91影视

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.

The change in length due to tension is given by,

$鈭�L_F=\frac{F{L}_0}{AY}$

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,

$鈭�L=鈭�L_F+鈭�L_T$

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

$鈭�L_T=L_0伪鈭�T$

Substitute values in the above equation.

$$\begin{gathered} 鈭�L=\frac{F{L}_0}{AY}+L_0伪鈭�T \\ \frac{鈭�L}{L_0}=\frac{F}{AY}+伪鈭�T \\ \frac{F}{AY}=\frac{鈭�L}{L_0}-伪鈭�T \end{gathered}$$

Solve the above equation for tensile stress.

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

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

Tensile stress of the steel wire is given by,

$\frac{F}{A}=Y_{steel}\left({伪}_{brass}-{伪}_{steel}\right)鈭�T$

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

Substitute values in the above,

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

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