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1. Stress
CHAPTER OBJECTIVES
• Review important principles of
statics
• Use the principles to determine
internal resultant loadings in a
body
• Introduce concepts of normal
and shear stress
• Discuss applications of analysis and design
members subjected to an axial load or direct shear
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1. Stress
CHAPTER OUTLINE
1.Introduction
2.Equilibrium of a deformable body
3.Stress
4.Average normal stress
5.Average shear stress
6.Allowable stress
7.Design of Simple Connections
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1. Stress
1.1 Mechanics
BRANCHES OF MECHANICS
Statics
Dynamics
Rigid Bodies
(Things that do not change shape)
Deformable Bodies
(Things that do change shape)
Incompressible
Compressible
Fluids
Mechanics
Type title here
• Statics – Equilibrium of bodies
At rest
Move with constant velocity
• Dynamics – Accelerated motion of bodies
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1. Stress
Mechanics of materials
• A branch of mechanics
• It studies the relationship of
– External loads applied to a deformable body,
and
– The intensity of internal forces acting within the
body
• are used to compute deformations of a body
• Study body’s stability when external forces are
applied to it
1.1 INTRODUCTION
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1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
External loads
• Surface forces
– Area of contact
– Concentrated force
– Linear distributed force
– Centroid C
• Body force (e.g., weight)
Center of gravity
A review of some main principles of static
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1. Stress
Support reactions
• for 2D problems
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Equations of equilibrium
• For equilibrium
– balance of forces
– balance of moments
• Draw a freebody diagram to account for all
forces acting on the body
• Apply the two equations to achieve equilibrium
state
∑
F = 0
∑
M
O
= 0
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Internal resultant loadings
• Define resultant force (F
R
) and moment (M
Ro
) 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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1. Stress
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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1. Stress
Internal resultant loadings
• For coplanar loadings:
– Apply ∑ F
x
= 0 to solve for N
– Apply ∑ F
y
= 0 to solve for V
– Apply ∑ M
O
= 0 to solve for M
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Procedure for Analysis
• Support Reactions: Method of sections
1.Choose segment to analyze
2.Determine Support Reactions
3.Draw freebody diagram for whole body
4.Apply equations of equilibrium
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Procedure for analysis
• Freebody diagram
1.Keep all external loadings in exact locations
before “sectioning”
2.Indicate unknown resultants, N, V, M, and T
at the section, normally at centroid C of
sectioned area
3.Coplanar system of forces only include N, V,
and M
4.Establish x, y, z coordinate axes with origin at
centroid
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
Procedure for analysis
• Equations of equilibrium
1.Sum moments at section, about each
coordinate axes where resultants act
2.This will eliminate unknown forces N and V,
with direct solution for M (and T)
3.Resultant force with negative value implies
that assumed direction is opposite to that
shown on freebody diagram
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
EXAMPLE 1.1
Determine resultant loadings acting on cross
section at C of beam.
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1. Stress
EXAMPLE 1.1 (SOLN)
Support Reactions
• Consider segment CB
FreeBody Diagram:
• Keep distributed loading exactly where it is on
segment CB after “cutting” the section.
• Replace it with a single resultant force, F.
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1. Stress
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 = ½ (180 N/m)(6 m) = 540 N
F acts 1/3(6 m) = 2 m from C.
FreeBody Diagram:
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1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
∑ F
x
= 0;
∑ F
y
= 0;
∑ M
c
= 0;
− N
c
= 0
N
c
= 0
V
c
− 540 N = 0
V
c
= 540 N
−M
c
− 504 N (2 m) = 0
M
c
= −1080 N∙m
+
+
+
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1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
Negative sign of M
c
means it acts in the opposite
direction to that shown below
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1. Stress
EXAMPLE 1.5
Mass of pipe = 2 kg/m,
subjected to vertical
force of 50 N and couple
moment of 70 N∙m at
end A. It is fixed to the
wall at C.
Determine resultant internal loadings acting on cross
section at B of pipe.
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1. Stress
EXAMPLE 1.5 (SOLN)
Support Reactions:
• Consider segment AB,
which does not involve
support reactions at C.
FreeBody Diagram:
• Need to find weight of
each segment.
