In the previous section, we saw the basic details about stress-strain diagram. In this section we will see Young's modulus
• We have already seen the basic details about the portion OA of the stress-strain curve
• Here we will discuss a few more details about this portion. It can be written in 8 steps:
1. We know that, 'point A' is the point of proportional limit
■ OA is a straight line because, Hooke’s law is satisfied in this portion
♦ That is., $\mathbf\small{\rm{stress = k \times strain}}$ is satisfied in this portion
♦ That is., $\mathbf\small{\rm{\frac{stress}{strain}=k}}$ is satisfied in this portion
2. We have seen three different types of stresses and strains. They are:
(i) Compressive stress which causes compressive strain
and Tensile stress which causes tensile strain
• These two can be combined as:
Longitudinal stress which causes longitudinal strain
(ii) Shear stress which causes shear strain
(iii) Hydraulic stress which causes volumetric strain
3. All items mentioned in (2) are stresses and strains
■ We can draw a separate stress-strain curve for each of them
That is.,
(i) We can draw a longitudinal stress - longitudinal strain curve
♦ We have already seen this curve
(ii) We can draw a shear stress- shear strain curve
♦ Some examples can be seen here
(iii) We can draw a hydraulic stress - hydraulic strain curve
♦ Some examples can be seen here
4. Each of the curves mentioned in (3), will have a proportional limit
• We have seen that, this 'proportional limit' is marked as point A
• Let us elaborate this a little more. It can be done in 3 steps:
(i) Consider the longitudinal stress - longitudinal strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}=k_1}}$
✰ Where k1 is a constant
(ii) Consider the shear stress - shear strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}=k_2}}$
✰ Where k2 is a constant
(iii) Consider the hydraulic stress - volumetric strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}=k_3}}$
✰ Where k3 is a constant
5. So we have to learn about three constants: k1, k2 and k3
■ These constants are commonly known as moduli of elasticity or elastic moduli
♦ Modulus is singular
♦ Moduli is plural
6. So we can write:
♦ k1 is the Elastic modulus related to longitudinal stress and longitudinal strain
♦ k2 is the Elastic modulus related to shear stress and shear strain
♦ k3 is the Elastic modulus related to hydraulic stress and volumetric strain
7. For differentiating between the three elastic moduli, each one is given a special name:
♦ The constant k1 is called Young's Modulus
♦ The constant k2 is called Shear Modulus
♦ The constant k3 is called Bulk Modulus
• So we can write:
♦ $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}}}$ = Young's modulus
♦ $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}}}$ = Shear modulus
♦ $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}}}$ = Bulk modulus
8. So the items given in (7) are the three elastic moduli
■ Note that:
• The word Elastic or Elasticity is specially mentioned along with moduli
• This is because, these moduli are applicable only when the material is within the proportional limit (That is., within the portion OA)
• At higher parts of the curve, the material is no longer elastic and so, the moduli are not applicable at those parts
• First we will discuss about Young’s modulus. It is represented by the letter Y
• We have seen that,Young’s modulus is related to longitudinal stress and strain
• Let us see an interesting feature related to longitudinal stress and strain. We will write it in 7 steps:
1. Fig.9.17(a) below shows the original length L of a steel cylinder
• The cylinder is resting on a rigid platform
• We see that, due to the compressive force F, the length of the cylinder decreases by ΔLc
♦ So we can write: Change in length due to compression = ΔLc
♦ Then, strain (𝜺c) due to compression will be given by: $\mathbf\small{\rm{\epsilon_c=\frac{\Delta L_c}{L}}}$
2. Fig.9.17(b) above shows the same steel cylinder mentioned in (1)
• This time, the cylinder is suspended from a rigid ceiling
• We see that, due to the stretching force F, the length of the cylinder increases by ΔLt
[This 'F' is same in magnitude as the 'F' in (1)]
♦ So we can write: Change in length due to tension = ΔLt
♦ Then, strain (𝜺t) due to tension will be given by: $\mathbf\small{\rm{\epsilon_t=\frac{\Delta L_t}{L}}}$
3. Remember that:
• The same steel cylinder is used in both the cases
