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CHAPTER OUTLINE
1. Introduction
2. Equilibrium of a deformable body
3. Stress
4. Average normal stress in an axially loaded bar
5. Average shear stress
6. Allowable stress
7. Design of simple connections
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Historical development
Beginning of 17th century (Galileo) Early 18th century (Saint-Venant, Poisson, Lam
and Navier)
In recent times, with advanced mathematical and
computer techniques, more complex problemscan be solved
1.1 INTRODUCTION
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1.2 EQUILIBRIUM OF A DEFORMABLE BODY
External loads
Surface forces
Area of contact
Concentrated force
Linear distributed force Centroid C(or
geometric center)
Body force (e.g., weight)
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Support reactions
for 2D problems
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Equations of equilibrium
For equilibrium balance of forces
balance of moments
Draw a free-body diagram to account for allforces acting on the body
Apply the two equations to achieve equilibriumstate
F= 0
MO= 0
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Internal resultant loadings
Define resultant force (FR) and moment (MRo) in 3D:
Normal force, N
Shear force, V
Torsional moment or torque, T
Bending moment, M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Internal resultant loadings
For coplanar loadings:
Normal force, N
Shear force, V
Bending moment, M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Internal resultant loadings
For coplanar loadings:
Apply Fx = 0 to solve for N
Apply Fy = 0 to solve for V
Apply MO = 0 to solve for M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Procedure for Analysis
Method of sections
1. Choose segment to analyze
2. Determine Support Reactions
3. Draw free-body diagram for whole body
4. Apply equations of equilibrium
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Procedure for analysis
Free-body diagram
1. Keep all external loadings in exact locations
before sectioning
2. Indicate unknown resultants, N, V, M, and Tat the section, normally at centroid Cof
sectioned area
3. Coplanar system of forces only include N, V,and M
4. Establishx,y,zcoordinate axes with origin at
centroid
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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Procedure for analysis
Equations of equilibrium
1. Sum moments at section, about each
coordinate axes where resultants act
2. This will eliminate unknown forces Nand V,with direct solution for M(and T)
3. Resultant force with negative value implies
that assumed direction is opposite to thatshown on free-body diagram
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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EXAMPLE 1.1
Determine resultant loadings acting on cross
section at Cof beam.
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EXAMPLE 1.1 (SOLN)
Support Reactions
Consider segment CB
Free-Body Diagram:
Keep distributed loading exactly where it is on
segment CBaftercutting the section.
Replace it with a single resultant force,F.
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EXAMPLE 1.1 (SOLN)
Intensity (w) of loading at C(by proportion)
w/6 m = (270 N/m)/9 m
w= 180 N/m
F= (180 N/m)(6 m) = 540 N
Facts 1/3(6 m) = 2 m from C.
Free-Body Diagram:
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1 St
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EXAMPLE 1.1 (SOLN)
Equilibrium equations:
Negative sign of Mcmeans it acts in the opposite
direction to that shown below
1 St
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EXAMPLE 1.5
Mass of pipe = 2 kg/m,
subjected to vertical
force of 50 N and couplemoment of 70 Nm at
endA. It is fixed to the
wall at C.
Determine resultant internal loadings acting on cross
section at Bof pipe.
1 St
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EXAMPLE 1.5 (SOLN)
Support Reactions:
Consider segmentAB,
which does not involve
support reactions at C.
Free-Body Diagram:
Need to find weight of
each segment.
