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 of
members subjected to an axial load or direct shear
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1. Stress
CHAPTER OUTLINE
1.
2.
3.
4.
5.
6.
7.
Introduction
Equilibrium of a deformable body
Stress
Average normal stress in an axially loaded bar
Average shear stress
Allowable stress
Design of simple connections
CEN223 Strength of Materials
2
1. Stress
1.1 INTRODUCTION
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
CEN223 Strength of Materials
3
1. Stress
1.1 INTRODUCTION
Historical development
• Beginning of 17th century (Galileo)
• Early 18th century (Saint-Venant, Poisson, Lamé
and Navier)
• In recent times, with advanced mathematical and
computer techniques, more complex problems
can be solved
CEN223 Strength of Materials
4
1. Stress
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)
CEN223 Strength of Materials
5
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CEN223 Strength of Materials
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1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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1. Stress
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1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Equations of equilibrium
• For equilibrium
– balance of forces
– balance of moments
• Draw a free-body diagram to account for all
forces acting on the body
• Apply the two equations to achieve equilibrium
state
∑F=0
∑ MO = 0
CEN223 Strength of Materials
7
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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
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CEN223 Strength of Materials
8
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Internal resultant loadings
• For coplanar loadings:
– Normal force, N
– Shear force, V
– Bending moment, M
CEN223 Strength of Materials
9
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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
CEN223 Strength of Materials
10
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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
CEN223 Strength of Materials
11
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
Procedure for analysis
• Free-body 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
CEN223 Strength of Materials
12
1. Stress
1.2 EQUILIBRIUM OF A DEFORMABLE BODY
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 free-body diagram
CEN223 Strength of Materials
13
1. Stress
EXAMPLE 1.1
Determine resultant loadings acting on cross
section at C of beam.
CEN223 Strength of Materials
14
1. Stress
EXAMPLE 1.1 (SOLN)
Support Reactions
• Consider segment CB
Free-Body Diagram:
• Keep distributed loading exactly where it is on
segment CB after “cutting” the section.
• Replace it with a single resultant force, F.
CEN223 Strength of Materials
15
1. Stress
EXAMPLE 1.1 (SOLN)
Free-Body Diagram:
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.
CEN223 Strength of Materials
16
1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
+
∑ Fx = 0;
− Nc = 0
Nc = 0
+
∑ Fy = 0;
Vc − 540 N = 0
Vc = 540 N
+ ∑ Mc = 0;
CEN223 Strength of Materials
−Mc − 504 N (2 m) = 0
Mc = −1080 N·m
17
1. Stress
EXAMPLE 1.1 (SOLN)
Equilibrium equations:
Negative sign of Mc means it acts in the opposite
direction to that shown below
CEN223 Strength of Materials
18
1. Stress
EXAMPLE 1.5
Determine resultant internal loadings acting on cross
section at B of pipe.
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.
CEN223 Strength of Materials
19
1. Stress
EXAMPLE 1.5 (SOLN)
Support Reactions:
• Consider segment AB,
which does not involve
support reactions at C.
Free-Body Diagram:
• Need to find weight of
each segment.
