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Stress and Strain
Stress
Stress is defined as force per unit area. It has the same units as pressure, and in fact pressure
is one special variety of stress. When a body is subjected to a set of forces and is in
equilibrium, then the internal force developed per unit area is called the stress.
������ =�������
=
Types of Stresses:
Stresses are three types such as normal (tensile or compressive), shearing (direct or torsional),
and bearing stress.
Normal Stress:
Stress that acts perpendicular to a surface. At equilibrium, the intensity of the force
perpendicular to or normal to the section point of the body is called the normal stress at a
point. It can be either tension or compression. It is denoted by σ.
Tension
Stress that acts to lengthen an object is termed as tensional stress.
Compression
Stress that acts to shorten an object is termed as compressional stress.
Shearing Stress:
Stress that acts parallel to a surface is termed as shearing stress. It can cause one object to
slide over another. It also tends to deform originally rectangular objects into parallelograms.
The most general definition is that shear acts to change the angles in an object. Shearing
stress is denoted by τ (tao). It may be direct or torsional.
Bearing stress:
Bearing stress is the stress due to the force that acts perpendicular to the common plane. It is
the contact pressure between separate bodies. As for example, soil pressure, pressure between
rivet and the contact surface.
Unit of Stress:
(a) SI Unit: kg/cm2, ton/cm
2, N/mm
2 (MPa), kN/m
2
(b) British Unit: Ib/in2
(psi), lb/ft2, kip/in
2 (ksi).
2
Strain:
Deformation of anybody per unit length is termed as strain. Strain is defined as the amount of
deformation an object experiences compared to its original size and shape. For example, if a
block L cm on a side is deformed so that it becomes l cm long, then the elongation is ∆ = (L-
l) and the strain is =∆/L. Note that strain is dimensionless.
Strain= ����������
�������� ������= ∆
�
Types of Strain:
Strain is two types:
i) Linear Strain
ii) Shearing Strain
Linear or Longitudinal Strain:
Strain that changes the length of a line without changing its direction is termed as linear
strain. It can be either compressional or tensional. Linear strain is denoted by ε.
Compression
Longitudinal or linear strain that shortens an object.
Tension
Longitudinal or linear strain that lengthens an object.
Shearing strain
Strain that changes the angles of an object. Shear causes lines to rotate. Shearing strain is
associated with the change in angles. It is denoted by γ.
Thermal strain
Strain due to change in temperature.
A typical stress-strain diagram:
The load per unit area or stress was plotted against the elongation per unit length or strain is
termed as stress-strain diagram. Figure shown in below is a typical stress-strain diagram.
Proportional Limit
In the stress-strain diagram, proportional limit up to that point where linear variation can be
obtained i.e. stress is proportional to strain.
3
Figure: A typical stress-strain diagram
Elastic Limit
The elastic limit is the stress beyond which the material will not return to its original shape
when unloaded.
Permanent Set
It is the point after elastic limit at which the material does not return in its original position.
There is a permanent deformation called permanent set.
Yield Point
The yield point is the point at which there is an appreciable elongation or yielding of the
material without any corresponding increase of load; indeed, the load may actually decrease
while the yielding occurs. However, the phenomenon of yielding is peculiar to structural
steel; other grades of steel and steel alloys or other materials do not possess it, as is indicated
by the typical stress-strain curves of these materials shown in the following figures.
Figure: Stress-strain diagram of different materials
4
Yield Strength
It is closely associated with yield point. In which load yield point created that is called yield
strength of the materials. For materials like timber, aluminum, which does not have a well-
defined yield point, yield strength is determined by the 0.2% offset method. This consists of
drawing a line parallel to the initial tangent of the stress-strain curve, this line being started at
an arbitrary offset- strain, usually of 0.2% (0.002 mm/mm or 0.002 in/in.). As shown in the
following figure, the intersection of that line with the stress-strain curve is called the yield
strength.
Figure: Offset method to determine yield strength
Ultimate Strength
The ultimate stress, or ultimate strength as it is more commonly called, is the highest ordinate
on the stress-strain curve.
