Chapter 9.3 - Shear modulus and Bulk modulus

In the previous section, we saw the basic details about Young's modulus. In this section we will see Shear modulus and Bulk modulus

Shear modulus

Basic details about shear modulus can be written in 4 steps:

1. We have already seen the basic details about shear stress and shear strain

(See figs.9.7 and 9.8 of section 9)

• We saw the expression for shearing stress: $\mathbf\small{\rm{Shearing \;stress=\frac{F}{A}}}$

• We saw two expressions for shear strain: $\mathbf\small{\rm{Shearing \;strain =\frac{\Delta x}{L}}}$ OR $\mathbf\small{\rm{Shearing \;strain =\theta}}$

• Thus we can easily find the shearing stress and corresponding shearing strain

2. Now consider the ratio: $\mathbf\small{\rm{\frac{Shearing \; stress}{Shearing \;strain}}}$

■ If the applied shearing stresses are less than or equal to the proportional limit, this ratio will be a constant

■ This constant is called shear modulus of the material

• So we can write: $\mathbf\small{\rm{Shear \;modulus =\frac{Shearing \; stress}{Shearing \;strain}}}$

• The symbol for shear modulus is: G

• Another name for shear modulus is: Modulus of rigidity

3. Now we can write an expression for shear modulus:

$\mathbf\small{\rm{G=\frac{Shearing \; stress}{Shearing \; strain}=\frac{\frac{F}{A}}{\frac{\Delta x}{L}}}}$

• Thus we get: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

• We can use the other expression for strain also:

$\mathbf\small{\rm{G=\frac{Shearing \; stress}{Shearing \; strain}=\frac{\frac{F}{A}}{\theta}}}$

• Thus we get: $\mathbf\small{\rm{G=\frac{F}{A \theta}}}$

4. We know that, strain has no unit

• So the unit of G is Nm^-2

 

We will see some solved examples after completing the discussion on bulk modulus

 

Bulk modulus

1. We have already seen the basic details about hydraulic stress and volume strain

(See figs.9.9 and 9.10 of section 9)

• We saw that, hydraulic stress is the force per unit area

• 'unit area' in this case is:

One unit of the total surface area of the body

• So 'force per unit area' can be replaced by pressure (p)

• We saw the expression for volume strain: $\mathbf\small{\rm{Volume \;strain =\frac{\Delta V}{V}}}$

• Thus we can easily find the hydraulic stress and corresponding volume strain

2. Now consider the ratio: $\mathbf\small{\rm{\frac{Hydraulic \; stress}{Volume \;strain}}}$

■ If the applied hydraulic stresses are less than or equal to the proportional limit, this ratio will be a constant

■ This constant is called bulk modulus of the material

• So we can write: $\mathbf\small{\rm{Bulk \;modulus =\frac{hydraulic \; stress}{Volume \;strain}}}$

• The symbol for bulk modulus is: B

3. Now we can write an expression for bulk modulus:

$\mathbf\small{\rm{B=\frac{Hydraulic \; stress}{Shearing \; strain}=\frac{p}{\frac{-\Delta V}{V}}}}$

• Thus we get: $\mathbf\small{\rm{B=-\frac{p \times V}{\Delta V}}}$

■ The -ve sign is provided because, when pressure increases, volume decreases. This can be elaborated in 3 steps:

(i) When a pressure p is applied, the volume decreases from V₁ to V₂

(ii) V₂ will be smaller than V₁

(iii) So the change in volume (ΔV = V₂-V₁) will be negative

4. We know that, strain has no unit

• So the unit of B is Nm^-2

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Compressibility

Basic details of compressibility can be understood using an example. It can be written in 7 steps: 

1. Let us take two spheres of the same volume V

    ♦ One is made of iron and the other is made of glass

2. We want to decrease the volume of the steel sphere by ΔV

• We want the same decrease (ΔV) to happen to the glass sphere also

3. Let us calculate the required pressures:

    ♦ Let P_I be the pressure required to reduce the volume of the iron sphere by ΔV

    ♦ Let B_I be the bulk modulus of iron

• Then we can write: $\mathbf\small{\rm{B_I=-\frac{p_I \times V}{\Delta V}}}$

    ♦ Let P_G be the pressure required to reduce the volume of the glass sphere by ΔV

