Lesson 03, simple stress and strain

Simple Stress and Strain
Lecturer;
Dr. Dawood S. Atrushi
PP'
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1
2
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␴max
␴ave
November 2014

Content
¢  Elastic deformations of axially
loaded members
¢  Statically determinate and
indeterminate members loaded
axially
¢  Average shear Stress
¢  Stress concentrations
¢  Allowable stress
¢  IS Units
November, 142 Strenght of Materials I - DAT

Elastic Deformation of
Axially Loaded Members
Consider a bar with
¢  gradually varying cross section along
its length, and
¢  subjected to concentrated loads at its
right end and also a variable external
load distributed along its length
November, 14Strenght of Materials I - DAT3

Mechanical deformation
(SI&4th
p.120-127;3rd
p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement of one end with respect to the
other.
x
P(x)
dx dx (original length)
P(x)+dP
d (elongation of dx)
P
FBD
a
a
View a-a
A(x)
L
Fig. 2.7 Thermal and mechanical deformation
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be
)(
)(
xA
xP
x
d
Generalized bar with gradually varying
cross-sectional along its length
(SI&4th
p.120-127;3rd
p.122-129)
Now we are going to find the elastic deformation of a member subjected to axial loads. Let’s
consider a generalized bar shown in Fig. 2.7, which has a gradually varying cross-sectional
are along its length L. For a more general case, the bar is subjected to concentrated loads at its
right end and also a variable external load distributed along its length (such as a distributed
load could be for example, to represent the weight of a vertical bar or friction forces acting on
bar surface). Here we wish to find the relative displacement of one end with respect to the
other.
x
P(x)
P(x)+dP
d (elongation of dx)
P
FBD
a
a
View a-a
A(x)
L
We pick a differential element of length dx and cross-sectional area A(x). FBD can be drawn
as middle of Fig. 2.7. Assume that resultant internal axial force is represented as P(x). The
load P(x) will deform the element into the shape indicated by the dashed outline.
The average stress in the cross-sectional area would be
)(
)(
xA
xP
x
d

¢  The average stress in the cross-
sectional area;
element of length dx and cross-sectional area A(x). FBD can b
Assume that resultant internal axial force is represented as P
he element into the shape indicated by the dashed outline.
he cross-sectional area would be
)(
)(
xA
xP
x
he cross-sectional area would be
dx
d
x
ties do not exceed the proportional limit, we can relate the
lement of length dx and cross-sectional area A(x). FBD ca
Assume that resultant internal axial force is represented a
e element into the shape indicated by the dashed outline.
e cross-sectional area would be
)(
)(
xA
xP
x
cross-sectional area would be
dx
d
x
es do not exceed the proportional limit, we can relate
¢  The average strain in the cross-
sectional area;

Using Hook’s law, i.e. σ = Eε;
Provided these quantities do not exceed the proportional limit, we
Hook’s law, i.e.
E
Therefore
dx
d
xE
xA
xP
)(
)(
Re-organize the equation, we have
dx
xExA
xP
d
)(
)(
For the entire length L of the bar, we must integrate this expression
displacement
L
dx
xExA
xP
0
)(
)(
Where: = displacement between two points
L = distance between the points
P(x) = Internal axial force distribution
A(x) = Cross-sectional area
E(x) = Young’s modulus
dx
d
xE
xA
xP
)(
)(
dx
xExA
xP
d
)(
)(
For the entire length L of the bar, we m
displacement
L
dx
xExA
xP
0
)(
)(
Where: = displacement be
L = distance between
P(x) = Internal axial for
A(x) = Cross-sectional
E(x) = Young’s modulu
dx
d
xE
xA
xP
)(
)(
dx
xExA
xP
d
)(
)(
For the entire length L of the bar, we must
displacement
L
dx
xExA
xP
0
)(
)(
Where: = displacement betwe
L = distance between th
P(x) = Internal axial force
Where
δ = displacement between two points
L = distance between the points
P(x) = Internal axial force distribution
For the entire length L of the bar;

Constant Load and Cross-
Sectional Area
For
P(x) = P = constant (no axially distributed load)
A(x) = A = constant (uniform area)
E(x) = E = constant (homogeneous material)
the displacement is;
Constant Load and Cross-Sectional Area
In many engineering cases, the structural
constant cross-sectional area and made of on
P(x) = P = constant (no axially d
A(x) = A = constant (uniform are
E(x) = E = constant (homogeneo
From Eq. (2.16), we have
EA
PL
Multi-Segment Bar
If the bar is subjected to several different

