Let's break down the solution step-by-step: --- # **Step 1: Analyze Loads** Given- Factored (Pu) 200 k (k) Dead load =75% of Pu - Live load = 25% Pu **Load Factors (per AISC):** - Dead Load (DL): 1.2 - Live Load (LL): 1.6 ### **Factored Load Calculation** Let: - \( D = .75 \times 200 = 150\,k \) - \( L = .25 \times 200 = 50\,k \) Factored load: \[ \text{Required Strength (} P_u \text{)} = 1.2 D + 1.6 L \] \[ P_u = 1.2(150) + 1.6(50) = 180 + 80 = 260\,k \] --- # **Step 2: Select a Pair of Channels** ## **a) A36 Steel (Fy = 36 ksi)** ### **Required Strength (Pn):** \[ P_n = \frac{P_u}{\phi} \] For compression members (per AISC), \( \phi = .9 \): \[ P_n = \frac{260}{.9} = 288.9\,k \] ### **Channel Selection** Each channel must carry half the load: \[ \frac{288.9}{2} = 144.45\,k \text{ per channel} \] Look up AISC tables for standard channels (assume C12x30, C12x25, etc.). **Example:** Try **C12x30** (AISC Steel Manual): - Area, \( A_g \) = 8.82 in² - \( P_n = A_g \times F_y = 8.82 \times 36 = 317.52\,k \) - For two channels: \( 317.52 \times 2 = 635.04\,k \) > 288.9 k **C12x25** (A_g = 7.37 in²): - \( 7.37 \times 36 = 265.32\,k \) - \( \times 2 = 530.64\,k \) > 288.9 k **Conclusion:** C12x25 is adequate (verify slenderness, but for this, assume adequate for direct compression). --- ## **b) A572 Gr 60 (Fy = 60 ksi)** \[ P_n = A_g \times F_y \] For same load: \[ P_n = 288.9\,k \rightarrow \frac{288.9}{2} = 144.45\,k \text{ per channel} \] Try **C10x20** (A_g = 5.87 in²): \[ 5.87 \times 60 = 352.2\,k \text{ (per channel)} \] \[ 352.2 \times 2 = 704.4\,k > 288.9\,k \] **C8x18.75** (A_g = 5.51 in²): \[ 5.51 \times 60 = 330.6\,k \] \[ 330.6 \times 2 = 661.2\,k > 288.9\,k \] **Conclusion:** C8x18.75 is adequate for A572 Gr 60. --- # **Step 3: Design Fillet Welds** ## **Weld Strength Calculation** - Gusset plate thickness = 5/8" = .625" - Use fillet weld on both sides of each channel web (assume effective weld length \( L \)) - Strength of fillet weld (per AISC): \( \phi R_n = .75 \times .707 \times w \times F_{EXX} \times L \) Where: - \( w \) = weld size (assume 5/16" for example) - \( F_{EXX} \) = 70 ksi (E70XX electrode) - \( L \) = length of weld on one side \[ \text{Strength per unit length} = .75 \times .707 \times \frac{5}{16}" \times 70\,ksi = .75 \times .707 \times .3125 \times 70 = 11.6\,kips/in \] ### **Required Weld Length** Total load per channel = 144.45 k Weld on both sides: \( 2L \) \[ 2L \times 11.6 = 144.45 \implies L = \frac{144.45}{2 \times 11.6} = 6.23\,in \] **Use weld length \( L = 6.5 \) in per side for safety.** --- # **Step 4: Summary Table** | Case | Channel Size | Area (in²) | Yield Strength (ksi) | Weld Size (in) | Weld Length/side (in) | |---------------|---------------|------------|----------------------|----------------|-----------------------| | **A36** | 2C12x25 | 7.37 | 36 | 5/16 | 6.5 | | **A572 Gr 60**| 2C8x18.75 | 5.51 | 60 | 5/16 | 6.5 | --- # **Final Answers** ### **a) For A36:** - Use two C12x25 channels. - Provide 5/16" fillet welds at least 6.5" long on each side. ### **b) For A572 Gr 60:** - Use two C8x18.75 channels. - Provide 5/16" fillet welds at least 6.5" long on each side. --- **Note:** Always verify slenderness and local buckling for final design. Weld size may be increased if minimum fillet weld requirements (per AISC) are higher due to plate thickness. Adjust channel selection based on available section tables and actual design constraints.