Bending Stress Check: Timber Garage Shelf Beam
1. Verdict
❌ Errors found. Your moment and your mm³ conversion are both correct. The error is in the final division, which is off by a factor of 10: 4,000,000 ÷ 491,610 is 8.14, not 81.4.
Corrected answer: σ ≈ 8.14 MPa ≈ 1,180 psi. This is about 81% of your stated 10 MPa allowable, not 8× over it.
2. What I understood
| Quantity | Symbol | Value | Unit | Source |
|---|---|---|---|---|
| Uniform distributed load | w | 2 | kN/m | given |
| Span | L | 4 | m | given |
| Elastic section modulus | S | 30 | in³ | given |
| Allowable bending stress | σ_allow | 10 | MPa | given (not independently verified) |
| Support condition | n/a | Simply supported | n/a | given |
Goal: Find the maximum bending stress in a simply supported timber shelf beam under uniform load, report it in MPa and psi, and compare it with the allowable stress.
3. Assumptions
- L = 4 m is the clear span between supports.
- w = 2 kN/m is a force per metre (not a mass) and already includes everything you want to check, such as shelf self-weight.
- S = 30 in³ is the elastic section modulus about the bending axis, with the beam oriented as intended (e.g. on edge).
- The load is static, and simple elastic bending (σ = M/S) applies.
- The 10 MPa allowable comes from your own source. I checked only the comparison, not the value.
4. Unit conversion table
| From | To | Factor (exact/rounded) | Result |
|---|---|---|---|
| 1 in | mm | 25.4 (exact) | 25.4 mm |
| 1 in³ | mm³ | 25.4³ = 16,387.064 (exact) | 16,387.064 mm³ |
| S = 30 in³ | mm³ | × 16,387.064 (exact) | 491,611.92 mm³ |
| M = 4 kN·m | N·mm | × 10⁶ (exact: 10³ for kN→N, 10³ for m→mm) | 4,000,000 N·mm |
| M = 4 kN·m | lbf·in | 1 lbf·in = 4.4482216152605 × 0.0254 = 0.112984829 N·m (exact-derived) | 35,402.8 lbf·in |
| σ = 8.1365 MPa | psi | 1 psi = 6,894.757 Pa (rounded) | 1,180.1 psi |
| σ_allow = 10 MPa | psi | 1 psi = 6,894.757 Pa (rounded) | 1,450.4 psi |
| Total load wL = 8 kN | lbf | 1 lbf = 4.4482216152605 N (exact) | 1,798.5 lbf |
| Total load 8 kN | equivalent mass | ÷ g₀ = 9.80665 m/s² (exact) | ≈ 815.8 kg |
5. Step-by-step check
| Step | Customer's working | My recomputation | Status | Note |
|---|---|---|---|---|
| 1. Max moment | M = 2 × 4² / 8 = 4 kN·m | 2 × 16 = 32; 32 ÷ 8 = 4 kN·m | ✅ | wL²/8 is the correct midspan moment for a simply supported beam under uniform load. |
| 2. Convert S | 30 × 16,387 = 491,610 mm³ | 30 × 16,387.064 = 491,611.92 mm³ | ✅ (minor) | Using the exact factor changes S by only 0.0004%. Negligible. |
| 3. Convert M | 4 × 1000 = 4000 N·mm, then ×1000 = 4,000,000 N·mm | 4 kN·m × 10³ = 4,000 N·m; × 10³ = 4,000,000 N·mm | ⚠️ | The final value is correct. The intermediate label is wrong: 4,000 is in N·m, not N·mm. |
| 4. Stress | σ = 4,000,000 / 491,610 = 81.4 MPa | 4,000,000 / 491,611.92 = 8.1365 N/mm² = 8.1365 MPa | ❌ | Decimal slip, off by a factor of 10. |
| 5. Compare to allowable | 81.4 > 10, "fails badly" | 8.14 / 10 = 0.81 (81% of allowable) | ❌ | The conclusion reverses: the computed stress is below the stated allowable. |
Arithmetic for Step 4:
- 491,611.92 × 8 = 3,932,895.36
- 4,000,000 − 3,932,895.36 = 67,104.64
- 67,104.64 ÷ 491,611.92 = 0.13650
- σ = 8.1365 MPa
Cross-check in US units:
- M = 4,000 N·m ÷ 0.112984829 N·m per lbf·in = 35,402.8 lbf·in
- σ = 35,402.8 ÷ 30 = 1,180.1 psi
- This agrees with 8.1365 MPa ÷ 6,894.757 Pa/psi = 1,180.1 psi. ✅
6. Errors and fixes
Error 1 (❌ main error): division off by ×10
- What went wrong: 4,000,000 ÷ 491,610 was evaluated as 81.4 instead of 8.14.
- Why: This is a decimal-place slip in the division. Quick check: 491,610 × 81.4 ≈ 40,017,000, which is 10× too big.
- Corrected working: σ = 4,000,000 N·mm ÷ 491,611.92 mm³ = 8.1365 N/mm² = 8.14 MPa
- Effect: The stress drops by a factor of 10 (−90%). The utilisation changes from "814%" to 81%, which reverses your pass/fail conclusion.
Issue 2 (⚠️ presentation): intermediate unit label
- What went wrong: "4 × 1000 = 4000 N·mm" should read 4000 N·m.
- Why it matters: The final value was still right. A mislabelled intermediate value is a common source of 10³ errors on later steps.
- Cleaner way: Write 4 kN·m × 10⁶ = 4,000,000 N·mm in one step.
- Effect on result: None.
Issue 3 (minor): rounded conversion factor
- What happened: You used 16,387 instead of the exact 16,387.064.
- Effect: 0.0004%. Negligible, but the exact value costs nothing.
7. Dimensional analysis
- Moment: M = wL²/8 gives (kN/m) × m² = kN·m ✅
- Stress in SI: σ = M/S gives N·mm ÷ mm³ = N/mm² = MPa ✅ (since 1 N/mm² = 10⁶ N/m² = 1 MPa)
- Stress in US units: lbf·in ÷ in³ = lbf/in² = psi ✅
8. Sanity check
Order of magnitude:
- σ ≈ 4 × 10⁶ ÷ 5 × 10⁵ ≈ 8 MPa.
- 81 MPa would need S to be about 50,000 mm³, which is 10× smaller than you have. That flags the slip immediately.
Section size:
- 30 in³ is roughly the section modulus of a nominal 2×12 lumber joist on edge.
- Actual size is 1.5 × 11.25 in, so S = bh²/6 = 1.5 × 11.25² / 6 ≈ 31.6 in³.
- That is a substantial member, so a single-digit MPa stress is plausible.
Load size:
- 2 kN/m over 4 m is 8 kN total, roughly the weight of 816 kg.
- That is a heavy, conservative load for a garage shelf. If your real load is lighter, the stress scales down in direct proportion.
Material range:
- Typical allowable bending stresses for common structural softwood are approximately on the order of 5–15 MPa (about 700–2,200 psi).
- This varies strongly with species, grade, size, moisture and design code.
- 8.1 MPa sits inside this typical range. Your 10 MPa allowable is not independently verified; confirm it against your source.
9. Final answer
| Quantity | SI | US customary |
|---|---|---|
| Max bending moment | 4 kN·m (4,000,000 N·mm) | 35,400 |