Question

The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
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Answer 1

Explanation:

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 × 109 Pa

Shear modulus, η = Shear stress / Shear strain = (F/A) / (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 m2

ΔL = Vertical deflection of the cube

∴ ΔL = FL / Aη

= 980 × 0.1 / [ 10-2 × (25 × 109) ]

= 3.92 × 10–7 m

The vertical deflection of this face of the cube is 3.92 ×10–7 m.

Answer 2

$\huge{\red{\underline{\underline{\sf{......Answer......}}}}}$

$\star{\blue{\underline{\sf{Given:-}}}}$

$\tt{\implies Length\;(l)=10\;cm=0.10\;m}$

$\tt{\implies Mass\;(m)=100\;kg}$

$\tt{\implies Shear\;modulus\;(\eta)=25\;GPa=25\times 10^{9}\;Pa}$

$\tt{\implies Acceleration\;due\;to\;gravity\;(g)=10\;m/s^{2}}$

$\star{\blue{\underline{\sf{To\;Find:-}}}}$

ToFind:−

⇒ Vertical Deflection.

$\star{\blue{\underline{\sf{Formula\;used:-}}}}$

$\star{\boxed{\blue{\tt{\eta = \dfrac{Fl}{A.\theta}}}}}$

$\textsf{Now, first we have to find area,}$

$\tt{\implies Area=(length)^{2}}$

$\tt{\implies Area=0.01\;m^{2}}$

$\textsf{Now, put the values in the formula, we get,}$

$\tt{\implies \eta=\dfrac{Fl}{A.\theta}}$

$\tt{\implies \theta=\dfrac{mg.l}{l^{2}\times \eta}}$

$\tt{\implies \dfrac{100\times 10\times 0.1}{0.01\times 25\times 10^{9}}}$

$\tt{\implies \dfrac{100}{0.25\times 10^{9}}}$

$\tt{\implies \dfrac{1\times 10^{2}}{0.25\times 10^{9}}}$

$\tt{\implies 4\times 10^{-7}\;m}$

Hence, vertical deflection = 4 × 10⁻⁷

$\huge{\underline{\underline{\sf{Extra\;Information:}}}}$

Shear modulus of elasticity (η): In an elastic limit the ratio of tangential stress to tangential strain is called shear modulus of elasticity.

$\sf{\eta=\dfrac{Tangential\;stress}{Tangential\;strain}}$

$\sf{\eta=\dfrac{F}{A\times \theta}}$