Rupture Strength
The rupture strength is the stress at failure. For structural steel it is somewhat lower than
ultimate strength because the rupture strength is computed by dividing the rupture load by the
original cross-sectional area, which, although convenient, is incorrect. The error is caused by
a phenomenon known as necking. As failure occurs, the material stretches very rapidly and
simultaneously narrows down as shown in the following figure.
5
Hooke's Law:
For homogeneous isotropic materials i.e. materials having the same properties in all
directions, stress is directly proportional to strain up to proportional limit or elastic limit.
i.e. σ α ε or σ = ε.E ------------------- (i)
Where, E = Modulus of elasticity
Now, if ε = 1 or unit strain then stress, σ = E
So modulus of elasticity may be defined as the stress of a body for its unit strain. It can also
be defined as the slope of straight-line portion of the stress-strain diagram.
The unit of E is the unit of σ.
A convenient variation of Hooke's law is obtained by replacing o by its equivalent � = �
�
and replacing ε by ∆
�, so that equation (i) becomes-
=∆�
× !
Or, ∆ = ��
��= "�
�--------------- (ii)
Equation (ii) expresses the relation among the total deformation ∆, the applied load P, the
total length L, the cross-sectional area A, and the modulus of elasticity E. The unit
deformation ∆ has the same unit as length L, since the units of σ and E, being equivalent.
Note that, equation (ii) is subject to the restrictions mentioned below:
1. The load must be axial.
2. The bar must have a constant cross-section.
3. Material must be homogeneous and isotropic.
4. The stress must not exceed the proportional limit.
Modulus of Rigidity
For homogeneous, isotropic materials shearing stress is directly proportional to shearing
strain up to proportional limit.
i.e. # $ %
Or, # = &%, Where, G = Modulus of rigidity
Now, if γ = 1 or unit shearing strain then τ = G,
So modulus of rigidity may be defined as the shearing stress of a body for its unit shearing
strain. The unit of G is the unit of τ .
6
Poisson's Ratio
Experiment show that if a bar is lengthened by axial tension, there is a reduction in the
transverse dimensions. Simeon D. Poisson showed that the ratio of the unit deformations or
strains in these directions is constant for stresses within the proportional limit. Accordingly,
this ration is named after him is Poisson's ratio. Poisson's ratio is the ratio of the lateral strain
to axial strain. It is denoted by ν (neu).
Mathematically, ν = − ������� (�����
�)��� (�����= −
*+
,-= − *.
,-
Where, /) is axial strain in X-direction and 01 or 02 is lateral strain in Y-direction and Z-
direction.
Common values of Poisson's ratio are 0.20 for concrete, 0.25 to 0.30 for steel and 0.33 for
most other materials.
Bi-axial and tri-axial strain
Poisson's ratio permits to extend Hooke's law of uniaxial stress to the case of bi-axial and tri-
axial stresses. Following figure shows that an element is subjected simultaneously to tensile
stress in the x and y direction. The strain in the x-direction due to tensile stress �) is "-
�,
simultaneously the tensile stress �1 will produce lateral contraction in the x-direction of the
amount -ν"+
� or -νε1.
+ =
So, the resultant unit deformation or strain in the x-direction will be, 0) = 0)3 + 0)5
But 0)3 = "-
� and 0)5 =-ν
"+
� ∴ 0)=
"-
�−ν
"+
�
Similarly the resultant unit deformation or strain in the y-direction will be, 01 = 013 + 015
But 013 = −ν "-
� and 015 =
"+
� ∴ 01=
"+
�−ν
"-
�
∴ for two dimensional body, 0)= "-
�−ν
"+
� =
7
�8�) − νσ:; and 01=
"+
�−ν
"-
� =
7
�8�1 − νσ<;
A further extension of this discussion results in the following expressions for strain caused by
simultaneous action of triaxial tensile stresses:
0) = 7
�8�) − ν(σ: + σ>);, 01 = 7
�8�1 − ν(σ> + σ<);, and 02 = 7
�8�2 − ν(σ< + σ:);
�1 = 0
�1 = 0
�) �) �) = 0 �) = 0
�1
�1
�1
�1
�) �)
7
Problem: A piece of 25 cm by 5 cm by 12 mm steel plate is subjected to uniformly
distributed stresses along its edges shown in the following Figure. (a) If Px = 10t and Py = 20t,
what change in thickness will occur due to the application of these forces? (b) To cause the
same change in thickness as in (a) by Px alone, what must be its magnitude? Let, E = 21 × 105
kg/cm2 and ν = 0.25.
