Simply supported beam: reactions, maximum moment and bending stress

A simply supported steel beam AB spans 6 m. It carries a uniformly distributed load of 12 kN/m over its full length (self-weight included) and a concentrated load of 30 kN at 2 m from support A. The beam is a W310×39 of ASTM A992 steel. Taking the allowable bending stress as 0.6·Fy, find the reactions, the maximum bending moment and where it occurs, and whether the beam is adequate in bending.

Given

Solution

1. 1.Moments about A give the reaction at B
   $$R_B=\frac{w{L}^2/2+Pa}{L}$$
   $$R_B=46\mathrm{kN}$$
2. 2.Vertical equilibrium gives the reaction at A
   $$R_A=wL+P-R_B$$
   $$R_A=56\mathrm{kN}$$
3. 3.Shear is zero beyond the point load, at
   $$x_m=\frac{R_A-P}{w}$$
   $$x_m=2.167\mathrm{m}$$
4. 4.The moment there is the maximum
   $$M_{max}=R_A{x}_m-\frac{w{x}_m^2}{2}-P(x_m-a)$$
   $$M_{max}=88.17\mathrm{kN}\cdot \mathrm{m}$$
5. 5.Bending stress from the elastic section modulus
   $${\sigma }_{max}=\frac{M_{max}}{S_x}$$
   $${\sigma }_{max}=161.2\mathrm{MPa}$$
6. 6.Allowable bending stress
   $${\sigma }_{allow}=0.6F_y$$
   $${\sigma }_{allow}=207\mathrm{MPa}$$

Worked in ProveCalc

The free body diagram node solves the same reactions and draws the shear and moment diagrams, so the zero-shear location is checked two ways. Units carry through every line: kN·m divided by mm³ comes out in MPa without a conversion step to get wrong.

The results above were evaluated by ProveCalc's compute engine (SymPy and Pint), the same engine the desktop app runs on your machine.

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