    ♦ Let B_G be the bulk modulus of glass

• Then we can write: $\mathbf\small{\rm{B_G=-\frac{p_G \times V}{\Delta V}}}$

4. Taking ratios, we get: $\mathbf\small{\rm{\frac{B_I}{B_G}=-\frac{p_I \times V}{\Delta V}\times -\frac{\Delta V}{p_G \times V}=\frac{P_I}{P_G}}}$

• From the data book, we have the B of iron and glass:

    ♦ B_I is 100 × 10⁹ N m^-2

    ♦ B_G is 37 × 10⁹ N m^-2

• Substituting the values of B, we get:

$\mathbf\small{\rm{\frac{P_I}{P_G}=\frac{B_I}{B_G}=\frac{100 \times 10^9}{37 \times 10^9}=2.7}}$

    ♦ That means, the pressure required in the case of iron is greater

    ♦ To be precise, the pressure required by iron is 2.7 times that required by glass

5. We can think about the 'reverse of the result in (4)':

• Pressure required by glass is lesser than that required by iron

    ♦ That means, it is easier to compress glass than steel

    ♦ That means, glass has greater compressibility

6. Compressibility is the reciprocal of bulk modulus

• The symbol of compressibility is k

• So we can write: $\mathbf\small{\rm{k=\frac{1}{B}}}$

• But $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

■ So we get: $\mathbf\small{\rm{k=\frac{1}{B}=-\frac{\Delta V/V}{p}}}$

7. A comparison can be written as follows:

    ♦ If a material has a high value for B, that material is difficult to compress

    ♦ If a material has high value for k, that material is easy to compress

• Solids have high values for B

    ♦ So they cannot be compressed easily

• Gases have high values of k

    ♦ So they can be compressed easily

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Now we will see some solved examples on shear and bulk moduli:

 

Solved example 9.17

A square lead slab of side 50 cm and thickness 10 cm is subject to a shearing force (on its narrow face) of 9.0 × 10⁴ N. The lower edge is riveted to the floor. How much will the upper edge be displaced?

Solution:

• Fig.9.20(a) below shows a 3D view

    ♦ The bottom face is riveted to the platform

    ♦ It is clear that, the shearing force F will be acting on an area of (0.5× 0.1) = 0.05 m²

• The deformed shape is shown in fig.9.20(b)

Fig.9.20

• Fig.9.21 below shows the 2D view

    ♦ The displacement Δx can be clearly seen in fig.9.21(b)

Fig.9.21

Data given is:

• F = 9.0 × 10⁴ N

• L = 50 cm = 0.50 m

• A = (0.5 × 0.1) m² = 0.05 m²

• G = 5.6 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

$\mathbf\small{\rm{\Rightarrow \Delta x=\frac{F \times L}{A \times G}}}$

2. Substituting the known values, we get:

Δx = 1.6 × 10^-4 m

 

Solved example 9.18

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Solution:

• Fig.9.22(a) below shows a 3D view

    ♦ The aluminium cube is firmly fixed on the vertical side of a wall

Fig.9.22

• In fig.9.22(b), a plate (shown in red color) is glued firmly on to the free face of the cube

    ♦ A wt of 100 kg (shown in cyan color) is attached to the plate through a rod. The rod is shown in yellow color

    ♦ Due to the 100 kg wt, the free face will deflect in the vertical direction

• Fig.9.23 below shows the 2D view
    ♦ The vertical deflection Δx can be clearly seen in fig.9.23(b)

Fig.9.23

Data given is:

• F = 100g N

• L = 10 cm = 0.10 m

• A = (0.1 × 0.1) m² =  0.01 m²

• G = 25 × 10⁹ N m^-2 

• g = 10 ms^-2 

1. We have: $\mathbf\small{\rm{G=\frac{F \times L}{A \times \Delta x}}}$

$\mathbf\small{\rm{\Rightarrow \Delta x=\frac{F \times L}{A \times G}}}$

2. Substituting the known values, we get:

Δx = 4 × 10^-7 m

 

