Simply Supported Beams

Simply Supported Beam with Symmetric Triangular Distributed Load: Reactions, Shear and Moment Formulas

A simply supported beam rests on a pin at one end and a roller at the other. It is statically determinate: the reactions follow from equilibrium alone, and the bending moment is zero at both supports. Below are the closed-form formulas for this load case: support reactions, maximum shear, maximum bending moment and the shear and moment equations along the beam.

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Load case and diagrams

Simply Supported Beam with Symmetric Triangular Distributed Load

Formulas

Reactions

R=wL4R = \frac{wL}{4}

Maximum shear force

−Vmax=−R-V_{max} = -R
+Vmax=+R+V_{max} = +R

Maximum bending moment

Mmax=wL212M_{max} = \frac{wL^2}{12}

Shear and moment equations along the beam

Shear force, V(x)

V(x)={−w4L(L2−4x2)ifx<L2w4L(L2−4(L−x)2)ifx≥L2V_{(x)}= \left\{ \begin{array}{lcc} -\frac{w}{4L}(L^2-4x^2) & \text{if} & x < \frac{L}{2} \\ \\ \frac{w}{4L}(L^2-4(L-x)^2) & \text{if} & x \geq \frac{L}{2} \end{array} \right.

Bending moment, M(x)

M(x)={wLx2(12−2x23L2)ifx<L2wL(L−x)2(12−2(L−x)23L2)ifx≥L2M_{(x)}= \left\{ \begin{array}{lcc} \frac{wLx}{2}(\frac{1}{2}-\frac{2x^2}{3L^2}) & \text{if} & x < \frac{L}{2} \\ \\ \frac{wL(L-x)}{2}(\frac{1}{2}-\frac{2(L-x)^2}{3L^2}) & \text{if} & x \geq \frac{L}{2} \end{array} \right.

Variables

SymbolMeaning
wwdistributed load value
LLspan between supports
xxdistance along the beam measured from the left end
RRsupport reaction

Sign convention: positive moment is sagging (tension at the bottom); negative moment is hogging (tension at the top), as in cantilevers and over interior supports.

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