Multi-Segment Bar
¢  If the bar is subjected to several different axial
forces or cross-sectional areas or Young’s
moduli, the equation can be used for each
segment. The total displacement can be
computed from algebraic addition as;
constant cross-sectional area and made of one homogeno
P(x) = P = constant (no axially distributed loa
A(x) = A = constant (uniform area)
E(x) = E = constant (homogeneous material)
From Eq. (2.16), we have
EA
PL
Multi-Segment Bar
If the bar is subjected to several different axial forces
moduli, the above equation can be used for each segm
computed from algebraic addition as
i ii
ii
EA
LP
Example 2.4: The composite bar shown in the figure is
having cross-sectional areas of AAB=200mm2
and ABC=
EAB=100GPa and EBC=210GPa respectively. Find the tota

Example 4
The composite bar shown in the figure is made of
two segments, AB and BC, having cross-sectional
areas of AAB=200 mm2 and ABC=100 mm2. Their
Young’s moduli are EAB=100 GPa and EBC=210
GPa respectively. Find the total displacement at
the right end.
Multi-Segment Bar
If the bar is subjected to several different axial forces or cross-
moduli, the above equation can be used for each segment. The
computed from algebraic addition as
i ii
ii
EA
LP
Example 2.4: The composite bar shown in the figure is made of
and ABC=100mm2
.
EAB=100GPa and EBC=210GPa respectively. Find the total displace
Step 1 FBD
Assume th
tension.
Step 2 Equil
Internal forc
F
(Opposite to
4m 4.2m
F1=10kN
F2=40kN
A B
C
E1A1 E2A2
F2=40kN

and ABC=100mm
EAB=100GPa and EBC=210GPa respectively. Find the total displa
Step 1 FB
Assume
tension.
Step 2 Eq
Internal fo
(Opposite
so Segme
Internal fo
PBC
Step 3 Compute the total deformation by using Eq. (2.18)
3
4m 4.2m
F1=10kN
F2=40kN
A B
C
E1A1 E2A2
F1=10kN
F2=40kN
C
E1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
Solution
Example 4
Step 1, Free body diagram

Step 2, Equilibriums
Internal force in the whole bar, ABC;
Solution
Example 4
=200mm and ABC=100mm . Their Young’s moduli are
ctively. Find the total displacement at the right end.
Step 1 FBDs for Segments AB and BC.
Assume the internal forces are in
tension.
Step 2 Equilibriums
Internal force in AB
00 12 FFPF ABx
kNPAB 30
(Opposite to our assumption of tension,
so Segment AB is in compression)
Internal force in BC
00 1FPF BCx
kNFP 10 (in tension)
F1=10kN
C
F1=10kN
C
F1=10kN
Step 1 FBDs for Segments AB and BC.
Assume the internal forces are in
tension.
Step 2 Equilibriums
Internal force in AB
00 12 FFPF ABx
kNPAB 30
(Opposite to our assumption of tension,
00 1FPF BCx
kNFPBC 101 (in tension)
F1=10kN
C
F1=10kN
C
F1=10kN
C
Internal force in the segment BC;

Step 3, Deformation
The total deformation;Step 3 Compute the total deformation by using Eq. (2.
9
3
2010100
1030
BCBC
BCBC
ABAB
ABAB
BCABAC
AE
LP
AE
LP
mAC 004.0002.0006.0
F1=10kN
C
E1A1 E2A2
B
PAB
F1=10kN
C
PBC
FBD1
FBD2
00 12 FFPF ABx
kNPAB 30
(Opposite to our assumption of tensio
00 1FPF BCx
kNFPBC 101 (in tension)
he total deformation by using Eq. (2.18)
69
3
69
3
1010010210
2.41010
1020010100
41030
BCBC
BCBC
ABAB
ABAB
AE
LP
AE
LP
mmmAC 4004.0002.0006.0 (towards left)
F1=10kN
F2=40kN
C
A1 E2A2
B
F1=10kN
C
PBC
0 PF ABx
kNPAB 30
(Opposite to our assumptio
so Segment AB is in compr
0 FPF BCx
kNFPBC 101 (in
ompute the total deformation by using Eq. (2.18)
9
3
69
3
10010210
2.41010
1020010100
41030
BCBC
BCBC
ABAB
ABAB
BCB
AE
LP
AE
LP
mmmAC 4004.0002.0006.0 (towards left)
4m 4.2m
F1=10kN
F2=40kN
C
E1A1 E2A2
B
B
F1=10kN
C
PBC

