A round steel bar of overall length 40 cm consists of two equal p
A. <span class="math-tex">\(\frac{3}{{10\pi }}\left[ {\frac{1}{{20}} + \frac{1}{{15}}} \right]\)</span>
B. <span class="math-tex">\(\frac{8}{{10\pi }}\left[ {\frac{1}{{10}} + \frac{1}{{15}}} \right]\)</span>
C. <span class="math-tex">\(\frac{2}{{10\pi }}\left[ {\frac{1}{{25}} + \frac{1}{{30}}} \right]\)</span>
D. <span class="math-tex">\(\frac{1}{{10\pi }}\left[ {\frac{1}{{25}} + \frac{1}{{16}}} \right]\)</span>
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Right Answer is: D
SOLUTION
Concept:
The deflection in any bar due to axial load is given by
Δ = (P x L)/ (A x E)
Where P = Axial load
L= Length of the bar
A = Cross sectional area of bar
E = Modulus of elasticity
E = 2 × 106 kg / cm2
Tensile load = 10 tonnes = 10 × 103 kg
\({\rm{\Delta }}L = \frac{{P{L_1}}}{{{A_1}{E_1}}} + \frac{{P{L_2}}}{{{A_2}{E_2}}}\)
\( = \frac{{10 \times {{10}^3} \times 20}}{{\frac{\pi }{4} \times {{10}^2} \times 2 \times {{10}^6}}} + \frac{{10 \times {{10}^3} \times 20}}{{\frac{\pi }{4} \times {8^2} \times 2 \times {{10}^6}}}\)
\(= \frac{1}{{10\pi }}\left[ {\frac{1}{{25}} + \frac{1}{{16}}} \right]\)