Solution:
We have, 02 = 7
�8�2 − ν(σ< + σ:);
We get, �) = �-
�-= 7A×7AAA
B×7.D= 1666.67 HI/KD, �1 =
�+
�+= DA×7AAA
DB×7.D= 666.67 HI/KD,
and �2 = 0
∴ (�) 02 = 7
�8�2 − ν(σ< + σ:);=
7
D7×7AL M0 − 0.25 (1666.67 + 666.67)P
= −27.77 × 10QB K/K
∴ Change in thickness =∆2= 02 × � = −27.77 × 10QB × 1.2
= 33.33 × 10QB (contraction) (Ans)
(b) We have, 02 = −27.77 × 10QB, �1 = �2 = 0, ) =?
We get, 02 = 7
�8�2 − ν(σ< + σ:);= −27.77 × 10QB =
7
D7×7AL M0 − 0.25 (�) + 0)P
∴ �) = 2332.68 HI/KD
Again we get, �) = �-
�-= 2332.68 ∴ ) = 2332.68 × 5 × 1.2 = 14000 HI = 14 � (Ans)
Bulk Modulus:
The ratio of the hydrostatic compressive stress to the decrease in volume is called modulus of
compression or bulk modulus.
8
Relation between bulk modulus, modulus of elasticity and Poisson's ratio:
Considering an infinitesimal elements whose sides are dx, dy and dz. So, the initial volume of
the element will be dxdydz. After straining the sides will become VW + VW0) , VX + VX01, and
VY + VY02. They may be written as dx(1+0)), VX(1 + 01), dz(1+02) respectively. After
subtracting the initial volume of the strained element, the change in volume is determined.
The change in volume is = dx(1+0))VX[1 + 01\ dz(1+02)- dxdydz
⇒ (1 + 0102+020) + 0)01 + 0) 0102 + 0)+01+02)VWVXVY − VWVXVY
⇒ (1 + 0)+01+02)VWVXVY − VWVXVY
[Note: the product of strain 0102+020) + 0)01 + 0) 0102 are too small are neglected]
⇒ (1 + 0) +01+02)VWVXVY − VWVXVY
⇒ VWVXVY + (0)+01+02)VWVXVY − VWVXVY
⇒ (0)+01+02 )VWVXVY
So, the change in volume per unit volume will be (*-]*+]*.)^)^1^2
^)^1^2= (0)+01+02)
Therefore, in the infinitesimal strain theory, the change in volume per unit volume, often
referred as dilatation, e and is defined as e = [0)+01+02\.
Based on the generalized Hooke's law, the dilatation can be found in terms of stresses and
material constants. For this purpose, the following three equations must be added together.
0) = 7
�8�) − ν(σ: + σ>);, 01 = 7
�8�1 − ν(σ> + σ<);, and 02 = 7
�8�2 − ν(σ< + σ:);
They yields, � = [0) +01 +02\ = (7QDν)
�(σ< + σ: + σ>).
Which means that dilatation is proportional to the algebraic sum of all normal stresses.
If an elastic body is subjected to hydrostatic pressure of uniform intensity p, so that σ< =
σy=σz=−a, then we get �=1−2ν!(−a−a−a) ⇒ �= −31−2ν!(a)
⇒ Qb
�= H = �
c(7QDν)
The quantity k represents the ratio of the hydrostatic compressive stress to the decrease in
volume and is called bulk modulus. So, the relation yields, H = �
c(7QDν).
Assignment: Show that for any material, relation between modulus of elasticity, modulus of
rigidity, and Poisson's ratio is, & = �
D(7]ν ).
