## Step-by-Step Solution Let's break down the problem and solve it step by step. --- ### **Given:** - **Tie rod** of length \( L \), loaded in tension by force \( F \) - **Objective:** Minimize mass \( m \) - **Constraints:** - Must not fail by **yielding** (plastic deformation) - Must not fail by **fast fracture** (brittle failure) - \( L \) and \( F \) are specified - **Free variables:** Section area \( A \), choice of material --- ## **Step 1: Express the Mass** The **mass** of the tie rod: \[ m = \rho \cdot A \cdot L \] where \( \rho \) is the density of the material. --- ## **Step 2: Performance Equation for Yielding Constraint** **Yielding constraint:** The maximum stress must not exceed the yield strength (\( \sigma_y \)) of the material. \[ \text{Stress} = \frac{F}{A} \leq \sigma_y \implies A \geq \frac{F}{\sigma_y} \] **Substitute into mass equation:** \[ m_y = \rho \cdot A \cdot L \geq \rho \cdot \frac{F}{\sigma_y} \cdot L \] **Performance equation for yielding:** \[ m_y \geq \frac{\rho F L}{\sigma_y} \] --- ## **Step 3: Performance Equation for Fast Fracture Constraint** **Fast fracture constraint:** The maximum stress must not exceed the fracture strength (\( \sigma_f \)). \[ \frac{F}{A} \leq \sigma_f \implies A \geq \frac{F}{\sigma_f} \] **Substitute into mass equation:** \[ m_f = \rho \cdot A \cdot L \geq \rho \cdot \frac{F}{\sigma_f} \cdot L \] **Performance equation for fracture:** \[ m_f \geq \frac{\rho F L}{\sigma_f} \] --- ## **Step 4: Define Material Indices** - **For yielding:** \( M_y = \frac{\sigma_y}{\rho} \) (maximize this to minimize mass) - **For fracture:** \( M_f = \frac{\sigma_f}{\rho} \) (maximize this to minimize mass) --- ## **Step 5: Coupling Equation** The **actual required area** must satisfy **both** constraints, so take the more stringent (i.e., larger area): \[ A \geq \max\left(\frac{F}{\sigma_y}, \frac{F}{\sigma_f}\right) \] Similarly, the **minimum mass** is: \[ m \geq \rho L \cdot \max\left(\frac{F}{\sigma_y}, \frac{F}{\sigma_f}\right) \] --- ## **Step 6: Coupling Between Indices** The **coupling equation** is: \[ m \geq \frac{\rho F L}{\min(\sigma_y, \sigma_f)} \] - The relevant material index is: \( M = \frac{\min(\sigma_y, \sigma_f)}{\rho} \) - To minimize mass, **maximize** this combined index. --- ## **Summary Table** | Constraint | Limiting Area | Performance Equation | Material Index (\( M \)) | |----------------------|-------------------|-------------------------------------|--------------------------| | Yielding | \( F/\sigma_y \) | \( m_y \geq \frac{\rho F L}{\sigma_y} \) | \( \frac{\sigma_y}{\rho} \) | | Fast Fracture | \( F/\sigma_f \) | \( m_f \geq \frac{\rho F L}{\sigma_f} \) | \( \frac{\sigma_f}{\rho} \) | | **Coupled** | \( F/\min(\sigma_y, \sigma_f) \) | \( m \geq \frac{\rho F L}{\min(\sigma_y, \sigma_f)} \) | \( \frac{\min(\sigma_y, \sigma_f)}{\rho} \) | --- ### **Final Answers:** - **Performance equations:** - Yielding: \( m_y \geq \frac{\rho F L}{\sigma_y} \) - Fracture: \( m_f \geq \frac{\rho F L}{\sigma_f} \) - **Material indices:** \( \frac{\sigma_y}{\rho} \) and \( \frac{\sigma_f}{\rho} \) - **Coupling equation:** \( m \geq \frac{\rho F L}{\min(\sigma_y, \sigma_f)} \) - **Relation:** The minimum mass is determined by the lower of \( \sigma_y \) or \( \sigma_f \), so the **critical material index** is \( \frac{\min(\sigma_y, \sigma_f)}{\rho} \). --- ### **Conclusion** - Both constraints are checked, but the **smaller** of the two strengths (yield or fracture) limits the design. - To **minimize mass**, choose a material that **maximizes** \( \frac{\min(\sigma_y, \sigma_f)}{\rho} \).