Solved example 9.19

The average depth of Indian Ocean is about 3000 m. Calculate the fractional compression, ΔV/V, of water at the bottom of the ocean, given that the bulk modulus of water is 2.2 × 10⁹ N m^-2. (Take g = 10 m s^-2 )

Solution:

Data given is:

• Depth h = 3000 m

    ♦ So pressure p = 𝝆gh = 1000 × 10 × 3000 =  3 × 10⁷ N

    ♦ 𝝆 is the density of water in kg m^-3

• B = 2.2 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the known values, we get:

ΔV/V = 1.36 × 10^-2

 

Solved example 9.20

Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 10⁵ Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Solution:

Data given is:

• Initial volume V₁ = 100 litre

• Final volume V₂ = 100.5 litre

    ♦ So ΔV = V₂-V₁ = 0.5 litre

          ✰ 1 litre = 10^-3 m3

          ✰ So V₁ = 100 litre = (100 × 10^-3) = 0.1 m³

          ✰ So ΔV = 0.5 litre = (0.5 × 10^-3) = 5 × 10^-4 m³

• Pressure increase = Δp = 100 atm = 1.013 × 10⁷ Pa = 1.013 × 10⁷ N m^-2

1. We have: $\mathbf\small{\rm{B=-\frac{\Delta p \times V_1}{\Delta V}}}$

2. Substituting the values, we get: B of water = 2.026 × 10⁹ N m^-2 

3. We have: B of air = 1 × 10⁵ N m^-2 

• Taking ratios, we get: $\mathbf\small{\rm{\frac{B\;of\;water}{B\;of\;air}}}$ = 2.026 × 10⁴

    ♦ That means, 'B of water' is very large when compared to 'B of air'

    ♦ 'B of water is 2.026 × 10⁴ times the 'B of air'

• This is because, when compared to air, it is very difficult to compress water

    ♦ When compared to air, the ‘space available between the molecules’ is very small in water

 

Solved example 9.21

What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 10³ kg m^-3  ?

Solution:

Data given is:

• Initial pressure p₁ = 1 atm (∵ the pressure at the surface is 1 atm)    

• Final pressure p₂ = 80.0 atm

    ♦ So Δp = p₂-p₁ = 79 atm = 79 × 1.013 × 10⁷ N m^-2

• Density at the surface = 1.03 × 10³ kg m^-3

• B of water = 2.026 × 10⁹ N m^-2 

 1. Given that, density at the surface = 1.03 × 10³ kg m^-3

    ♦ So 1 m³ of water at the surface will have a mass of 1.03 × 10³ kg

    ♦ So V₁ = 1 m³

2. We take this 'same mass' and 'same volume' to a depth, where the pressure is 80 atm

    ♦ Due to the high pressure at that depth, the volume will be reduced to a new value (V₂) which will be less than 1 m³

    ♦ But mass will remain the same

■ We have to find V₂

3. We have: $\mathbf\small{\rm{B=-\frac{\Delta p \times V_1}{\Delta V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V=-\frac{\Delta p \times V_1}{B}}}$

4. Substituting the values, we get: ΔV = -0.00395 m³

• So (V₂ - V₁) = -0.00395

⇒ (V₂ - 1) = -0.00395

⇒ V₂ = (1 - 0.00395) = 0.99605 m³

5. So new density = $\mathbf\small{\rm{\frac{Mass}{New\;volume}=\frac{1.03 \times 10^3}{0.99605}}}$ = 1.034 × 10³ kg m^-3 

 

Solved example 9.22

Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Solution:

Data given is:

• Pressure p = 10 atm = 10 × 1.013 × 10⁵ N m^-2

• B of glass = 37 × 10⁹ N m^-2 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 2.74 × 10^-5

 

Solved example 9.23

Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 × 10⁶ Pa.

Solution:

Data given is:

• Pressure p = 7 × 10⁶ N m^-2

• B of copper = 140 × 10⁹ N m^-2 

• Initial volume V = (0.1)³ = 0.001 m³ 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 5.0 × 10^-5

So ΔV = V × 5.0 × 10^-5 = 5.0 × 10^-8 m³

 

Solved example 9.24

How much should the pressure on a litre of water be changed to compress it by 0.10%?

