A steel rail, rigidly fixed at its ends is assumed to be stress free at 20°C. If the stress required to cause the buckling of the rail is −75 Mpa, at what temperature will the rail buckle? (E = 200 GPa and α = 12.5 × 10^-6/°C)

Concept:

Thermal stress in a material that is rigidly fixed and unable to expand or contract is given by:

\( \sigma = E \alpha \Delta T \)

Where:

σ is the thermal stress

&E; is the Young's modulus of the material

α is the coefficient of linear thermal expansion

ΔT is the change in temperature

Calculation:

Given:

Initial temperature, \(T_i = 20^\circ\text{C}\)

Stress required to cause buckling, \(\sigma = -75 \text{ MPa}\)

Young's modulus, \(E = 200 \text{ GPa} = 200 \times 10^3 \text{ MPa}\)

Coefficient of linear thermal expansion, \(\alpha = 12.5 \times 10^{-6} / ^\circ\text{C}\)

Rearranging the formula to solve for ΔT:

\( \Delta T = \frac{\sigma}{E \alpha} \)

Substituting the given values:

\( \Delta T = \frac{-75 \text{ MPa}}{(200 \times 10^3 \text{ MPa})(12.5 \times 10^{-6} / ^\circ\text{C})} \)

Calculate the denominator:

\( E \alpha = (200 \times 10^3) \times (12.5 \times 10^{-6}) \)

\( E \alpha = 200 \times 12.5 \)

\( E \alpha = 2500 \times 10^{-3} \)

\( E \alpha = 2.5 \)

Now, calculate ΔT:

\( \Delta T = \frac{-75}{2.5} \)

\( \Delta T = -30^\circ\text{C} \)

Since the change in temperature is negative, indicating a decrease for compressive stress:

\( T_f = T_i + \Delta T \)

\( T_f = 20^\circ\text{C} + 30^\circ\text{C} \)

\( T_f = 50^\circ\text{C} \)

The temperature at which the rail will buckle is 50°C.

The correct answer is:
2) 50°C