A brass wire (d = 2.0 mm, E = 110 GPa) is pretensioned to T = 85 N. The coefficient of thermal expan- sion for the wire is $ 19.5 \times 10^{-6} /{ }^{\circ} \mathrm{C} $ . The temperature change at which the wire goes slack is approximately:

Question

Quizplus · Accepted Answer

The formula for thermal stress is:

$$\sigma_t = \alpha E \Delta T$$

where $\alpha$ is the coefficient of thermal expansion, E is the Young's modulus, and $\Delta T$ is the change in temperature.

We know that the wire is pretensioned to T = 85 N. When the thermal stress exceeds this tension, the wire will go slack. The maximum thermal stress that the wire can withstand before going slack is:

$$\sigma_{max} = \frac{T}{A}$$

where A is the cross-sectional area of the wire.

Combining these two equations yields:

$$\frac{T}{A} = \alpha E \Delta T$$

Solving for $\Delta T$, we get:

$$\Delta T = \frac{T}{A \alpha E}$$

Substituting the given values, we get:

$$\Delta T = \frac{85 \mathrm{N}}{\frac{\pi}{4}(2 \mathrm{mm})^2 19.5 	imes 10^{-6}/{ }^{\circ} \mathrm{C} (110 \mathrm{GPa})} \approx +12.6 ^ { \circ } \mathrm { C }$$

Since the coefficient of thermal expansion is positive, the temperature change that causes the wire to go slack is also positive. Therefore, the answer is (C).

A Brass Wire (D = 2 19.5×10−6/∘C 19.5 \times 10^{-6} /{ }^{\circ} \mathrm{C} 19.5×10−6/∘C

A Brass Wire (D = 2 $19.5 \times 10^{-6} /{ }^{\circ} \mathrm{C}$