Question

Cylindrical bars P and Q have identical lengths and radii, but are composed of different linear elastic materials. The Young's modulus and coefficient of thermal expansion of Q are twice the corresponding values of P. Assume the bars to be perfectly bonded at the interface, and their weights to be negligible. The bars are held between rigid supports as shown in the figure and the temperature is raised by \(\Delta T\). Assume that the stress in each bar is homogeneous and uniaxial. Denote the magnitudes of stress in P and Q by \(\sigma_1\) and \(\sigma_2\), respectively. Which of the statement(s) given is/are CORRECT?

• A. The interface between P and Q moves to the left after heating
• B. The interface between P and Q moves to the right after heating
• C. \(\sigma_1<\sigma_2\)
• D. \(\sigma_1 = \sigma_2\)

Hint:

For statically indeterminate problems like this, always start with the two fundamental conditions: 1. {Equilibrium:} Sum of forces = 0. This often gives a relationship between stresses. 2. {Compatibility:} Deformations must fit the geometric constraints. This gives a relationship between strains/displacements. Solving these two simultaneously gives the unknown forces/stresses.

Answer 1

The Correct Option is A, D

Step 1: Understanding the Concept:
This is a problem of thermal stresses in a composite bar fixed between two rigid supports. When the temperature is raised, both bars will try to expand. Since they are constrained by the rigid supports, a compressive stress will be induced in both bars. The bars are also bonded together, so they must have the same final deformation and the forces must be in equilibrium.
Step 2: Key Formula or Approach:
Let \(L_P = L_Q = L\), \(A_P = A_Q = A\).
Let the properties of bar P be \(E_P, \alpha_P\).
Let the properties of bar Q be \(E_Q, \alpha_Q\).
Given: \(E_Q = 2E_P\) and \(\alpha_Q = 2\alpha_P\).
1. Equilibrium Condition: Since there are no external forces, the compressive force in bar P must be equal to the compressive force in bar Q. \[ F_P = F_Q \implies \sigma_1 A = \sigma_2 A \implies \sigma_1 = \sigma_2 \] This immediately tells us that statement (D) is correct and (C) is incorrect. 2. Compatibility Condition: The total change in length of the composite bar must be zero because the supports are rigid. \[ \Delta L_{\text{total}} = \Delta L_P + \Delta L_Q = 0 \] The change in length of each bar is the sum of its free thermal expansion and its mechanical contraction due to the compressive stress. Let the compressive force be \(F\). \[ \Delta L = (\text{Thermal Expansion}) - (\text{Mechanical Contraction}) = L\alpha\Delta T - \frac{FL}{AE} \] So the compatibility equation is: \[ (L\alpha_P\Delta T - \frac{FL}{AE_P}) + (L\alpha_Q\Delta T - \frac{FL}{AE_Q}) = 0 \] 3. Interface Movement: Let the interface move by an amount \(\delta\) to the left. The final length of bar P is \(L-\delta\) and the final length of bar Q is \(L+\delta\). The change in length of P is \(-\delta\) and the change in length of Q is \(+\delta\).
- For bar P: \(\Delta L_P = L\alpha_P\Delta T - \frac{\sigma_1 L}{E_P} = -\delta\)
- For bar Q: \(\Delta L_Q = L\alpha_Q\Delta T - \frac{\sigma_2 L}{E_Q} = +\delta\)
We can solve for \(\delta\) to determine the direction of movement.
Step 3: Detailed Calculation:
1. Stress Equality: From the equilibrium of forces at the interface, \(F_P = F_Q\). Since the areas are identical, \(\sigma_1 = \sigma_2\). So, (D) is correct.
2. Interface Movement Analysis:
From the two compatibility equations for \(\delta\): \[ \delta = - (L\alpha_P\Delta T - \frac{\sigma_1 L}{E_P}) = \frac{\sigma_1 L}{E_P} - L\alpha_P\Delta T \] \[ \delta = L\alpha_Q\Delta T - \frac{\sigma_2 L}{E_Q} \] Since \(\sigma_1 = \sigma_2 = \sigma\), let's substitute this into the second equation for \(\delta\): \[ \delta = L(2\alpha_P)\Delta T - \frac{\sigma L}{2E_P} \] Now we have two expressions for \(\delta\). Let's first solve for the stress \(\sigma\) using the total compatibility equation: \[ L\alpha_P\Delta T - \frac{FL}{AE_P} + L\alpha_Q\Delta T - \frac{FL}{AE_Q} = 0 \] Divide by \(L\): \[ (\alpha_P + \alpha_Q)\Delta T = \frac{F}{A} \left(\frac{1}{E_P} + \frac{1}{E_Q}\right) \] \[ ( \alpha_P + 2\alpha_P )\Delta T = \sigma \left(\frac{1}{E_P} + \frac{1}{2E_P}\right) \] \[ 3\alpha_P \Delta T = \sigma \left(\frac{3}{2E_P}\right) \implies \sigma = 2\alpha_P E_P \Delta T \] Now substitute this stress \(\sigma\) into the expression for \(\delta\): \[ \delta = \frac{\sigma L}{E_P} - L\alpha_P\Delta T = \frac{(2\alpha_P E_P \Delta T)L}{E_P} - L\alpha_P\Delta T \] \[ \delta = 2L\alpha_P\Delta T - L\alpha_P\Delta T = L\alpha_P\Delta T \] Since \(L, \alpha_P, \Delta T\) are all positive, \(\delta>0\). We defined \(\delta\) as a movement to the left. Therefore, the interface moves to the left. Statement (A) is correct and (B) is incorrect. Step 4: Final Answer:
The correct statements are (A) and (D).
Step 5: Why This is Correct:
Equilibrium of internal forces on the interface requires the stresses to be equal as the areas are equal. The compatibility condition, combined with the material properties, shows that the interface must move to accommodate the different thermal expansions and mechanical contractions. Since bar Q has a much stronger tendency to expand (due to higher \(\alpha_Q\)) but is also much stiffer (higher \(E_Q\)), the final calculation shows that the expansion tendency dominates, causing the interface to shift leftwards into bar P.