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1. Stress
EXAMPLE 1.5 (SOLN)
W
BD
= (2 kg/m)(0.5 m)(9.81 N/kg)
= 9.81 N
W
AD
= (2 kg/m)(1.25 m)(9.81 N/kg)
= 24.525 N
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1. Stress
EXAMPLE 1.5 (SOLN)
Equilibrium equations:
∑ F
x
= 0;
∑ F
y
= 0;
(F
B
)
x
= 0
(F
B
)
y
= 0
∑ F
z
= 0;
(F
B
)
z
− 9.81 N − 24.525 N − 50 N = 0
(F
B
)
z
= 84.3 N
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1. Stress
EXAMPLE 1.5 (SOLN)
Equilibrium Equations:
∑ (M
B
)
x
= 0;
(M
c
)
x
+ 70 N∙m− 50 N (0.5 m) − 24.525 N (0.5 m)
− 9.81 N (0.25m) = 0
(M
B
)
x
= − 30.3 N∙m
∑ (M
B
)
y
= 0;
(M
c
)
y
+ 24.525 N (0.625∙m) + 50 N (1.25 m) = 0
(M
B
)
y
= − 77.8 N∙m
∑(M
B
)
z
= 0;
(M
c
)
z
= 0
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1. Stress
EXAMPLE 1.5 (SOLN)
N
B
= (F
B
)
y
= 0
V
B
= √ (0)
2
+ (84.3)
2
= 84.3 N
T
B
= (M
B
)
y
= 77.8 N∙m
M
B
= √ (30.3)
2
+ (0)
2
= 30.3 N∙m
The direction of each moment is determined
using the righthand rule: positive moments
(thumb) directed along positive coordinate axis
Equilibrium Equations:
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1. Stress
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)
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1. Stress
1.3 STRESS
Concept of stress
• Consider ∆A in figure below
• Small finite force, ∆F acts on ∆A
• As ∆A →0, ∆ F →0
• But stress (∆F / ∆A) →finite limit (∞)
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1. Stress
Normal stress
• Intensity of force, or force per unit area, acting
normal to ∆A
• Symbol used for normal stress,is σ (sigma)
• Tensile stress: normal force “pulls” or “stretches”
the area element ∆A
• Compressive stress: normal force “pushes” or
“compresses” area element ∆A
1.3 STRESS
σ
z
=
lim
∆A →0
∆F
z
∆A
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1. Stress
Shear stress
• Intensity of force, or force per unit area, acting
tangent to ∆A
• Symbol used for normal stress is τ (tau)
1.3 STRESS
τ
zx
=
lim
∆A →0
∆F
x
∆A
τ
zy
=
lim
∆A →0
∆F
y
∆A
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1. Stress
General state of stress
• Figure shows the state of stress
acting around a chosen point in a
body
Units (SI system)
• Newtons per square meter (N/m
2
)
or a pascal (1 Pa = 1 N/m
2
)
• kPa = 10
3
N/m
2
(kilopascal)
• MPa = 10
6
N/m
2
(megapascal)
• GPa = 10
9
N/m
2
(gigapascal)
1.3 STRESS
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1. Stress
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
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1. Stress
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 P be
applied along centroidal axis of cross section
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Average normal stress distribution
σ = average normal stress at any
point on cross sectional area
P = internal resultant normal force
A = xsectional area of the bar
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
F
Rz
= ∑ F
xz
∫ dF = ∫
A
σ dA
P = σA
+
P
A
σ =
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1. Stress
Equilibrium
• Consider vertical equilibrium of the element
∑ F
z
= 0
σ (∆A) − σ’ (∆A) = 0
σ = σ’
Above analysis
applies to members
subjected to tension
or compression.
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Maximum average normal stress
• For problems where internal force P and x
sectional A were constant along the longitudinal
axis of the bar, normal stress
σ
= P/A is also
constant
• If the bar is subjected to several external loads
along its axis, change in xsectional area may
occur
• Thus, it is important to find the maximum
average normal stress
• To determine that, we need to find the location
where ratio P/A is a maximum
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Maximum average normal stress
• Draw an axial or normal force diagram (plot of
P vs. its position x along bar’s length)
• Sign convention:
– P is positive (+) if it causes tension in the
member
– P is negative (−) if it causes compression
• Identify the maximum average normal stress
from the plot
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Procedure for Analysis
Average normal stress
• Use equation of σ = P/A for xsectional area of a
member when section subjected to internal
resultant force P
Axially loaded members
• Internal Loading:
• Section member perpendicular to its longitudinal
axis at pt where normal stress is to be determined
• Draw freebody diagram
• Use equation of force equilibrium to obtain
internal axial force P at the section
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
Procedure for Analysis
Axially loaded members
• Average Normal Stress:
• Determine member’s xsectional area at the
section
• Compute average normal stress σ = P/A
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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1. Stress
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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1. Stress
EXAMPLE 1.6 (SOLN)
Internal loading
Normal force diagram
By inspection, largest
loading area is BC,
where P
BC
= 30 kN
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1. Stress
EXAMPLE 1.6 (SOLN)
Average normal stress
σ
BC
=
P
BC
A
30(10
3
) N
(0.035 m)(0.010 m)
=
= 85.7 MPa
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1. Stress
EXAMPLE 1.8
Specific weight γ
st
= 80 kN/m
3
Determine average compressive stress acting at
points A and B.
