CE 224-MECHANICS OF DEFORMABLE BODIES

MODULE 2: SIMPLE STRAIN

INTRODUCTION

You may have noticed that certain objects can stretch easily, but stretching
objects like an iron rod seems impossible. This module will help you understand how
deformation takes place in a material. The focus of this module would be the relation
between stress and strain.

OBJECTIVES

• Learn and understand the concepts of strain.

• Utilize Hooke’s Law to calculate unknown stresses and strain.
• Determine material parameters from stress-strain curve.
• Recognize elastic and plastic components of deformation in a material.
• Identify the regions of behavior on a stress-strain diagram.

DISCUSSION PROPER

Simple Strain

Also known as unit deformation, strain is the ratio of the change in length caused by the
applied force, to the original length.

ε=δ/L

where δ is the deformation and L is the original length, thus ε is dimensionless.

Suppose that a metal specimen be placed in tension-compression-testing

machine. As the axial load is gradually increased in increments, the total elongation over
the gauge length is measured at each increment of the load and this is continued until
failure of the specimen takes place. Knowing the original cross-sectional area and length
of the specimen, the normal stress σ and the strain ε can be obtained. The graph of these
quantities with the stress σ along the y-axis and the strain ε along the x-axis is called the
stress-strain diagram. The stress-strain diagram differs in form for various materials.
The diagram shown below is that for a medium-carbon structural steel.

Metallic engineering materials are classified as either ductile or brittle materials.

A ductile material is one having relatively large tensile strains up to the point of rupture
like structural steel and aluminum, whereas brittle materials has a relatively small strain
up to the point of rupture like cast iron and concrete. An arbitrary strain of 0.05
mm/mm is frequently taken as the dividing line between these two classes.

Stress-strain diagram of a medium-carbon structural steel

Proportional Limit (Hooke's Law)
From the origin O to the point called proportional limit, the
stress-strain curve is a straight line. This linear relation
between elongation and the axial force causing was first
noticed by Sir Robert Hooke in 1678 and is called Hooke's Law
that within the proportional limit, the stress is directly
proportional to strain or

σ∝ε or σ=kε

The constant of proportionality k is called the Modulus of

Elasticity or Young’s Modulus and is equal to the slope of the stress-strain diagram from
O to P. Then,

σ=Eε
Elastic Limit
The elastic limit is the limit beyond which the material will no longer go back to its
original shape when the load is removed, or it is the maximum stress that may e
developed such that there is no permanent or residual deformation when the load is
entirely removed.

Elastic and Plastic Ranges

The region in stress-strain diagram from O to E is called the elastic range. The region
from E to R is called the plastic range.

Yield Point
Yield point is the point at which the material will have an appreciable elongation or
yielding without any increase in load.

Ultimate Strength
The maximum ordinate in the stress-strain diagram is the ultimate strength or tensile
strength.

Rapture Strength
Rapture strength is the strength of the material at rupture. This is also known as the
breaking strength.

Modulus of Resilience
Modulus of resilience is the work done on a unit volume of material as the force is
gradually increased from O to P, in N•m/m3. This may be calculated as the area under
the stress-strain curve from the origin O to up to the elastic limit E (the shaded area in
the figure). The resilience of the material is its ability to absorb energy without creating
a permanent distortion.

Modulus of Toughness
Modulus of toughness is the work done on a unit volume of material as the force is
gradually increased from O to R, in N•m/m3. This may be calculated as the area under
the entire stress-strain curve (from O to R). The toughness of a material is its ability to
absorb energy without causing it to break.
Working Stress, Allowable Stress, and Factor of Safety
Working stress is defined as the actual stress of a material under a given loading. The
maximum safe stress that a material can carry is termed as the allowable stress. The
allowable stress should be limited to values not exceeding the proportional limit.
However, since proportional limit is difficult to determine accurately, the allowable tress
is taken as either the yield point or ultimate strength divided by a factor of safety. The
ratio of this strength (ultimate or yield strength) to allowable strength is called the factor
of safety.

In the linear portion of the stress-strain diagram, the tress is proportional to strain and
is given by

σ=Eε

since σ=P/A and ε=δ/L, then P/A=Eδ/L

𝑃𝐿 𝜎𝐿
𝛿= =
𝐴𝐸 𝐸
To use this formula, the load must be axial, the bar must have a uniform cross-sectional
area, and the stress must not exceed the proportional limit.

