Question

tangential force of 2.5×10^5 N applied on a surface of area 4cm^2 displaces it sideways with respect to the opposite fixed layer so that the angle of shear is 2° . Find the modulus of rigidity of the material

Answer only if you know it​

Answer 1

Answer:

17.857 Gpa or 17857.14 Mpa

Explanation:

shear stress = 250000/400 (N/ mm²) = 625 Mpa

shear strain = 2° = 2*π/180 radian = .035 rad

G or modulus of rigidity = 625/.035 = 17.857 Gpa

Answer 2

The modulus of rigidity of the material is 3.125×10¹⁰.

Given:-

Tangential force = 2.5 × 10⁵N

Area of surface = 4cm²

Angle of shear = 2°

To Find:-

The modulus of rigidity of the material.

Solution:-

We can easily find the value of modulus of rigidity of the material by using these simple steps.

As

Tangential force (f) = 2.5 × 10⁵N

Area of surface (A) = 4cm²

Angle of shear (a) = 2°

Modulus of rigidity (¥) =?

According to the formula of modulus of rigidity of the material.

$¥ = \frac{f}{A \times a}$

on putting the values, we get

$¥ = \frac{2.5 \times {10}^{5} }{4 \times {10}^{ - 4} \times 2 }$

$¥ = \frac{2.5 \times {10}^{5} }{8 \times {10}^{ - 4} }$

$¥ = 0.3125 \times {10}^{(5 + 4)}$

$¥ =0.3125 \times {10}^{9}$

$¥ =3.125 \times {10}^{10}$

Hence, The modulus of rigidity of the material is 3.125×10¹⁰.

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