Statically Determinate
Members Loaded Axially
Statically determinate
¢  System with the same
number of unknown
reactions as equations of
statics.
¢  Known reactions can be
determined strictly from
equilibrium equations.
(SI&4th
Ed p. 134-139; 3rd
Ed p. 137-
Statically Determinate and Indetermin
When a bar is supported at one end and s
in Fig. 2.8(a), there is only one unknown
the unknown reaction can easily be dete
unknown reactions as equations of sta
reactions can be determined strictly from
FA
P
(a) Statically determinate
A
B
Fig. 2.8 Statically deter

Statically Indeterminate
Members Loaded Axially
Statically indeterminate
The system that has more
unknown forces than
equations of statics.
The reactions can not be
determined only from
equilibrium equations.
(SI&4th
Ed p. 134-139; 3rd
Ed p. 137-142)
Statically Determinate and Indeterminate
When a bar is supported at one end and subjected to an axial force P at the other end as
in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of
the unknown reaction can easily be determined. So such a system with the same num
unknown reactions as equations of statics is called statically determinate. – i.e.
reactions can be determined strictly from equilibrium equations.
FA
P
P
(b) Statically indeterminate
FA
FB
C
B
AA
B
LAC
LCB
L
Fig. 2.8 Statically determinate and indeterminate structures

Compatibility Conditions
¢  What we need is an
additional equation that
specifies how the
structure is displaced due
to the applied loading.
Such an equation is
usually termed the
compatibility equation.
(SI&4th
Ed p. 134-139; 3rd
Ed p. 137-142)
Statically Determinate and Indeterminate
When a bar is supported at one end and subjected to an axial force P at the other end as
in Fig. 2.8(a), there is only one unknown reaction force FA. By using the equations of
the unknown reaction can easily be determined. So such a system with the same num
unknown reactions as equations of statics is called statically determinate. – i.e.
reactions can be determined strictly from equilibrium equations.
FA
P
P
(b) Statically indeterminate
FA
FB
C
B
AA
B
LAC
LCB
L
Fig. 2.8 Statically determinate and indeterminate structures

Compatibility Condition
FA
C
B
A
=0
C’
= AC + CB=0
FA
AC
C
C’
PAC
FA
FBD
B
C
C’
Elongated
Contracted
CB
FB
PCB
FBD
Fig. 2.9 Compatibility condition
Let’s respectively look at the free body diagram for segment AC and CB as in Fig. 2.9.
(indeed FBD can be in any level of structural system or structural members).
For segment AC,
The amount that length AC elongates CB contracts

¢  Now, since both ends of the bar are
fully fixed, then the total change in
length between A and B must be zero.
¢  The amount that length AC elongates
CB contracts; so the equation can be
written as:
δAC +δCB =0

From the free body diagram for segment AC
and CB;
B
= AC + CB=0
B
Let’s respectively look at the free body diagram for segment AC and CB as i
For segment AC,
00 ACAy
PFF AAC FP Tension (+)
ACAC
ACA
ACAC
ACAC
AC
EA
LF
EA
LP
elongation (+)
For segment CB,
00 CBBy
PFF BCB FP Compression ( )
CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( )
Compatibility condition:
LFLF
B
= AC + CB=0
B
Elongated
Let’s respectively look at the free body diagram for segment AC and CB as in Fig.
For segment AC,
00 ACAy
PFF AAC FP Tension (+)
ACAC
ACA
ACAC
ACAC
AC
EA
LF
EA
LP
elongation (+) (
For segment CB,
00 CBBy
CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( ) (
0CBBACA
CBAC
LFLF
(

CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( )
0
CBCB
CBB
ACAC
ACA
CBAC
EA
LF
EA
LF
Combining Compatibility equation (2.23) with the equation of
solve for the two unknowns FA and FB as,
PFF
EA
LF
EA
LF
BA
CBCB
CBB
ACAC
ACA
0
i.e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B
A
L
F
A
A
If ConstEAEA CBCBACAC , we have
P
L
L
F AC
B and P
L
L
F CB
A
00 CBBy
PFF BCB FP Compressio
CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( )
0
CBCB
CBB
ACAC
ACA
CBAC
EA
LF
EA
LF
Combining Compatibility equation (2.23) with the equation of static
PFF
EA
LF
EA
LF
BA
CBCB
CBB
ACAC
ACA
0
i.e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B
EA
L
A
L
F
ACAC
AC
CB
A
For segment CB,
00 CBBy
CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( )
0
CBCB
CBB
ACAC
ACA
CBAC
EA
LF
EA
LF
Combining Compatibility equation (2.23) with the equation of statics (2.19
PFF
EA
LF
EA
LF
BA
CBCB
CBB
ACAC
ACA
0
i.e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B
EA
L
EA
L
EA
L
F
CB
CB
ACAC
AC
CBCB
CB
A
LAC LCB
or segment CB,
00 CBBy
CBCB
CBB
CBCB
CBCB
CB
EA
LF
EA
LP
contraction ( ) (2.22)
ompatibility condition:
0
CBCB
CBB
ACAC
ACA
CBAC
EA
LF
EA
LF
(2.23)
ombining Compatibility equation (2.23) with the equation of statics (2.19), we now
olve for the two unknowns FA and FB as,
PFF
EA
LF
EA
LF
BA
CBCB
CBB
ACAC
ACA
0
(2.24)
e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
CBCB
CB
A
ConstEAEA CBCBACAC , we have
LAC LCB
Combining compatibility equation with the equation of statics,
we now can solve for the two unknowns FA and FB as;

Lecture N
PFF BA
i.e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B FA
P
L
L
F AC
B and P
L
L
F CB
A
Lecture
PFF BA
i.e. P
EA
L
EA
L
EA
L
F
CBCB
CB
ACAC
AC
ACAC
AC
B F
P
L
L
F AC
B and P
L
L
F CB
A

Example 5
Two bars made of Copper and Aluminum
are fixed to the rigid abutments. Originally,
there is a gap of 5mm between the ends as
shown in the figure. Determine average
normal stress in both bars if increase the
temperature from 10°C to 210°C.
Example 2.5: Two bars made of Copper and Aluminum are fixed to the rigid abutmen
Originally, there is a gap of 5mm between the ends as shown in the figure. Determine avera
normal stress in both bars if increase the temperature from 10 C to 210 C.
Copper Aluminumcu=17 × 10-6
Ecu=110GPa
al=23×10-6
Eal=69GPa
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
F
Thermally expanded T,cu due to T
Mechanical force push it back by F,cu
Copper
Copper
d = 0.01m
T
Ecu=110GPa
al=23×10-6
Eal=69GPa
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
F
Copper
Copper
d = 0.01m
T
Ecu=110GPa
al=23×10-6
Eal=69GPa
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
F
Copper
Copper
d = 0.01m
T

Ecu=110GPa
al=23×10-6
Eal=69GPa
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01m
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210
Originally, there is a gap of 5mm between the ends as shown in the figure. Determine aver
Ecu=110GPa
al=23×10-6
Eal=69GPa
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01m
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210
Ecu=110GPa
al=23×10-6
Eal=69GPa
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01m
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210
Ecu=110GPa
al=23×10-6
Eal=69GPa
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01m
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210
Ecu=110GPa
al=23×10-6
Eal=69GPa
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01m
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210
Ecu=110GPa
al=23×1
Eal=69GP
T0=10oC
T=210oC
T= 200oC
0.4m 0.8m
0.005m
d = 0.01m
T,cu
F,cu
T,Al
F,Al
F
F
T
Copper
Copper
Aluminum
Aluminum
d = 0.01
T
522
10857010
4
143
4
..
.
dA m2
and CT o
20010210

