Question

Question 19
A rectangle block of aluminium is 45 cm by 45 cm by 15 cm when unstressed. A force of 1273 Nis applied
tangentially to the upper surface causing a 2.7 cm displacement relative to the lower surface. The block is
placed such that 45x45 comes on the lower and upper surface.
Calculate the shear modulus​

Answer 1

Answer:

1482................

Answer 2

Given :

l = 45cm

b = 45cm

h = 15cm

F = 1273N

x = 2.7cm

To find :

Shear Modulus (G)

Solution :

$Shearing Stress = \frac{F}{A}$

$Shearing Strain = \frac{x}{h}$

From Hooke's Law

Stress ∝ Strain

Shearing Stress ∝ Shearing Strain

Shearing Stress = G * Shearing Strain

$G = \frac{Stress}{Strain}$

$G = \frac{\frac{F}{A} }{\frac{x}{h} }$

$G = \frac{F}{A} *\frac{h}{x}$

$G = \frac{1273}{45 * 45} *\frac{15}{2.7} *10^{4}$

$G = 3.49 * 10^{4} N/m^{2}$

Therefore, Shear Modulus (G) is 3.49*10^4 N/m^2.