CEN223 Strength of Materials
20
1. Stress
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
CEN223 Strength of Materials
21
1. Stress
EXAMPLE 1.5 (SOLN)
Equilibrium equations:
∑ Fx = 0;
(FB)x = 0
∑ Fy = 0;
(FB)y = 0
∑ Fz = 0;
CEN223 Strength of Materials
(FB)z − 9.81 N − 24.525 N − 50 N = 0
(FB)z = 84.3 N
22
1. Stress
EXAMPLE 1.5 (SOLN)
Equilibrium Equations:
∑ (MB)x = 0;
(Mc)x + 70 N·m − 50 N (0.5 m) − 24.525 N (0.5 m)
− 9.81 N (0.25m) = 0
(MB)x = − 30.3 N·m
∑ (MB)y = 0;
(Mc)y + 24.525 N (0.625·m) + 50 N (1.25 m) = 0
(MB)y = − 77.8 N·m
∑(MB)z = 0;
CEN223 Strength of Materials
(Mc)z = 0
23
1. Stress
EXAMPLE 1.5 (SOLN)
Equilibrium Equations:
NB = (FB)y = 0
VB = √ (0)2 + (84.3)2 = 84.3 N
TB = (MB)y = 77.8 N·m
MB = √ (30.3)2 + (0)2 = 30.3 N·m
The direction of each moment is determined
using the right-hand rule: positive moments
(thumb) directed along positive coordinate axis
CEN223 Strength of Materials
24
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)
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CEN223 Strength of Materials
25
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1. Stress
1.3 STRESS
Concept of stress
• Consider ΔA in figure below
• Small finite force, ΔF acts on ΔA
• As ΔA → 0, Δ F → 0
• But stress (ΔF / ΔA) → finite limit (∞)
CEN223 Strength of Materials
26
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1. Stress
1.3 STRESS
Normal stress
• Intensity of force, or force per unit area, acting
normal to ΔA
• Symbol used for normal stress, is σ (sigma)
σz =
lim ΔFz
ΔA →0
ΔA
• Tensile stress: normal force “pulls” or “stretches”
the area element ΔA
• Compressive stress: normal force “pushes” or
“compresses” area element ΔA
CEN223 Strength of Materials
27
1. Stress
1.3 STRESS
Shear stress
• Intensity of force, or force per unit area, acting
tangent to ΔA
• Symbol used for normal stress is τ (tau)
τzx =
τzy =
CEN223 Strength of Materials
lim ΔFx
ΔA →0
ΔA
lim ΔFy
ΔA →0
ΔA
28
1. Stress
1.3 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/m2)
or a pascal (1 Pa = 1 N/m2)
• kPa = 103 N/m2 (kilo-pascal)
• MPa = 106 N/m2 (mega-pascal)
• GPa = 109 N/m2 (giga-pascal)
1
CEN223 Strength of Materials
MPa
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29
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
CEN223 Strength of Materials
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1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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
CEN223 Strength of Materials
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1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Average normal stress distribution
+ FRz = ∑ Fxz
∫ dF = ∫A σ dA
P = σA
P
σ=
A
σ = average normal stress at any
point on cross sectional area
P = internal resultant normal force
A = x-sectional area of the bar
CEN223 Strength of Materials
32
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Equilibrium
• Consider vertical equilibrium of the element
∑ Fz = 0
σ (ΔA) − σ’ (ΔA) = 0
σ = σ’
Above analysis
applies to members
subjected to tension
or compression.
CEN223 Strength of Materials
33
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Maximum average normal stress
• For problems where internal force P and xsectional 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 x-sectional 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
CEN223 Strength of Materials
34
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
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
CEN223 Strength of Materials
35
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Procedure for Analysis
Average normal stress
• Use equation of σ = P/A for x-sectional area of a
member when section subjected to internal
resultant force P
CEN223 Strength of Materials
36
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Procedure for Analysis
Axially loaded members
• Internal Loading:
• Section member perpendicular to 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 P at the section
CEN223 Strength of Materials
37
1. Stress
1.4 AVERAGE NORMAL STRESS IN AXIALLY LOADED BAR
Procedure for Analysis
Axially loaded members
• Average Normal Stress:
• Determine member’s x-sectional area at the
section
• Compute average normal stress σ = P/A
CEN223 Strength of Materials
38
1. Stress
EXAMPLE 1.6
Bar width = 35 mm, thickness = 10 mm
Determine max. average normal stress in bar when
subjected to loading shown.
CEN223 Strength of Materials
39
1. Stress
EXAMPLE 1.6 (SOLN)
Internal loading
Normal force diagram
By inspection, largest
loading area is BC,
where PBC = 30 kN
CEN223 Strength of Materials
40
1. Stress
EXAMPLE 1.6 (SOLN)
Average normal stress
PBC
30(103) N
= 85.7 MPa
σBC =
=
(0.035 m)(0.010 m)
A
CEN223 Strength of Materials
41
1. Stress
EXAMPLE 1.8
Specific weight γst = 80 kN/m3
Determine average compressive stress acting at
points A and B.
