1

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