Let's break down the problem step-by-step: ### **Given Data:** - **Material**: Steel, \( E = 29 \times 10^6 \) psi - **Maximum force in BD**: \( 0.016P \) - **Maximum allowable stress in BD**: \( \sigma_{max} = 34 \) ksi \( = 34,000 \) psi - **Maximum change in length of BD**: \( \Delta L_{max} = 0.0008 \) times the length of AB - **Length AB**: \( AB = 72 + 72 = 144 \) in. - **Load \( P \)**: 100 kips = 100,000 lbs - **Length BC**: 54 in (for reference) - **Find**: Allowable diameter \( d \) for rod BD. --- ### **Step 1: Maximum Force in BD** Given: \[ F_{BD} = 0.016P = 0.016 \times 100,000 = 1,600 \text{ lbs} \] --- ### **Step 2: Stress Limitation** The maximum stress in BD should not exceed 34 ksi: \[ \sigma = \frac{F_{BD}}{A} \leq 34,000 \text{ psi} \] Area of rod (circular): \[ A = \frac{\pi}{4} d^2 \] So, \[ \frac{1,600}{\frac{\pi}{4} d^2} \leq 34,000 \] \[ 1,600 \leq 34,000 \times \frac{\pi}{4} d^2 \] \[ 1,600 \leq 26,703.5 d^2 \] \[ d^2 \geq \frac{1,600}{26,703.5} \] \[ d^2 \geq 0.0599 \] \[ d \geq \sqrt{0.0599} \] \[ d \geq 0.245 \text{ in} \] So, the diameter must be at least **0.245 in** to satisfy the stress limit. --- ### **Explanation:** - We calculated the force in BD from the given relation. - We used the formula for normal stress (\(\sigma = \frac{F}{A}\)) and set it equal to the maximum allowed stress. - Solved for the minimum diameter that meets the stress requirement. --- ### **Summary Table** | Step | Formula | Calculation | Result | |-----------------|--------------------------------|-------------------------|------------| | Force in BD | \(0.016 \times 100,000\) | | 1,600 lbs | | Area | \(\frac{\pi}{4} d^2\) | | | | Stress | \(\sigma = \frac{F}{A}\) | \(\frac{1,600}{A} \leq 34,000\) | | | Diameter | \(d^2 \geq \frac{1,600}{26,703.5}\) | \(d \geq 0.245\) in | 0.245 in | --- **If you want to check the deformation (elongation) limit, let me know and we can work through that as well!**