Simply supported beam of span L with triangular load (zero at one end, maximum at the other) A simply supported prismatic beam of span L is subjected to a linearly varying (triangular) load that is zero at the left support and increases to a maximum at the right support. If W denotes the total load on the span, the maximum sagging bending moment M_max is:
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AM_max = (2 W L) / (9√3)
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BM_max = W L / 8
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CM_max = W L / 6
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DM_max = W L / 9
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EM_max = (W L) / (3√3)
Answer
Correct Answer: M_max = (2 W L) / (9√3)
Explanation
Introduction / Context:Linearly varying (triangular) loads are common idealizations of wind, soil, or hydrostatic pressures. For a simply supported beam, the position of maximum bending moment is not necessarily midspan; it is located where the shear force becomes zero. Expressing the answer in terms of total load W avoids ambiguity between W (total) and w (intensity).
Given Data / Assumptions:
- Span L, triangular load intensity w(x) increasing from 0 at x = 0 to w0 at x = L.
- Total load W = (1/2) w0 L.
- Small deflection linear theory; supports are simple (pin and roller).
Concept / Approach:
Let w(x) = k x where k = w0/L. Determine reactions from equilibrium, write the shear V(x) by integrating the load, find x where V(x) = 0, then evaluate M(x) there to obtain M_max. Finally express the result in terms of W and L.
Step-by-Step Solution:
Total load W = ∫₀ᴸ k x dx = k L² / 2 ⇒ k = 2W / L².Moments about A: R_B L = ∫₀ᴸ k x · x dx = k L³ / 3 ⇒ R_B = k L² / 3.Hence R_A = W − R_B = k L² / 2 − k L² / 3 = k L² / 6.Shear: V(x) = R_A − ∫₀ˣ k ξ dξ = k L² / 6 − k x² / 2.Set V = 0 ⇒ x² = L² / 3 ⇒ x = L / √3.Moment: M(x) = R_A x − ∫₀ˣ k ξ (x − ξ) dξ = (k/6)(L² x − x³).At x = L/√3: M_max = (k L³) / (9√3) = (2W/L² · L³) / (9√3) = (2 W L) / (9√3).Verification / Alternative check:
Numerically, (2/(9√3)) ≈ 0.1283, a well-known coefficient for this loading case. The location x = 0.577L matches the zero-shear condition from standard beam tables.
Why Other Options Are Wrong:
- W L / 8 and W L / 6 are for other load patterns (e.g., UDL or central point load, respectively).
- W L / 9 and (W L)/(3√3) do not satisfy equilibrium/compatibility for this load shape.
Common Pitfalls:
- Confusing W (total) with w (intensity). Always convert consistently.
- Assuming maximum moment at midspan; here it occurs at x = L/√3 from the zero-load end.
Final Answer:
M_max = (2 W L) / (9√3).