Beam Formulas: Reactions, Shear and Moment for 37 Load Cases

Closed-form beam formulas for the most common support and load configurations: support reactions, maximum shear force and maximum bending moment, plus the shear V(x) and moment M(x) equations along the beam. The tables cover simply supported beams, propped cantilevers, fixed-fixed beams, cantilevers, overhanging beams and two- and three-span continuous beams.

Open any case for its full set of equations, or press Calculate to solve it with your own values.

Simply Supported Beams

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.

Load case	Reactions	Max. shear	Max. moment
Uniform Distributed Load	$$R = \frac{wL}{2}$$	$$-V_{max} = -R$$ $$+V_{max} = +R$$	$$M_{max} = \frac{wL^2}{8}$$	Calculate →
Triangular Distributed Load	$$R_A = \frac{wL}{6}$$ $$R_B = \frac{wL}{3}$$	$$V_1 = -R_A$$ $$V_{max} = -R_B$$	$$M_{max} = \frac{wL^2}{9\sqrt{3}}$$	Calculate →
Symmetric Triangular Distributed Load	$$R = \frac{wL}{4}$$	$$-V_{max} = -R$$ $$+V_{max} = +R$$	$$M_{max} = \frac{wL^2}{12}$$	Calculate →
Central Uniform Distributed Load	$$R_A = \frac{wb}{2L}(2c + b)$$ $$R_B = \frac{wb}{2L}(2a + b)$$	$$V_1 = -R_A$$ $$V_2 = R_B$$	$$M_{max} = R_A(a+\frac{R_A}{2w})$$	Calculate →
Uniform Load at One End	$$R_A = \frac{wa}{2L}(2L-a)$$ $$R_B = \frac{wa^2}{2L}$$	$$V_1 = -R_A$$ $$V_2 = R_B$$	$$M_{max} = \frac{R_A^2}{2w}$$	Calculate →
Uniform Load at Each End	$$R_A = \frac{w_1a(2L-a)+w_2 c^2}{2L}$$ $$R_B = \frac{w_2c(2L-c)+w_1 a^2}{2L}$$	$$V_1 = -R_A$$ $$V_2 = R_B$$	$$M_{max} = \left\{ \begin{array}{lcc} \frac{R_A^2}{2w_1} & \text{if} & R_A < w_1 a \\ \\ \frac{R_B^2}{2w_2} & \text{if} & R_B < w_2 c \end{array} \right.$$	Calculate →
Central Point Load	$$R = \frac{P}{2}$$	$$V = R$$	$$M_{max} = \frac{PL}{4}$$	Calculate →
Point Load	$$R_A = \frac{Pb}{L}$$ $$R_B = \frac{Pa}{L}$$	$$V_1 = -R_A$$ $$V_2 = R_B$$	$$M_{max} = \frac{Pab}{L}$$	Calculate →
Two Equal Point Loads Uniformly Distributed	$$R = P$$	$$V1 = -R$$ $$V2 = R$$	$$M_{max} = Pa$$	Calculate →
Two Equal Point Loads	$$R_A = \frac{P(L-a+b)}{L}$$ $$R_B = \frac{P(L-b+a)}{L}$$	$$V_1 = -R_A$$ $$V_2 = -R_A+P$$ $$V_3 = R_B$$	$$M_1 = R_A a$$ $$M_2 = R_B b$$	Calculate →
Two Different Point Loads	$$R_A = \frac{P_1(L-a)+P_2b}{L}$$ $$R_B = \frac{P_1a+P_2(L-b)}{L}$$	$$V_1 = -R_A$$ $$V_2 = -R_A+P_1$$ $$V_3 = R_B$$	$$M_1 = R_A a$$ $$M_2 = R_B b$$	Calculate →

Beams Fixed at One End

A propped cantilever is fixed at one end and simply supported (roller) at the other. It is statically indeterminate to the first degree, so the fixed end develops a moment and the reactions are not symmetric.

Load case	Reactions	Max. shear	Max. moment
Distributed Load	$$R_1 = \frac{3wL}{8}$$ $$R_2 = \frac{5wL}{8}$$	$$V_1 = -R_1$$ $$V_{max} = R_2$$	$$M_1 = \frac{9wL^2}{128}$$ $$M_{max} = -\frac{wL^2}{8}$$	Calculate →
Central Point Load	$$R_1 = \frac{5P}{16}$$ $$R_2 = \frac{11P}{16}$$	$$V_1 = -R_1$$ $$V_{max} = R_2$$	$$M_1 = \frac{5PL}{32}$$ $$M_{max} = \frac{3PL}{16}$$	Calculate →
Point Load	$$R_A = \frac{Pb^2}{2L^3}(a+2L)$$ $$R_B = \frac{Pa}{2L^3}(3L^2-a^2)$$	$$V_1 = -R_A$$ $$V_2 = +R_B$$	$$M_1 = aR_A$$ $$M_{max} = -\frac{Pab}{2L^2}(a+L)$$	Calculate →

Beams Fixed at Both Ends

A fixed-fixed beam is clamped at both ends. Both ends resist rotation, so negative (hogging) moments appear at the supports and the midspan moment is smaller than in a simply supported beam.

