Question

Question:(II) An aluminum bar has the desired length when at 12°C. How much stress is required to keep it at this length if the temperature increases to 35°C? (See Table 9–1.)

Short answer

The thermal stress of the aluminum bar is \(4.02 \times {10^7}\;{\rm{N/}}{{\rm{m}}^2}\).

Step-by-step solution

Identification of given data

• The coefficient of thermal expansion of aluminum bar is\({\alpha _{{\rm{alum}}}} = 25 \times {10^{ - 6}}\;{\rm{/^\circ C}}\).
• The Young modulus of the aluminum bar is\({E_{{\rm{alum}}}} = 70 \times {10^9}\;{\rm{N/}}{{\rm{m}}^2}\).
• The initial temperature of the aluminum bar is\({T_{\rm{i}}} = 12^\circ {\rm{C}}\).
• The final temperature of the aluminum bar is \({T_{\rm{f}}} = 35^\circ {\rm{C}}\).

Understanding the thermal stress of the aluminum bar

• The thermal stress can be obtained by the analysis of thermal expansion.The thermal stress of aluminum is the product of its coefficient of thermal expansion, Young’s modulus, and the change in the temperature of the bar.

Determination of the thermal stress of the aluminum bar

The thermal stress can be expressed as shown below:

\(\begin{aligned}{c}{\sigma _{{\rm{therm}}}} &= {\alpha _{{\rm{alum}}}}{E_{{\rm{alum}}}}\Delta T\\ &= {\alpha _{{\rm{alum}}}} \times {E_{{\rm{alum}}}} \times \left( {{T_{\rm{f}}} - {T_{\rm{i}}}} \right)\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}{\sigma _{{\rm{therm}}}} &= \left( {25 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right) \times \left( {70 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^2}} \right) \times \left( {35^\circ {\rm{C}} - 12^\circ {\rm{C}}} \right)\\ &= 1.75 \times {10^6}\;{\rm{N/}}{{\rm{m}}^2} \cdot {\rm{^\circ C}} \times 23^\circ {\rm{C}}\\ &= 4.02 \times {10^7}\;{\rm{N/}}{{\rm{m}}^2}\end{aligned}\)

Thus, the thermal stress of the aluminum bar is \(4.02 \times {10^7}\;{\rm{N/}}{{\rm{m}}^2}\).