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Give step-by-step solution with explanation and final answer:Safe Load Tables: Problem 3: Design a simply supported steel bear to carry a unformiy distributed ive load of 4 Kips! ft and a superimposed unifomly distributed dead load of 2.5 kip! ft on a span of 16 f. Your choice should be the ightest shage.

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Give step-by-step solution with explanation and final answer:Uploaded ImageSafe Load Tables: Problem 3: Design a simply supported steel bear to carry a unformiy distributed ive load of 4 Kips! ft and a superimposed unifomly distributed dead load of 2.5 kip! ft on a span of 16 f. Your choice should be the ightest shage.

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Asked by: varrisanju5@gmail.com
Created at: 2025-10-29 06:05:33

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# Step-by-Step Solution **Given:** - Simply supported steel beam - Live load, \( w_L = 4 \) kips/ft - Dead load, \( w_D = 2.5 \) kips/ft - Span, \( L = 16 \) ft - Find: The lightest steel shape --- ## **Step 1: Calculate Total Uniform Load** \[ w_{total} = w_L + w_D = 4 + 2.5 = 6.5\ \text{kips/ft} \] --- ## **Step 2: Maximum Moment for Simply Supported Beam** The maximum moment for a uniformly distributed load on a simply supported beam is: \[ M_{max} = \frac{w_{total}L^2}{8} \] Substitute the values: \[ M_{max} = \frac{6.5 \times 16^2}{8} \] \[ M_{max} = \frac{6.5 \times 256}{8} \] \[ M_{max} = \frac{1664}{8} \] \[ M_{max} = 208\ \text{kip-ft} \] Convert to **kip-in** (1 ft = 12 in): \[ M_{max} = 208 \times 12 = 2496\ \text{kip-in} \] --- ## **Step 3: Select a Trial Section (AISC Table)** We must ensure the beam's nominal moment capacity \((M_n)\) is **greater than** the required moment (with safety factor considerations). Assume LRFD design (Load and Resistance Factor Design): \[ \text{Required } \phi M_n \geq M_u \] - \(\phi\) (steel flexure) = .9 - \(M_u\) = factored moment ### **Factored Load (LRFD):** \[ w_u = 1.2w_D + 1.6w_L = 1.2(2.5) + 1.6(4) = 3 + 6.4 = 9.4\ \text{kips/ft} \] \[ M_u = \frac{w_u L^2}{8} = \frac{9.4 \times 16^2}{8} = \frac{9.4 \times 256}{8} = \frac{2406.4}{8} = 300.8\ \text{kip-ft} \] \[ M_u = 300.8 \times 12 = 3609.6\ \text{kip-in} \] \[ \phi M_n \geq 3609.6\ \text{kip-in} \] \[ M_n \geq \frac{3609.6}{.9} = 401.7\ \text{kip-in} \] --- ## **Step 4: Use AISC Table to Select Lightest Shape** Look for the lightest W-shape (wide flange) in the AISC Steel Manual with a nominal flexural strength \( M_n \geq 401.7\ \text{kip-in} \). - **\( M_n = F_y S_x \)** - \( F_y \) = 50 ksi (typical for ASTM A992) - \( S_x = \frac{M_n}{F_y} = \frac{401.7}{50} = 80.21\ \text{in}^3 \) Check AISC table for the lightest W-shape with \( S_x \geq 80.21 \) in³: - **W16x26:** \( S_x = 80.2 \) in³ - Weight = 26 lb/ft (lightest at or above required \( S_x \)) --- ## **Step 5: Final Answer** ### **The lightest suitable section is:** \[ \boxed{\text{W16} \times 26} \] - \( S_x = 80.2 \) in³ - Weight: 26 lb/ft - Satisfies the required moment capacity. --- ### **Summary Table** | Step | Value / Section | |-----------------------|-----------------------| | Total load | 6.5 kips/ft | | Factored moment (\(M_u\)) | 3609.6 kip-in | | Required \( S_x \) | 80.21 in³ | | Lightest shape | W16x26 | --- **Always double-check with the most current AISC Manual for official values.**

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