Beam Analysis Calculator for simply supported beam with point load

Using this calculator you can visualise the shear force, bending moment and deflection of a simply supported beam when a point load is applied at a distance 'a' from the left support.
CalculatorApplied force is negative (-) in the downwards direction.
InputsGeometry and Loading
Length of beam,﻿L
:10.00 m​
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Distance from left support to Point load,﻿a
:1.00 m​
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Magnitude of force applied, ﻿F
:-50.00 kN​
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Beam Properties
Elastic Modulus,﻿E
:200 GPa​
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Second Moment of Inertia,﻿I
:1.00e-4 m^4​
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Simply Supported Beam with Point Load
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Free body diagram
OutputsNote, self-weight loading is excluded.
﻿﻿Max Shear
:45.0 kN​
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﻿﻿Max Moment
:45.0 kN m​
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﻿﻿Max Deflection
:-15.8 mm​
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﻿﻿Ra
:45.0 kN​
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﻿﻿Rb
:5.0 kN​
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Beam Analysis EquationsUsing Macaulay's Theorem and the Double Integration Method, we can create the equations for shear force, bending moment and deflection as follows:
Shear Force
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V(x)=Ra<x−0>0+F<x−a>0V(x) = R_a<x-0>^{0} + F<x-a>^{0}V(x)=Ra​<x−0>0+F<x−a>0
Bending Moment 
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M(x)=Ra<x−0>1+F<x−a>1M(x) = R_a<x-0>^{1} + F<x-a>^{1}M(x)=Ra​<x−0>1+F<x−a>1
Deflection
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Y(x)=1EI[Ra6<x−0>3+F6<x−a>3−(Fb36L+RaL26)x]Y(x) =\frac{1}{EI}[  \frac{R_a}{6}<x-0>^{3} + \frac{F}{6}<x-a>^{3} - (\frac{Fb^{3}}{6L} + \frac{R_{a}L^{2}}{6})x]Y(x)=EI1​[6Ra​​<x−0>3+6F​<x−a>3−(6LFb3​+6Ra​L2​)x]
Want to know how to derive these formulas? Keep reading!
DerivationStep 1: Find the beam support reactions by taking moments at each end.
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Free body diagram
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ΣM@x=L=0Ra×L=−F×bRa=−FbL\Sigma M_{@x=L} = 0 \\ R_a \times L = -F\times b \\ R_a = \frac{-Fb}{L}ΣM@x=L​=0Ra​×L=−F×bRa​=L−Fb​
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ΣM@x=0=0Rb×L=−F×aRb=−FaL\Sigma M_{@x=0} = 0 \\ R_b \times L = -F\times a \\ R_b = \frac{-Fa}{L}ΣM@x=0​=0Rb​×L=−F×aRb​=L−Fa​
Step 2: Find the shear force ﻿V(x)
 and bending moment ﻿M(x)
 equations by using the table of Macaulay's Singularity Functions on the homepage. There will be two terms in both ﻿V(x)
 and ﻿M(x)
 equations since there is ﻿Ra​
 reaction force at ﻿a=0
 and ﻿F
 applied force at ﻿a
. 
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V(x)=Ra<x−0>0+F<x−a>0V(x) = R_a<x-0>^{0} + F<x-a>^{0}V(x)=Ra​<x−0>0+F<x−a>0
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M(x)=Ra<x−0>1+F<x−a>1M(x) = R_a<x-0>^{1} + F<x-a>^{1}M(x)=Ra​<x−0>1+F<x−a>1
Step 3: Perform the Double Integration Method to find the deflection equation.
Integrate the Bending Moment Equation once to get the Slope Equation. 
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θ(x)=1EI∫M(x)dxθ(x)=1EI[Ra2<x−0>2+F2<x−a>2+C1]\theta(x) = \frac{1}{EI}\int M(x) \hspace{0.1cm} dx  \\ \theta(x) = \frac{1}{EI} [\frac{R_a}{2}<x-0>^{2} + \frac{F}{2}<x-a>^{2} +\hspace{0.1cm} C_{1}]θ(x)=EI1​∫M(x)dxθ(x)=EI1​[2Ra​​<x−0>2+2F​<x−a>2+C1​]
Integrate the Slope Equation to get the Deflection Equation.
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Y(x)=1EI∫θ(x)dxY(x)=1EI[Ra6<x−0>3+F6<x−a>3+C1x+C2]Y(x) = \frac{1}{EI}\int \theta(x) \hspace{0.1cm} dx  \\ Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{F}{6}<x-a>^{3} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]Y(x)=EI1​∫θ(x)dxY(x)=EI1​[6Ra​​<x−0>3+6F​<x−a>3+C1​x+C2​]
Apply the Boundary Conditions to find the constants ﻿C1​
 and ﻿C2​
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BC 1: @ x=0, Y(x)=00=1EI[Ra6<0−0>3+F6<0−a>3+C1(0)+C2]0=1EI[0+0+0+C2]→C2=0\text{BC 1: @ x=0, Y(x)=0} \\ 0 =\frac{1}{EI}[ \frac{R_a}{6}<0-0>^{3} + \frac{F}{6}<0-a>^{3} +\hspace{0.1cm} C_{1}(0) +\hspace{0.1cm} C_{2}] \\ 0 = \frac{1}{EI}[0+0+0+C_2] \\\rightarrow C_2 = 0BC 1: @ x=0, Y(x)=00=EI1​[6Ra​​<0−0>3+6F​<0−a>3+C1​(0)+C2​]0=EI1​[0+0+0+C2​]→C2​=0
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BC 2: @ x=L, Y(x)=00=1EI[Ra6<L−0>3+F6<L−a>3+C1(L)]C1L=−Ra6L3−F6b3→C1=−[Fb36L+Ra6L2]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[ \frac{R_a}{6}<L-0>^{3} + \frac{F}{6}<L-a>^{3} +\hspace{0.1cm} C_{1}(L)] \\ C_1L = -\frac{R_a}{6}L^{3}-\frac{F}{6}b^{3} \\ \rightarrow C_1 = -[\frac{Fb^{3}}{6L} + \frac{R_a}{6}L^{2}]BC 2: @ x=L, Y(x)=00=EI1​[6Ra​​<L−0>3+6F​<L−a>3+C1​(L)]C1​L=−6Ra​​L3−6F​b3→C1​=−[6LFb3​+6Ra​​L2]
After substituting the constant ﻿C1​
, your final Deflection equation becomes:
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Y(x)=1EI[Ra6<x−0>3+F6<x−a>3−(Fb36L+RaL26)x]Y(x) =\frac{1}{EI}[  \frac{R_a}{6}<x-0>^{3} + \frac{F}{6}<x-a>^{3} - (\frac{Fb^{3}}{6L} + \frac{R_{a}L^{2}}{6})x]Y(x)=EI1​[6Ra​​<x−0>3+6F​<x−a>3−(6LFb3​+6Ra​L2​)x]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a simply supported beam with a point load!
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