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