A balcony is supported by two wood cantilever beams, fixed within the wall. The schematic below shows one of the cantilever beams which sustains a distributed load q=2 kN/m and a concentrated load P=5 kN at the free end of the beam. The allowable strength of the timber used in the structure is []=10 MPa. The cross section of the beam has a rectangular shape: width 0.12 m; the height-to-width ratio h/b=2.

Calculate the maximum normal stress in the beam and determine if the beam would fail.

Question

A balcony is supported by two wood cantilever beams, fixed within the wall. The schematic below shows one of the cantilever beams which sustains a distributed load q=2 kN/m and a concentrated load P=5 kN at the free end of the beam. The allowable strength of the timber used in the structure is []=10 MPa. The cross section of the beam has a rectangular shape: width 0.12 m; the height-to-width ratio h/b=2.

Calculate the maximum normal stress in the beam and determine if the beam would fail.

BlackTom AI · Accepted Answer

Question restatement and given data:
- A balcony is supported by two fixed-end cantilever beams embedded in the wall. One cantilever carries a distributed load q = 2 kN/m and a concentrated load P = 5 kN at the free end. The timber has allowable normal stress σ_allow = 10 MPa. Cross-section is rectangular with width b = 0.12 m and height-to-width ratio h/b = 2, so h = 0.24 m. The length of the cantilever beam is not explicitly given in the prompt.

What you must determine:
- Maximum normal (bending) stress in the beam, using σ = M c / I, and compare with the allowable strength to decide whether the beam would fail.

Important missing information:
- The beam length L is required to compute the bending moment at the fixed end. Without L, you cannot obtain M, and thus cannot determine σ = M c / I.
- The prompt also doesn’t include the full set of answer options, so we cannot evaluate alternatives. We can nonetheless show how the calculation would proceed once L is known.

Cross-section properties calculation:
- For a rectangle with width b and height h, second moment of area I = b h^3 / 12.
  Here, h = 0.24 m, b = 0.12 m, so h^3 = 0.24^3 = 0.013824 m^3, and I = 0.12 * 0.013824 / 12 ≈ 0.00013824 m^4.
- The distance from the neutral axis to the outer fiber is c = h/2 = 0.24 / 2 = 0.12 m.
- Therefore, yielding σ = M c / I depends on the fixed-end moment M at the wall.

Moment at the fixed end for a cantilever with both a uniform (distributed) load and a tip load:
- For a cantilever of length L with distributed load q (kN/m) along its length and a concentrated load P at the free end, the end reactions produce a fixed-end bending moment:
  M = q L^2 / 2 + P L.
  Plugging q = 2 kN/m and P = 5 kN gives M = (2) L^2 / 2 + 5 L = L^2 + 5 L  (in kN·m).

Stress calculation:
- With I ≈ 1.3824e-4 m^4 and c = 0.12 m, the bending stress is σ = M c / I.
- Convert M to N·m for consistency: M (kN·m) × 1000 = M_Nm.
  So σ = (M_Nm × 0.12) / 1.3824e-4.
- To compare with the allowable 10 MPa, we can determine the maximum M that the section can carry before σ reaches 10 MPa:
  Set σ_allow = 10e6 Pa = M c / I, so M_allow = σ_allow I / c.
  M_allow = (10e6 Pa) × (1.3824e-4 m^4) / (0.12 m) ≈ 11,520 N·m = 11.52 kN·m.

Critical_length condition for safety:
- The actual M = L^2 + 5L (in kN·m). To be safe, M ≤ 11.52 kN·m.
- Solve the inequality: L^2 + 5L ≤ 11.52.
  Solve L^2 + 5L − 11.52 ≤ 0. The positive root of L^2 + 5L − 11.52 = 0 is
  L = [−5 + sqrt(5^2 + 4 × 11.52)] / 2 = [−5 + sqrt(25 + 46.08)] / 2 = [−5 + sqrt(71.08)] / 2 ≈ [−5 + 8.433] / 2 ≈ 1.7165 m.
- Therefore, if the cantilever length L ≤ about 1.72 m, the maximum bending stress will be ≤ 10 MPa and the beam would be considered safe under the given loading; if L > 1.72 m, the beam would exceed the allowable stress and could fail.

Notes on the provided answer candidate:
- The prompt includes a single candidate statement: 'The maximum normal stress σ = 8.46 MPa, so the beam is safe.' This aligns with a scenario where L is such that σ ≈ 8.46 MPa (below 10 MPa). However, since the actual length L is not specified in the question, we cannot independently verify that σ = 8.46 MPa without L. The safe conclusion depends on L; if L is short enough (≤ ~1.72 m, as shown), σ could be below 10 MPa. If L is larger, σ would exceed 10 MPa.

Summary of reasoning flow (without asserting a final verdict):
- Identify the need for L to compute M = q L^2 / 2 + P L.
- Compute I and c from the given cross-section dimensions.
- Express M_allow for the given material strength from σ_allow = M c / I, yielding a numeric capacity for M in N·m (about 11.52 kN·m).
- Translate the condition M = L^2 + 5 L (in kN·m) being ≤ 11.52 kN·m into a maximum safe length L ≈ 1.72 m.
- Conclude that the beam would be safe only if L ≤ ~1.72 m; otherwise, it would fail according to the 10 MPa allowable stress.

If you provide the beam length L or the full set of answer choices, I can finalize the assessment and show the exact numerical σ and the verdict for each option.

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