This slide introduces the concept of simple strain, a term used in mechanics to describe the deformation of a material under an applied force. The slide includes a diagram illustrating the deformation of a rectangular object under a tensile force, as well as a formula for calculating strain. Simple strain is a fundamental concept in the study of materials and mechanics, and understanding it is essential for many engineering applications

1. Mechanics of Solid
ME 2129
Credit: 3.00
Pranto Karua
Assistant Professor
Department of Mechanical Engineering
Khulna University of Engineering & Technology
Simple Strain
Stress-strain Diagram, Hook’s Law, Thermal Stress

2. Simple Strain
2
 Introduction to Strain
 Different Types of Strain
 Stress-Strain diagram
 Hook’s law: Axial And Shearing Deformations
 Poisson’s Ratio
 Statically Indeterminate Members
 Thermal Stresses
 Strength of Materials (4th Edition)
-Andrew Pytel, Ferdinand L. Singer
Reference Books
 Mechanics of Materials (10th Edition)
-Russell C. Hibbeler

3. Strain
3
 When a body is subjected to some external force, there is some change in the
dimension of the body.
 The ratio of change in dimension of body to its original dimension is called as strain.
𝜀 =
𝛿
𝐿
where δ is the deformation and L is the original length, thus ε is dimensionless.
 Simple strain can be classified as tensile strain, compressive strain, shear strain,
volumetric strain.

4. Tensile Strain
4
Tensile strain is the ratio of the increase in length to its original length.
Tensile strain = Increase in length, (l – lo) / Original length, lo

5. Compressive Strain
5
Compressive strain is the ratio of the decrease in length to its original length.
Compressive strain = Decrease in length, (lo – l) / Original length, lo

6. Stress-Strain Diagram
6
 Suppose a metal specimen is placed in a tension-compression testing machine. As the
axial load is gradually increased in increments, the total elongation 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.
 Stress-Strain diagram differs in form for various materials.
 Metallic engineering materials are classified as either ductile or brittle materials.

7. 7
A ductile material is one having relatively large tensile strains up to the point of
rupture like structural steel and aluminum.
And brittle materials has a relatively small strain up to the point of rupture like cast
iron and concrete.
Fig 1.1: S.S.D for ductile material (Mild Steel) Fig 1.2: S.S.D for brittle material (Cast Iron)
Stress-Strain Diagram

8. Stress-Strain Diagram
In figure 1.1, there exists a linear relationship between the elongation and the axial force
up to the point P, which is termed as proportional limit.
Hooke's Law states that within the proportional limit, the stress is directly
proportional to strain.
This relationship can be expressed as
where σ is the stress, ε is the strain and the constant of
proportionality, k is called the Modulus of Elasticity, E
or Young’s Modulus and is equal to the slope of the
stress-strain diagram from O to P.
 Thus, the relationship can be written,
σ = E ε
8
σ ∝ ε or σ = k ε

9. Stress-Strain Diagram: Hook’s Law
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 be
developed such that there is no permanent or residual deformation when the load is
entirely removed.
 The region in stress-strain diagram from P to E is called the elastic range. The region
from E to Y 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.
9

10. Stress-Strain Diagram: Hook’s Law
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.
10

11. Stress-Strain Diagram: Mild Steel
Figure 1.3: S.S.D. for mild steel 11

12. Hook’s Law: Axial and Shearing Stress
Axial Deformations
 Axial forces cause axial deformations.
From Hook’s Law, we get
If σ is replaced by its equivalent P/A and ε is replaced by 𝛿/L, then
σ = E ε
𝑃
𝐴
= 𝐸
𝛿
𝐿
Hence, we get
𝛿
=
𝜎𝐿
𝐸
Where,
=
𝑃𝐿
𝐴𝐸
𝑃 = Axial load
𝐴 = Cross sectional area
𝐸 = Modulus of elasticity
𝛿 = Axial deformations
𝐿 = Length 12