1 St
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EXAMPLE 1.5 (SOLN)
WBD = (2 kg/m)(0.5 m)(9.81 N/kg)= 9.81 N
WAD = (2 kg/m)(1.25 m)(9.81 N/kg)
= 24.525 N
1 St
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EXAMPLE 1.5 (SOLN)
Equilibrium equations:
Fx= 0;
Fy= 0;
(FB)x= 0
(FB
)y= 0
Fz= 0; (FB)z 9.81 N 24.525 N 50 N = 0(FB)z= 84.3 N
1 St
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EXAMPLE 1.5 (SOLN)
Equilibrium Equations:
(MB)x= 0;
(Mc)x+ 70 Nm 50 N (0.5 m) 24.525 N (0.5 m)
9.81 N (0.25m) = 0(MB)x= 30.3 Nm
(MB)y= 0;
(Mc)y+ 24.525 N (0.625m) + 50 N (1.25 m) = 0(MB)y= 77.8 Nm
(MB)z= 0; (Mc)z= 0
1 Stress
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EXAMPLE 1.5 (SOLN)
NB= (FB)y= 0
VB= (0)2+ (84.3)2= 84.3 N
TB= (MB)y= 77.8 Nm
MB= (30.3)2+ (0)2= 30.3 Nm
The direction of each moment is determinedusing the right-hand rule: positive moments
(thumb) directed along positive coordinate axis
Equilibrium Equations:
1 Stress
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1.3 STRESS
Concept of stress
To obtain distribution of force acting over a
sectioned area
Assumptions of material:
1. It is continuous (uniform distribution of matter)2. It is cohesive (all portions are connected
together)
1 Stress
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1.3 STRESS
Concept of stress
Consider Ain figure below
Small finite force, Facts on A
As A 0, F 0
But stress (F /A) finite limit ()
1 Stress
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Normal stress
Intensityof force, or force per unit area, acting
normalto A
Symbol used for normal stress,is (sigma)
Tensile stress: normal force pulls or stretchesthe area element A
Compressive stress: normal force pushes or
compresses area element A
1.3 STRESS
z=lim
A 0
FzA
1 Stress
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Shear stress
Intensityof force, or force per unit area, acting
tangenttoA
Symbol used for normal stress is (tau)
1.3 STRESS
zx=lim
A 0
FxA
zy=lim
A 0
FyA
1 Stress
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General state of stress
Figure shows the state of stressacting around a chosen point in abody
Units (SI system)
Newtons per square meter (N/m2)or apascal(1 Pa = 1 N/m2)
kPa = 103
N/m2
(kilo-pascal) MPa = 106N/m2(mega-pascal)
GPa = 109N/m2(giga-pascal)
1.3 STRESS
1 Stress
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1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Examples of axially loaded bar
Usually long and slender structural members
Truss members, hangers, bolts
Prismatic means all the cross sections are the same
1 Stress
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Assumptions
1. Uniform deformation: Bar remains straight before
and after load is applied, and cross section
remains flat or plane during deformation
2. In order for uniform deformation, force Pbeapplied along centroidal axis of cross section
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
1 Stress
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Average normal stress distribution
= average normal stress at anypoint on cross sectional area
P= internal resultant normal force
A= x-sectional area of the bar
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
FRz= Fxz dF = AdA
P= A
+
P
A=
1 Stress
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Equilibrium
Consider vertical equilibrium of the element
Fz= 0 (A) (A) = 0=
Above analysis
applies to members
subjected to tensionor compression.
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
1 Stress
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Maximum average normal stress
For problems where internal force Pand x-sectional A were constantalong the longitudinalaxis of the bar, normal stress =P/Ais alsoconstant
If the bar is subjected to several external loadsalong its axis, change in x-sectional area mayoccur
Thus, it is important to find the maximum
average normal stress To determine that, we need to find the locationwhere ratioP/Ais a maximum
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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Procedure for Analysis
Average normal stress Use equation of = P/Afor x-sectional area of a
member when section subjected to internal
resultant force P
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1 Stress
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Procedure for Analysis
Axially loaded members Internal Loading:
Section memberperpendicularto its longitudinal
axis at pt where normal stress is to be
determined
Draw free-body diagram
Use equation of force equilibrium to obtain
internal axial force Pat the section
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
1. Stress
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Procedure for Analysis
Axially loaded members Average Normal Stress:
Determine members x-sectional area at the
section
Compute average normal stress =P/A
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
1. Stress
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EXAMPLE 1.6
Bar width = 35 mm, thickness = 10 mm
Determine max. average normal stress in bar when
subjected to loading shown.
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EXAMPLE 1.6 (SOLN)
Internal loading
Normal force diagram
By inspection, largest
loading area isBC,wherePBC= 30 kN
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EXAMPLE 1.6 (SOLN)
Max. Average normal stress,
BC=PBC
A
30(103) N
(0.035 m)(0.010 m)= = 85.7 MPa
1. Stress
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EXAMPLE 1.8
Specific weight st= 80 kN/m3
Determine average compressive stress acting at
pointsAandB.
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EXAMPLE 1.8 (SOLN)
Internal loading
Based on free-body diagram,
weight of segmentABdetermined from
Wst= stVst
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EXAMPLE 1.8 (SOLN)
Average normal stress
+ Fz= 0; P Wst= 0
P (80 kN/m3)(0.8 m)(0.2 m)2= 0
P= 8.042 kN
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EXAMPLE 1.8 (SOLN)
Average compressive stress
Cross-sectional area at section is:
A= (0.2)m2
8.042 kN
(0.2 m)2
P
A==
= 64.0 kN/m2
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1.5 AVERAGE SHEAR STRESS
Shear stress is the stress component that act
in the plane of the sectioned area.
