The strain energy stored in a hollow circular shaft of outer diameter (D) and inner diameter (d) subjected to shear stress is

A $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
B $$\frac{{{\tau ^2}}}{{2{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
C $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$
D $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$

Correct Answer: $$\frac{{{\tau ^2}}}{{4{\text{C}}}}\left( {\frac{{{{\text{D}}^2} + {{\text{d}}^2}}}{{\text{D}}}} \right) \times {\text{Volume of shaft}}$$

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#Mechanical Engineering #Strength Of Materials In ME

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The strain energy stored in a body due to shear stress, is (where $$\tau $$ = Shear stress, C = Shear modulus and V = Volume of the body)
A $$\frac{{\tau {\text{V}}}}{{2{\text{C}}}}$$
B $$\frac{{2{\text{C}}}}{{\tau {\text{V}}}}$$
C $$\frac{{{\tau ^2}{\text{V}}}}{{2{\text{C}}}}$$
D $$\frac{{2{\text{C}}}}{{{\tau ^2}{\text{V}}}}$$
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D $$\frac{\pi }{{64}}\left( {{{\text{D}}^4} - {{\text{d}}^4}} \right)$$
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A $$\frac{\pi }{4} \times \tau \left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right)$$
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D $$\frac{\pi }{{32}} \times \tau \left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
The polar modulus for a hollow shaft of outer diameter (D) and inner diameter (d) is
A $$\frac{\pi }{4}\left( {\frac{{{{\text{D}}^2} - {{\text{d}}^2}}}{{\text{D}}}} \right)$$
B $$\frac{\pi }{{16}}\left( {\frac{{{{\text{D}}^3} - {{\text{d}}^3}}}{{\text{D}}}} \right)$$
C $$\frac{\pi }{{16}}\left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
D $$\frac{\pi }{{32}}\left( {\frac{{{{\text{D}}^4} - {{\text{d}}^4}}}{{\text{D}}}} \right)$$
A shaft is subjected to bending moment M and a torque T simultaneously. The ratio of the maximum bending stress to maximum shear stress developed in the shaft, is
A $$\frac{{\text{M}}}{{\text{T}}}$$
B $$\frac{{\text{T}}}{{\text{M}}}$$
C $$\frac{{2{\text{M}}}}{{\text{T}}}$$
D $$\frac{{2{\text{T}}}}{{\text{M}}}$$
When a body is subjected to biaxial stress i.e. direct stresses ($${\sigma _{\text{x}}}$$) and ($${\sigma _{\text{y}}}$$) in two mutually perpendicular planes accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), then maximum shear stress is
A $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{1}{2}\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + 4\tau _{{\text{xy}}}^2} $$
C $$\sqrt {{{\left( {{\sigma _{\text{x}}} - {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
D $$\sqrt {{{\left( {{\sigma _{\text{x}}} + {\sigma _{\text{y}}}} \right)}^2} + \tau _{{\text{xy}}}^2} $$
When a body is subjected to a direct tensile stress ($${\sigma _{\text{x}}}$$) in one plane accompanied by a simple shear stress ($${\tau _{{\text{xy}}}}$$ ), the maximum shear stress is
A $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
B $$\frac{{{\sigma _{\text{x}}}}}{2} - \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
C $$\frac{{{\sigma _{\text{x}}}}}{2} + \frac{1}{2}\sqrt {\sigma _{\text{x}}^2 - 4\tau _{{\text{xy}}}^2} $$
D $$\frac{1}{2}\sqrt {\sigma _{\text{x}}^2 + 4\tau _{{\text{xy}}}^2} $$
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B T
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C $$\frac{{{\text{8F}}}}{{\pi {{\text{d}}^2}}}$$
D $$\frac{{{\text{16F}}}}{{3\pi {{\text{d}}^2}}}$$
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A Minimum at the center to maximum at the circumference
B Maximum at the center to minimum at the circumference
C Zero at the center to maximum at the circumference
D Maximum at the center to zero at the circumference

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