♦ So the 'cross sectional area on which force is applied' will be the same in both cases
• The same force F is applied in both the cases
• Since forces and areas are the same, we get:
Compressive stress (𝜎c) applied in (1) = Tensile stress (𝜎t) applied in (2)
• Also, since the same steel cylinder is used in both cases, the initial length L is same
4. Experiments show that:
If all the conditions in (3) are satisfied, ΔLc will be equal to ΔLt
• From this we get:
$\mathbf\small{\rm{\frac{\Delta L_c}{L}=\frac{\Delta L_t}{L}}}$
⇒ $\mathbf\small{\rm{\epsilon_c=\epsilon_t}}$
• That is: Compressive strain (𝜺c) = Tensile strain (𝜺t)
• Also, from (3), we have: Compressive stress (𝜎c) = Tensile stress (𝜎t)
5. Now let us calculate Y:
• Let us calculate Y using the compressive test
♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_c}{\epsilon_c}}}$
• Next, let us calculate Y using the tensile test
♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_t}{\epsilon_t}}}$
6. Equality of the two fractions:
♦ Both the numerators in (5) are equal
♦ Both the denominators in (5) are also equal
• So both the results in (5) are the same
7. So we can write:
■ If the body is made of a material like steel and if all the conditions in (3) are satisfied, we can determine Y using either compressive method or tensile method. Both methods will give the same result
• We have already seen the basic details about the portion OA of the stress-strain curve
• Here we will discuss a few more details about this portion. It can be written in 8 steps:
1. We know that, 'point A' is the point of proportional limit
■ OA is a straight line because, Hooke’s law is satisfied in this portion
♦ That is., $\mathbf\small{\rm{stress = k \times strain}}$ is satisfied in this portion
♦ That is., $\mathbf\small{\rm{\frac{stress}{strain}=k}}$ is satisfied in this portion
2. We have seen three different types of stresses and strains. They are:
(i) Compressive stress which causes compressive strain
and Tensile stress which causes tensile strain
• These two can be combined as:
Longitudinal stress which causes longitudinal strain
(ii) Shear stress which causes shear strain
(iii) Hydraulic stress which causes volumetric strain
3. All items mentioned in (2) are stresses and strains
■ We can draw a separate stress-strain curve for each of them
That is.,
(i) We can draw a longitudinal stress - longitudinal strain curve
♦ We have already seen this curve
(ii) We can draw a shear stress- shear strain curve
♦ Some examples can be seen here
(iii) We can draw a hydraulic stress - hydraulic strain curve
♦ Some examples can be seen here
4. Each of the curves mentioned in (3), will have a proportional limit
• We have seen that, this 'proportional limit' is marked as point A
• Let us elaborate this a little more. It can be done in 3 steps:
(i) Consider the longitudinal stress - longitudinal strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}=k_1}}$
✰ Where k1 is a constant
(ii) Consider the shear stress - shear strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}=k_2}}$
✰ Where k2 is a constant
(iii) Consider the hydraulic stress - volumetric strain curve
♦ The initial portion OA will be a straight line
♦ Within this OA, we will get: $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}=k_3}}$
✰ Where k3 is a constant
5. So we have to learn about three constants: k1, k2 and k3
■ These constants are commonly known as moduli of elasticity or elastic moduli
♦ Modulus is singular
♦ Moduli is plural
6. So we can write:
♦ k1 is the Elastic modulus related to longitudinal stress and longitudinal strain
♦ k2 is the Elastic modulus related to shear stress and shear strain
♦ k3 is the Elastic modulus related to hydraulic stress and volumetric strain
7. For differentiating between the three elastic moduli, each one is given a special name:
♦ The constant k1 is called Young's Modulus
♦ The constant k2 is called Shear Modulus
♦ The constant k3 is called Bulk Modulus
• So we can write:
♦ $\mathbf\small{\rm{\frac{Longitudinal\;stress}{Longitudinal\;strain}}}$ = Young's modulus
♦ $\mathbf\small{\rm{\frac{Shear\;stress}{Shear\;strain}}}$ = Shear modulus
♦ $\mathbf\small{\rm{\frac{Hydraulic\;stress}{Volumetric\;strain}}}$ = Bulk modulus
8. So the items given in (7) are the three elastic moduli
■ Note that:
• The word Elastic or Elasticity is specially mentioned along with moduli
