Question

A round bar made of same material consists of 3 parts each of 100 mm length having diameters of 40mm, 50 mm and 60 mm, respectively. If the bar is subjected to an axial load of 10 kN, the total elongation of the bar in mm would be (E is the modulus of elasticity in kN/mm${}^2$)

• A. $\frac{0.4}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})$
• B. $\frac{4}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})$
• C. $\frac{4\sqrt{2}}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})$
• D. $\frac{4D}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})$

Answer 1

The correct option is D $\frac{4D}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})$

Total elongation
$=\frac{4PL}{\pi E}(\frac{1}{d_1^2}+\frac{1}{d_2^2}+\frac{1}{d_3^2})$
$=\frac{4\times 10\times 100}{\pi E\times 100}[\frac{1}{16}+\frac{1}{25}+\frac{1}{36}]mm$
$=\frac{4D}{\pi E}(\frac{1}{16}+\frac{1}{25}+\frac{1}{36})mm$