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
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 PI be the pressure required to reduce the volume of the iron sphere by ΔV
♦ Let BI be the bulk modulus of iron
• Then we can write: $\mathbf\small{\rm{B_I=-\frac{p_I \times V}{\Delta V}}}$
♦ Let PG be the pressure required to reduce the volume of the glass sphere by ΔV
♦ Let BG 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:
♦ BI is 100 × 109 N m-2
♦ BG 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)
♦ 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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