Maximum Bending Moment Beam

Maximum bending moment in a beam

On a simply supported beam carrying a uniformly distributed load, the largest bending moment lands at midspan and works out to M_max = wL² / 8. Swap that for a concentrated load P at midspan and you get M_max = PL / 4. Plug in w = 12 kN/m and L = 6 m and the moment comes out to 54 kN·m. On the resistance side, a section can carry M_R = W·f_y, where W is the elastic section modulus and f_y the yield strength (around 250 MPa for ASTM A36 or steel grade C-25).

When you design, M_max is what fixes the section you need. Divide it by f_y/γ to find W_req, then grab the smallest I, W, H or channel profile that clears that value. Run the ultimate-limit-state checks under NBR 6118 for concrete and NBR 8800 for steel, keeping the load factors γ_g and γ_q in the mix.

Applications

Sizing floor beams, lintels, roof purlins and pergolas. Quick sanity checks before you open FEM software like SAP2000, Robot or RFEM. Picking steel profiles I, W and H straight from Gerdau or ArcelorMittal tables. Laying out reinforcement in reinforced-concrete beams. And early-stage cost estimates per metre of beam.

FAQ

Why does M_max occur at midspan for UDL? The shear V = wL/2 − wx hits zero exactly at x = L/2, and the moment peaks right where the shear flips sign.

What about cantilevers? A cantilever of length L under UDL reaches M_max = wL²/2 at the fixed end, which is four times what the simply supported beam sees.

How do I get the section modulus W? Use W = I / y_max. A rectangle b×h gives W = b·h²/6, and for rolled profiles the catalogue prints W for you.

Is the formula valid for continuous beams? Not really. Once you have two or more spans you need the three-moment theorem or matrix methods. The midspan moment falls to about wL²/14, and a hogging moment of wL²/8 shows up over the interior supports.