Beam Analysis Calculator for simply supported beam with UDL

Using this calculator you can visualise the shear force, bending moment and deflection of a simply supported beam when a uniformly distributed load (UDL) is applied spanning distance 'a' to 'b' from the left support.
CalculationsApplied 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 start of UDL,﻿a
:5.00 m​
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Distance from left support to end of UDL,﻿b
:7.00 m​
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Magnitude of UDL,﻿F
:-15.00 kN / m​
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Beam Properties
Elastic Modulus,﻿E
:200.00 GPa​
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Second Moment of Inertia, ﻿I
:3.54e+7 mm^4​
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Simply Supported Beam with UDL
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Free body diagram
Outputs Note, self-weight loading is excluded.
Max Forces and Deflection
﻿﻿Max Shear
:-18.00 kN​
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﻿﻿Max Moment
:64.80 kN m​
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﻿﻿Max Deflection
:-82.29 mm​
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Reactions
﻿﻿Ra
:12.00 kN​
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﻿﻿Rb
:18.00 kN​
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ExplanationUsing 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>1−F<x−b>1V(x) = R_a<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} V(x)=Ra​<x−0>0+F<x−a>1−F<x−b>1
Bending Moment 
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M(x)=Ra<x−0>1+F2<x−a>2−F2<x−b>2M(x) = R_a<x-0>^{1} + \frac{F}{2}<x-a>^{2} - \frac{F}{2}<x-b>^{2}M(x)=Ra​<x−0>1+2F​<x−a>2−2F​<x−b>2
Deflection 
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Y(x)=1EI[Ra6<x−0>3+F24<x−a>4−F24<x−b>4+C1x]whereC1=−[RaL26+F(L−a)424L−F(L−b)424L]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{F}{24}<x-a>^{4} - \frac{F}{24}<x-b>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_aL^2}{6} + \frac{F(L-a)^4}{24L} - \frac{F(L-b)^4}{24L} ]Y(x)=EI1​[6Ra​​<x−0>3+24F​<x−a>4−24F​<x−b>4+C1​x]whereC1​=−[6Ra​L2​+24LF(L−a)4​−24LF(L−b)4​]
Want to know how to derive the equations? 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=0Rb×L=−w×eRa=−w×eL\Sigma M_{@x=L} = 0 \\ R_b\times L= -w\times e \\ R_a = \frac{-w\times e}{L}ΣM@x=L​=0Rb​×L=−w×eRa​=L−w×e​
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ΣM@x=0=0Ra×L=−w×fRb=−w×fL\Sigma M_{@x=0} = 0 \\ R_a\times L= -w\times f \\ R_b = \frac{-w\times f}{L}ΣM@x=0​=0Ra​×L=−w×fRb​=L−w×f​
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. Because the load is not applied up to the right end of the beam, there are a few extra steps to consider:
Change the FBD so that the UDL does a wrap-around the beam and stops at the ﻿b
 from the wall on the underside of the beam
Cut a section 'A' just before the right support reaction
Apply Macaulay's Theorem as normal to get the ﻿V(x)
 and ﻿M(x)
 equations considering only the forces present on the left side of the section cut
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Free body diagram adjusted for Macaulay's Theorem
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V(x)=Ra<x−0>0+F<x−a>1−F<x−b>1V(x) = R_a<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} V(x)=Ra​<x−0>0+F<x−a>1−F<x−b>1
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M(x)=Ra<x−0>1+F2<x−a>2−F2<x−b>2M(x) = R_a<x-0>^{1} + \frac{F}{2}<x-a>^{2} - \frac{F}{2}<x-b>^{2}M(x)=Ra​<x−0>1+2F​<x−a>2−2F​<x−b>2
Step 3: Perform the Double Integration Method to find the deflection equation.
Integrate the Bending moment equations once to get the Slope Equation.
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θ(x)=1EI∫M(x)dxθ(x)=1EI[Ra2<x−0>2+F6<x−a>3−F6<x−b>3+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}{6}<x-a>^{3} - \frac{F}{6}<x-b>^{3} +\hspace{0.1cm} C_{1}]θ(x)=EI1​∫M(x)dxθ(x)=EI1​[2Ra​​<x−0>2+6F​<x−a>3−6F​<x−b>3+C1​]
Integrate the Slope Equation to find the Deflection Equation.
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Y(x)=1EI∫θ(x)dxY(x)=1EI[Ra6<x−0>3+F24<x−a>4−F24<x−b>4+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}{24}<x-a>^{4} - \frac{F}{24}<x-b>^{4} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]Y(x)=EI1​∫θ(x)dxY(x)=EI1​[6Ra​​<x−0>3+24F​<x−a>4−24F​<x−b>4+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+F24<0−a>4−F24<0−b>4+C1(0)+C2]0=1EI[0+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}{24}<0-a>^{4} - \frac{F}{24}<0-b>^{4} +\hspace{0.1cm} C_{1}(0) +\hspace{0.1cm} C_{2}] \\0=\frac{1}{EI}[ 0 + 0 + 0 + 0 +C_{2}] \\ C_{2}= 0BC 1: @ x=0, Y(x)=00=EI1​[6Ra​​<0−0>3+24F​<0−a>4−24F​<0−b>4+C1​(0)+C2​]0=EI1​[0+0+0+0+C2​]C2​=0
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BC 2: @ x=L, Y(x)=00=1EI[Ra6<L−0>3+F24<L−a>4−F24<L−b>4+C1(L)]0=1EI[RaL36+F(L−a)424+F(L−b)424+C1L]C1=−[RaL26+F(L−a)424L+F(L−b)424L]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[\frac{R_a}{6}<L-0>^{3} + \frac{F}{24}<L-a>^{4} - \frac{F}{24}<L-b>^{4} +\hspace{0.1cm} C_{1}(L)] \\ 0=\frac{1}{EI}[\frac{R_aL^3}{6} + \frac{F(L-a)^4}{24} + \frac{F(L-b)^4}{24} +C_{1}L] \\ C_{1} = -[\frac{R_aL^2}{6} + \frac{F(L-a)^4}{24L} + \frac{F(L-b)^4}{24L} ]BC 2: @ x=L, Y(x)=00=EI1​[6Ra​​<L−0>3+24F​<L−a>4−24F​<L−b>4+C1​(L)]0=EI1​[6Ra​L3​+24F(L−a)4​+24F(L−b)4​+C1​L]C1​=−[6Ra​L2​+24LF(L−a)4​+24LF(L−b)4​]
So you final Deflection equation becomes:
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Y(x)=1EI[Ra6<x−0>3+F24<x−a>4−F24<x−b>4+C1x]whereC1=−[RaL26+F(L−a)424L+F(L−b)424L]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{F}{24}<x-a>^{4} - \frac{F}{24}<x-b>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_aL^2}{6} + \frac{F(L-a)^4}{24L} + \frac{F(L-b)^4}{24L} ]Y(x)=EI1​[6Ra​​<x−0>3+24F​<x−a>4−24F​<x−b>4+C1​x]whereC1​=−[6Ra​L2​+24LF(L−a)4​+24LF(L−b)4​]
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 UDL!
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