.

Apply equilibrium analyses on

resulting free-bodies to determineinternal shear forces and bending

couples

SOLUTION


130/142

SOLUTION:

Treating the entire beam as a rigid body, determine

the reaction forces kN14kN40:0from DBBy RRMF

Section the beam and apply equilibrium analyses

on resulting free-bodies

00m0kN200

kN200kN200

111

11

MMM

VVFy

mkN500m5.2kN200

kN200kN200

222

22

MMM

VVFy

0kN14

mkN28kN14

mkN28kN26

mkN50kN26

66

55

44

33

MV

MV

MV

MV

Id tif th i h d b di


131/142

5 - 131

Identify the maximum shear and bending-

moment from plots of their distributions.

mkN50kN26 Bmm MMV


determine the corresponding

maximum normal stress.

36

3

36

2

612

61

m1033.833

mN1050

m1033.833

m250.0m080.0

S

M

hbS

B

m

Pa100.60 6m



132/142

The structure shown is constructed

of a W10x112 rolled-steel beam. (a)

Draw the shear and bending-

moment diagrams for the beam and

the given loading. (b) determinenormal stress in sections just to the

right and left of point D.

SOLUTION:

Replace the 10 kip load with an

equivalent force-couple system atD.Find the reactions atBby considering

the beam as a rigid body.

Section the beam at points near the

support and load application points.Apply equilibrium analyses on

resulting free-bodies to determine

internal shear forces and bending

couples.

Apply the elastic flexure formulas todetermine the maximum normal

stress to the left and right of pointD.

SOLUTION:


133/142

SOLUTION:

Replace the 10 kip load with equivalent force-

couple system atD. Find reactions atB.

Section the beam and apply

equilibrium analyses on resulting free-

bodies.

ftkip5.1030kips3030

:

2

21

1

xMMxxM

xVVxFCtoAFrom

y

ftkip249604240

kips240240

:

2

xMMxM

VVF

DtoCFrom

y

ftkip34226kips34

:

xMV

BtoDFrom



134/142


determine the maximum normal stress to

the left and right of pointD.From Appendix C for a W10x112 rolled

steel shape, S= 126 in3about theX-X

axis.

3

3

in126

inkip1776

:

in126

inkip2016

:

S

M

DofrighttheTo

S

MDoflefttheTo

m

m

ksi0.16m

ksi1.14m

Sample Problem


135/142

A simply supported steel beam is

to carry the distributed andconcentrated loads shown.

Knowing that the allowable normal

stress for the grade of steel to be

used is 160 MPa, select the wide-

flange shape that should be used.

SOLUTION:

Considering the entire beam as a free-body, determine the reactions atAand

D.

Develop the shear diagram for the

beam and load distribution. From the

diagram, determine the maximumbending moment.

Determine the minimum acceptable

beam section modulus. Choose the

best standard section which meets thiscriteria.

Considering the entire beam as a free-body


136/142

Considering the entire beam as a free-body,

determine the reactions atAandD.

kN0.52

kN50kN60kN0.580

kN0.58m4kN50m5.1kN60m50

y

yy

A

A

AF

DDM

Develop the shear diagram and determine the

maximum bending moment.

kN8

kN60

kN0.52

B

AB

yA

V

curveloadunderareaVV

AV

Maximum bending moment occurs atV= 0 orx= 2.6 m.

kN6.67

,max

EtoAcurveshearunderareaM

i h i i bl b


137/142

Determine the minimum acceptable beam

section modulus.

3336

maxmin

mm105.422m105.422

MPa160

mkN6.67

all

MS

Choose the best standard section which

meets this criteria.

4481.46W200

5358.44W250

5497.38W310

4749.32W360

63738.8W410

mm, 3

SShape

Problem


138/142

Determine the minimum width b of thebeam loaded as shown if the flexural

stress is not to exceed 10 MPa.

200mm

b

A

2000N/m

5000N

B

C

D

5000N

C

Solution


139/142

MD= 03R1 = 500(1) + 2000(4)(2)

R1 = 7000N

R2 = 5000 + 2000(4)7000

R2 = 6000N

From V & M diagram

Mmax = 5000 N-m

Sb = Mc/I; c = d/2, I = bd3/12

10x106N/m2= 5000 N-m (0.2m/2) / b(0.23m)/12

b = 0.075mb = 75mm

-2000

5000

-4000

1000

-6000

-1000 N-m

5000 N-m

2000N/m

B

C

D1m 2m 1m

Problem


140/142

A cantilever beam, 60mm x 200mm highand 6m long carries a load that varies

uniformly from zero at the free end to 1000

N/m at the wall. Compute the magnitudeand location of the maximum flexural

stress. Determine the type and magnitude

of the stress in a fiber 40mm from the top

of the beam at a section 3m from the wall.


141/142

Max Moment is at fixed endM = (1000N/m)(6m)(2m)

M = 6000N-m

Fb = 6M / bd2

Sb = 6 (6000)(1000) N-mm / 60mm(200mm)2

Sb = 15 N/mm2

Sb = 15 MPa

Stress at 3m from the free end

M = (500N/m)(3m)(1m)

M = 750 N-m

I = bh3/12 = 60 (2003)/12

I = 40x106mm4

Sb = My / I = 750 (1000) N/mm (60) / 40x106mm4

Sb = 1.125N/mm2(Tensile)

200mm

60mm

40

60

1000N/m

6m

M

T

C


142/142

Thanks!

END OF TOPIC

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