If however, the cross-sectional area is not uniform, the axial deformation can be
determined by considering a differential length and applying integration.

𝑃 𝐿 𝑑𝑥
𝛿= ∫
𝐸 0 𝐿
where A = ty, and y and t if variable, must be expressed in terms of x.
For a rod of unit mass ρ suspended vertically from one end, the total elongation due to
its own weight is

𝜌𝑔𝐿2 𝑀𝑔𝐿
𝛿= =
2𝐸 2𝐴𝐸
where ρ is in kg/m , L is the length of the rod in mm, M is the total mass of the rod in
3

kg, A is the cross-sectional area of the rod in mm2, and g = 9.81 m/s2.

Stiffness, k
Stiffness is the ratio of the steady force acting on an elastic body to the resulting
displacement. It has the unit of N/mm.

𝑃
𝑘=
𝛿

Example 1:
The following data were recorded during the tensile test of a 14-mm-diameter mild steel
rod. The gage length was 50 mm.

Load (N) Elongation (mm) Load (N) Elongation (mm)

0 0 46 200 1.25
6 310 0.010 52 400 2.50
12 600 0.020 58 500 4.50
18 800 0.030 68 000 7.50
25 100 0.040 59 000 12.5
31 300 0.050 67 800 15.5
37 900 0.060 65 000 20.0
40 100 0.163 65 500 Fracture
41 600 0.433

Plot the stress-strain diagram and determine the following mechanical properties: (a)
proportional limits; (b) modulus of elasticity; (c) yield point; (d) ultimate strength; and
(e) rupture strength.
Solution:

Area, A = 0.25π(142) = 49π mm2

Length, L = 50 mm
Strain = Elongation/Length
Stress = Load/Area

Load Elongation Strain Stress

(N) (mm) (mm/mm) (MPa)
0 0 0 0
6 310 0.010 0.0002 40.99
12 600 0.020 0.0004 81.85
18 800 0.030 0.0006 122.13
25 100 0.040 0.0008 163.05
31 300 0.050 0.001 203.33
37 900 0.060 0.0012 246.20
40 100 0.163 0.0033 260.49
41 600 0.433 0.0087 270.24
46 200 1.250 0.025 300.12
52 400 2.500 0.05 340.40
58 500 4.500 0.09 380.02
68 000 7.500 0.15 441.74
59 000 12.500 0.25 383.27
67 800 15.500 0.31 440.44
65 000 20.000 0.4 422.25
61 500 Failure 399.51
 From stress-strain diagram:
a. Proportional Limit = 246.20 MPa
b. Modulus of Elasticity
E = slope of stress-strain diagram within proportional limit
E = 246.20/0.0012 = 205 166.67 MPa = 205.2 GPa
c. Yield Point = 270.24 MPa
d. Ultimate Strength = 441.74 MPa
e. Rupture Strength = 399.51 MPa

Example 2:
A steel rod having a cross-sectional area of 300 mm2 and a length of 150 m is suspended
vertically from one end. It supports a tensile load of 20 kN at the lower end. If the unit
mass of steel is 7850 kg/m3 and E = 200 × 103 MN/m2, find the total elongation of the
rod.

Solution:

Elongation due to its own weight,

δ1=PL/AE

Where:
P = W = 7850(1/1000)3(9.81)[300(150)(1000)]
P = 3465.3825 N
L = 75(1000) = 75 000 mm
A = 300 mm2
E = 200 000 MPa
Thus,
δ1=3465.3825(75000)/300(200000)
δ1=4.33 mm

Elongation due to applied load,

δ2=PL/AE

Where:
P = 20 kN = 20 000 N
L = 150 m = 150 000 mm
A = 300 mm2
E = 200 000 MPa

Thus,
δ2=20000(150000)300(200000)
δ2=50 mm

Total elongation:
δ=δ1+δ2
δ=4.33+50=54.33 mm answer

Example 3:
A steel wire 30 ft long, hanging vertically, supports a load of 500 lb. Neglecting the
weight of the wire, determine the required diameter if the stress is not to exceed 20 ksi
and the total elongation is not to exceed 0.20 in. Assume E = 29 × 10 6 psi.