• This is because, these moduli are applicable only when the material is within the proportional limit (That is., within the portion OA)
• At higher parts of the curve, the material is no longer elastic and so, the moduli are not applicable at those parts
Young's modulus
• We have seen that,Young’s modulus is related to longitudinal stress and strain
• Let us see an interesting feature related to longitudinal stress and strain. We will write it in 7 steps:
1. Fig.9.17(a) below shows the original length L of a steel cylinder
 |
| Fig.9.17 |
• We see that, due to the compressive force F, the length of the cylinder decreases by ΔLc
♦ So we can write: Change in length due to compression = ΔLc
♦ Then, strain (𝜺c) due to compression will be given by: $\mathbf\small{\rm{\epsilon_c=\frac{\Delta L_c}{L}}}$
2. Fig.9.17(b) above shows the same steel cylinder mentioned in (1)
• This time, the cylinder is suspended from a rigid ceiling
• We see that, due to the stretching force F, the length of the cylinder increases by ΔLt
[This 'F' is same in magnitude as the 'F' in (1)]
♦ So we can write: Change in length due to tension = ΔLt
♦ Then, strain (𝜺t) due to tension will be given by: $\mathbf\small{\rm{\epsilon_t=\frac{\Delta L_t}{L}}}$
3. Remember that:
• The same steel cylinder is used in both the cases
♦ So the 'cross sectional area on which force is applied' will be the same in both cases
• The same force F is applied in both the cases
• Since forces and areas are the same, we get:
Compressive stress (𝜎c) applied in (1) = Tensile stress (𝜎t) applied in (2)
• Also, since the same steel cylinder is used in both cases, the initial length L is same
4. Experiments show that:
If all the conditions in (3) are satisfied, ΔLc will be equal to ΔLt
• From this we get:
$\mathbf\small{\rm{\frac{\Delta L_c}{L}=\frac{\Delta L_t}{L}}}$
⇒ $\mathbf\small{\rm{\epsilon_c=\epsilon_t}}$
• That is: Compressive strain (𝜺c) = Tensile strain (𝜺t)
• Also, from (3), we have: Compressive stress (𝜎c) = Tensile stress (𝜎t)
5. Now let us calculate Y:
• Let us calculate Y using the compressive test
♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_c}{\epsilon_c}}}$
• Next, let us calculate Y using the tensile test
♦ We have: $\mathbf\small{\rm{Y=\frac{\sigma_t}{\epsilon_t}}}$
6. Equality of the two fractions:
♦ Both the numerators in (5) are equal
♦ Both the denominators in (5) are also equal
• So both the results in (5) are the same
7. So we can write:
■ If the body is made of a material like steel and if all the conditions in (3) are satisfied, we can determine Y using either compressive method or tensile method. Both methods will give the same result
Next we will derive a formula to find Y. It can be written in 3 steps:
1. We have: $\mathbf\small{\rm{Y=\frac{Longitudinal\;stress\;(\sigma)}{Longitudinal\;strain\;(\epsilon)}}}$
2. But $\mathbf\small{\rm{\sigma=\frac{F}{A}\;\;and\;\;\epsilon=\frac{\Delta L}{L}}}$
3. So the result in (1) becomes: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}}}$
■ Thus we get: $\mathbf\small{\rm{Y=\frac{F \times L}{A \times \Delta L}}}$
■ Note that, since strain has no unit, the 'unit of Y' is same as that of stress, which is: Nm-2
1. We have: $\mathbf\small{\rm{Y=\frac{Longitudinal\;stress\;(\sigma)}{Longitudinal\;strain\;(\epsilon)}}}$
2. But $\mathbf\small{\rm{\sigma=\frac{F}{A}\;\;and\;\;\epsilon=\frac{\Delta L}{L}}}$
3. So the result in (1) becomes: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}}}$
■ Thus we get: $\mathbf\small{\rm{Y=\frac{F \times L}{A \times \Delta L}}}$
■ Note that, since strain has no unit, the 'unit of Y' is same as that of stress, which is: Nm-2
Now we will see two important properties related to Y. They can be written in 7 steps:
1. We have: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}}}$
• This can be rearranged as: $\mathbf\small{\rm{\sigma=Y \times \epsilon}}$
2. The '=' sign implies that '𝜎' and '(Y × 𝜺)' are equal
• So if Y is large, 𝜺 will become small. Then only the equality between '𝜎' and '(Y×𝜺)' can be maintained
3. So we can write:
■ If the 'material of large Y' is used to make a cylinder, even a large 𝜎 will produce only a small strain
• In other words:
■ A 'material having large Y' will be stronger than a 'material having smaller Y'
4. Let us see a comparison:
• From the data book, we have:
♦ YAluminium = 70 ×109 N m-2
♦ YCopper = 120 ×109 N m-2
♦ YSteel = 200 ×109 N m-2
5. Let us apply a same load on three different cylinders, made of aluminium, copper and steel