November, 14Strenght of Materials I - DAT
23
T,Al
522
10857010
4
143
4
..
.
dA m2
and
Let’s firstly look at the copper bar. When the bar system i
copper bar expand towards right by Cu,T . After the copp
mechanical force F will develop, which will prevent the
We assume that due to such a mechanical force, the copp
real total deformation of copper bar will be computed as
CuFCuTCu ,, (elongation +, Contractio
Similarly, we have
Al,FAl,TAl (elongation +, Contraction
Because these two expanding bars should fill the gap, we p
0050.CuAl
T,Al
Aluminum
5
1085 m2
and CT o
20010210
hen the bar system is heated up from 10 C to 210 C, the
Cu,T . After the copper bar touch to the aluminum bar, a
ch will prevent the copper bar from expanding further.
ical force, the copper bar is pressed back by Cu,F . The
ll be computed as
ion +, Contraction )
on +, Contraction )
Total deformation of copper;
T,Al
522
10857010
4
143
4
..
.
dA m2
and T 210
Let’s firstly look at the copper bar. When the bar system is heated u
copper bar expand towards right by Cu,T . After the copper bar tou
mechanical force F will develop, which will prevent the copper ba
We assume that due to such a mechanical force, the copper bar is p
CuFCuTCu ,, (elongation +, Contraction )
Similarly, we have
Al,FAl,TAl (elongation +, Contraction )
Because these two expanding bars should fill the gap, we prescribe a c
0050.CuAl
From these two equations, we have
0050.Al,FAl,TCu,FCu,T
T= 200oC
T,Al
T Aluminum
522
10857010
4
143
4
..
.
dA m2
and T 210
Let’s firstly look at the copper bar. When the bar system is heated u
copper bar expand towards right by Cu,T . After the copper bar tou
mechanical force F will develop, which will prevent the copper ba
We assume that due to such a mechanical force, the copper bar is p
Similarly, we have
Because these two expanding bars should fill the gap, we prescribe a
0050.CuAl
Similarly, we have;

real total deformation of copper bar will be computed
CuFCuTCu ,, (elongation +, Contr
Similarly, we have
Al,FAl,TAl (elongation +, Contra
Because these two expanding bars should fill the gap,
0050.CuAl
i.e.
E
F
TL
AE
LF
TL
A
AlAl
CuCu
Cu
CuCu
LL
TLTL
F
AlCu
AlAlCuCu 1017005.0 6
CuFCuTCu ,, (elongation +, Contraction
Similarly, we have
Because these two expanding bars should fill the gap, we presc
0050.CuAl
i.e.
AE
LF
TL
AE
LF
TL
AlAl
Al
AlAl
CuCu
Cu
CuCu
AE
L
AE
L
TLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
85.710110
4.0
02001017005.0
9
6
These two expanding bars should fill the
gap, we prescribe a compatibility condition
as;
copper bar expand towards right by Cu,T . After the copper bar touch to
mechanical force F will develop, which will prevent the copper bar from
We assume that due to such a mechanical force, the copper bar is pressed
Similarly, we have
Because these two expanding bars should fill the gap, we prescribe a compa
0050.CuAl
i.e. 0050.
AE
LF
TL
AE
LF
TL
AlAl
Al
AlAl
CuCu
Cu
CuCu
N
AE
L
AE
L
TLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
691085.710110
4.0
10234.02001017005.0
59
6
Solve for F;
F = 206.2 N

The average normal stress can be
computed as;
Lectu
i.e.
AE
LF
TL
AE
LF
TL
AAl
AlAl
CuCu
Cu
CuCu
N
AE
L
AE
L
TLTL
F
AlAl
Al
CuCu
Cu
AlAlCuCu
2.206
10110
0
21017005.0
9
6
The average normal stress can be computed as
MPa.
.
.
A
F
632
10857
2206
5

Average Shear Stress
Shear Stress
The intensity or force per unit area
acting tangentially to A is called Shear
Stress, (tau), and it is expressed as;
m
AE
LF
TL
AlAl
Al
AlAlAlFAlTAl
1065.31005.31068.3 363
,,
2.9 AVERAGE SHEAR STRESS (S
The intensity or force per unit area acting ta
is expressed as in Eq. (2.3) as:
A
F
lim t
A 0
In order to show how the shear stress can de
an example. The block is supported by two
vertically as shown in Fig. 2.10. If the force

Let’s take a block supported by two rigid
bodies and an external force F as an
example;
A
lim
A 0
(2.3)
In order to show how the shear stress can develop in a structural member, let’s take a block as
an example. The block is supported by two rigid bodies and an external force F is applied
vertically as shown in Fig. 2.10. If the force is large enough, it will cause the material of the
block to deform and fail along the vertical planes as shown.
A FBD of the unsupported center segment indicate that shear force V=F/2 must be applied at
each section to hold the segment in equilibrium.
F
F
F
V
V
A
B
C
D
Sectioned area A
Block
Rigid Rigid
Fig. 2.10 Average shear stress
020 FVF y
A
lim
A 0
(2.3)
F
F
F
V
V
A
B
C
D
Sectioned area A
Block
Rigid Rigid
020 FVF y
A
lim
A 0
(2.3)
F
F
F
V
V
A
B
C
D
Sectioned area A
Block
Rigid Rigid
020 FVF y

