Influence-line application: a simply supported beam of span 20 m carries a moving uniformly distributed load (UDL) of intensity 4 t/m over a loaded length of 5 m. What is the absolute maximum bending moment and where does it occur?

Introduction / Context:For moving loads on simply supported beams, the absolute maximum bending moment is found using the influence line for bending moment at a section. For a partial uniformly distributed load (UDL) shorter than the span, the maximum bending moment at a given section occurs when the load is placed symmetrically about that section. This problem applies the influence-line method to compute the peak value and confirm its location.

Given Data / Assumptions:

• Span L = 20 m (simply supported).
• UDL intensity w = 4 t/m acting over a length a = 5 m.
• Beam is prismatic, linear-elastic; self-weight neglected compared to moving load.
• We seek the absolute maximum bending moment under the moving UDL.

Concept / Approach:The influence line (IL) ordinate for bending moment at midspan of a simply supported beam is given by: ordinate at a distance s from midspan equals L/4 − s^2 / L. For a UDL of length a centered at midspan, the bending moment is the integral of the IL over the loaded length multiplied by w.

Step-by-Step Solution:Place the 5 m UDL symmetrically about the midspan to maximize the moment there.Use IL ordinate at distance s from midspan: y(s) = L/4 − s^2 / L.Compute M_max at midspan: M = w ∫ from s = −a/2 to +a/2 of (L/4 − s^2 / L) ds.Evaluate integral: ∫(L/4) ds = (L a)/4; ∫(s^2/L) ds = (a^3)/(12 L).Thus, M = w * [ (L a)/4 − (a^3)/(12 L) ].Substitute L = 20 m, a = 5 m, w = 4 t/m → M = 4 * [ (20*5)/4 − (125)/(12*20) ] t·m.Compute numerically: (20*5)/4 = 25; (125)/(240) ≈ 0.520833; bracket ≈ 24.479167.Therefore M ≈ 4 * 24.479167 = 97.916668 t·m ≈ 97.9 t·m at midspan.

Verification / Alternative check:Using “equivalent concentrated load” intuition, total load w a = 20 t placed at midspan gives 100 t·m; accounting for load spread slightly reduces to ≈ 97.9 t·m, matching the integral.

Why Other Options Are Wrong:

• 87.5 t·m variants underestimate the moment; correct IL integration yields ≈ 97.9 t·m.
• 3.5 t·m is far too low for the given load and span.
• 125.0 t·m exceeds the point-load upper bound (20 * 20 / 4 = 100 t·m).

Common Pitfalls:Placing the UDL off-center; using point-load IL directly without integrating; arithmetic slips in a^3/(12L) term.

Final Answer:97.9 t·m at the midpoint