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 × 10 9 Pa
Shear modulus, η =\( \frac{\text{Shear stress}}{\text{ Shear strain }}= \frac{\frac{F}{A}}{\frac{L}{ ΔL}}\)
Where, 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 m 2
ΔL = Vertical deflection of the cube
\(∴ ΔL = \frac{FL }{ Aη} \)
=\( \frac{980 × 0.1 }{ 10 ^{- 2} × (25 × 10 ^9) }\)
= 3.92 × 10 - 7 m
The vertical deflection of this face of the cube is 3.92 ×10 - 7 m.
The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5×10–4 is ____ kJ/m3. Assume that material is elastic up to the linear strain of 5×10–4
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?