A steel rod of 50 mm diameter is 4 m long. If E = 2 × 10 6 kg/cm
A. Impact factor is 2
B. max. stress is 1.018 tonne/cm<sup style="">2</sup>
C. Elongation is 0.203 cm
D. Height through which the weight falls is 0 cm
Please scroll down to see the correct answer and solution guide.
Right Answer is:
SOLUTION
Concept:
\({\rm{Impact\;factor\;}}\left( {IF} \right) = L \times \sqrt {L + \frac{{24}}{{{\delta _{st}}}}} \)
For suddenly applied load, h = 0
\(Stress = \frac{{Load}}{{Area}} \times IF\)
\(Elongation = \frac{{Stress}}{E} \times L\)
Calculation:
Given:
Diameter = 50 mm, length = 4 m, E = 2 × 106 kg/cm2, Load = 10 tonnes
\(Area = \frac{\pi }{4}{d^2} = \frac{\pi }{4} \times 25 = \frac{{25\pi }}{4}\;c{m^2}\)
Now,
Stress \(= 2 \times \frac{P}{A}\)
\({\rm{Stress}} = \frac{{2 \times 10}}{{25\pi }} \times 4 = 1.018\;tonne/c{m^2}\)
Now,
\({\rm{Elongation}} = Stress \times \frac{L}{E}\)
\({\rm{Elongation}} = 1.018 \times \frac{{400}}{{2 \times {{10}^3}}}\)
∴ Elongation = 0.2037 cm