|Class 11 Physics Mechanical Properties of Solids||Shear Modulus|
Shear Modulus (Modulus of Rigidity)
Relation between Young’s Modulus and Shear Modulus
Problem:- A box shaped piece of wax has a top area of 10cm2 and height of 2cm.When a shearing force of 0.5N is applied to the upper surface, displaces 4mm relative to the bottom surface. What are the shearing stress, shearing strain and shear modulus for wax?
Answer:- Area = 10cm2, height =2cm, (displacement) x=4mm
=0.5/10x10-4 = 500Pa
= (4x10-3)/ (2x10-2) = 0.2
=500/0.2 = 2500 Pa.
Problem:- 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 are displaced?
Answer: - The lead slab is fixed and the force is applied parallel to the narrow face as shown in Fig. The area of the face parallel to which this force is applied is
A = 50 cm × 10 cm
= 0.5 m × 0.1 m
= 0.05 m2
Therefore, the stress applied is
= (9.4 × 104 N/0.05 m2)
= 1.80 × 106Nm–2
We know that shearing strain = (Δx/L) = Stress /G. Therefore the displacement Δx = (Stress × L)/G = (1.8 × 106 N m–2 × 0.5m)/ (5.6 × 109 N m–2)
= 1.6 × 10–4 m = 0.16 mm
Problem:- 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?
Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η = Shear stress/Shear Strain = (F/A)/L/ ΔL
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
ΔL = FL/A η = (980x0.1)/10-2x (25x109)
= 3.92 × 10–7 m
The vertical deflection of this face of the cube is 3.92 ×10–7 m.