13. Hook’s Law: Axial and Shearing Stress
Shearing Deformations
 Shearing forces cause shearing deformations.
 In shearing deformations, an element undergoes a change in its shape such as from a
rectangle to a parallelogram without any change in its length.
Assuming Hook’s Law to apply shear,
𝜏 = 𝐺𝛾
Where,
If 𝜏 is replaced by its equivalent V/As and 𝛾 is replaced
by 𝛿𝑠/L, then
𝑉 = Axial load
As = Cross sectional area
𝐺 = Modulus of rigidity
𝛿𝑠 = Shearing deformations
𝐿 = Length
𝑉
As
= 𝐺
𝛿𝑠
𝐿
Hence, we get
𝛿𝑠 =
𝑉𝐿
𝐴𝑠𝐺
13

14. Problems
14
Problem 211 A bronze bar is fastened between a steel bar and an aluminum bar as shown in figure.
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 Est = 200 GPa, Eal = 70 GPa, and Ebr = 83 GPa.

15. Problems
15
Problem 212 The rigid bar ABC shown in figure is hinged at A and supported by a steel rod at
B. Determine the largest load P that can be applied at C if the stress in the steel rod is
limited to 30 ksi and the vertical movement of end C must not exceed 0.10 in.
Try yourself

16. Problems
16
Problem 213 The rigid bar AB, attached to two vertical rods as shown in figure, is horizontal
before the load P is applied. Determine the vertical movement of P if its magnitude is 50
kN.

17. Problems
17
Problem 214
Try yourself
The rigid bars AB and CD shown in figure are supported by pins at A and C and
the two rods. Determine the maximum force P that can be applied as shown if its vertical
movement is limited to 5 mm. Neglect the weights of all members.

18. Problems
18
Problem 215 A uniform concrete slab of total weight W is to be attached, as shown in figure,
to two rods whose lower ends are on the same level. Determine the ratio of the areas of the
rods so that the slab will remain level.

19. Poisson’s Ratio
19
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.
The elongation in the x direction is accompanied by a contraction in the other directions.

20. Poisson’s Ratio
20
Poisson’s ratio can be expressed as
𝑣 = −
𝜀𝑦
𝜀𝑥
= −
𝜀𝑧
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.
 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 tensile stress σy 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
𝜀𝑥 =
𝜎𝑥
𝐸
− 𝑣
𝜎𝑦
𝐸

21. Poisson’s Ratio
21
Bulk Modulus of Elasticity, 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
𝐾 =
𝐸
3(1 − 2𝑣
=
𝜎
Δ 𝑉 𝑉
where V is the volume and ΔV is change in volume and the ratio ΔV/V is called volumetric
strain.
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 + 𝑣
 Similarly, the resulting strain in the y direction will be
𝜀𝑦 =
𝜎𝑦
𝐸
− 𝑣
𝜎𝑥
𝐸

22. Problems
22
Problem 222 A solid cylinder of diameter d carries an axial load P. Show that its change in
diameter is 4Pν / πEd.

23. Problems
23
Problem 227 A 150-mm-long bronze tube, closed at its ends, is 80 mm in diameter and has a
wall thickness of 3 mm. It fits without clearance in an 80-mm hole in a rigid block. The tube
is then subjected to an internal pressure of 4.00 MPa. Assuming ν = 1/3 and E = 83 GPa,
determine the tangential stress in the tube.
Try yourself

24. Problems
24
Problem 233 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.
Try yourself

25. Problems
25
Problem 236 A rigid block of mass M is supported by three symmetrically spaced rods as
shown in figure. Each copper rod has an area of 900 mm2; E = 120 GPa; and the allowable
stress is 70 MPa. The steel rod has an area of 1200 mm2; E = 200 GPa; and the allowable
stress is 140 MPa. Determine the largest mass M which can be supported.
Try yourself

26. Problems
26
Problem 239 The rigid platform in figure has negligible mass and rests on two steel bars, each
250.00 mm long. The center bar is aluminum and 249.90 mm long. Compute the stress in
the aluminum bar after the center load P = 400 kN has been applied. For each steel bar, the
area is 1200 mm2 and E = 200 GPa. For the aluminum bar, the area is 2400 mm2 and E = 70
GPa.
Try yourself

27. Problems
27
Problem 243 A homogeneous rod of constant cross section is attached to unyielding
supports. It carries an axial load P applied as shown in figure. Prove that the reactions are
given by R1 = Pb/L and R2 = Pa/L.
Try yourself