Solution:
Based on maximum allowable stress,
σ=P/A
20000=500(1/4πd2)
d=0.1784 in

Based on maximum allowable deformation,

δ=PL/AE
500(30×12)
0.20 = 1 2 6
𝜋𝑑 (29×10 )
4

d=0.1988 in

Use the bigger diameter, d = 0.1988 in answer

Example 4:
An aluminum bar having a cross-sectional area of 0.5 in2 carries the axial loads applied
at the positions shown. Compute the total change in length of the bar if E = 10 × 10 6 psi.
Assume the bar is suitably braced to prevent lateral buckling.
Solution:

P1 = 6000 lb tension
P2 = 1000 lb compression
P3 = 4000 lb tension

δ=PL/AE

δ=δ1−δ2+δ3
δ=6000(3×12)0.5(10×106)−1000(5×12)0.5(10×106)+4000(4×12)0.5(10×106)
δ=0.0696 in. (lengthening) answer

Shearing Deformation

Shearing forces cause shearing deformation. An element subject to shear does not
change in length but undergoes a change in shape.
The change in angle at the corner of an original rectangular element is called the shear
strain and is expressed as

𝛿𝑠
𝛾=
𝐿

The ratio of the shear stress τ and the shear strain γ is called the modulus of elasticity in
shear or modulus of rigidity and is denoted as G, in MPa.

𝜏
𝐺=
𝛾

The relationship between the shearing deformation and the applied shearing force is

𝑉𝐿 𝜏𝐿
𝛿𝑠 = =
𝐴𝑠𝐺 𝐺

where V is the shearing force acting over an area A s.

Poisson's Ratio
When a bar is subjected to a tensile loading there is an increase in length of the bar in
the direction of the applied load, but there is also a decrease in a lateral dimension
perpendicular to the load. The ratio of the sidewise deformation (or strain) to the
longitudinal deformation (or strain) is called the Poisson's ratio and is denoted by ν. For
most steel, it lies in the range of 0.25 to 0.3, and 0.20 for concrete.

𝜀𝑦 𝜀𝑧
𝜈=− =−
𝜀𝑥 𝜀𝑥
where εx is strain in the x-direction and εy and εz are the strains in the perpendicular
direction. The negative sign indicates a decrease in the transverse dimension when εx is
positive.

Biaxial Deformation
If an element is subjected simultaneously by tensile stresses, σx and σy, in the x and y
directions, the strain in the x direction is σx/E and the strain in the y direction is σy/E.
Simultaneously, the stress in the y direction will produce a lateral contraction on the x
direction of the amount -ν εy or -ν σy/E. The resulting strain in the x direction will be

𝜎𝑥 𝜎𝑦 (𝜀𝑥 + 𝜈𝜀𝑦)𝐸
𝜀𝑥 = −𝑣 𝑜𝑟 𝜎𝑥 =
𝐸 𝐸 1 − 𝑣2
and
𝜎𝑦 𝜈𝜎𝑥 (𝜀𝑦 + 𝜈𝜀𝑥 )𝐸
𝜀𝑦 = − 𝑜𝑟 𝜎𝑦 =
𝐸 𝐸 1 − 𝑣2

Triaxial Deformation
If an element is subjected simultaneously by three mutually perpendicular normal
stresses σx, σy, and σz, which are accompanied by strains εx, εy, and εz, respectively,
1
𝜀𝑥 = [𝜎𝑥 − 𝜈(𝜎𝑦 + 𝜎𝑧)]
𝐸
1
𝜀𝑦 = [𝜎𝑦 − 𝜈(𝜎𝑥 + 𝜎𝑧)]
𝐸
1
𝜀𝑧 = [𝜎𝑧 − 𝜈(𝜎𝑥 + 𝜎𝑦)]
𝐸
Tensile stresses and elongation are taken as positive. Compressive stresses and
contraction are taken as negative.