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1. Stress
EXAMPLE 1.8 (SOLN)
Internal loading
Based on freebody diagram,
weight of segment AB determined from
W
st
= γ
st
V
st
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1. Stress
EXAMPLE 1.8 (SOLN)
Average normal stress
+
∑ F
z
= 0;
P − W
st
= 0
P − (80 kN/m
3
)(0.8 m)π(0.2 m)
2
= 0
P = 8.042 kN
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1. Stress
EXAMPLE 1.8 (SOLN)
Average compressive stress
Crosssectional area at section is:
A = π(0.2)m
2
8.042 kN
π(0.2 m)
2
P
A
=
σ =
σ = 64.0 kN/m
2
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1. Stress
1.5 AVERAGE SHEAR STRESS
• Shear stress is the stress component that act
in the plane of the sectioned area.
• Consider a force F acting to the bar
• For rigid supports, and F is large enough, bar
will deform and fail along the planes identified
by AB and CD
• Freebody diagram indicates that shear force,
V = F/2 be applied at both sections to ensure
equilibrium
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1. Stress
1.5 AVERAGE SHEAR STRESS
Average shear stress over each
section is:
τ
avg
= average shear stress at
section, assumed to be same
at each pt on the section
V =internal resultant shear force at
section determined from
equations of equilibrium
A = area of section
P
A
τ
avg
=
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1. Stress
1.5 AVERAGE SHEAR STRESS
• Case discussed above is example of simple or
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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1. Stress
Single shear
• Steel and wood joints shown below are
examples of singleshear connections, also
known as lap joints.
• Since we assume members are thin, there are
no moments caused by F
1.5 AVERAGE SHEAR STRESS
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1. Stress
Single shear
• For equilibrium, xsectional area of bolt and
bonding surface between the two members are
subjected to single shear force, V = F
• The average shear stress equation can be
applied to determine average shear stress
acting on colored section in (d).
1.5 AVERAGE SHEAR STRESS
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1. Stress
1.5 AVERAGE SHEAR STRESS
Double shear
• The joints shown below are examples of double
shear connections, often called double lap joints.
• For equilibrium, xsectional area of bolt and
bonding surface between two members
subjected to double shear force, V = F/2
• Apply average shear stress equation to
determine average shear stress acting on
colored section in (d).
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1. Stress
1.5 AVERAGE SHEAR STRESS
Procedure for analysis
Internal shear
1.Section member at the pt where the τ
avg
is to be
determined
2.Draw freebody diagram
3.Calculate the internal shear force V
Average shear stress
1.Determine sectioned area A
2.Compute average shear stress τ
avg
= V/A
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1. Stress
EXAMPLE 1.10
Depth and thickness = 40 mm
Determine average normal stress and average
shear stress acting along (a) section planes aa,
and (b) section plane bb.
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Internal loading
Based on freebody diagram, Resultant loading
of axial force, P = 800 N
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Average stress
Average normal stress,
=
P
A
800 N
(0.04 m)(0.04 m)
= 500 kPa
=
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Internal loading
No shear stress on section, since shear force at
section is zero.
τ
avg
= 0
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
+
∑ F
x
= 0;
− 800 N + N sin 60° + V cos 60° = 0
+
∑ F
y
= 0;
V sin 60° − N cos 60° = 0
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
Or directly using x’, y’ axes,
∑ F
x’
= 0;
∑ F
y’
= 0;
+
+
N − 800 N cos 30° = 0
V − 800 N sin 30° = 0
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average normal stress
=
N
A
692.8 N
(0.04 m)(0.04 m/sin 60)
= 375 kPa
=
30
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1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average shear stress
τ
avg
=
V
A
400 N
(0.04 m)(0.04 m/sin 60)
= 217 kPa
=
Stress distribution shown below
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1. Stress
1.6 ALLOWABLE STRESS
• When designing a structural member or
mechanical element, the stress in it must be
restricted to safe level
• Choose an allowable load that is less than the
load the member can fully support
• One method used is the factor of safety (F.S.)
