Let’s break down the solution into **3 steps** as requested, explaining each step and providing the final answer at the end. --- ### **Step 1: Find the Maximum Buckling Load (Euler's Formula)** For a pin-pin column, the **critical (buckling) load** is: \[ P_{cr} = \frac{\pi^2 E I}{(L_{eff})^2} \] Where: - \(E = 30\, \text{Msi} = 30 \times 10^6\, \text{psi}\) - \(I =\) Minimum moment of inertia of cross-section - \(L_{eff} = L\) (for pin-pin: \(L_{eff} = L\)) - Length \(L = 3\, \text{ft} = 36\, \text{in}\) **Cross-section:** \(1\, \text{in} \times 1.5\, \text{in}\) The smaller width (1 in) will be the axis about which buckling is most likely. **Moment of inertia about weaker axis:** \[ I = \frac{1 \times (1.5)^3}{12} = \frac{1 \times 3.375}{12} = 0.281\, \text{in}^4 \] **Plugging values:** \[ P_{cr} = \frac{\pi^2 \times (30 \times 10^6) \times 0.281}{(36)^2} \] \[ = \frac{9.8696 \times 30 \times 10^6 \times 0.281}{1296} \] \[ = \frac{83.243 \times 10^6 \times 0.281}{1296} \] \[ = \frac{23.39 \times 10^6}{1296} \] \[ = 18,050\, \text{lb} \] **Explanation:** *We have calculated the maximum compressive force \(P\) that the rod AB can carry without buckling, using Euler’s formula for a pin-pin column.* --- ### **Step 2: Find the Corresponding Shrinkage (Axial Shortening) \(\delta\)** The **axial deformation** under maximum load (if applied): \[ \delta = \frac{PL}{AE} \] Where: - \(P = 18,050\, \text{lb}\) - \(L = 36\, \text{in}\) - \(A = 1 \times 1.5 = 1.5\, \text{in}^2\) - \(E = 30 \times 10^6\, \text{psi}\) \[ \delta = \frac{18,050 \times 36}{1.5 \times 30 \times 10^6} \] \[ = \frac{649,800}{45,000,000} \] \[ = 0.0144\, \text{in} \] **Explanation:** *This step computes the expected shrinkage (shortening) of the rod if it is loaded to the maximum safe value.* --- ### **Step 3: Final Answers** - **Maximum force \(P\) that can be applied to handle without buckling:** \(\boxed{18,050\, \text{lb}}\) - **Shrinkage \(\delta\) of rod AB under this load:** \(\boxed{0.0144\, \text{in}}\) **Explanation:** *The first answer ensures safety against buckling. The second answer tells how much the rod will shorten if that maximum safe load is applied.* --- > **Final Answers:** > - **Maximum \(P\):** \(18,050\, \text{lb}\) > - **Shrinkage \(\delta\):** \(0.0144\, \text{in}\)