Let's solve the problem step by step: --- ## **Step 1: Gather Section Properties for W14x48** From the AISC Steel Manual: - **W14x48** (A992 steel) - \( F_y = 50 \) ksi - Depth \( d =13.90 \) in - \( A = 14.1 \) in² - \( I_x = 426 \) in⁴ - \( S_x = 61.4 \) in³ - \( r_x = 5.49 \) in --- ## **Step 2: Factored Loads** - **Axial Load, \( P_u \):** 25 kips - **Moment, \( M_u \):** 100 kip-ft = 120 kip-in --- ## **Step 3: Flexural Strength Check** ### **a. Nominal Moment Capacity (\( M_n \))** Since bending is about strong axis: #### **Plastic Moment:** \[ M_p = F_y \cdot Z_x \] From AISC, \( Z_x = 68.2 \) in³ for W14x48 \[ M_p = 50 \text{ ksi} \times 68.2 \text{ in}^3 = 341 \text{ kip-in} = 284.2 \text{ kip-ft} \] #### **LRFD Strength:** \[ \phi_b = .9 \] \[ \phi_b M_n = .9 \times 284.2 = 256 \text{ kip-ft} \] --- ### **b. Compare Demand to Capacity** \[ M_u = 100 \text{ kip-ft} < 256 \text{ kip-ft} \] **Flexure OK** --- ## **Step 4: Axial Strength Check** ### **a. Slenderness Ratio** \[ K = 1., \quad L = 30' = 360\text{ in}, \quad r_x = 5.49\text{ in} \] \[ \frac{KL}{r} = \frac{1. \times 360}{5.49} = 65.6 < 200 \] OK --- ### **b. Critical Buckling Stress (\( F_{cr} \))** \[ F_{cr} = .658^{\lambda_c^2} F_y \] Where: \[ \lambda_c = \frac{1}{\pi} \frac{KL}{r} \sqrt{\frac{F_y}{E}} \] \[ \sqrt{\frac{F_y}{E}} = \sqrt{\frac{50}{29000}} = .0416 \] \[ \lambda_c = \frac{1}{\pi} \times 65.6 \times .0416 = .870 \] \[ F_{cr} = .658^{(.870)^2} \times 50 = .658^{.757} \times 50 = .735 \times 50 = 36.8 \text{ ksi} \] ### **c. Axial Capacity** \[ \phi_c = .9 \] \[ P_n = F_{cr} \times A = 36.8 \times 14.1 = 518 \text{ kips} \] \[ \phi_c P_n = .9 \times 518 = 466 \text{ kips} \] \[ P_u = 25 \text{ kips} < 466 \text{ kips} \] **Axial OK** --- ## **Step 5: Combined Bending and Axial Check (Interaction Equation)** For members subjected to both axial compression and bending (AISC 360-16 Eq. H1-1): \[ \frac{P_u}{\phi_c P_n} + \frac{M_u}{\phi_b M_n} \leq 1. \] \[ \frac{25}{466} + \frac{100}{256} = .054 + .391 = .445 < 1. \] --- ## **Step 6: Direct Analysis Method & Appendix 8 Considerations** - All calculations above are compatible with the Direct Analysis Method. - No stability or nonlinear issues are expected with \( K = 1. \) and the given loads. --- ## **Conclusion** ### **Final Answer** \[ \boxed{\text{Yes, a W14x48 is sufficient for this setup using the Direct Analysis Method (Appendix 8).}} \] All strength and interaction checks are satisfied.