# Step-by-Step Solution: Selecting the Lightest Section for Axial Load ## Given Data - **Dead Load (DL):** 600 kips - **Live Load (LL):** 300 kips - **Total Axial Load, \(P_u\):** (to be calculated) - **Element Length (\(L)):** 28 ft = 336 in - **Bracing in \(y\) direction:** 14 ft from base = 168 in - **End Conditions:** Fixed at base, pinned at top - **Steel Grade:** ASTM A572 Gr. 60 (\(F_y = 60\) ksi) - **Check:** Slenderness, Local Buckling --- ## 1. Factored Axial Load (\(P_u\)) ASD and LRFD have different load combinations. Assume **LRFD** (most common for strength design): \[ P_u = 1.2 \times DL + 1.6 \times LL = 1.2 \times 600 + 1.6 \times 300 = 720 + 480 = \boxed{120\ \text{kips}} \] \[ P_u = 1,200,000\ \text{lbs} \] --- ## 2. Effective Length (\(K\)), Unbraced Length (\(L_{KL}\)) - **Length, \(L\):** 28 ft = 336 in - **Bracing:** Only in \(y\), so use full length for \(x\) axis, 14 ft for \(y\) axis. - **End Conditions:** - **Fixed-Pinned:** \(K \approx .7\) (AISC Table C-C2.2) - For y-y axis (braced at 14 ft): \(L_y = 14\) ft, \(K_y = 1.\) (assuming conservative value for braced axis) --- ## 3. Slenderness Ratio (\(\lambda\)) \[ \lambda = \frac{K L}{r} \] Where \(r\) is radius of gyration. --- ## 4. Required Nominal Strength (\(P_n\)) \[ \phi P_n \geq P_u \] \[ \phi = .9\ \text{(for compression, AISC)} \] \[ P_n = \frac{P_u}{\phi} = \frac{1,200,000}{.9} = 1,333,333\ \text{lbs} = 1,333\ \text{kips} \] --- ## 5. Select Lightest Section ### a. Try a Wide Flange (W) Section Let's check what area is theoretically needed: \[ P_n = F_{cr} \cdot A_g \] Where \(F_{cr} =\) critical buckling stress (AISC E3; Euler or inelastic buckling). --- ### b. Calculate Slenderness for Trial Section For a preliminary trial, let's try **W14x90** (just for calculation; will check lighter sections). From AISC Steel Manual (values for W14x90): - \(A_g = 26.5\ \text{in}^2\) - \(r_x = 5.32\ \text{in}\) - \(r_y = 2.06\ \text{in}\) #### Slenderness about x-x axis (unbraced, 28 ft): \[ K_x = .7,\ L_x = 336\ \text{in} \] \[ \lambda_x = \frac{.7 \times 336}{5.32} = \frac{235.2}{5.32} \approx 44.2 \] #### Slenderness about y-y axis (braced, 14 ft): \[ K_y = 1.,\ L_y = 168\ \text{in} \] \[ \lambda_y = \frac{1. \times 168}{2.06} = \frac{168}{2.06} \approx 81.6 \] **Use larger slenderness:** \(\lambda_y = 81.6\) --- ## 6. Calculate \(F_{cr}\) (Critical Stress) AISC E3: \[ \lambda_c = \frac{K L}{r} \cdot \sqrt{\frac{F_y}{\pi^2 E}} \] But more practically: \[ \lambda_c = \frac{1}{\pi} \cdot \frac{K L}{r} \cdot \sqrt{\frac{F_y}{E}} \] Let’s use the AISC formula: \[ \lambda_c = \frac{1}{\pi} \cdot \frac{K L}{r} \cdot \sqrt{\frac{F_y}{E}} \] Where: - \(F_y = 60\) ksi - \(E = 29000\) ksi \[ \lambda_c = \frac{81.6}{\pi} \sqrt{\frac{60}{29000}} = \frac{81.6}{3.142} \times \sqrt{.00207} \] \[ = 25.98 \times .0455 = 1.18 \] Since \(\lambda_c > 1.5\), use: \[ F_{cr} = .877 \frac{\pi^2 E}{(K L/r)^2} \] \[ F_{cr} = .877 \frac{\pi^2 \cdot 29000}{81.6^2} \] \[ = .877 \cdot \frac{286,870}{6,658} \] \[ = .877 \cdot 43.12 = 37.84\ \text{ksi} \] --- ## 7. Find Required Area \[ P_n = F_{cr} \cdot A_g \] \[ A_g = \frac{P_n}{F_{cr}} = \frac{1,333,333}{37,840} = 35.2\ \text{in}^2 \] But **W14x90** only has \(A_g = 26.5\ \text{in}^2\), **NOT ENOUGH**. --- ## 8. Try a Heavier Section Let’s try **W14x132**: - \(A_g = 38.8\ \text{in}^2\) - \(r_x = 6.44\ \text{in}\) - \(r_y = 2.18\ \text{in}\) \[ \lambda_{y} = \frac{168}{2.18} = 77.1 \] \[ F_{cr} = .877 \frac{286,870}{77.1^2} = .877 \cdot \frac{286,870}{5,946} = .877 \cdot 48.26 = 42.33\ \text{ksi} \] \[ A_g = \frac{1,333,333}{42,330} = 31.5\ \text{in}^2 \] **W14x132** provides \(A_g = 38.8 \text{ in}^2\), which is adequate. --- ## 9. Try a Slightly Lighter Section **W14x120:** - \(A_g = 35.4\ \text{in}^2\) - \(r_y = 2.14\ \text{in}\) - \(\lambda_y = 168/2.14 = 78.5\) - \(F_{cr} = .877 \cdot 286,870 / 78.5^2 = .877 \cdot 46.53 = 40.8\) ksi \[ A_g = \frac{1,333,333}{40,800} = 32.7\ \text{in}^2 \] **W14x120** has \(A_g = 35.4\ \text{in}^2\), which is adequate. --- ## 10. Check Even Lighter Section **W14x109:** - \(A_g = 32.2\ \text{in}^2\) - \(r_y = 2.11\ \text{in}\) - \(\lambda_y = 168/2.11 = 79.6\) - \(F_{cr} = .877 \cdot 286,870 / 79.6^2 = .877 \cdot 45.29 = 39.7\) ksi \[ A_g = \frac{1,333,333}{39,700} = 33.6\ \text{in}^2 \] **W14x109** only has \(A_g = 32.2\), which is slightly less than required. --- ## 11. **Final Selection** The lightest section that meets the required area is **W14x120**. --- ## 12. **Check Local Buckling (AISC Table 4-17)** For W14x120: - \(b_f/2t_f = 14.3/2 / 1.26 = 5.67\) - \(h/t_w = 12.97/.685 = 18.9\) Both are **less than λr** limits for A572 Gr. 60 (per AISC Table B4.1b). Thus, **no local buckling concern**. --- # **Final Answer** ## **Use W14x120 (A572 Gr. 60) as the lightest section.** - **Area:** 35.4 in² - **Critical Buckling Stress:** ≈ 40.8 ksi - **Factored Axial Strength:** φPn = .9 × 35.4 × 40.8 = 1,298 kips (> 1,200 kips required) --- ## **Summary Table** | Section | Area (in²) | φPn (kips) | Adequate? | |-----------|------------|------------|-----------| | W14x90 | 26.5 | 973 | No | | W14x109 | 32.2 | 1,151 | No | | **W14x120** | **35.4** | **1,298** | **Yes** | | W14x132 | 38.8 | 1,480 | Yes | --- ## **Key Checks** - **Slenderness:** OK (\(< 200\)) - **Local Buckling:** OK (per AISC limits) --- ## **Conclusion** **Select W14x120 (A572 Gr. 60) for the column.** --- **Note:** If you are allowed to use built-up sections or HSS, you can check those for lighter options. If using ASD, multiply required strength by .6 for required area. **Always confirm with current AISC Steel Manual and local codes.** --- **[No image included; table and calculations provided.]**