Beam Analysis Calculator for simply supported beam with point moment

Using this calculator you can visualise the shear force, bending moment and deflection of a simply supported beam when a point moment is applied at a distance 'a' from the left support.
CalculatorApplied moment is positive (+) in the clockwise direction.
InputsGeometry and Loading
﻿﻿Length of beam, L
:8.00m​
 
﻿﻿Distance from left support to Moment, a
:2.00m​
 
﻿﻿Magnitude of applied moment, M0
:-6.00kN m​
 
Beam Properties
﻿﻿Elastic Modulus, E
:200.00GPa​
 
﻿﻿Second Moment of Inertia, I
:142,000,000.00mm4​
 
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Simply Supported Beam with Point Moment
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Free body diagram
Outputs﻿﻿MaxShear
:0.50kN​
 
﻿﻿MaxMoment
:-3.00kN m​
 
﻿﻿MaxDeflection
:0.424mm​
 
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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+M0<x−a>−1=Ra<x−0>0V(x) = R_a<x-0>^{0} + M_0<x-a>^{-1}\\= R_a<x-0>^{0} V(x)=Ra​<x−0>0+M0​<x−a>−1=Ra​<x−0>0
Bending Moment 
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M(x)=Ra<x−0>1+M0<x−a>0M(x) = R_a<x-0>^{1} + M_0<x-a>^{0}M(x)=Ra​<x−0>1+M0​<x−a>0
Deflection
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Y(x)=1EI[Ra6<x−0>3+M02<x−a>2+C1x]whereC1=−[Ra6L2+M0(L−a)22L]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{M_0}{2}<x-a>^{2} +\hspace{0.1cm} C_{1}x]  \\ \text{where}\hspace{0.3cm} C_1 = -[ \frac{R_a}{6}L^{2} + \frac{M_0(L-a)^{2}}{2L}]Y(x)=EI1​[6Ra​​<x−0>3+2M0​​<x−a>2+C1​x]whereC1​=−[6Ra​​L2+2LM0​(L−a)2​]
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 for Macaulay's Theorem
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ΣM@x=L=0Ra×L=−M0Ra=−M0L\Sigma M_{@x=L} = 0 \\ R_a \times L = -M_0 \\ R_a = \frac{-M_0}{L}ΣM@x=L​=0Ra​×L=−M0​Ra​=L−M0​​
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ΣM@x=0=0Rb×L=M0Rb=MoL\Sigma M_{@x=0} = 0 \\ R_b \times L = M_0 \\ R_b = \frac{M_o}{L}ΣM@x=0​=0Rb​×L=M0​Rb​=LMo​​
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 ﻿M0​
 applied force at ﻿a
. 
❗Note: ﻿ifn<0→⟨x−a⟩n=0
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﻿
V(x)=Ra<x−0>0+M0<x−a>−1=Ra<x−0>0V(x) = R_a<x-0>^{0} + M_0<x-a>^{-1}\\=R_a<x-0>^{0}V(x)=Ra​<x−0>0+M0​<x−a>−1=Ra​<x−0>0
﻿
M(x)=Ra<x−0>1+M0<x−a>0M(x) = R_a<x-0>^{1} + M_0<x-a>^{0}M(x)=Ra​<x−0>1+M0​<x−a>0
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+M0<x−a>1+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} + {M_0}<x-a>^{1} +\hspace{0.1cm} C_{1}]θ(x)=EI1​∫M(x)dxθ(x)=EI1​[2Ra​​<x−0>2+M0​<x−a>1+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+M02<x−a>2+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{M_0}{2}<x-a>^{2} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]Y(x)=EI1​∫θ(x)dxY(x)=EI1​[6Ra​​<x−0>3+2M0​​<x−a>2+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+M02<0−a>2+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{M_0}{2}<0-a>^{2} +\hspace{0.1cm} C_{1}(0) +\hspace{0.1cm} C_{2}] \\ 0 = \frac{1}{EI}[0+0+0+C_2] \\ C_2 = 0BC 1: @ x=0, Y(x)=00=EI1​[6Ra​​<0−0>3+2M0​​<0−a>2+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+M02<L−a>2+C1L]C1=−[Ra6L2+M0(L−a)22L]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[  \frac{R_a}{6}<L-0>^{3} + \frac{M_0}{2}<L-a>^{2} +\hspace{0.1cm} C_{1}L] \\ C_1 = -[ \frac{R_a}{6}L^{2} + \frac{M_0(L-a)^{2}}{2L}]BC 2: @ x=L, Y(x)=00=EI1​[6Ra​​<L−0>3+2M0​​<L−a>2+C1​L]C1​=−[6Ra​​L2+2LM0​(L−a)2​]
After substituting the constant ﻿C1​
, your final Deflection equation becomes:
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Y(x)=1EI[Ra6<x−0>3+M02<x−a>2+C1x]whereC1=−[Ra6L2+M0(L−a)22L]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{M_0}{2}<x-a>^{2} +\hspace{0.1cm} C_{1}x]  \\ \text{where}\hspace{0.3cm} C_1 = -[ \frac{R_a}{6}L^{2} + \frac{M_0(L-a)^{2}}{2L}]Y(x)=EI1​[6Ra​​<x−0>3+2M0​​<x−a>2+C1​x]whereC1​=−[6Ra​​L2+2LM0​(L−a)2​]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a simply supported beam with an applied moment!
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