4MA008 – Engineering Science
Stress and Strain
Learning Outcomes
By the end of the session, students should be able to;
• Define compressive and tensile stress.
• Calculate the stress, strain, and modulus of elasticity for different
bodies.
• Explain the relationship between stress, strain, and modulus of
elasticity.
Stress, σ
• We will be looking at compressive and
tensile stress.
• These are stresses that are induced as a
result of pushing (compressive) forces or
pulling (tensile) forces.
• Tensile stress is indicated by a positive
number whereas compressive stress is
negative.
• Stress can be found using the following
formula;
Where;
𝐹
σ = Stress (Pa or 𝑁𝑚−2 )
σ=
F= Force (N)
𝐴
2
A= Area (𝑚 )
Tensile Stress
Compressive Stress
Example 1
If a circular shaft has a diameter of 30mm, is 1m in length, and has a
force of 150N pulling on it’s end then calculate the tensile stress in the
beam?
Cross-Section
150N
1m
ø30mm
Solution 1
𝐹
σ=
𝐴
𝐹
σ=
𝐴
𝐹 = 150𝑁
𝐴 = π𝑟 2 𝑜𝑟
π × 30 ×
𝐴=
4
π𝑑 2
4
10−3 2
𝐴 = 706.858 × 10−6 𝑚2
150
σ=
706.858 × 10−6
σ = 212.207 × 103 𝑃𝑎 𝑜𝑟 212.207𝑘𝑃𝑎
Example 2
If a beam has a rectangular cross-section of 30mm width and 40mm
height, is 1m in length, and has a force of 275N pushing on it’s end
then calculate the compressive stress in the beam?
Cross-Section
30mm
275N
40mm
1m
Solution 2
𝐹
σ=
𝐴
𝐹 = −275𝑁
𝐴=W×𝐻
𝐴 = 30 × 10−3 × 40 × 10−3
𝐴 = 1.2 × 10−3 𝑚2
𝐹
σ=
𝐴
−275
σ=
1.2 × 10−3
σ = −229.167 × 103 𝑃𝑎 𝑜𝑟 − 229.167𝑘𝑃𝑎
Strain, ε
Tensile
F
• When a force is applied to a body it
will produce a deformation, 𝑥.
• Strain is defined as deformation per
unit of original length.
• It is unitless as it is a ratio of
deformed length to original length.
• Strain can be found using the
formula;
Where;
𝑥
ε = Strain (Unitless)
ε=
𝑥 = Deformation (m)
𝐿
L = Original Length (m)
𝒙
L
Compressive
F
𝒙
L
Example 3
A beam has length of 300mm before a force is applied.
When a force is applied the beam deforms by 40mm,
calculate the strain of the beam.
F
300mm
300mm
40mm
Solution 3
𝑥
ε=
𝐿
𝐿 = 300 × 10−3 𝑚
𝑥 = 40 × 10−3 𝑚
40 × 10−3
ε=
300 × 10−3
ε = 0.133 (𝑢𝑛𝑖𝑡𝑙𝑒𝑠𝑠)
Example 4
A beam has length of 500mm before a force is applied.
When a force is applied the beam the new length of the
beam is 575mm, calculate the strain of the beam.
F
500mm
575mm
Solution 4
ε=
𝑥
𝐿
𝐿 = 500 × 10−3 𝑚
𝑥 = 𝐹𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ − 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ
𝑥 = 575 × 10−3 − 500 × 10−3
𝑥 = 75 × 10−3 𝑚
75 × 10−3
ε=
500 × 10−3
ε = 0.15 (𝑢𝑛𝑖𝑡𝑙𝑒𝑠𝑠)
Modulus of Elasticity, E
• Defines the elastic behaviour of a
material.
• Linear
elastic
materials
are
described using Hooke’s law.
• Also known as Young’s Modulus.
• Specific values for materials can be
found using a table.
• Modulus of Elasticity can be found
using the formula;
Where;
σ
E = Modulus of Elasticity (Pa)
𝐸=
ε
σ = Stress (Pa)
ε = Strain (Unitless)
Example 5
If a beam has a ultimate tensile strength of 200MPa and a maximum
strain of 32 × 10−3 then what is the Modulus of Elasticity of the
material.
σ
E
ε
Solution 5
σ
𝐸=
ε
σ = 200 × 106 𝑃𝑎
ε = 32 × 10−3
200 × 106
𝐸=
32 × 10−3
𝐸 = 6.25 × 109 𝑃𝑎 𝑜𝑟 6.25𝐺𝑃𝑎
Stress, Strain, and Modulus of Elasticity
𝐹
σ
If, 𝐸 =
and, σ =
𝐴
ε
Then,
𝐹𝐿
𝐸=
𝐴𝑥
𝑥
ε=
𝐿
Where;
ε = Strain (Unitless)
F = Force (N)
L = Original Length (m)
A = Area (𝑚2 )
𝑥 = Deformation (m)
Example 6
A circular shaft has a diameter of 35mm and is 500mm in length. If a
tensile force of 1kN is applied and the shaft deforms by 10mm then
calculate the Young’s Modulus of the material.
1kN
500mm
500mm
Cross-Section
ø35mm
10mm
Solution 6
𝐹L
E=
𝐴x
𝐹 = 1 × 103 𝑁
𝐿 = 500 × 10−3 𝑚
𝑥 = 10 × 10−3 𝑚
𝐴=
π𝑟 2
π𝑑
𝑜𝑟
4
𝐹L
E=
𝐴x
(1 × 103 ) × (500 × 10−3 )
E=
(962.113 × 10−6 )(10 × 10−3 )
2
π × 35 × 10−3
𝐴=
4
2
𝐴 = 962.113 × 10−6 𝑚2
E = 51.969 × 106 𝑃𝑎 𝑜𝑟 51.969M𝑃𝑎
Exercise
1. A bar has a length of 800mm and is stretched to 810mm when
subjected to a force of 25kN. If the bar has a cross-sectional
diameter of 16mm then calculate the stress and strain and
therefore the Young’s Modulus of the material.
2. A bar is used to support a load of 1000Kg, has a modulus of
elasticity of 185GPa, and is 1.5m in length. If the bar cannot
compress more than 0.25mm then calculate the cross-sectional
area and therefore the diameter of the bar.
3. A steel bar has a Young’s modulus of 215GPa, has a length of 2.5m,
and a diameter of 50mm. If the bar is then subjected to a tensile
force of 50kN then how much will the bar deform.