Consider a force Facting to the bar
For rigid supports, and Fis large enough, bar
will deform and fail along the planes identifiedbyABand CD
Free-body diagram indicates that shear force,
V=F/2be applied at both sections to ensure
equilibrium
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1.5 AVERAGE SHEAR STRESS
Average shear stress over each
section is:
avg= average shear stress atsection, assumed to be sameat each pt on the section
V= internal resultant shear force at
section determined from
equations of equilibrium
A= area of section
V
Aavg=
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1.5 AVERAGE SHEAR STRESS
Case discussed above is example of simpleor
direct shear Caused by the direct action of applied load F
Occurs in various types of simple connections,e.g., bolts, pins, welded material
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Single shear
Steel and wood joints shown below are
examples of single-shear connections, also
known as lap joints.
Since we assume members are thin, there areno moments caused byF
1.5 AVERAGE SHEAR STRESS
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Single shear
For equilibrium, x-sectional area of bolt and
bonding surface between the two members are
subjected to single shear force, V=F
The average shear stress equation can beapplied to determine average shear stress
acting on colored section in (d).
1.5 AVERAGE SHEAR STRESS
1. Stress
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1.5 AVERAGE SHEAR STRESS
Double shear
The joints shown below are examples of double-shear connections, often called double lap joints.
For equilibrium, x-sectional area of bolt andbonding surface between two members
subjected to double shear force, V=F/2 Apply average shear stress equation todetermine average shear stress acting oncolored section in (d).
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1.5 AVERAGE SHEAR STRESS
Procedure for analysis
Internal shear
1. Section member at the pt where the avgis to be
determined
2. Draw free-body diagram3. Calculate the internal shear force V
Average shear stress
1. Determine sectioned areaA2. Compute average shear stress avg= V/A
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EXAMPLE 1.10
Depth and thickness = 40 mm
Determine average normal stress and average
shear stress acting along (a) section planes a-a,
and (b) section plane b-b.
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EXAMPLE 1.10 (SOLN)
Part (a)
Internal loadingBased on free-body diagram, Resultant loadingof axial force,P= 800 N
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EXAMPLE 1.10 (SOLN)
Part (a)
Average stressAverage normal stress,
=P
A
800 N
(0.04 m)(0.04 m) = 500 kPa=
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EXAMPLE 1.10 (SOLN)
Part (a)
Internal loadingNo shear stress on section, since shear force atsection is zero.
avg= 0
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EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
+
Fx= 0; 800 N +Nsin 60+ Vcos 60 = 0+
Fy= 0; Vsin 60 Ncos 60= 0
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EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
Or directly usingx,yaxes,
Fx= 0;
Fy= 0;
+
+
N 800 N cos 30= 0
V 800 N sin 30= 0
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EXAMPLE 1.10 (SOLN)
Part (b)
Average normal stress
=N
A
692.8 N
(0.04 m)(0.04 m/sin 60)= 375 kPa=
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EXAMPLE 1.10 (SOLN)
Part (b)
Average shear stress
avg =V
A
400 N
(0.04 m)(0.04 m/sin 60)= 217 kPa=
Stress distribution shown below
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1.6 ALLOWABLE STRESS
When designing a structural member or
mechanical element, the stress in it must berestricted to safe level
Choose an allowable load that is less than theload the member can fully support
One method used is the factor of safety (F.S.)
F.S.=Ffail
Fallow
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1.6 ALLOWABLE STRESS
If load applied is linearly related to stress
developed within member, then F.S. can alsobe expressed as:
F.S.=fail
allowF.S.=
fail
allow
In all the equations, F.S. is chosen to be greater than 1,to avoid potential for failure
Specific values will depend on types of material usedand its intended purpose
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1.7 DESIGN OF SIMPLE CONNECTIONS
To determine area of section subjected to a
normal force, use
A=P
allow
A=
V
allow
To determine area of section subjected to a shearforce, use
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1.7 DESIGN OF SIMPLE CONNECTIONS
Cross-sectional area of a tension member
Condition:
The force has a line of action that passes
through the centroid of the cross section.
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1.7 DESIGN OF SIMPLE CONNECTIONS
Cross-sectional area of a connecter subjected to
shear
Assumption:
If bolt is loose or clamping force of bolt is unknown,assume frictional force between plates to be
negligible.
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Assumptions:
1. (b)allowof concrete