Strength of Materials Test 2

Result

Strength of Materials Test 2

Score
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Rank
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TIME Taken - -

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Question 1
1 / -0

A bumper consists of two helical springs of circular section brings to rest a weight of 2000 kg and moving at 2 m/sec. While doing so, the springs are compressed by 150 mm. Then, the maximum force in N on each spring (assuming gradually increasing load) is

A
26665-26668

B
266654-266681

C
266657-266686

D
266659-266682

Solution

Concept:
The kinetic energy of the moving mass is converted into the potential energy of the spring.
Kinetic energy = \(\frac{1}{2}m{v^2}\;\)
Potential energy of spring = \(\frac{1}{2}k{x^2}\)
Stiffness: The force required to produce unit deflection in the spring is called as stiffness of the spring.
Calculation:
Deflection (x) = 150 mm = 0.15 m
\(\begin{array}{l} \frac{1}{2}k{x^2} = \frac{1}{2}m{v^2}\\ \Rightarrow k = m{\left( {\frac{v}{x}} \right)^2} = 2000 × {\left( {\frac{2}{{15}}} \right)^2} = 355.556\ kN/m \end{array}\)
Stiffness of one spring (k) \(= \frac{{355.556}}{2} = 177.78\ kN/m\)
Maximum force = k × x \(= 177.78 × {10^3} × 0.15 = 26666.67\ N\)
Question 2
1 / -0

If a shaft is simultaneously subjected to torque T/2 and a bending moment of 2M, then the ratio of maximum bending stress to maximum shear stress will be______

A
4M/T

B
8M/T

C
4T/M

D
8T/M

Solution

Concept:
Maximum bending across shaft occurs at either top or bottom fiber and it is given by the bending formula:
\({{\rm{\sigma }}_{{\rm{max}}}} = \frac{{\rm{M}}}{{\rm{Z}}} = \frac{{2{\rm{M}}}}{{\frac{{\rm{\pi }}}{{64}}{{\rm{d}}^4}}} \times \frac{{\rm{d}}}{2} = \frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)
Maximum shear stress is given by the twisting formula:
\({{\rm{\tau }}_{{\rm{max}}}} = \frac{{{\rm{Tr}}}}{{\rm{I}}} = \frac{{\left( {\frac{1}{2}} \right) \times \left( {\frac{{\rm{d}}}{2}} \right)}}{{\frac{{\rm{\pi }}}{{32}}{{\rm{d}}^4}}} = \frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}\)
\(\frac{{{{\rm{\sigma }}_{{\rm{\;max}}}}}}{{{{\rm{\tau }}_{{\rm{max}}}}}} = \frac{{\frac{{64{\rm{M}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}}{{\frac{{8{\rm{T}}}}{{{\rm{\pi }}{{\rm{d}}^3}}}}} = \frac{{8{\rm{M}}}}{{\rm{T}}}\)