CEN223 Strength of Materials
42
1. Stress
EXAMPLE 1.8 (SOLN)
Internal loading
Based on free-body diagram,
weight of segment AB determined from
Wst = γstVst
CEN223 Strength of Materials
43
1. Stress
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
CEN223 Strength of Materials
44
1. Stress
EXAMPLE 1.8 (SOLN)
Average compressive stress
Cross-sectional area at section is:
A = (0.2)m2
σ=
P
A
=
8.042 kN
(0.2 m)2
σ = 64.0 kN/m2
CEN223 Strength of Materials
45
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
Free-body diagram indicates that shear force,
V = F/2 be applied at both sections to ensure
equilibrium
CEN223 Strength of Materials
46
1. Stress
1.5 AVERAGE SHEAR STRESS
Average shear stress over each
section is:
P
τavg =
A
τavg = 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
CEN223 Strength of Materials
47
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
CEN223 Strength of Materials
48
1. Stress
1.5 AVERAGE SHEAR STRESS
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 are
no moments caused by F
CEN223 Strength of Materials
49
1. Stress
1.5 AVERAGE SHEAR STRESS
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 be
applied to determine average shear stress
acting on colored section in (d).
A
=
Zalloa
CEN223 Strength of Materials
50
1. Stress
1.5 AVERAGE SHEAR STRESS
Double shear
• The joints shown below are examples of doubleshear connections, often called double lap joints.
• For equilibrium, x-sectional 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).
CEN223 Strength of Materials
51
1. Stress
1.5 AVERAGE SHEAR STRESS
Procedure for analysis
Internal shear
1. Section member at the pt where the τavg is to be
determined
2. Draw free-body diagram
3. Calculate the internal shear force V
Average shear stress
1. Determine sectioned area A
2. Compute average shear stress τavg = V/A
CEN223 Strength of Materials
52
1. Stress
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.
CEN223 Strength of Materials
53
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Internal loading
Based on free-body diagram, Resultant loading
of axial force, P = 800 N
CEN223 Strength of Materials
54
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Average stress
Average normal stress, σ
P
800 N
= 500 kPa
σ=
=
A
(0.04 m)(0.04 m)
CEN223 Strength of Materials
55
1. Stress
EXAMPLE 1.10 (SOLN)
Part (a)
Internal loading
No shear stress on section, since shear force at
section is zero.
τavg = 0
CEN223 Strength of Materials
56
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
+
∑ Fx = 0; − 800 N + N sin 60° + V cos 60° = 0
+
∑ Fy = 0;
CEN223 Strength of Materials
V sin 60° − N cos 60° = 0
57
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Internal loading
Or directly using x’, y’ axes,
+
∑ Fx’ = 0;
N − 800 N cos 30° = 0
+
∑ Fy’ = 0;
V − 800 N sin 30° = 0
CEN223 Strength of Materials
58
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average normal stress
N
692.8 N
σ=
= 375 kPa
=
(0.04 m)(0.04 m/sin 60°)
A
CEN223 Strength of Materials
59
1. Stress
EXAMPLE 1.10 (SOLN)
Part (b)
Average shear stress
V
400 N
τavg =
= 217 kPa
=
A (0.04 m)(0.04 m/sin 60°)
Stress distribution shown below
CEN223 Strength of Materials
60
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ข้
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.)
:-
Ffail
F.S. =
Fallow
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CEN223 Strength of Materials
61
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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:
σfail
F.S. = σ
allow
•
•
τfail
F.S. = 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
CEN223 Strength of Materials
62
1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
•
To determine area of section subjected to a
normal force, use
P
A=
σallow
•
To determine area of section subjected to a shear
force, use
V
A=
τallow
CEN223 Strength of Materials
63
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1. Stress
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.
CEN223 Strength of Materials
64
1. Stress
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.
CEN223 Strength of Materials
65
1. Stress
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.
Assumptions:
1. (σb)allow of concrete <
(σb)allow of base plate
2. Bearing stress is
uniformly distributed
between plate and
concrete
CEN223 Strength of Materials
66
1. Stress
1.7 DESIGN OF SIMPLE CONNECTIONS
Required area to resist shear caused by axial load
•
•
Although actual shear-stress distribution along rod
difficult to determine, we assume it is uniform.
Thus use A = V / τallow to calculate l, provided d and
τallow is known.
CEN223 Strength of Materials
67
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 x-sectional area
2. Draw a free-body diagram of segment of
member
3. Use equations of equilibrium to determine
internal resultant force
CEN223 Strength of Materials
68
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/τallow or A = V/τallow
CEN223 Strength of Materials
69
* *
EXAMPLE 1.13
The two members pinned together at B. If the pins
have an allowable shear stress of τallow = 90 MPa,
and allowable tensile stress of rod CB is
(σt)allow = 115 MPa
Determine to nearest
mm the smallest
diameter of pins A
and B
and the
ให หาเ น าน น
diameter of rod CB
necessary to support
the load.
