Understanding the Problem: Thermal Expansion in a Composite Bar

The problem describes a composite bar made of a steel bar placed between two copper bars. All bars have the same initial length and cross-sectional area at 20°C. They are rigidly connected at both ends. When the temperature increases to 320°C, the total length of the composite bar increases by a specific amount (1.5 mm). We are given the Young's moduli (E) and coefficients of thermal expansion (α) for both steel and copper, and we need to find the original length of the bars.

Since the bars are rigidly connected, they must expand together by the same total amount. However, steel and copper have different coefficients of thermal expansion. Copper (αc) has a higher coefficient than steel (αs). This means that if they were free to expand, copper would expand more than steel for the same temperature increase.

Because they are forced to expand together, the material that wants to expand more (copper) will be compressed, and the material that wants to expand less (steel) will be stretched. This introduces internal stresses: compressive stress in copper and tensile stress in steel.

Applying Principles of Thermal Expansion and Stress

Let the original length of each bar at 20°C be \(L\). The temperature change is \(\Delta T = 320^\circ\text{C} - 20^\circ\text{C} = 300^\circ\text{C}\).

The free thermal expansion for steel would be:

\(\Delta L_{\text{thermal, s}} = L \alpha_s \Delta T\)

The free thermal expansion for copper would be:

\(\Delta L_{\text{thermal, c}} = L \alpha_c \Delta T\)

Since \(\alpha_c > \alpha_s\), \(\Delta L_{\text{thermal, c}} > \Delta L_{\text{thermal, s}}\).

Due to the rigid connection, the bars experience stresses. Let \(\sigma_s\) be the tensile stress in steel and \(\sigma_c\) be the compressive stress in copper. The strain due to stress is given by Hooke's Law, \(\epsilon = \sigma / E\). The change in length due to stress is \(\Delta L_{\text{stress}} = \text{Strain} \times L = (\sigma / E) L\).

The actual change in length for steel is the free thermal expansion plus the expansion due to tensile stress:

\(\Delta L_{\text{actual, s}} = \Delta L_{\text{thermal, s}} + \Delta L_{\text{stress, s}} = L \alpha_s \Delta T + \frac{\sigma_s}{E_s} L\)

The actual change in length for copper is the free thermal expansion minus the contraction due to compressive stress:

\(\Delta L_{\text{actual, c}} = \Delta L_{\text{thermal, c}} - \Delta L_{\text{stress, c}} = L \alpha_c \Delta T - \frac{\sigma_c}{E_c} L\)

Since the bars are rigidly connected, their actual final length must be the same, so the actual change in length for each bar must be equal to the total given expansion, \(\Delta L_{\text{total}} = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}\).

\(\Delta L_{\text{actual, s}} = \Delta L_{\text{actual, c}} = \Delta L_{\text{total}}\)

Thus, \(L \alpha_s \Delta T + \frac{\sigma_s}{E_s} L = L \alpha_c \Delta T - \frac{\sigma_c}{E_c} L\). Dividing by \(L\) (assuming \(L > 0\)):

\(\alpha_s \Delta T + \frac{\sigma_s}{E_s} = \alpha_c \Delta T - \frac{\sigma_c}{E_c}\)

Rearranging terms:

\(\frac{\sigma_s}{E_s} + \frac{\sigma_c}{E_c} = (\alpha_c - \alpha_s) \Delta T\)

Now consider the forces. Let \(A\) be the area of each bar. The total tensile force in the steel bar must balance the total compressive force in the two copper bars. The force in steel is \(F_s = \sigma_s A\). The total force in copper is \(F_c = 2 \sigma_c A\). For equilibrium, \(F_s = F_c\):

\(\sigma_s A = 2 \sigma_c A\)

\(\sigma_s = 2 \sigma_c\)

Substitute \(\sigma_s = 2 \sigma_c\) into the previous equation:

\(\frac{2 \sigma_c}{E_s} + \frac{\sigma_c}{E_c} = (\alpha_c - \alpha_s) \Delta T\)

\(\sigma_c \left( \frac{2}{E_s} + \frac{1}{E_c} \right) = (\alpha_c - \alpha_s) \Delta T\)

We can now solve for \(\sigma_c\).

