A steel plate of face-area 4 cm^2 and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied
A steel plate of face-area 4 cm2 and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4x1010 N/m2.
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Let the lateral displacement of the upper surface with respect to lower surface =Δx
The thickness of the plate, d = 0.5 cm = 0.005 m
Shearing Strain = Δx/d = Δx/0.005 =200*Δx
Shearing stress = Force /area =10/(4/10000) N/m² =25000 N/m²
Shearing stress/shearing strain = Rigidity modulus
→25000/(200*Δx) =8.4x10¹⁰ N/m²
→Δx =25000/(200*8.4x10¹⁰) m
→Δx ≈ 1.5x10⁻⁹ m
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