Vertical equilibrium;
V
Fig. 2.10 Average shear str
020 FVF y
2/FV
The average shear stress distributed over each sectioned are
defined by
A
V
avg
avg = assume to be the same at each point over the sect
V = Internal shear force
A = Area at the section
Average shear stress distributed over each
section;
Fig. 2.10 Averag
020 FVF y
2/FV
The average shear stress distributed over each sec
defined by
A
V
avg
avg = assume to be the same at each point ov
Assume to be uniform over the section
Internal shear force
Area at the section
V
Fig. 2.10 Average shear s
020 FVF y
2/FV
The average shear stress distributed over each sectioned a
defined by
A
V
avg
avg = assume to be the same at each point over the sec

Stress concentrations
¢  For a uniform cross-
sectional bar that is
applied an axial
force, both
experiment and
theory of elasticity
find that the normal
stress will be
uniformly distributed
over the cross-
section.
intensity of the forces distributed
the stress on that section and is
igma). The stress in a member of
to an axial load P (Fig. 1.8) is
he magnitude P of the load by the
5
P
A
(1.5)
dicate a tensile stress (member in
dicate a compressive stress (mem-
used in this discussion, with P ex-
square meters (m2
), the stress s
unit is called a pascal (Pa). How-
an exceedingly small quantity and
s unit must be used, namely, the
(MPa), and the gigapascal (GPa). Fig. 1.8 Member with an axial load.
(a) (b)
A
P
A
P' P'
P
␴ ϭ

Stress Concentration
l  If we drill a hole for some reasons in the
component, the typical example is to
build a connection with other structural
elements.
avg
K max
in which avg=P/A’ is the assumed average stress
from the figures or tables (as in Fig. 2.11(c)), and
from avg=P/A’, where A’ is the smallest cross-sec
maximum stress at the cross section can be comput
'A
P
KK avgmax
(a)
(b)
P
P
A’
w 2r
l  In engineering practice, the actual stress
distribution does not need. Instead, only
the maximum stress at these sections
must be known.

e points of
larger than
a structural
den change
ear the dis-
stresses in
Figure 2.58
ss distribu-
Figure 2.59
widths con-
rowest part
the use of
who has to
h an analy-
PP'
P'
r
D
d
1
2
d
1
2
␴max
␴ave
Fig. 2.58 Stress distribution near circular
hole in flat bar under axial loading.

Stress Concentration
If we drill a hole for some reasons in the
elements.
In engineering practice, though, the actual stress dis
Instead, only the maximum stress at these sections
designed to resist this highest stress when the axial
maximum normal stress at the critical section can be
advanced mathematical techniques using the the
investigations are usually reported in graphical for
Concentration Factor K.
avg
K max
in which avg=P/A’ is the assumed average stress as
from the figures or tables (as in Fig. 2.11(c)), and the
from avg=P/A’, where A’ is the smallest cross-sectio
maximum stress at the cross section can be computed a
'A
P
KK avgmax
from avg=P/A’, where A’ is the smallest cross-sec
maximum stress at the cross section can be comput
'A
P
KK avgmax
(a)
(b)
P
P
A’
w 2r
Fig. 2.11 Stress c
Stress concentration occurs in the case that there
By observing Fig. 2.11(c), it is interesting to no
sectional area, the higher the stress concentration
A´ is the smallest cross section

Stress Concentration Factor
If we drill a hole for some reasons in the
elements.
Stress Concentration Factor
In engineering practice, though, the actual stress distribution
Instead, only the maximum stress at these sections must be
designed to resist this highest stress when the axial load is a
maximum normal stress at the critical section can be determin
advanced mathematical techniques using the theory of
investigations are usually reported in graphical form (as in
Concentration Factor K.
avg
K max
in which avg=P/A’ is the assumed average stress as in Fig. 2
from the figures or tables (as in Fig. 2.11(c)), and the average
from avg=P/A’, where A’ is the smallest cross-sectional area.
maximum stress at the cross section can be computed as:
P
KK
avg
K max
(2.28)
which avg=P/A’ is the assumed average stress as in Fig. 2.11(b). Provided K has been known
om the figures or tables (as in Fig. 2.11(c)), and the average normal stress has been calculated
om avg=P/A’, where A’ is the smallest cross-sectional area. Then from the above equation the
aximum stress at the cross section can be computed as:
'A
P
KK avgmax (2.29)
(a)
(b)
(c)
P
P
A’ StressConcentrationFactorK
w 2r
r/w