Relationship between E, G, and ν

The relationship between modulus of elasticity E, shear modulus G and Poisson's
ratio ν is:

𝐸
𝐺=
2(1 + 𝜈 )

Bulk Modulus of Elasticity or Modulus of Volume Expansion, K

The bulk modulus of elasticity K is a measure of a resistance of a material to change in
volume without change in shape or form. It is given as

𝜎
𝐸
𝐾= = 𝛥𝑉
3(1 − 2𝜈) 𝑉

where V is the volume and ΔV is change in volume. The ratio ΔV/V is called volumetric
strain and can be expressed as

𝛥𝑉 𝜎 3(1 − 2𝜈)
= =
𝑉 𝐾 𝐸

Example 1:
A solid cylinder of diameter d carries an axial load P. Show that its change in diameter is
4Pν / πEd.

Solution:
ν=−εy/εx
εy=−νεx
εy=−νσx/E
𝛿𝑦 (−𝑃)
= −𝜈 𝐴𝐸
𝑑
𝑃𝑣𝑑
𝛿𝑦 = 1
𝜋𝑑2 𝐸
4
𝟒𝑷𝝂
𝜹𝒚 = 𝝅𝑬𝒅 (okay!)
Example 2:
A welded steel cylindrical drum made of a 10-mm plate has an internal diameter of 1.20
m. Compute the change in diameter that would be caused by an internal pressure of 1.5
MPa. Assume that Poisson's ratio is 0.30 and E = 200 GPa.

Solution:
σy = longitudinal stress
σy=pD/4t=1.5(1200)/4(10)
σy=45MPa

σx = tangential stress
σy=pD/2t=1.5(1200)/2(10)
σy=90MPa
εx=σx/E−νσy/E
εx=90/200000−0.3(45/200000)
εx=3.825×10−4

εx=ΔD/D
ΔD=εxD=(3.825×10−4)(1200)
ΔD=0.459 mm answer

Example 3:
A rectangular steel block is 3 inches long in the x direction, 2 inches long in the y
direction, and 4 inches long in the z direction. The block is subjected to a triaxial loading
of three uniformly distributed forces as follows: 48 kips tension in the x direction, 60
kips compression in the y direction, and 54 kips tension in the z direction. If ν = 0.30
and E = 29 × 106 psi, determine the single uniformly distributed load in the x direction
that would produce the same deformation in the y direction as the original loading.

Solution:

For triaxial deformation (tensile triaxial stresses),

(compressive stresses are negative stresses)
1
𝜀𝑦 = [𝜎𝑦 − 𝜈(𝜎𝑥 + 𝜎𝑧)]
𝐸

σx=Px/Ayz=48/4(2)=6.0 ksi (tension)

σy=Py/Axz=60/4(3)=5.0 ksi (compression)
σz=Pz/Axy=54/2(3)=9.0ksi (tension)

Thus
1
𝜀𝑦 = 29𝑥106 [−5000 − 0.30(6000 + 9000)]
εy=−3.276×10−4
 εy is negative, thus, tensile force is required in the x-direction to produce the same
deformation in the y-direction as the original forces.

For equivalent single force in the x-direction,

(uniaxial stress)
ν=−εy/εx
−νεx=εy
−νσx/E=εy
−0.30(σx/29×106)=−3.276×10−4
σx=31666.67 psi
σx=Px/4(2)=31666.67
Px=253333.33 lb
Px=253.33 kips (tension) answer

Statically Indeterminate Members

When the reactive forces or the internal resisting forces over a cross section exceed the
number of independent equations of equilibrium, the structure is called statically
indeterminate. These cases require the use of additional relations that depend on the
elastic deformations in the members.

Example 1:
A steel bar 50 mm in diameter and 2 m long is surrounded by a shell of a cast iron 5 mm
thick. Compute the load that will compress the combined bar a total of 0.8 mm in the
length of 2 m. For steel, E = 200 GPa, and for cast iron, E = 100 GPa.

Solution:

δ=PL/AE
δ=δcastiron=δsteel=0.8mm

𝑃𝑐𝑎𝑠𝑡𝑖𝑟𝑜𝑛(2000)
δcastiron= 1 = 0.8
[4𝜋(602 −502 )](100000)
Pcastiron=11000π N

𝑃𝑠𝑡𝑒𝑒𝑙(2000)
δsteel= 1 = 0.8
[4𝜋(502 )](200000)

Psteel=50000π N

ΣFV=0
P=Pcastiron+Psteel
P=11000π+50000π
P=61000π N
P=191.64 kN answer

Thermal Stress

Temperature changes cause the body to expand or contract. The amount δT, is given by

𝛿𝑇 = 𝛼𝐿(𝑇𝑓 − 𝑇𝑖 ) = 𝛼𝐿𝛥𝑇
where α is the coefficient of thermal expansion in m/m°C, L is the length in meter,
Ti and Tf are the initial and final temperatures, respectively in °C. For steel, α = 11.25 ×
10-6 m/m°C.

