Chapter 19, Problem 19.54AP

Two metal bars are made of invar and a third bar is made of aluminum. At 0°C, each of the three bars is drilled with two holes 40.0 cm apart. Pins are put through the holes to assemble the bars into an equilateral triangle as in Figure P18.31. (a) First ignore the expansion of the invar. Find the angle between the invar bars as a function of Celsius temperature. (b) Is your answer accurate for negative as well as positive temperatures? (c) Is it accurate for 0°C? (d) Solve the problem again, including the expansion of the invar. Aluminum melts at 660°C and invar at 1 427°C. Assume the tabulated expansion coefficients are constant. What are (e) the greatest and (f) the smallest attainable angles between the invar bars?

Figure P18.31

To determine

The angle between the invar bars as a function of Celsius temperature.

Answer to Problem 19.54AP

The angle between the invar bars as a function of Celsius temperature is $2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2}\right)$.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle is,

$\begin{array}{c}\sin \left(\frac{\theta }{2}\right)=\frac{\frac{L_0\left(1+{\alpha }_{\text{Al}}\Delta T\right)}{2}}{L_0}\\ =\frac{L_0\left(1+{\alpha }_{\text{Al}}\Delta T\right)}{2L_0}\\ =\frac{\left(1+{\alpha }_{\text{Al}}{T}_{\text{C}}\right)}{2}\end{array}$

Here,

$L_0$ is the length of each bar at $0°\text{C}$.

${\alpha }_{\text{Al}}$ is the average coefficient of linear expansion of the aluminum.

$\Delta T$ is the temperature difference.

$T_{\text{C}}$ is the Celsius temperature.

Rearrange the above equation for $\theta$.

$\begin{array}{c}\sin \left(\frac{\theta }{2}\right)=\frac{\left(1+{\alpha }_{\text{Al}}{T}_{\text{C}}\right)}{2}\\ \theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2}\right)\end{array}$

Conclusion:

Therefore, the angle between the invar bars as a function of Celsius temperature is $2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2}\right)$.

To determine

Whether the answer is accurate for negative as well as positive temperature or not.

Answer to Problem 19.54AP

Yes the answer is accurate for negative as well as positive temperature.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

If the temperature drops, the Celsius temperature becomes negative. The negative value of the Celsius temperature describes the contraction of the bars. So the answer is accurate for the negative temperature same as the positive temperature.

Conclusion:

Therefore, yes the answer is accurate for negative as well as positive temperature.

To determine

Whether the answer is accurate at $0°\text{C}$ or not.

Answer to Problem 19.54AP

Yes, the answer is accurate at $0°\text{C}$.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

The expression for the angle between the invar bars is,

$\theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2}\right)$

Substitute $0°\text{C}$ for $T_{\text{C}}$ in above equation.

$\begin{array}{l}\theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}\left(0°\text{C}\right)}{2}\right)\\ =2{\sin }^{-1}\left(\frac{1}{2}\right)\\ =60°\end{array}$

The value of each angle of the right angle triangle is $60°$ so this is accurate.

Conclusion:

Therefore, Yes, the answer is accurate at $0°\text{C}$.

To determine

The angle between the invar bars and the aluminum bar as a function of Celsius temperature.

Answer to Problem 19.54AP

The angle between the invar bars and the aluminum bar as a function of Celsius temperature is $2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\right)$.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

In the given diagram consider the right triangle that each invar bar makes with one half of the aluminum bar.

Then, the angle between invar bars and aluminum bar is,

$\begin{array}{c}\sin \left(\frac{\theta }{2}\right)=\frac{\frac{L_0\left(1+{\alpha }_{\text{Al}}\Delta T\right)}{2}}{L_0\left(1+{\alpha }_{\text{invar}}\Delta T\right)}\\ =\frac{L_0\left(1+{\alpha }_{\text{Al}}\Delta T\right)}{2L_0\left(1+{\alpha }_{\text{invar}}\Delta T\right)}\\ =\frac{\left(1+{\alpha }_{\text{Al}}{T}_{\text{C}}\right)}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\end{array}$

Here,

${\alpha }_{\text{invar}}$ is the average coefficient of linear expansion of the invar.

Rearrange the above equation for $\theta$.

$\begin{array}{c}\sin \left(\frac{\theta }{2}\right)=\frac{\left(1+{\alpha }_{\text{Al}}{T}_{\text{C}}\right)}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\\ \theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\right)\end{array}$

Conclusion:

Therefore, the angle between the invar bars and the aluminum bar as a function of Celsius temperature is $2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\right)$.

To determine

The greatest attainable angle between the invar bars.

Answer to Problem 19.54AP

The greatest attainable angle between the invar bars is $61.0°$.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

The average coefficient of linear expansion of the aluminum is $24\times {10}^{-6}°{\text{C}}^{-1}$.

The average coefficient of linear expansion of the invar is $0.9\times {10}^{-6}°{\text{C}}^{-1}$.

The greatest angle is occur at $660°\text{C}$.

The equation for the angel between the invar bars is,

$\theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\right)$

Substitute $24\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_{\text{Al}}$, $0.9\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_{\text{invar}}$ and $660°\text{C}$ for $T_{\text{C}}$ in above equation to find $\theta$.

$\begin{array}{c}\theta =2{\sin }^{-1}\left(\frac{1+\left(24\times {10}^{-6}°{\text{C}}^{-1}\right)\left(660°\text{C}\right)}{2\left(1+\left(0.9\times {10}^{-6}°{\text{C}}^{-1}\right)\left(660°\text{C}\right)\right)}\right)\\ =2{\sin }^{-1}\left(\frac{1.01584}{2.001188}\right)\\ =2{\sin }^{-1}\left(0.5076\right)\\ =61.0°\end{array}$

Conclusion:

Therefore, the greatest attainable angle between the invar bars is $61.0°$.

To determine

The smallest attainable angle between the invar bars.

Answer to Problem 19.54AP

The smallest attainable angle between the invar bars is $59.6°$.

Explanation of Solution

Given info: The distance between two holes drilled at $0°\text{C}$ is $40.0\text{cm}$, the melting temperature of the aluminum is $660°\text{C}$ and the melting temperature of invar is $1427°\text{C}$.

The average coefficient of linear expansion of the aluminum is $24\times {10}^{-6}°{\text{C}}^{-1}$.

The average coefficient of linear expansion of the invar is $0.9\times {10}^{-6}°{\text{C}}^{-1}$.

The smallest angle is occur at $-273°\text{C}$.

The equation for the angel between the invar bars is,

$\theta =2{\sin }^{-1}\left(\frac{1+{\alpha }_{\text{Al}}{T}_{\text{C}}}{2\left(1+{\alpha }_{\text{invar}}{T}_{\text{C}}\right)}\right)$

Substitute $24\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_{\text{Al}}$, $0.9\times {10}^{-6}°{\text{C}}^{-1}$ for ${\alpha }_{\text{invar}}$ and $-273°\text{C}$ for $T_{\text{C}}$ in above equation to find $\theta$.

$\begin{array}{c}\theta =2{\sin }^{-1}\left(\frac{1+\left(24\times {10}^{-6}°{\text{C}}^{-1}\right)\left(-273°\text{C}\right)}{2\left(1+\left(0.9\times {10}^{-6}°{\text{C}}^{-1}\right)\left(-273°\text{C}\right)\right)}\right)\\ =2{\sin }^{-1}\left(\frac{0.9934}{1.9995}\right)\\ =2{\sin }^{-1}\left(0.497\right)\\ =59.6°\end{array}$

Conclusion:

Therefore, the smallest attainable angle between the invar bars is $59.6°$.