Allowable Stress
¢  When the stress (intensity of force) of
an element exceeds some level, the
structure will fail.
¢  We usually adopt allowable force or
allowable stress to measure the
threshold of safety in engineering.

Uncertainties that we must take
into account in engineering:
¢  The load for design may be different
from the actual load.
¢  Size of structural member may not be
very precise due to manufacturing and
assembly.
¢  Various defects in material due to
manufacturing processing.

To consider such uncertainties a Factor
of Safety, F.S., is usually introduced;
Various defects in material due to manufacturing proce
One simple method to consider such uncertainties is to us
Safety, F.S., which is a ratio of failure load Ffail (found from
the allowable one Fallow
allow
fail
F
F
.S.F
If the applied load is linearly related to the stress developed
using A/F , then we can define the factor of safety as
the allowable stress allow
allow
fail
SF ..
Usually, the factor of safety is chosen to be greater than
failure. This is dependent on the specific design case. For
safety for some of its components may be as high as 3. For
allow
fail
F
F
.S.F
If the applied load is linearly rela
using A/F , then we can defi
the allowable stress allow
allow
fail
SF ..
Usually, the factor of safety is ch
failure. This is dependent on the
safety for some of its components
F.S. (safer), the heavier the structu
to balance the safety and cost.
The value of F.S. can be found in
use Eq. (2.6) to compute the allow
allow
Usually, the factor of safety i
failure. This is dependent on t
safety for some of its compon
F.S. (safer), the heavier the str
to balance the safety and cost.
The value of F.S. can be foun
use Eq. (2.6) to compute the al
..SF
fail
allow
Example 2.2: In Example
Cu,allow=50MPa, please deter
strength point of view.
F
Fallow : Allowable load
Ffail : Failure load
σallow : Allowable stress
The factor of safety is chosen to be greater than 1

Example 6
¢  If the maximum
allowable stress for
copper in Example 1
is σCu,allow=50 MPa,
please determine the
minimum size of the
wire/cable from the
material strength
point of view.
f Stress
its in the SI system is the Newton per square meter or Pascal, i.e. : Pa = N/m2
.
neering, Pa seems too small, so we usually use:
Kilo Pascal KPa (=Pa 103
) e.g. 20,000Pa=20kPa
Mega Pascal MPa (=Pa 106
) e.g. 20,000,000Pa=20MPa
Giga Pascal GPa (=Pa 109
) e.g. 20,000,000,000Pa=20GPa
le 2.1: An 80 kg lamp is supported by a single electrical
cable of diameter d = 3.15 mm. What is the stress carried
cable.
ermine the stress in the wire/cable as Eq. (2.4), we need
ss sectional area A of the cable and the applied internal
:
26
22
10793.7
4
00315.0
4
m
d
A
N.mgF 7848980
MPa.
.A
F
6100
107937
784
6
ble Stress (SI&4th
p. 48-49, 3rd
p. 51-52)
Example 2.1, we may concern whether or not 80kg would be too heavy, or say
MPa stress would be too high for the wire/cable, from the safety point of view. Indeed,
80kg
F
a a
Section
a-a
A
d
FBD
F