• Assume that the cylinders have the same length and diameter
• Since the loads and diameters are the same, the stress (𝜎) will be the same
• Then we get:
𝜎 = (YAluminium × 𝜺Aluminium) = (YCopper × 𝜺Copper) = (YSteel × 𝜺Steel)
• Comparing the values in (4), we can write:
♦ 𝜺Steel will be the smallest
♦ 𝜺Aluminium will be the largest
6. We see that, large forces are required to cause large strains in steel
• That means, very large forces will be required to make steel elongate like plastic
• That means, if we do not apply such very large forces, the steel will always return to it's original length
• That means, steel is more elastic than aluminium and copper
■ That is why, steel is used in the construction of buildings, transmission towers etc.,
• The engineer will specify the maximum load that is allowed on a building or tower
♦ When we apply that load, the pillars and beams will deform a little
♦ When that load is removed, the pillars and beams will regain their original dimensions
♦ This is because, steel is elastic even when large loads are applied
7. Thus we see two important properties related to Y:
(i) A material having larger Y will be stronger
(ii) A material having larger Y will be more elastic
1. We have seen that:
• Stress = k × Strain (or Longitudinal stress = Y × Longitudinal strain)
• is similar to
• the equation y = mx
♦ We often this equation in coordinate geometry classes
■ So the slope m corresponds to the young's modulus Y
2. Consider two graphs shown in the fig.9.18 below:
• The graph in fig.a has a steep slope
• The graph in fig.b has a gentle slope
3. Forming a right triangle:
(i) Mark any two convenient points P and Q in the graph in fig.a
(ii) Draw horizontal and vertical dashed lines through P and Q
♦ Let them meet at R
(iii) We will get a right triangle PQR with PQ as the hypotenuse
(iv) We see that:
The height (QR) of the triangle is large and the base (PR) is small
4. Slope of the graph is obtained as: $\mathbf\small{\rm{Slope=\frac{Height}{Base}}}$
♦ The numerator (Height) is larger
♦ The denominator (Base) is smaller
♦ So the slope of the graph in fig.a is high
5. But the slope corresponds to Y
■ So we can write:
If the portion OA in the stress-strain curve is steep (high slope), the material will have a high value for Y
6. The opposite happens in fig.b
• We see that:
(i) The height of the triangle is small and the base is large
(ii) That is:
♦The numerator (Height) is smaller
♦ The denominator (Base) is larger
(iii) So the slope of the graph in fig.a is low
■ So we can write:
If the portion OA in the stress-strain curve has a gentle slope, the material will have a lower value for Y
7. Note that, if we want to compare two materials in this way, their graphs must be drawn to the same scale
Solved examples 9.5 to 9.10
1. We have: $\mathbf\small{\rm{Y=\frac{\sigma}{\epsilon}}}$
• This can be rearranged as: $\mathbf\small{\rm{\sigma=Y \times \epsilon}}$
2. The '=' sign implies that '𝜎' and '(Y × 𝜺)' are equal
• So if Y is large, 𝜺 will become small. Then only the equality between '𝜎' and '(Y×𝜺)' can be maintained
3. So we can write:
■ If the 'material of large Y' is used to make a cylinder, even a large 𝜎 will produce only a small strain
• In other words:
■ A 'material having large Y' will be stronger than a 'material having smaller Y'
4. Let us see a comparison:
• From the data book, we have:
♦ YAluminium = 70 ×109 N m-2
♦ YCopper = 120 ×109 N m-2
♦ YSteel = 200 ×109 N m-2
5. Let us apply a same load on three different cylinders, made of aluminium, copper and steel
• Assume that the cylinders have the same length and diameter
• Since the loads and diameters are the same, the stress (𝜎) will be the same
• Then we get:
𝜎 = (YAluminium × 𝜺Aluminium) = (YCopper × 𝜺Copper) = (YSteel × 𝜺Steel)
• Comparing the values in (4), we can write:
♦ 𝜺Steel will be the smallest
♦ 𝜺Aluminium will be the largest
6. We see that, large forces are required to cause large strains in steel
• That means, very large forces will be required to make steel elongate like plastic
• That means, if we do not apply such very large forces, the steel will always return to it's original length
• That means, steel is more elastic than aluminium and copper
■ That is why, steel is used in the construction of buildings, transmission towers etc.,
• The engineer will specify the maximum load that is allowed on a building or tower
