A cube has side length 5 cm and modulus of rigidity 105 N/m2. The displacement produced by a force of 10 N in the upper face of cube is mm.
Answer
2
Solution
Area of the face: A=L2=(0.05)2=2.5 10-3 m2
Shear stress = FA= 102.5 10-3=4 103 N/m2
Shear strain = stress = 4 103105=0.04
∵ Shear strain = xL ⇒ x=0.04 5 cm=0.2 cm
∴ x=2 mm
Answer: 2