Solution
November, 14
Strenght of Materials I - DAT
38
Example 2.2: In Example 2.1, if the max
Cu,allow=50MPa, please determine the minimum
Mathematically, allowCu
d
mg
A
F
,2
4
Therefore:
mg
d
allowCu
10469.4
4
,
Obviously, the lower the allowable stress, the bigg
structural strength and elemental size.
In engineering, there are two significant problems
Problem (1) Stress Analysis: for a specific str
..SF
fail
allow
mple 2.2: In Example 2.1, if the maximum allowable
allow=50MPa, please determine the minimum size of the wire
ngth point of view.
hematically, allowCu
d
mg
A
F
,2
4
refore: mm
mg
d
allowCu
469.410469.4
4 3
,
iously, the lower the allowable stress, the bigger the cable size.
ctural strength and elemental size.
ngineering, there are two significant problems associated with st
..SF
fail
allow
Example 2.2: In Example 2.1, if the maxi
Cu,allow=50MPa, please determine the minimum
Mathematically, allowCu
d
mg
A
F
,2
4
Therefore:
mg
d
allowCu
10469.4
4
,
Obviously, the lower the allowable stress, the bigge
structural strength and elemental size.
In engineering, there are two significant problems as
value of F.S. can be found in design codes and engineering hand
Eq. (2.6) to compute the allowable stress:
..SF
fail
allow
ample 2.2: In Example 2.1, if the maximum allowable s
allow=50MPa, please determine the minimum size of the wire/ca
ngth point of view.
hematically, allowCu
d
mg
A
F
,2
4
refore: mm
mg
d
allowCu
469.410469.4
4 3
,
viously, the lower the allowable stress, the bigger the cable size. St
ctural strength and elemental size.

IS UNITS
from
MECHANICS OF MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. Dewolf
David F. Mazurek

SI Prefixes
SI Prefixes
Multiplication Factor Prefix† Symbol
1 000 000 000 000 5 1012
tera T
1 000 000 000 5 109
giga G
1 000 000 5 106
mega M
1 000 5 103
kilo k
100 5 102
hecto‡ h
10 5 101
deka‡ da
0.1 5 1021
deci‡ d
0.01 5 1022
centi‡ c
0.001 5 1023
milli m
0.000 001 5 1026
micro m
0.000 000 001 5 1029
nano n
0.000 000 000 001 5 10212
pico p
0.000 000 000 000 001 5 10215
femto f
0.000 000 000 000 000 001 5 10218
atto a
† The first syllable of every prefix is accented so that the prefix will retain its identity.
Thus, the preferred pronunciation of kilometer places the accent on the first syllable, not
the second.
‡ The use of these prefixes should be avoided, except for the measurement of areas and vol-

Principal SI Units Used in Mechanics
Principal SI Units Used in Mechanics
Quantity Unit Symbol Formula
Acceleration Meter per second squared p m/s2
Angle Radian rad †
Angular acceleration Radian per second squared p rad/s2
Angular velocity Radian per second p rad/s
Area Square meter p m2
Density Kilogram per cubic meter p kg/m3
Energy Joule J N ? m
Force Newton N kg ? m/s2
Frequency Hertz Hz s21
Impulse Newton-second p kg ? m/s
Length Meter m ‡
Mass Kilogram kg ‡
Moment of a force Newton-meter p N ? m
Power Watt W J/s
Pressure Pascal Pa N/m2
Stress Pascal Pa N/m2
Time Second s ‡
Velocity Meter per second p m/s
Volume, solids Cubic meter p m3
Liquids Liter L 1023
m3
Work Joule J N ? m
† Supplementary unit (1 revolution 5 2p rad 5 3608).
‡ Base unit.

U.S.CustomaryUnitsandTheirSI
Equivalents
Quantity U.S. Customary Units SI Equivalent
Acceleration ft/s2
0.3048 m/s2
in./s2
0.0254 m/s2
Area ft2
0.0929 m2
in2
645.2 mm2
Energy ft ? lb 1.356 J
Force kip 4.448 kN
lb 4.448 N
oz 0.2780 N
Impulse lb ? s 4.448 N ? s
Length ft 0.3048 m
in. 25.40 mm
mi 1.609 km
Mass oz mass 28.35 g
lb mass 0.4536 kg
slug 14.59 kg
ton 907.2 kg
Moment of a force lb ? ft 1.356 N ? m
lb ? in. 0.1130 N ? m
Moment of inertia
Of an area in4
0.4162 3 106
mm4
Of a mass lb ? ft ? s2
1.356 kg ? m2
Power ft ? lb/s 1.356 W
hp 745.7 W
Pressure or stress lb/ft2
47.88 Pa
lb/in2
(psi) 6.895 kPa
Velocity ft/s 0.3048 m/s
in./s 0.0254 m/s
mi/h (mph) 0.4470 m/s
mi/h (mph) 1.609 km/h
Volume, solids ft3
0.02832 m3
in3
16.39 cm3
Liquids gal 3.785 L
qt 0.9464 L
Work ft ? lb 1.356 J

Thank You

Lesson 03, simple stress and strain

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