9
Deflection (deformation) of an Axially Loaded Member:
When the deflection of an axially loaded member is a design parameter, it is necessary to
determine the deformations. Axial deformations are also required in the analysis of statically
indeterminate bars. The deflection characteristics of bars also provide necessary information
for determining the stiffness of systems in mechanical vibration analysis.
Consider the axially loaded bar shown in the following figure for deriving a relation for
axial bar deformation. The applied forces P1, P2, and P3 are held in equilibrium by the force
P4. The cross-sectional area A of the bar is permitted to change gradually. The change in
length that takes place in the bar between points B and D due to the applied force is to be
determined. An arbitrary element cut from the bar is also shown in figure. From free-body
considerations, this element is subjected to a pull P(x), which, in general, is a variable
quantity.
The infinitesimal deformation dδ that takes place within this element upon application of the
forces is equal to the strain xε multiplied by the length dx. The total deformation between any
two given points on a bar is simply the sum of the element deformations.
Now, dxddx
dxx εδε
δ=⇒=
∫ ∫=⇒=∴x
o
x
o
xdxd εδδδ
The magnitude of the strain εx depends on the magnitude of the stress σx. The σx is found in
general by dividing the variable force P(x) by the corresponding area A(x) i.e. ( )( )xA
xPx =σ . For
linearly elastic materials, according to Hooke's law for uniaxial stress, E
xx
σε =
( )( )∫=⇒
x
oExA
dxxPδ ∫=∴
x
o
x
E
σδ
10
A = 0.01028
sq.in.
160′ dy
Problem: Find the deflection of the free end caused by the application of a concentrated
force P. The elastic modulus of the material is E (neglect the weight of the bar). The cross-
section area of the bar is A.
Solution :
We get, ( )( ) AE
P
Eand
A
P
xA
xP xxx ===
σεσ
We know,( )( ) 1C
ExA
dxxPx
o
+= ∫δ ∫=⇒x
o
dxEA
Pδ
1C
AE
Px+=⇒ δ
0,,0 1 === CthenoxAt δ AE
Px=∴δ
( )., AnsAE
PLLxAt == δ
Problem: A copper aerial wire 160 ft. long, weighting 0.0396 lb per ft. and having a cross
sectional area of 0.01028 sq. in. is suspended vertically from its upper end; E =15 x 106 psi (i)
Calculate the total amount of elongation of this wise because of its own weight (ii) Find the
total elongation if a weight of 50 lb is attached to its lower end (iii) Determine what
maximum weight W, this wire can safely support at its lower en if the unit tress in it must not
exceed its elastic limit of 10,000 psi.
B
A
P P
P
P
dx
x
11
Solution:
Here,
( ) ..01028.0,,1015,0396.00396.0 6insqAanddyLpsiElbylbyP ==×==×=
(i) We get, ∫=L
oAE
PL1δ ∫ ××
=⇒160
0
61101510258.0
096.0 ydyδ
( )∫ ×
×=
×=
×=⇒
160
0
2
6
160
0
2
6612
160
1015
852.3
21015
852.3
1015
852.3 yydyδ
( ).0394.010287.3 3
1 Ansinchft =×=⇒ −δ
(ii) Here, P = 50 lb, L = 160′, A = 0.01028 sq. in., E = 15×166 psi
.6225.00518.010151028.0
1605062 inchft
AE
PL==
××
×==δ
∴ Total elongation = δ1+δ2=0.394+0.6225=0.662 inch (Ans.)
(iii) Stress due to self wt = psiA
W616
1028.0
1600396.011 =
×==σ
Maximum stress, σ =10,000 psi.
∴σ2 = σ - σ1 = 10,-616 = 9384 psi
( ).96501028.0
93842 AnslbWW
A
W=⇒=⇒=σ
Assignment: A steel bar 500 ft long, suspended vertically from its one end, supports a load P
at its lower end. The maximum stress in the bar is 8000 psi caused by the load and by its own
weight. If E = 30 x 106 psi and steel weigh 490 lb/ft, find the total elongation of the bar. (Ans
δ = 1.43 inch).
Assignment: A steel rod having a cross-sectional area 1/2 in2 and a length of 600 ft is
suspended vertically. It supports a load of 5000 at its lower end. If steel weigh 490 lb/ft3 and
E = 30 x 106 psi and, find the total elongation in the rod. (Ans δ = 2.64 inch).