Load case	Reactions	Max. shear	Max. moment
Distributed Load	$$R = \frac{wL}{2}$$	$$V_{max} = R$$	$$M_1 = \frac{wL^2}{24}$$ $$M_{max} = \frac{wL^2}{12}$$	Calculate →
Central Point Load	$$R = \frac{P}{2}$$	$$V = R$$	$$M_{max} = \frac{PL}{8}$$	Calculate →
Point Load	$$R_A = \frac{Pb^2}{L^3}(3a+b)$$ $$R_B = \frac{Pa^2}{L^3}(a+3b)$$	$$V_1 = -R_A$$ $$V_2 = +R_B$$	$$M_1 = -\frac{Pab^2}{L^2}$$ $$M_2 = \frac{2Pa^2b^2}{L^3}$$ $$M_3 = -\frac{Pa^2b}{L^2}$$	Calculate →

Cantilever Beams

A cantilever beam is fixed at one end and free at the other. All load is carried back to the fixed support, where both the shear and the bending moment are largest.

Load case	Reactions	Max. shear	Max. moment
Triangular Distributed Load	$$R = \frac{wL}{2}$$	$$V_{max} = R$$	$$M_{max} = \frac{wL^2}{6}$$	Calculate →
Uniform Distributed Load	$$R = wL$$	$$V_{max} = R$$	$$M_{max} = \frac{wL^2}{2}$$	Calculate →
Point Load	$$R = P$$	$$V = R$$	$$M_{max} = -Pb$$	Calculate →
End Point Load	$$R = P$$	$$V = R$$	$$M_{max} = -PL$$	Calculate →

Single Span and Cantilever Beams

An overhanging beam rests on two supports and extends past one or both of them. Loads on the overhang produce negative moments over the support and can reduce the positive moment in the span.

Load case	Reactions	Max. shear	Max. moment
Uniform Distributed Load	$$R_A = \frac{w}{2L}(L^2-a^2)$$ $$R_B = \frac{w}{2L}(L+a)^2$$	$$V_1 = R_A$$ $$V_2 = \frac{w}{2L}(L^2+a^2)$$ $$V_3 = wa$$	$$M_1 = \frac{w}{8L^2}(L+a)^2(L-a)^2$$ $$M_2 = \frac{wa^2}{2}$$	Calculate →
End Uniform Distributed Load	$$R_1 = -\frac{w a^2}{2L}$$ $$R_2 = \frac{w a}{2L}(2L+a)$$	$$V_1 = -R_1$$ $$V_2 = -wa$$	$$M_{max} = -\frac{w a^2}{2}$$	Calculate →
End Point Load	$$R_1 = -\frac{Pa}{L}$$ $$R_2 = \frac{P}{L}(L+a)$$	$$V_1 = -R_1$$ $$V_2 = -P$$	$$M_{max} = -Pa$$	Calculate →
Central Uniform Distributed Load in Supported Zone	$$R = \frac{wL}{2}$$	$$V = R$$	$$M_{max} = \frac{wL^2}{8}$$	Calculate →
Point Load	$$R_1 = \frac{Pb}{L}$$ $$R_2 = \frac{Pa}{L}$$	$$V_1 = -R_1$$ $$V_2 = R_2$$	$$M_{max} = \frac{Pab}{L}$$	Calculate →
Uniform Distributed Load Cantilever on Both Sides	$$R_A = \frac{wL(L-2c)}{2b}$$ $$R_B = \frac{wL(L-2a)}{2b}$$	$$V_1 = wa$$ $$V_2 = -R_A+wa$$ $$V_3 = R_B-wc$$ $$V_4 = -wc$$	$$M_1 = -\frac{wa^2}{2}$$ $$M_2 = R_A(\frac{R_A}{2w}-a)$$ $$M_3 = -\frac{wc^2}{2}$$	Calculate →

Two-Span Beams

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.

