Question

(a) Equation (17.12) gives the stress required to keep the length of a rod constant as its temperature changes. Show that if the length is permitted to change by an amount ∆L when its temperature changes by ∆T, the stress is equal to$\frac{\mathbf{F}}{\mathbf{A}}\mathbf{=}\mathbf{Y}\left(\frac{∆L}{L_0}‐\alpha ∆T\right)$

where F is the tension on the rod, ${\mathbf{L}}_{\mathbf{0}}$ is the original length of the rod, A its cross-sectional area, its coefficient of linear expansion, and Y its Young’s modulus. (b) A heavy brass bar has projections at its ends (Fig. P17.79). Two fine steel wires, fastened between the projections, are just taut (zero tension) when the whole system is at 20°C. What is the tensile stress in the steel wires when the temperature of the system is raised to 140°C? Make any simplifying assumptions you think are justified, but state them.

Short answer

a) it is proved that the change in length and temperature the stress is equal to $\frac{F}{A}=Y\left(\frac{∆L}{L_0}-\alpha ∆T\right)$.

b) the tensile stress of steel wire is $1.92\times {10}^8\mathrm{Pa}$.

Step-by-step solution

Concept/Significance of young modulus

Young's modulus is a feature of materials that indicates how stiff or elastic they are. It is the stress-to-strain ratio in mathematics.

verification of the stress equation

The change in length due to tension is given by,

$∆L_F=\frac{F{L}_0}{AY}$

Here, F is the force on the rod, A is the cross-sectional area of the rod, and Yis the young modulus of the rod

The total change in length is also given by,

$∆L=∆L_F+∆L_T$

Here, $∆L_F$is the change in length due to force on rod and $∆L_T$is the change in length due to heating whose value is

$∆L_T=L_0\alpha ∆T$

Substitute values in the above equation.

$$\begin{gathered} ∆L=\frac{F{L}_0}{AY}+L_0\alpha ∆T \\ \frac{∆L}{L_0}=\frac{F}{AY}+\alpha ∆T \\ \frac{F}{AY}=\frac{∆L}{L_0}-\alpha ∆T \end{gathered}$$

Solve the above equation for tensile stress.

localid="1664360539848" $$\begin{gathered} \frac{F}{AY}=\frac{∆L}{L_0}-\alpha ∆T \\ \frac{F}{A}=Y\left(\frac{∆L}{L_0}-\alpha ∆T\right) \end{gathered}$$

Thus, it is proved by above expression that the change in length and temperature the stress is equal to $\frac{F}{A}=Y\left(\frac{∆L}{L_0}-\alpha ∆T\right)$.

Determination of tensile stress of steel wire

Tensile stress of the steel wire is given by,

$\frac{F}{A}=Y_{steel}\left({\alpha }_{brass}-{\alpha }_{steel}\right)∆T$

Here, $Y_{steel}$is the young modulus for brass whose values is $20\times {10}^{10}Pa,{\alpha }_{steel}$is the coefficient of linear expansion for brass whose value is $2\times {10}^{-5}{\left(°C\right)}^{-1},{\alpha }_{steel}$is the coefficient of linear expansion for steel whose value is $1.2\times {10}^{-5}{\left(°C\right)}^{-1}$and $∆T$is the change in temperature.

Substitute values in the above,

$$\begin{gathered} \frac{F}{A}=\left(20\times {10}^{10}Pa\right)\left(2\times {10}^{-5}{\left(°C\right)}^{-1}-1.2\times 10{\left(°C\right)}^{-1}\right)\left(120°C\right) \\ =1.92\times {10}^{8}Pa \end{gathered}$$

Thus, the tensile stress of steel wire is $1.92\times {10}^8\mathrm{Pa}$.