Problem : Two bars are to be cut from a 1 cm thick metal plate so that both bars have a
constant thickness of 1 cm. Bar A is to have a constant width of 4 cm throughout its entire
length. Bar B is to be 6 cm in wide at the top and 2 cm wide at the bottom. Each bar is to be
subjected to the same load P. Determine the ratio B
A
L
Lso that both bars will stretch the same
amount. Neglect the weight of the bar.
W=?
12
Solution:
For Bar A, ( )
( )1.......842 E
PL
E
PL
AE
PLL AAA
A =×
==∆
From Bar B, 23
16
21
1
1 B
B
LX
LX
X=⇒==
+
Again, BB L
xy
Lxx
y 4
2
22
1
=∴== ; Area = ( ) ( )BB L
x
L
xyxA
8422 =×==
( )2............3log8
23
2/e
BL
LB
E
PLdx
AE
PL
B
B
==∆ ∫
Now, 3log88
e
BA
BAE
PL
E
PLLL ×=⇒∆=∆
.)(1.13log Ans
L
Le
B
A ==∴
Problem: The dimensions of a frustum of a right circular supported at the large end on a rigid
base are shown in the following figure. Determine the deflection of the top due to the weight
of the body. The unit weight of material is γ; the elastic modulus is E.
4cm
A
P
LA
B
6cm
P
LB
2cm
B
LB
y
x
2cm
X1
6 cm
24 cm
2 cm
6 cm
24 cm
2 cm
X
dx
X1
13
Solution :
cmxxx
123
24
11
11 =⇒+
= and 1212
1 xr
x
rx
x =⇒=
( )14412
22
2 xxrA xx
πππ =
== , and γππ
××−
= 121
123
1 2
2
xx
Px
−= 121443
3xπγ
∫ ∫∫
×+=
−
===∆36
12
36
12
36
12
2
2
3
0
14412
23
144
121443
x
x
Edx
Ex
x
EA
dxP
AE
PL
x
x
L γ
π
πγ
( ) ( ).160AnsShorening
E
γ=
Problem: The composite rod shown in the following figure is firmly attached to unyielding
supports. Compute the stress in each material caused by the application of axial load P = 200
kN. (Area of aluminum bar is 9 cm2 and EA = 70 GPa. Area of steel bar is 12 cm
2 and E =
200 GPa.)
Solution:
( )iRR BA −−−−−−−−−−=+ 200
000,2012
30
70009
20
×
×=∆
×
×==∆ BA R
sandR
AaEa
PaLaa
Here, 000,2012
30
70009
20
×
×=
×
×∴∆=∆ BA RR
sa ⇒ RB = 2.539 RA - (ii)
Solving (i) and (ii)
2/28.69
5.565.56 cmkNkNR aA === σ
2/96.1112
5.1435.1435.56200 cmkNkNR sB ==−= σ
14
Problem: A composite bar consists of an aluminum section rigidly fastened between a bronze
section and a steel section as shown in following figure. Axial loads are applied at the
position indicated. Determine the stress in each section.
Solution:
To calculate the stresses, we must first determine the axial load n each section. The
appropriate free-body diagrams are shown in the following figure.
From the figure we can determine Pbr = 4000 lb (tension, Pal = 5000 lb (compression), and Pst
= 7000 lb (compression). The stresses in each section are-
( ).330,32.1
4000Anspsi
A
P
br
brbr ===σ
( ).780,28.1
5000Anspsi
A
P
al
alal ===σ
( ).380,46.1
7000Anspsi
A
P
st
stst ===σ
Problem: A rod of variable cross-section built in at one end is subjected to three axial forces
as shown in the following figure. Find the maximum normal stress.
