Question

2. A metallic cube of each side is 5cm is subjected to a
shearing force of 120 kgf. The top face is displaced
through 0.2 with respect to the buttom face.
Calculate the following.
1.Shearing stress
2.Shearing strain
3shear modulus

Answer 1

Explanation:

Given:-

• Each side of a metallic cube = 5cm.

• Shearing force = 120kgf

• Change in length = 0.2cm

To find:-

• Shearing stress

• Shearing strain

• Shear modulus.

Solution:-

$\sf Shearing \: force = 120kgf$

$\sf As \: we \: know \: that:-$

$\sf \dashrightarrow 1kgf = 10N$

$\sf \dashrightarrow 120kgf = 120 \times 10$

$\sf \dashrightarrow 1200N$

$\sf \therefore\underline {Shearing \: force\: (F) = 1200N}$

$\sf Side\: of \: cube = 5cm = 0.05m$

$\sf Area\: of \: cube (A)= 0.05 \times 0.05\:= 0.0025m^{2}$

$\sf \triangle l = 0.2cm = 0.002m$

$\sf Now,\: Shearing \: stress \: (\tau) = \dfrac{F}{A}$

$\sf \implies \tau= \dfrac{1200}{0.0025}$

$\sf \implies \tau= 480000$

$\sf \implies \tau = 4.8\times 10^{5} N/{m}^{2}$

$\sf \dashrightarrow \underline{\: \: \underline{ \: \pink{ Shearing \: stress = 4.8\times 10^{5}N/{m}^{2}}\: }\: \:}$

$\sf Shearing \: strain \: (\varepsilon) = \dfrac{\triangle l}{l}$

$\sf\implies \varepsilon = \dfrac{0.002}{0.05}$

$\sf \implies\varepsilon = 0.04$

$\sf \dashrightarrow \underline{\: \: \underline{ \: \orange{ Shearing \: strain = 0.04}\: }\: \:}$

$\sf Shearing \: modulus = \dfrac{Shearing \: stress}{Shearing \: strain}$

$\sf = \dfrac{4.8 \times 10^{5}}{0.04}$

$\sf = 1.2\times 10^{7}$

$\sf \dashrightarrow \underline{\: \: \underline{ \: \purple{ Shearing \: modulus = 1.2 \times 10^{7}N/{m}^{2}}\: }\: \:}$