Calculation Steps

Given values:

• \(\Delta L_{\text{total}} = 1.5 \text{ mm} = 1.5 \times 10^{-3} \text{ m}\)
• \(\Delta T = 300^\circ\text{C}\)
• \(E_s = 220 \text{ GN/m}^2 = 220 \times 10^9 \text{ N/m}^2\)
• \(E_c = 110 \text{ GN/m}^2 = 110 \times 10^9 \text{ N/m}^2\)
• \(\alpha_s = 0.000012 \text{ per } ^\circ\text{C} = 12 \times 10^{-6} \text{ per } ^\circ\text{C}\)
• \(\alpha_c = 0.0000175 \text{ per } ^\circ\text{C} = 17.5 \times 10^{-6} \text{ per } ^\circ\text{C}\)

Calculate \((\alpha_c - \alpha_s) \Delta T\):

\((\alpha_c - \alpha_s) \Delta T = (17.5 \times 10^{-6} - 12 \times 10^{-6}) \times 300 = (5.5 \times 10^{-6}) \times 300 = 1650 \times 10^{-6} = 1.65 \times 10^{-3}\)

Calculate \(\left( \frac{2}{E_s} + \frac{1}{E_c} \right)\):

\(\frac{2}{E_s} + \frac{1}{E_c} = \frac{2}{220 \times 10^9} + \frac{1}{110 \times 10^9} = \frac{1}{110 \times 10^9} + \frac{1}{110 \times 10^9} = \frac{2}{110 \times 10^9} = \frac{1}{55 \times 10^9}\)

Calculate \(\sigma_c\):

\(\sigma_c = \frac{(\alpha_c - \alpha_s) \Delta T}{\left( \frac{2}{E_s} + \frac{1}{E_c} \right)} = \frac{1.65 \times 10^{-3}}{1 / (55 \times 10^9)} = 1.65 \times 10^{-3} \times 55 \times 10^9 = (1.65 \times 55) \times 10^6 = 90.75 \times 10^6 \text{ N/m}^2\)

Calculate \(\sigma_s\):

\(\sigma_s = 2 \sigma_c = 2 \times 90.75 \times 10^6 = 181.5 \times 10^6 \text{ N/m}^2\)

Now use the actual expansion of the steel bar (or copper bar) to find the original length \(L\). We know \(\Delta L_{\text{actual, s}} = \Delta L_{\text{total}}\):

\(\Delta L_{\text{total}} = L \alpha_s \Delta T + \frac{\sigma_s}{E_s} L = L \left( \alpha_s \Delta T + \frac{\sigma_s}{E_s} \right)\)

Substitute the values:

\(1.5 \times 10^{-3} = L \left( (12 \times 10^{-6} \times 300) + \frac{181.5 \times 10^6}{220 \times 10^9} \right)\)

\(1.5 \times 10^{-3} = L \left( (3600 \times 10^{-6}) + \frac{181.5}{220} \times 10^{-3} \right)\)

\(1.5 \times 10^{-3} = L \left( 3.6 \times 10^{-3} + 0.825 \times 10^{-3} \right)\)

\(1.5 \times 10^{-3} = L (4.425 \times 10^{-3})\)

Solve for \(L\):

\(L = \frac{1.5 \times 10^{-3}}{4.425 \times 10^{-3}} = \frac{1.5}{4.425}\)

\(L \approx 0.33898 \text{ m}\)

Rounding to three decimal places, the original length is approximately 0.339 m.

Final Answer

The original length of the bar is approximately 0.339 m.

Using the formula \(L = \frac{\Delta L_{\text{total}}}{\alpha_s \Delta T + \frac{\sigma_s}{E_s}}\) and the relation \(\sigma_s = 2 \sigma_c\) and the equilibrium equation \(\frac{2 \sigma_c}{E_s} + \frac{\sigma_c}{E_c} = (\alpha_c - \alpha_s) \Delta T\), we found the value of \(L\).

The calculation led to \(L \approx 0.339\) m.