F.S.=
F
fail
F
allow
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1. Stress
1.6 ALLOWABLE STRESS
• If load applied is linearly related to stress
developed within member, then F.S. can also
be expressed as:
F.S.=
fail
allow
F.S.=
τ
fail
τ
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 used
and its intended purpose
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
• To determine area of section subjected to a
normal force, use
A =
P
allow
A =
V
τ
allow
• To determine area of section subjected to a shear
force, use
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Crosssectional 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. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Crosssectional 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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1. Stress
Assumptions:
1.(σ
b
)
allow
of concrete <
(σ
b
)
allow
of base plate
2.Bearing stress is
uniformly distributed
between plate and
concrete
1.7 DESIGN OF SIMPLE CONNECTIONS
Required area to resist bearing
• Bearing stress is normal stress produced by the
compression of one surface against another.
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
• Although actual shearstress distribution along rod
difficult to determine, we assume it is uniform.
• Thus use A = V/ τ
allow
to calculate l, provided d and
τ
allow
is known.
Required area to resist shear caused by axial load
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Procedure for analysis
When using average normal stress and shear stress
equations, consider first the section over which the
critical stress is acting
Internal Loading
1.Section member through xsectional area
2.Draw a freebody diagram of segment of
member
3.Use equations of equilibrium to determine
internal resultant force
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1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Procedure for Analysis
Required Area
• Based on known allowable stress, calculate
required area needed to sustain load from
A = P/τ
allow
or A = V/τ
allow
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1. Stress
EXAMPLE 1.13
The two members pinned together at B. If the pins
have an allowable shear stress of
τ
allow
= 90 MPa,
and allowable tensile stress of rod CB is
(σ
t
)
allow
= 115 MPa
Determine to nearest
mm the smallest
diameter of pins A
and B and the
diameter of rod CB
necessary to support
the load.
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1. Stress
EXAMPLE 1.13 (SOLN)
Draw freebody diagram:
=
P
A
800 N
(0.04 m)(0.04 m)
= 500 kPa
=
No shear stress on section, since shear force at
section is zero
τ
avg
= 0
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1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
d
A
= 6.3 mm
A
A
=
V
A
T
allow
2.84 kN
90 × 10
3
kPa
=
= 31.56 × 10
6
m
2
= π(d
A
2
/4)
d
B
= 9.7 mm
A
B
=
V
B
T
allow
6.67 kN
90 × 10
3
kPa
=
= 74.11 × 10
6
m
2
= π(d
B
2
/4)
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1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
d
A
= 7 mm d
B
= 10 mm
Choose a size larger to nearest millimeter.
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1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of rod:
d
BC
= 8.59 mm
A
BC
=
P
(
t
)
allow
6.67 kN
115 × 10
3
kPa
=
= 58 × 10
6
m
2
= π(d
BC
2
/4)
d
BC
= 9 mm
Choose a size larger to nearest millimeter.
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1. Stress
CHAPTER REVIEW
• Internal loadings consist of
1.Normal force, N
2.Shear force, V
3.Bending moments, M
4.Torsional moments, T
• Get the resultants using
1.method of sections
2.Equations of equilibrium
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1. Stress
CHAPTER REVIEW
• Assumptions for a uniform normal stress
distribution over xsection of member (σ = P/A)
1.Member made from homogeneous isotropic
material
2.Subjected to a series of external axial loads
that,
3.The loads must pass through centroid of
crosssection
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1. Stress
CHAPTER REVIEW
• Determine average shear stress by using
τ = V/A equation
– V is the resultant shear force on cross
sectional area A
– Formula is used mostly to find average
shear stress in fasteners or in parts for
connections
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1. Stress
CHAPTER REVIEW
• Design of any simple connection requires that
– Average stress along any crosssection not
exceed a factor of safety (F.S.) or
– Allowable value of σallow or τallow
– These values are reported in codes or
standards and are deemed safe on basis of
experiments or through experience
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