28. Problems
28
Problem 244 A homogeneous bar with a cross sectional area of 500 mm2 is attached to rigid
supports. It carries the axial loads P1 = 25 kN and P2 = 50 kN, applied as shown in figure.
Determine the stress in segment BC. (Hint: Use the results of Prob. 243, and compute the
reactions caused by P1 and P2 acting separately. Then use the principle of superposition to
compute the reactions when both loads are applied.).
Try yourself

29. Problems
29
Problem 247
Try yourself
The composite bar in figure is stress-free before the axial loads P1 and P2 are
applied. Assuming that the walls are rigid, calculate the stress in each material if P1 = 150
kN and P2 = 90 kN.

30. Problems
30
Problem 254
Try yourself
As shown in figure, a rigid bar with negligible mass is pinned at O and attached
to two vertical rods. Assuming that the rods were initially stress-free, what maximum load
P can be applied without exceeding stresses of 150 MPa in the steel rod and 70 MPa in the
bronze rod.

31. Problems
31
Problem 256
Try yourself
Three rods, each of area 250 mm2, jointly support a 7.5 kN load, as shown in
figure. Assuming that there was no slack or stress in the rods before the load was applied,
find the stress in each rod. Use Est = 200 GPa and Ebr = 83 GPa.

32. Thermal Stress
Temperature changes cause the body to expand or contract. The amount δT is given by
𝛿𝑇 = 𝛼𝐿(𝑇𝑓 − 𝑇𝑖 = 𝛼𝐿∆𝑇
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.
𝛿𝑇 = 𝛼𝐿∆𝑇
Deformation due to temperature changes for a
homogeneous rod mounted between unyielding
supports can be computed as:
32

33. Thermal Stress
From the sketch of deformations, we see that 𝛿𝑇 = 𝛿𝑃
 To reattach the rod to the wall will evidently require a pull P to produce the load
deformation, 𝛿𝑃.
𝛿𝑃 =
𝑃𝐿
𝐴𝐸
=
𝜎𝐿
𝐸
In equivalent terms, 𝛼𝐿∆𝑇 =
𝜎𝐿
𝐸
∴ 𝜎 = 𝛼𝐸∆𝑇
where σ is the thermal stress in MPa, E is the modulus of elasticity of the rod in MPa.
If the pull force P can’t able to reattach the wall, then free temperature contraction is
equal to the sum of the load deformation and the yield of the walls.
From the sketch of deformations, we see that 𝛿𝑇 = 𝛿𝑃 + Yield
𝛼𝐿∆𝑇 =
𝜎𝐿
𝐸
+ Yield
∴ 33

34. Problems
Problem 265 : A bronze bar 3 m long with a cross-sectional area of 320 mm2 is placed
between two rigid walls as shown in the figure. At a temperature of -20°C, the gap Δ = 2.5
mm. Find the temperature at which the compressive stress in the bar will be 35 MPa.
Use α = 18.0 × 10-6 m/(m·°C) and E = 80 GPa.
34

35. Problems
Problem 268 :
Try yourself
The rigid bar ABC in figure is pinned at B and attached to the two vertical
rods. Initially, the bar is horizontal and the vertical rods are stress-free. Determine the stress
in the aluminum rod if the temperature of the steel rod is decreased by 40°C. Neglect
the weight of bar ABC.
36

36. Problems
Problem 271 : A rigid bar of negligible weight is supported as shown in figure. If W = 80
kN, compute the temperature change that will cause the stress in the steel rod to be 55
MPa. Assume the coefficients of linear expansion are 11.7 µm/(m·°C) for steel and
18.9 µm/(m·°C) for bronze.
35

37. Problems
Problem 275 :
Try yourself
A rigid horizontal bar of negligible mass is connected to two rods as shown in
figure. If the system is initially stress-free. Calculate the temperature change that will
cause a tensile stress of 90 MPa in the brass rod. Assume that both rods are subjected to
the change in temperature.
37

38. THANK YOU!
Any Query
 prantokarua@me.kuet.ac.bd
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