Let's break down the problem step by step as per your image and text. Since the image contains a structural platform plan, let's interpret and solve the key questions: --- ## **Given Data** - **Platform Dimensions:** 24 ft × 8 ft - **Spacing:** Girders and columns at 6 ft intervals - **Height:** 20 ft steel pipe post - **Loading:** 100 psf dead load + 100 psf live load = **200 psf total load** - **Lumber Allowables:** Fb = 800 psi, Fv = 120 psi, E = 1.2 × 10⁶ psi - **Use 2× planks (actual size: 1.5" × 7.25" for 2×8, 1.5" × 9.25" for 2×10, etc.)** - **Girder Top Flange is continuously supported** - **Use C = 1.5 for bending and deflection** - **Deflection limit: L/240** --- ## **1. What size 2" joists are required?** ### **Step 1: Find Joist Span and Tributary Width** - **Span = 6 ft** (distance between girders) - **Tributary Width = 2 ft** (distance between joists, typical for platforms) ### **Step 2: Calculate Load per Joist** - **Uniform Load per Area:** 200 psf × 2 ft = **400 plf** (pounds/linear foot) ### **Step 3: Maximum Moment and Shear** - **Maximum Moment (simple span):** \[ M_{max} = \frac{wL^2}{8} \] \[ M_{max} = \frac{400 \times 6^2}{8} = \frac{400 \times 36}{8} = 180 \, \text{lb-ft} = 21,600 \, \text{lb-in} \] - **Maximum Shear:** \[ V_{max} = \frac{wL}{2} = \frac{400 \times 6}{2} = 120 \, \text{lb} \] ### **Step 4: Section Properties for 2×8, 2×10, 2×12** - **2×8:** \(d = 7.25"\), \(b = 1.5"\) - **2×10:** \(d = 9.25"\), \(b = 1.5"\) - **2×12:** \(d = 11.25"\), \(b = 1.5"\) - **Section Modulus \(S\):** \[ S = \frac{b d^2}{6} \] - 2×8: \(S = \frac{1.5 \times 7.25^2}{6} = 13.17 \, \text{in}^3\) - 2×10: \(S = \frac{1.5 \times 9.25^2}{6} = 21.4 \, \text{in}^3\) - 2×12: \(S = \frac{1.5 \times 11.25^2}{6} = 31.6 \, \text{in}^3\) - **Allowable Bending Stress:** \(F_b = 800 \, \text{psi} \), but use adjusted value with \(C = 1.5\): \(F_{b,adj} = 800 \times 1.5 = 120 \, \text{psi}\) ### **Step 5: Required Section Modulus:** \[ S_{req} = \frac{M_{max}}{F_{b,adj}} \] \[ S_{req} = \frac{21,600}{120} = 18 \, \text{in}^3 \] - **2×8:** 13.17 < 18 (NG) - **2×10:** 21.4 > 18 (OK) - **2×12:** 31.6 > 18 (OK) **Use 2×10 joists (minimum).** --- ## **2. Is a continuous 2" plank adequate as the deck?** - **Span = 2 ft (between joists)** - **Load = 200 psf × 2 ft = 400 plf** - **Maximum Moment:** \[ M_{max} = \frac{400 \times 2^2}{8} = 200 \, \text{lb-ft} = 2,400 \, \text{lb-in} \] - **Section modulus for 2×8 plank:** 13.17 in³ - **Allowable Moment:** \[ M_{allow} = F_{b,adj} \times S = 120 \times 13.17 = 15,804 \, \text{lb-in} \] - **Required: 2,400 < 15,804** (OK) **Yes, a continuous 2" plank is adequate.** --- ## **3. Choose an economical W shape for the girders.** - **Span = 8 ft (between columns)** - **Tributary width = 6 ft (from joists)** - **Load = 200 psf × 6 ft = 1,200 plf** - **Maximum Moment:** \[ M_{max} = \frac{wL^2}{8} = \frac{1,200 \times 8^2}{8} = 9,600 \, \text{lb-ft} = 115,200 \, \text{lb-in} \] - **Steel Allowable Stress (typical):** 24 ksi - **Required S:** \[ S_{req} = \frac{115,200}{24,000} = 4.8 \, \text{in}^3 \] - **Common W shapes** (from AISC tables): - W8×10: S = 10.3 in³ (OK, economical) - W8×13: S = 13. in³ (OK) **Use W8×10 or W8×13 (check local availability and deflection).** --- ## **4. Choose an economical size for the pipe columns.** - **Load per column:** - Each girder supports (8 ft × 6 ft = 48 ft²) at each column - Total load: 48 ft² × 200 psf = 9,600 lb per column - **20 ft tall column, check slenderness:** - Slenderness limit: \( \frac{KL}{r} < 200 \) (for steel) - Use a pipe (e.g., 4" Sch 40; r ≈ 1.24 in) \[ \frac{20 \times 12}{1.24} = 193.5 < 200 \] - **Capacity (approximate):** 4" Sch 40 pipe > 10,000 lb (OK) **Use 4" Sch 40 steel pipe (verify actual load capacity with detailed column design).** --- ## **5. What is the platform’s maximum deflection? Is it less than L/240?** - **For joists (span = 6 ft = 72 in):** - \(L/240 = 72/240 = .3" \) - **Deflection for uniform load:** \[ \Delta_{max} = \frac{5 w L^4}{384 E I} \] - \(w = 400 \, \text{plf} = 33.3 \, \text{plf-in}\) - \(E = 1.2 \times 10^6 \, \text{psi}\) - \(I_{2x10} = \frac{1.5 \times 9.25^3}{12} = 98.9 \, \text{in}^4\) - \(L = 72 \, \text{in}\) \[ \Delta_{max} = \frac{5 \times 33.3 \times 72^4}{384 \times 1.2 \times 10^6 \times 98.9} \] \[ = \frac{5 \times 33.3 \times 26,873,856}{45,584,640,000} \] \[ = \frac{5 \times 33.3 \times 26,873,856}{45,584,640,000} = \frac{4,468,044}{45,584,640,000} = .098 \text{ in} \] - \(.098" < .3"\) (OK) --- ## **Summary Table** | Component | Selected Size | Adequacy | |---------------------|-----------------|-------------------------| | Joists | 2×10 | OK | | Deck Plank | 2" Dimensional | OK | | Girder | W8×10 | OK, check deflection | | Pipe Column | 4" Sch 40 | OK, check actual design | | Max Deflection | .098" | < L/240 (OK) | --- ## **Final Answers** 1. **Use 2×10 joists.** 2. **A continuous 2" plank is adequate for the deck.** 3. **Use an economical W8×10 (or similar) girder.** 4. **Use a 4" Sch 40 steel pipe for columns.** 5. **Maximum deflection is about .1", which is less than L/240 = .3".** --- If you need calculations for another element (like the girders' deflection or detailed steel checks), let me know!