Question

A steel bar of 2 m length is fixed at both ends at 20°C. The coefficient of thermal expansion is 11 x 10-6/°C and the modulus of the elasticity is 2 x 106 kg/cm2. If the temperature is changed to 18°C, then the bar will be experience a stress of

A. 22 kg/cm2 (tensile) B. 22 kg/cm2 (compressive) C. 44 kg/cm2 (compressive) D. 44 kg/cm2 (tensile)

Answer 1

Answer:

A. 22 kg/cm2 (tensile)

Explanation:

If the temperature drops to 18C degrees (by 2 degrees), it would shrink by an amount of

$\Delta L = 11\times10^{-6}*2 = 22\times10^{-6} m$

The strain would be:

$\epsilon = \Delta L / L = 22\times10{-6} / 2 = 11\times10^{-6}$

We can calculate the stress from strain and modulus of the elasticity:

$\sigma = E\epsilon = 2\times10^6 * 11\times10^{-6} = 22 kg/cm^2$

This stress would be tensile because the bar shrinks

Answer 2

Answer:

C. 44 kg/cm2 (compressive).

Explanation:

The main difference between tensile and compressive stress is that tensile stress results in elongation whereas compressive stress results in shortening.

Below is an attachment containing the solution.