Let's break down the problem step by step: --- # **Generalized Hooke’s Law with Temperature** For a linear elastic isotropic material, the generalized Hooke’s Law including thermal effects is: \[ \epsilon_{ij} = \frac{1+\nu}{E}\sigma_{ij} - \frac{\nu}{E}\delta_{ij}\sigma_{kk} + \alpha \Delta T \delta_{ij} \] where: - \(\epsilon_{ij}\): Strain tensor components - \(\sigma_{ij}\): Stress tensor components - \(E\): Young's modulus - \(\nu\): Poisson’s ratio - \(\alpha\): Coefficient of linear expansion - \(\Delta T\): Temperature change - \(\delta_{ij}\): Kronecker delta (1 when \(i=j\), otherwise) - \(\sigma_{kk}\): Sum of normal stresses (trace of the stress tensor) --- # **Physical Setup** - At \(T_\), the block fits exactly between the rigid walls. - When heated to \(T_ + \Delta T\), the block wants to expand but is prevented by the walls, creating stress. - We want to find the induced stress \(\sigma_y\) in the block under two cases. --- ## **Case (a): Block is Thin in the \(z\)-Direction** ### **Assumption:** Plane Stress Condition (\(\sigma_z = \)) #### **Step 1: Constraints** - No expansion in the \(y\)-direction (between walls): \(\epsilon_y = \) - Free in the \(z\)-direction: \(\sigma_z = \) #### **Step 2: Write the Strain Equation** For the \(y\)-direction: \[ \epsilon_y = \frac{1}{E} (\sigma_y - \nu \sigma_x) + \alpha \Delta T \] But since the block is free along \(x\): \(\sigma_x = \). So, \[ \epsilon_y = \frac{\sigma_y}{E} + \alpha \Delta T \] But the wall doesn't allow expansion: \[ \epsilon_y = \] #### **Step 3: Solve for Stress** \[ = \frac{\sigma_y}{E} + \alpha \Delta T \] \[ \sigma_y = -E \alpha \Delta T \] **Interpretation**: The stress is compressive since the block wants to expand but is restrained. --- ## **Case (b): Block is Very Thick in the \(z\)-Direction** ### **Assumption:** Plane Strain Condition (\(\epsilon_z = \)) #### **Step 1: Constraints** - No expansion in the \(y\)-direction: \(\epsilon_y = \) - No expansion in the \(z\)-direction: \(\epsilon_z = \) #### **Step 2: Write Strain Equations** \[ \epsilon_y = \frac{1}{E}[\sigma_y - \nu (\sigma_x + \sigma_z)] + \alpha \Delta T \] \[ \epsilon_z = \frac{1}{E}[\sigma_z - \nu (\sigma_x + \sigma_y)] + \alpha \Delta T \] But block is free along \(x\): \(\sigma_x = \). So: \[ \epsilon_y = \frac{1}{E}(\sigma_y - \nu \sigma_z) + \alpha \Delta T \] \[ \epsilon_z = \frac{1}{E}(\sigma_z - \nu \sigma_y) + \alpha \Delta T \] #### **Step 3: Apply Constraints** \[ \epsilon_y = \implies \frac{\sigma_y - \nu \sigma_z}{E} + \alpha \Delta T = \] \[ \epsilon_z = \implies \frac{\sigma_z - \nu \sigma_y}{E} + \alpha \Delta T = \] #### **Step 4: Solve Simultaneous Equations** Write both: \[ \sigma_y - \nu \sigma_z = -E \alpha \Delta T \quad (1) \] \[ \sigma_z - \nu \sigma_y = -E \alpha \Delta T \quad (2) \] Multiply (1) by \(\nu\): \[ \nu \sigma_y - \nu^2 \sigma_z = -\nu E \alpha \Delta T \] Add to (2): \[ (\sigma_z - \nu \sigma_y) + (\nu \sigma_y - \nu^2 \sigma_z) = -E \alpha \Delta T - \nu E \alpha \Delta T \] \[ \sigma_z - \nu^2 \sigma_z = -(1+\nu) E \alpha \Delta T \] \[ \sigma_z (1 - \nu^2) = -(1+\nu) E \alpha \Delta T \] \[ \sigma_z = -\frac{(1+\nu)}{1-\nu^2} E \alpha \Delta T \] Now substitute \(\sigma_z\) into (1): \[ \sigma_y - \nu \left[-\frac{(1+\nu)}{1-\nu^2} E \alpha \Delta T\right] = -E \alpha \Delta T \] \[ \sigma_y + \frac{\nu (1+\nu)}{1-\nu^2} E \alpha \Delta T = -E \alpha \Delta T \] \[ \sigma_y = -E \alpha \Delta T - \frac{\nu (1+\nu)}{1-\nu^2} E \alpha \Delta T \] \[ \sigma_y = -E \alpha \Delta T \left[ 1 + \frac{\nu (1+\nu)}{1-\nu^2} \right] \] \[ \sigma_y = -E \alpha \Delta T \left[ \frac{1-\nu^2 + \nu(1+\nu)}{1-\nu^2} \right] \] \[ 1 - \nu^2 + \nu + \nu^2 = 1 + \nu \] \[ \sigma_y = -E \alpha \Delta T \left( \frac{1+\nu}{1-\nu} \right) \] --- # **Summary of Results** | Case | Stress, \(\sigma_y\) | |------|----------------------| | (a) Thin in \(z\) | \(-E \alpha \Delta T\) | | (b) Thick in \(z\) | \(-E \alpha \Delta T \left(\frac{1+\nu}{1-\nu}\right)\) | --- ## **Physical Meaning** - **Case (a):** Only direct restraint in the \(y\)-direction, so stress is simply the negative of what thermal strain would have been. - **Case (b):** Block is also restrained in \(z\), so Poisson effect adds to the restraint, increasing the stress. --- ## **If you need the full derivations or have questions about each step, just ask!**

# **Final Answer: Stress Calculations from Thermal Expansion** The stress \(\sigma_y\) induced in the block due to thermal expansion can be summarized as follows: ## **Results Summary** | Case | Description | Stress, \(\sigma_y\) | |------|---------------------------------|---------------------------------------------------------| | (a) | Thin in \(z\)-direction | \(\sigma_y = -E \alpha \Delta T\) | | (b) | Very thick in \(z\)-direction | \(\sigma_y = -E \alpha \Delta T \left(\frac{1+\nu}{1-\nu}\right)\) | ### **Key Variables:** - \(E\): Young's modulus - \(\nu\): Poisson's ratio - \(\alpha\): Coefficient of linear expansion - \(\Delta T\): Change in temperature --- ## **Interpretation of Results** - **For Case (a)**: The stress is directly related to the thermal expansion in the \(y\)-direction, resulting in simple compression. - **For Case (b)**: The stress is influenced by both the thermal expansion and the Poisson effect, leading to a more complex relationship due to the constraints in both \(y\) and \(z\) directions. If you need further clarification on any specific part, feel free to ask!