9000 lb 2000 lb
7000 lb 4000 lb
Bronze
(A = 1.2 in.2)
Aluminium
A = 1.8 in.2 Steel
A = 1.6 in.2
1.3 ft 1.6 ft 1.7 ft
4000 lb Pbr
9000 lb 4000 lb Pal
9000 lb 2000 lb 4000 lb Pst
A=25 cm2
32 t 9 t
18 t
A=12.5 cm2
15
Solution:
Reaction at A is Ra = 32-9+18 = 41t (left ward)
From the figure, Pab = 41t, Pbc = 9t and Pcd = 18t
( )+===∴ 2/64.125
41cmton
A
P
ab
ababσ
( )+===∴ 2/75.05.12
9cmton
A
P
bc
bcbcσ
( )+===∴ 2/44.15.12
18cmton
A
P
cd
cdcdσ
( )./64.1 2
max Anscmton=∴σ
Assignment:
A rod is composed of an aluminum section rigidly attached between steel and bronze
sections, as shown in the following figure. Axial loads are applied at the positions indicated.
If P = 3000 lb and the cross sectional area of the rod is 0.5 in.2, determine the stress in each
section.
Assignment:
A rod is composed of an aluminum section rigidly attached between steel and bronze
sections, as shown in the following figure. Axial loads are applied at the positions indicated.
Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum
of90 MPa, or in bronze of 100 MPa. (Ans: P = 10.0 kN)
32 t 9 t
18 t 41 t
a b c d
41 t 9 t 18 t
+ + +
Steel Aluminum Bronze
4P
2 ft 3 ft 2.5 ft
16
Elastic Strain energy:
In mechanics, energy is defined as the capacity to do work and work is the product of a force
and the distance in the direction that the force moves. According to the principal of
conservation of energy, the internal work done (i.e. internal force x deformation) is equal to
the external work done by the applied forces. In solid, deformable bodies, stresses multiplied
by their respective areas are forces and deformations are distances. The product of these two
quantities is the internal work done in a body by externally applied forces. This internal work
is stored in a body as the elastic strain energy or the elastic strain energy.
Let us consider an element of elastic material.
σx= normal stress along x – direction, and εx= linear strain in x - direction
At any instant, force acting on any force ABCD = (σx)(dy.dz) and elongation =εx.dx. The
element is made of a linearly elastic material, stress is proportional to strain. Therefore, if the
element is initially free of stress, the force, which acts on the element, increases linearly from
zero until it attains full value. So, the average force acting on the element, while deformation
is taking place dzdydzdy
x
x .2
12
.0σ
σ=
+=
Now, infinitesimal work done dxdydzdu xx .2
1εσ ×
==
= dvE
dvE
dvdzdydx xxxxxxx
22
.
2
...
2
. 2σσσεσεσ===
Where, dv = volume of the element
Steel
A = 500 mm2
Aluminum A = 500 mm
2
Bronze
A = 500 mm2
4P 2P
P
2.5 m 2.0 m 1.5 m
17
===∴2
.
2
2
xxx
Edv
du εσσ Strain energy per unit volume = Strain energy density
Total work done, ∫∫ ==v
o
xxv
o
x dvdvE
u .2
..
2
2 εσσ
Similarly, for shearing strain it can be shown that ∫=v
o
x dvE
U .2
2τ
Problem: Two bars made by linearly elastic materials whose proportions are shown in the
following figures to absorb the same amount of energy delivered by axial forces. Compare
the stresses in the two bars caused by the same input of energy.
Solution:
Let, σ1 and σ2 are stresses of member 1 and 2.
We know, ∫=v
o
x dvE
u2
2
1
σ
⇒ ( )ALE
VE
dvE
u
v
o222
2
1
2
1
2
11
σσσ=== ∫
Again, ∫∫
+==partupperpartlower
x dvE
dvE
dvE
u2
2
22
2
22
2
2
2
σ
σσ
⇒
=
+=
×+
= ∫ 8
5
28
3
424
32
842
2
2
2
2
2
2
2
22
AL
E
ALAL
E
LA
E
AL
Eu
v
o
σσσσ
Absorb same amount of energy
∴ 21 uu =∴
( ) ( ))522
2
2
2
1 ALE
ALE
σσ=⇒ 2
2
2
18
5σσ =⇒ 21 79.0 σσ =⇒
( ).265.11
2 Ans=⇒σ
σ
A
1
I 2A
L/4 A
3L/4
2
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