If temperature deformation is permitted to occur freely, no load or stress will be induced

in the structure. In some cases where temperature deformation is not permitted, an
internal stress is created. The internal stress created is termed as thermal stress.

For a homogeneous rod mounted between unyielding supports as shown, the thermal
stress is computed as:

deformation due to temperature changes;

𝛿𝑇 = 𝛼𝐿𝛥𝑇
deformation due to equivalent axial stress;
δP=PL/AE=σL/E

δT=δP
αLΔT=σL/E
σ=EαΔT
where σ is the thermal stress in MPa, E is the modulus of elasticity of the rod in MPa.

If the wall yields a distance of x as shown, the following calculations will be made:

δT=x+δP
αLΔT=x+σLE
where σ represents the thermal stress.

Take note that as the temperature rises above the normal, the rod will be in
compression, and if the temperature drops below the normal, the rod is in tension.

Example 1:
A steel rod with a cross-sectional area of 0.25 in2 is stretched between two fixed points.
The tensile load at 70°F is 1200 lb. What will be the stress at 0°F? At what temperature
will the stress be zero? Assume α = 6.5 × 10-6 in/(in·°F) and E = 29 × 106 psi.

Solution:

For the stress at 0°F:

 δ=δT+δst
𝜎𝐿 𝑃𝐿
= 𝛼𝐿(𝛥𝑇) +
𝐸 𝐴𝐸
σ=αE(ΔT)+P/A
σ=(6.5×10−6)(29×106)(70)+1200/0.25
σ=17995psi=18 ksi answer

For the temperature that causes zero stress:

α(ΔT)=P/AE
(6.5×10−6)(T−70)=1200/0.25(29×106)
T=95.46∘F answer

Example 2:
A steel rod is stretched between two rigid walls and carries a tensile load of 5000 N at
20°C. If the allowable stress is not to exceed 130 MPa at -20°C, what is the minimum
diameter of the rod? Assume α = 11.7 µm/(m·°C) and E = 200 GPa.

Solution:

σ=αE(ΔT) + P/A
130=(11.7×10−6)(200000)(40)+5000/A
A=500036.4=137.36 mm2

1/4πd2=137.36
d=13.22 mm answer
SAQ (SELF ASSESMENT EXAMINATION)

1. A bronze bar is fastened between a steel bar and an aluminum bar as shown in Fig. p-
211. Axial loads are applied at the positions indicated. Find the largest value of P that
will not exceed an overall deformation of 3.0 mm, or the following stresses: 140 MPa in
the steel, 120 MPa in the bronze, and 80 MPa in the aluminum. Assume that the
assembly is suitably braced to prevent buckling. Use E st = 200 GPa, Eal = 70 GPa, and
Ebr = 83 GPa.

2. A reinforced concrete column 200 mm in diameter is designed to carry an axial

compressive load of 300 kN. Determine the required area of the reinforcing steel if the
allowable stresses are 6 MPa and 120 MPa for the concrete and steel, respectively. Use
Eco = 14 GPa and Est = 200 GPa.

3. Steel railroad reels 10 m long are laid with a clearance of 3 mm at a temperature of

15°C. At what temperature will the rails just touch? What stress would be induced in the
rails at that temperature if there were no initial clearance? Assume α = 11.7 µm/(m·°C)
and E = 200 GPa.
SUMMARY

Strain is the ratio between the deformation of a body to its original length.
Sometimes it was too small that we could not notice. The relation between stress and
strain is called the stress-strain diagram. This diagram is very essential in studying
different types of materials. By knowing the material specifications, we would be able to
use the material as a structural member.

REFERENCES

1. Besavilla, Strength of Materials 2nd ed. VIB Publisher

2. A. Pytel and F. Singer, Strength of Materials 4th ed. Harper Collins Publishers

3. Mathalino.com

ISUE__ __ Syl ___

Revision: 02

Effectivity: August 1, 2020