♦ When we apply that load, the pillars and beams will deform a little
♦ When that load is removed, the pillars and beams will regain their original dimensions
♦ This is because, steel is elastic even when large loads are applied
7. Thus we see two important properties related to Y:
(i) A material having larger Y will be stronger
(ii) A material having larger Y will be more elastic
Let us now write a useful information that can be obtained from graphs. It can be written in 7 steps:
• Stress = k × Strain (or Longitudinal stress = Y × Longitudinal strain)
• is similar to
• the equation y = mx
♦ We often this equation in coordinate geometry classes
■ So the slope m corresponds to the young's modulus Y
2. Consider two graphs shown in the fig.9.18 below:
 |
| Fig.9.18 |
• The graph in fig.b has a gentle slope
3. Forming a right triangle:
(i) Mark any two convenient points P and Q in the graph in fig.a
(ii) Draw horizontal and vertical dashed lines through P and Q
♦ Let them meet at R
(iii) We will get a right triangle PQR with PQ as the hypotenuse
(iv) We see that:
The height (QR) of the triangle is large and the base (PR) is small
4. Slope of the graph is obtained as: $\mathbf\small{\rm{Slope=\frac{Height}{Base}}}$
♦ The numerator (Height) is larger
♦ The denominator (Base) is smaller
♦ So the slope of the graph in fig.a is high
5. But the slope corresponds to Y
■ So we can write:
If the portion OA in the stress-strain curve is steep (high slope), the material will have a high value for Y
6. The opposite happens in fig.b
• We see that:
(i) The height of the triangle is small and the base is large
(ii) That is:
♦The numerator (Height) is smaller
♦ The denominator (Base) is larger
(iii) So the slope of the graph in fig.a is low
■ So we can write:
If the portion OA in the stress-strain curve has a gentle slope, the material will have a lower value for Y
7. Note that, if we want to compare two materials in this way, their graphs must be drawn to the same scale
Let us see some solved examples:
Now we will see an experiment to determine the Young's modulus. It can be written in 12 steps:
1. We will be determining the ‘Y of the material’ with which, a wire is made. Consider fig.9.19 below:
 |
| Fig.9.19 |
• Two wires are suspended from a rigid support
• The two wires:
♦ Are long and straight
♦ Have the same length
♦ Have the same diameter
2. The wire on the left is called the reference wire
♦ The main scale is attached to this wire
3. The wire on the right is called the experimental wire
♦ The vernier scale is attached to this wire
4. The reference wire carries a pan at it’s bottom
♦ A fixed weight is placed in this pan
5. The experimental wire also carries a pan at it’s bottom
♦ The weight in this pan can be increased or decreased
6. An initial small mass is placed on both the pans
♦ Due to the weights of those masses, both the wires become straight
7. The initial reading (r1) is noted from the scale
8. The load in the experimental wire is gradually increased
♦ The additional mass (m) is noted
♦ Then the additional weight = mg
9. Due to the additional weight, the wire will experience tensile stress and will elongate
♦ The new reading (r2) is noted
♦ The difference (r2-r1) will give the elongation ΔL
• Let L be initial length of the experimental wire
♦ Then strain is given by: $\mathbf\small{\rm{\epsilon=\frac{\Delta L}{L}}}$
10. Let r be the radius of the experimental wire
• Then area of cross section of the experimental wire is given by: $\mathbf\small{\rm{A=\pi r^2}}$
• Then stress experienced by the experimental wire is given by: $\mathbf\small{\rm{\sigma=\frac{mg}{\pi r^2}}}$
11. So we get:
• Y of the material of the wire = $\mathbf\small{\rm{\frac{\sigma}{\epsilon}=\frac{\frac{mg}{\pi r^2}}{\frac{\Delta L}{L}}}}$
$\mathbf\small{\rm{\Rightarrow Y=\frac{mg\;L}{\pi r^2\;\Delta L}}}$
12. An advantage of using the reference wire can be written in 3 steps:
(i) Even when the room temperature is normal, a wire made of steel may elongate a little
(ii) So the elongation that we measure after applying the weight, will not be the 'true elongation due to tensile stress'
♦ It will include the 'elongation due to temperature increase' also
(iii) But if a reference wire is present, such a problem will not arise because, both wires will be 'expanding to the same amount' due to temperature increase
In the next section, we will see shear modulus
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