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

---

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

---

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

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Chapter 8.6 - Cavendish's Experiment

In the previous section, we saw the forces exerted by a spherical shell on a point mass inside or outside it. In the discussions so far in this chapter, we came across the universal gravitational constant (G) many times. We will now see how scientists determined the actual value of G

In 1798, the British scientist Henry Cavendish set up an apparatus called the torsion balance. His aim was to determine the density of the Earth. Though he succeeded in finding the density, he did not find it necessary to determine the value of G. However, the same setup was used by later scientists to calculate G accurately. We will write the procedure in steps:

1. Fig.8.22 below shows the schematic arrangement of the setup

Fig.8.22

• There are four balls
    ♦ The two small green balls are identical. They have the same diameter
    ♦ The two large magenta balls are identical. They have the same diameter
• All the four balls are made of lead
2. The green balls are attached to the ends of a rod
• The rod is suspended using a thin wire. This thin wire is shown in blue color
    ♦ The top end of this wire is attached to a rigid support
    ♦ The bottom end of this wire is attached at point O
3. The following 3 conditions should be satisfied:
(i) Consider an imaginary line joining the centers of the green balls
• Let the length of this line be l
• Point O must be the exact midpoint of this imaginary line
(ii) Consider an imaginary line joining the centers of the magenta balls
• Length of this imaginary line should be the same l mentioned in (i)
• Point O must be the exact midpoint of this imaginary line also
(iii) The green and magenta balls should be placed on alternate sides. This can be explained using the fig.8.22
• On the left side, the green ball is placed at the rear of the magenta ball
• On the right side, the green ball is placed at the front of the magenta ball
4. When the conditions 3(i) and 3(ii) are satisfied, the centers of all the four spheres will lie on an imaginary circle with center at O
• This imaginary circle is indicated by the dashed yellow curve
5. The setup is complete. Now we can begin the experiment
• Let r be the initial distance between any one magenta ball and it's corresponding green ball. This is shown in fig.8.22
• Due to the gravitational force of attraction, the green balls will move towards the magenta balls
6. The magnitude of the force acting on any one green ball is given by:
$\mathbf\small{|\vec{F}|=\frac{GMm}{r^2}}$
• Where:
    ♦ M is the mass of the larger magenta ball
    ♦ m is the mass of the smaller green ball
• This same magnitude acts on both the green balls. But in opposite directions
7. Since the forces are equal in magnitude but opposite in directions, a torque is created
• Magnitude of this torque = Magnitude of any one force × Perpendicular Distance between the forces
= $\mathbf\small{|\vec{F}|l=\frac{GMml}{r^2}}$
8. The green balls will not touch the magenta balls
• But that is unexpected. The torque must rotate the two green balls and bring them into contact with with the magenta balls
• So what is stopping the green balls ?
The answer can be written in 8 steps:
(i) When the green balls rotate, the rod attached to them also rotates
(ii) When the rod rotates, the blue wire also rotates
(iii) But the blue wire is fixed firmly at it's top end
So the wire is not 'free to rotate'
(iv) The bottom portions of the wire rotate. But the top portions resist rotation
• As a result, the wire is twisted
(v) When the wire is twisted, it tries to resist 'being twisted'
• This is called torsional resistance
• In our present case, 'torsional resistance' means, the torque which is resisting the rotation of the green balls   
• This torsional resistance is indicated by the white curved arrow in fig.8.23 below:

Fig.8.23

(vi) In the fig.8.23, note the two directions:
    ♦ Direction of rotation of the green balls
    ♦ Direction of the white curved arrow
• The above two directions are opposite to each other. This is indeed so because, the torsional resistance will be opposing the rotation of the green balls   
(vii) We calculated the torque created by $\mathbf\small{|\vec{F}|}$ in (7)
• At the initial stages, this torque is able to over come the torsional resistance offered by the wire
• So the green balls move towards the magenta balls
(viii) But as the rotation continues, the torsional resistance increases
• It increases to such a level that, it becomes equal to the value calculated in (7)
• At that stage, the rotation stops
• So at that stage, we can write:
Torsional resistance offered by the wire = $\mathbf\small{\frac{GMml}{r^2}}$
9. Consider the equation written in 8(viii) above
• If we can find the 'torsional resistance offered by the wire' at the final stage, we will be able to calculate G
• This is because, all others are known quantities
• So our next aim is to find this 'torsional resistance'
• For that, we have to perform a separate experiment. The following steps from (10) to (14) give a basic idea about that experiment
10. In fig.8.24 below, a blue cylinder is fixed at one of it's ends. The other end is free

Fig.2.24

11. In the fig.8.25 below, a torsion is applied at the free end
• This torsion is indicated by the cyan curved arrow

Fig.8.25

• Consider any fibre on the surface, say PQ 
• When torsion is applied, the end Q will move to a new position Q'
• But since the other end of the cylinder is fixed, P remains at the same position
12. So we get an angle Q'PQ
• This angle is called the angle of twist (θ)
• It is measured in radians
    ♦ If the cylinder is strong, we will get only a small θ even if we apply a large torsion
    ♦ If the cylinder is weak, we will get a large θ even if we apply a small torsion
• Thus, θ is a property of the object
13. A number of trials are done on the cylinder
• In each trial, a known torsion is applied and the corresponding θ is noted
■ From that data, we get an important information:
The exact torsion ($\mathbf\small{\tau}$) required to obtain a θ of 1 radian
■ Each object has it's own unique value of $\mathbf\small{\tau}$  
14. In our present case, the 'object of interest' is the blue wire used for suspending the green balls
• So the experiment is performed on the blue wire and it's $\mathbf\small{\tau}$ is determined
15. Now we get back to our main experiment
• We have the initial and final positions of any one of the two green balls
• From those positions, we can determine the angle (θ') through which the blue wire is twisted
• Also we have $\mathbf\small{\tau}$, which is the unique property of the wire
16. So, if we multiply θ' by $\mathbf\small{\tau}$, we will get the exact torsional resistance
• That means: Torsional resistance = $\mathbf\small{\tau \theta'}$
17. So the equation in 8(v) becomes: $\mathbf\small{\tau \theta'=\frac{GMml}{r^2}}$
• In this equation, G is the only unknown. So it can be easily calculated

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• So we have seen how scientists determined the value of G
• In the next section, we will see the acceleration due to gravity of the Earth

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