Load case	Reactions	Max. shear	Max. moment
Distributed Load in One Zone	$$R_1 = \frac{7wL}{16}$$ $$R_2 = \frac{5wL}{8}$$ $$R_3 = \frac{-wL}{16}$$	$$V_1 = \frac{7wL}{16}$$ $$V_2 = \frac{9wL}{16}$$ $$V_3 = -\frac{wL}{16}$$	$$M_{max} = -\frac{wL^2}{16}$$ $$M_1 = \frac{49wL^2}{512}$$	Calculate →
Central Point Load in One Zone	$$R_1 = \frac{13P}{32}$$ $$R_2 = \frac{11P}{16}$$ $$R_3 = -\frac{3P}{32}$$	$$V_1 = -\frac{13P}{32}$$ $$V_2 = \frac{19P}{32}$$ $$V_3 = -\frac{3P}{32}$$	$$M_{max} = \frac{13PL}{64}$$ $$M_1 = -\frac{3PL}{32}$$	Calculate →
Point Load	$$R_1 = \frac{Pb}{4L^3}(4L^2-a(L+a))$$ $$R_2 = \frac{Pa}{2L^3}(2L^2+b(L+a))$$ $$R_3 = -\frac{Pab}{4L^3}(L+a)$$	$$V_1 = -R_1$$ $$V_2 = \frac{Pa}{4L^3}(4L^2+b(L+a))$$ $$V_3 = R_3$$	$$M_1 = -\frac{Pab}{4L^2}(L+a)$$ $$M_{max} = \frac{Pab}{4L^3}(4L^2-a(L+a))$$	Calculate →
Distributed Load	$$R_A = \frac{3wL}{8}$$ $$R_B = \frac{10wL}{8}$$ $$R_C = \frac{3wL}{8}$$	$$V_1 = R_A$$ $$V_{max} = \frac{5wL}{8}$$	$$M_{max} = -\frac{wL^2}{8}$$ $$M_1 = \frac{9wL^2}{128}$$	Calculate →
Central Point Load in Both Zones	$$R_1 = \frac{5P}{16}$$ $$R_2 = \frac{11P}{8}$$	$$V_1 = \frac{5P}{16}$$ $$V_2 = \frac{11P}{16}$$	$$M_1 = -\frac{3PL}{16}$$ $$M_2 = \frac{5PL}{32}$$	Calculate →
Distributed Load with Different Span Lengths	$$R_A = \frac{M_1}{a}+\frac{wa}{2}$$ $$R_B = wa+wb-R_A-R_C$$ $$R_C = \frac{M_1}{b}+\frac{wb}{2}$$	$$V_1 = -R_A$$ $$V_2 = wa-R_A$$ $$V_3 = -wb+R_C$$ $$V_4 = R_C$$	$$M_1 = -\frac{wb^3+wa^3}{8(a+b)}$$ $$M_2 = R_Ax_1-\frac{wx_1^2}{2}$$ $$M_3 = R_Cx_2-\frac{wx_2^2}{2}$$	Calculate →
Point Loads with Different Span Lengths	$$R_1 = \frac{M_2}{a}+\frac{P_1}{2}$$ $$R_2 = P_1+P_2-R_1-R_3$$ $$R_3 = \frac{M_2}{b}+\frac{P_2}{2}$$	$$V_1 = -R_1$$ $$V_2 = P_1-R_1$$ $$V_3 = -P_2+R_3$$ $$V_4 = R_3$$	$$M_1 = R_1\frac{a}{2}$$ $$M_2 = -\frac{3}{16}(\frac{P_1 a^2+P_2 b^2}{a+b})$$ $$M_3 = R_3\frac{b}{2}$$	Calculate →

Three-Span Beams

A three-span continuous beam runs over four supports. The interior supports carry negative moments, and pattern loading (loading some spans only) can govern the design.

Load case	Reactions	Max. shear	Max. moment
Partial Distributed Load	$$R_A = 0.383wL$$ $$R_B = 1.200wL$$ $$R_C = 0.450wL$$ $$R_D = -0.033wL$$	$$V_1 = -R_A$$ $$V_2 = 0.617wL$$ $$V_3 = -0.583wL$$ $$V_4 = 0.417wL$$ $$V_5 = -0.033wL$$	$$M_1 = 0.0735wL^2$$ $$M_2 = -0.1170wL^2$$ $$M_3 = 0.0534wL^2$$ $$M_4 = -0.0330wL^2$$	Calculate →
End Distributed Load	$$R_1 = 0.450wL$$ $$R_2 = 0.550wL$$	$$V_1 = R_1$$ $$V_2 = R_2$$	$$M_1 = 0.1013wL^2$$ $$M_2 = -0.050wL^2$$	Calculate →
Distributed Load	$$R_1 = 0.400wL$$ $$R_2 = 1.100wL$$	$$V_1 = R_1$$ $$V_2 = 0.600wL$$ $$V_3 = 0.500wL$$	$$M_1 = 0.080wL^2$$ $$M_2 = 0.100wL^2$$ $$M_3 = 0.025wL^2$$	Calculate →

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