Solution: 

Data given is:

• B of water = 2.2 × 10⁹ N m^-2 

• ΔV/V = 0.10 % =$\mathbf\small{\rm{\frac{0.1}{100}}}$ =0.001

1. We have: $\mathbf\small{\rm{B=-\frac{\Delta p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta p=B \times \Delta V/V}}$

2. Substituting the values, we get: Δp = 2.2 × 10⁶ N m^-2

 

Solved example 9.25

Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 10⁷ Pa? Assume that each rivet is to carry one quarter of the load

Solution:

• Fig.9.24 below shows the two strips

    ♦ One is shown in cyan color. The other is shown in blue color

Fig.9.24

• The four hemispherical parts are the rivets

    ♦ There are four more hemispheres at the bottom. But they are not visible in this view

• The strips are pulled by F on either ends

    ♦ This will produce shearing stresses in the four rivets 

Data given is:

• Diameter of each rivet = 6 mm

    ♦ So cross sectional area of each rivet = $\mathbf\small{\rm{\frac{\pi \times 6^2}{4}}}$ = 28.26 mm² = 28.26 × 10^-6 m²  

• Maximum shearing stress allowable = 6.9 × 10⁷ N m^-2 

1. We have: Shearing stress = $\mathbf\small{\rm{Shearing \; stress=\frac{Force}{Area}}}$

• So maximum force that a rivet can take = (Allowable Shearing stress × Area) 

= (6.9 × 10⁷ × 28.26 × 10^-6) = 1949.94 N

2. So the four rivets together can take (4 × 1949.94) = 7800 N

3. That means, the maximum force (F) with which, we can pull from one end of the riveted strips is 7800 N

Solved example 9.26

The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Solution:

Data given is:

• Pressure p = 1.1 × 10⁸ N m^-2

• B of steel = 160 × 10⁹ N m^-2 

• Initial volume V = 0.32 m³ 

1. We have: $\mathbf\small{\rm{B=-\frac{p}{\Delta V/V}}}$

$\mathbf\small{\rm{\Rightarrow \Delta V/V=-\frac{p}{B}}}$

2. Substituting the values, we get: ΔV/V = 6.875 × 10^-4

3. So ΔV = V × 6.875 × 10^-4 = 2.148 × 10^-3 m³

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In the next chapter, we will see mechanical properties of fluids

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Chapter 9.2 - Young's Modulus

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

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Young's modulus

• 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

Fig.9.17

• The cylinder is resting on a rigid platform
• We see that, due to the compressive force F, the length of the cylinder decreases by ΔL_c
    ♦ So we can write: Change in length due to compression = ΔL_c
    ♦ Then, strain (𝜺_c) 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 ΔL_t
[This 'F' is same in magnitude as the 'F' in (1)]
    ♦ So we can write: Change in length due to tension = ΔL_t
    ♦ Then, strain (𝜺_t) 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 (𝜎_c) applied in (1) = Tensile stress (𝜎_t) 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, ΔL_c will be equal to ΔL_t 
• 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 (𝜺_c) = Tensile strain (𝜺_t)
• Also, from (3), we have: Compressive stress (𝜎_c) = Tensile stress (𝜎_t)
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

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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:
    ♦ Y_Aluminium = 70 ×10⁹ N m^-2
    ♦ Y_Copper = 120 ×10⁹ N m^-2
    ♦ Y_Steel = 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 = (Y_Aluminium × 𝜺_Aluminium) = (Y_Copper × 𝜺_Copper) = (Y_Steel × 𝜺_Steel)
• Comparing the values in (4), we can write:
    ♦ 𝜺_Steel will be the smallest
    ♦ 𝜺_Aluminium 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

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Let us now write a useful information that can be obtained from graphs. It can be written in 7 steps:

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:

Fig.9.18

• 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

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Let us see some solved examples:

Solved examples 9.5 to 9.10

Solved examples 9.11 to 9.16

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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 

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In the next section, we will see shear modulus

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