Two-Span Beams

Two-Span Continuous Beam with Central Point Load in One Zone: Reactions, Shear and Moment Formulas

A two-span continuous beam runs over three supports. Continuity over the middle support produces a negative moment there and reduces the span moments compared with two separate simple spans. 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

Two-Span Continuous Beam with Central Point Load in One Zone

Formulas

Reactions

R1=13P32R_1 = \frac{13P}{32}
R2=11P16R_2 = \frac{11P}{16}
R3=−3P32R_3 = -\frac{3P}{32}

Maximum shear force

V1=−13P32V_1 = -\frac{13P}{32}
V2=19P32V_2 = \frac{19P}{32}
V3=−3P32V_3 = -\frac{3P}{32}

Maximum bending moment

Mmax=13PL64M_{max} = \frac{13PL}{64}
M1=−3PL32M_1 = -\frac{3PL}{32}

Shear and moment equations along the beam

Shear force, V(x)

V(x)={−R1ifx<L2−R1+PifL2≤x≤LR3ifL<x≤2LV_{(x)}= \left\{ \begin{array}{lcc} -R_1 & \text{if} & x < \frac{L}{2} \\ \\ -R_1+P & \text{if} & \frac{L}{2} \leq x \leq L \\ \\ R_3 & \text{if} & L < x \leq 2L \end{array} \right.

Bending moment, M(x)

M(x)={R1xifx<L2(R1−P)(x−L2)+R1L2ifL2≤x≤L−R3(x−2L)ifL<x≤2LM_{(x)}= \left\{ \begin{array}{lcc} R_1x & \text{if} & x < \frac{L}{2} \\ \\ (R_1-P)(x-\frac{L}{2})+\frac{R_1L}{2} & \text{if} & \frac{L}{2} \leq x \leq L \\ \\ -R_3(x-2L) & \text{if} & L < x \leq 2L \end{array} \right.

Variables

SymbolMeaning
LLspan between supports
PPpoint load value
xxdistance along the beam measured from the left end

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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