Shear Modulus of Elasticity is one of the important mechanical properties of solids along with Young’s modulus and bulk modulus and it describes how a material responds to shear forces. It is defined as the ratio of shear stress to shear strain in a material. Consider a rectangular block whose lower face is fixed while a tangential force F is applied to its upper surface of area A. An equal and opposite reaction force acts on the lower fixed face forming a couple that produces torque. Due to this torque, the block undergoes shear deformation causing the upper surface to shift by a small distance Δx . As a result the rectangular block changes its shape into a parallelepiped and this change in shape represents the shear strain produced in the material.
Let AB = DC = 1 and ∠ABA' = θ
Tangential stress =\frac {F}{A} Shear Strain = θ = tan θ =AA'/AB
= \frac{Δx}{l} The modulus of rigidity is given by,
η = \frac{Tangential stress}{shear strain}
= \frac{F/A}{\theta}
= \frac{F}{A\theta}
= \frac{F}{A} \cdot \frac{l}{\Delta x}
Units and dimensions of η: The SI unit of modulus of rigidity is Nm-2 and its CGS unit is dyne/cm2. Its dimensional formula is [ML-1T-2].
Bulk modulus (B)
Bulk modulus of elasticity is the ratio of normal stress to the resulting volumetric strain within the elastic limit of a material. It measures a material’s resistance to uniform compression when pressure is applied equally in all directions. A higher bulk modulus indicates that the material is less compressible.
Consider a body of volume V and surface area A .suppose a force F acts uniformly over the whole surface of the body and it decreases the volume by ΔV as shown in the figure. Then bulk modulus of elasticity is given by
B = \frac{normal stress}{volumetric strain}
= -\frac{F/A}{\Delta V / V}
B = -\frac{F}{A} \cdot \frac{V}{\Delta V} where p (=F/A) is the normal pressure. A negative sign shows that the volume decreases with the increase in stress.
Units and dimensions of B: The SI unit of the bulk modulus is N/m2 or Pascal (Pa) and its CGS unit are done/cm2. Its dimensional formula is [ML-1 T-2]
Compressibility
The reciprocal of the bulk modulus of a material is called compressibility. It represents how much a material’s volume changes under applied pressure.
The SI unit of compressibility is Pa⁻¹ (or N⁻¹ m²), and the CGS unit is dyne⁻¹ cm².
Sample Problems
Question 1: A uniform wire of Steel of length 2.5 m and density 8.0 g/cm-3 weighs 50 g. When stretched by a force of 10 kg, the length increases by 2 mm. Calculate Young's modulus of steel.
Solution: Given l=2.5 m =250 cm
Δl =2 mm =0.2 cm
F=10 kgf =10 ×9.8 N =10 ×9.8 ×105 dyne
Mass =volume × density
A =Mass/l ×ρ =50/250 ×8 =0.025 cm2
Young's modulus =F/A . l/Δl
=10 ×9.8 ×105 ×250/0.025 ×0.2
=4.9 ×1011 dyne cm -2
Question 2: What is the percentage increase in the length of a wire of diameter 2.5 mm stretched by a force of 100 kg wt? Young's modulus of elasticity of the wire is 12.5 ×10^11 dyne cm -2
Solution : Given
r = 1.25 mm = 0.125 cm
F =100 ×9.8 = 980 N=98×10^6 dyne
Y = 12.5 ×10^11dyne cm ^- 2
Y = F/A.l/Δl or Δl/l =F/AY =F/πr2Y
The percentage increase in length is
Δl/l ×100 =F×100/πr2 Y
= 98× 106× 7 ×100 /22 ×(0.125)2 ×12.5×1011
= 1 5.965×10-2
= 0.16%
Question 3: Calculate the increase in pressure required to decrease the volume of a water sample by 0.01% . Given that the bulk modulus of water is 2.2 ×10 9 Pa.
Solution : Here, fractional decrease in volume
ΔV/V = -0.01% = -0.01/100 =-10-4
Bulk modulus of water B=2.2 ×10 9 Pa
Bulk modulus B =P/ΔV/V
P = -B.ΔV/V = -(2.2 ×109). (-10 -4)
=2.2 ×105 Pa
Question 4: Find the change in volume which 1m3 of water will undergo when taken from the surface to the bottom of a lake 100 m deep. Given volume elasticity of water is 22,000 atmosphere.
Solution : Here V =1m3, h=100 m, g=9.8 ms -2
ρ(water) =1000 kg m -3
P = hρh =100 ×1000 ×9.8 =9.8 ×10^5Nm^-2
B=22000 atm =22000 ×1.013×10^5 Nm^-2
=22.286×10^8Nm^-2
ΔV =pV/B =9.8×105 ×1/22.286 ×108
=4.4 ×10^-4 m3
Question 5: An Indian rubber cube of side 7 cm has one side fixed while tangential force equal to the weight of 200 kg is applied to the opposite face. Find the shearing strain produced and the distance through which the strained side moves. modulus of rigidity for rubber is 2 ×107 dyne cm-2
Solution : Here ,l=7 cm ,F =200 kg f=200 ×1000 ×981 dyne
η =2 ×20^7 dyne cm-2
Area of the free face,
A =l2 =7 cm × 7 cm =49 cm2
η =F/Aθ
θ =F/Aη =200 × 1000 × 981 /49 × 2 ×10^7
=0.2 radian
Δl = lθ =7 × 0.2 =1.4 cm
Unsolved Problems
Question 1: A material is subjected to a pressure of 5 × 106 Pa and its volume decreases by 0.2%. Calculate the bulk modulus of the material.
Question 2: A cube of side 0.5 m is subjected to a tangential force of 2000 N on its top surface. If the top surface is displaced by 0.001 m, find the shear modulus of the material.
Question 3: The shear modulus of a material is 4 × 1010 and its Poisson’s ratio is 0.25. Calculate Young’s modulus using E = 2G (1 + ν).
Question 4: A solid has a bulk modulus 1.5×1011 Pa and shear modulus 6×1010 Pa. Determine Young’s modulus using E = 9 KG/3 K + G.
Question 5: A material has Young’s modulus 2×1011 Pa and bulk modulus 1.6 × 1011 Pa. Find the shear modulus of the material.
Question 6: A cylindrical rod of length 2 m and cross sectional area 5 × 10-4 m2 is subjected to a tangential force of 4000 N producing a displacement of 0.002 m. Calculate the modulus of rigidity of the material.