ให้ห หาเ น า นย
้ห
์กล าง
กล างข
A : TU (
d
ศู
ศู
ผ่
ผ่
ส้
ส้
CEN223 Strength of Materials
ที่
พื
ผั
1. Stress
=
10
น ว
้
mm
70
1
)
^
FBB
si
3
∅
통
(
Ris
6 Ry
Ax
=
d
i)
'
RB -
·
통
>
Rz
1
.
B
AY
+
Reactions
I
Ma =
Fy
Ax
My
,
,
Double
:
T
shear
EsA
:
0
t
;
Ay
ほ☆
σ
:
号
“
-
6 RN
RB
や
:A
( 5 696 70 ^ )
,
R/e
:
O vos
5
「B
RA
寄
=
s
。
37
Mpa
.
bat
AzT Ay
=
=
14 I =
(
5
. 696
kγ
”
:
0
紫 に iHr )
0
号
-
↓= 4A = 4.3764D = 6.35 mm ( 0
ㅐ
k ^
Bu
A =
B
:
single shear
,
V
=
R
=N
A
Gallow
「 気
:
A
:
筐
〆
a
n 10 w
+10
A :
N
Las
Mpa)
115
=
A
cy
=
。 荒品 ( 品
=
に1
際
↑z
㎡
:
=
2
.
A
=
tTrz )
* 8 y 97
.
=
I
8
.
59
mm
I
สุ
=
ำ
(
x
A
mma
ต่
Ad
97
.
7H
.
87
smm
2
)
)
d = PA = 4.40) = 9.21 mm ( 0
57
=
do
w
=
90
Tallow
Pin
vod
A =
×
=
&N
2
; -A
③= As
√2
RN
湾
Kuwa
R
=
=
A =
:
U
s
RB ( emtem) = ·
Az :
3
등
A
B
,
Ri
,
0 1 - bkN( Im ) +
Ay =
を
pin
A
t
1 y
保何
t "
Ovospin
BlOred
allow
m
un
U
4)
2
=
+
+f
ente
‰
^
^
2)
}
"
:
{
ำ
ต่
Σ
一
Σ
สุ
小
^
ดของ pin
p
ดข ของ in
mma
1. Stress
EXAMPLE 1.13 (SOLN)
Draw free-body diagram:
Determine reactions at supports A and B:
S MA=0: RB = 6.67 kN
S FY=0: AY = 2.0 kN
S FX=0: AX = 5.32 kN
CEN223 Strength of Materials
71
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
2.84 kN
−6 m2 = (d 2/4)
AA =
=
31.56

10
=
A
tallow 90  103 kPa
VA
dA = 6.3 mm
6.67 kN
−6 m2 = (d 2/4)
AB =
=
74.11

10
=
B
tallow 90  103 kPa
dB = 9.7 mm
VB
CEN223 Strength of Materials
72
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of pins:
Choose a size larger to nearest millimeter.
dA = 7 mm
CEN223 Strength of Materials
dB = 10 mm
73
1. Stress
EXAMPLE 1.13 (SOLN)
Diameter of rod:
6.67 kN
P
−6 m2 = (d 2/4)
ABC =
=
58

10
=
BC
3 kPa
115

10
(st)allow
dBC = 8.59 mm
Choose a size larger to nearest millimeter.
dBC = 9 mm
CEN223 Strength of Materials
74
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
CEN223 Strength of Materials
75
1. Stress
CHAPTER REVIEW
•
Assumptions for a uniform normal stress
distribution over x-section 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
cross-section
CEN223 Strength of Materials
76
1. Stress
CHAPTER REVIEW
•
Determine average shear stress by using
τ = V/A equation
– V is the resultant shear force on crosssectional area A
– Formula is used mostly to find average
shear stress in fasteners or in parts for
connections
CEN223 Strength of Materials
77
1. Stress
CHAPTER REVIEW
•
Design of any simple connection requires that
– Average stress along any cross-section 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
CEN223 Strength of Materials
78