Beam Analysis Calculator for cantilever beam with point moment

Using this calculator you can visualise the shear force, bending moment and deflection of a cantilever beam when a point moment is applied at a distance 'a' from the wall support.
CalculationsApplied moment is positive (+) in the clockwise direction.
InputsGeometry and Loading
﻿﻿Length of beam, L
:10.00m​
 
﻿﻿Distance from wall to Moment, a
:4.00m​
 
﻿﻿Magnitude of Moment, M0
:-5.00kN m​
 
Beam Properties
﻿﻿Elastic Modulus, E
:200.00 GPa​
 
﻿﻿Second Moment of Inertia, SI
:142,000,000.00mm4​
  
﻿
Cantilever beam with Point Moment
﻿
Free body diagram
Outputs﻿﻿MaxShear
:0.00kN​
 
﻿﻿MaxMoment
:5.00kN m​
 
﻿﻿MaxDeflection
:5.634mm​
 
﻿
Can’t display the image because of an internal error. Our team is looking at the issue.
﻿
﻿
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
﻿
V(x)=−M<x−0>−1+V<x−0>0+M0<x−a>−1=V<x−0>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + M_0<x-a>^{-1} \\= V<x-0>^{0}V(x)=−M<x−0>−1+V<x−0>0+M0​<x−a>−1=V<x−0>0
Bending Moment 
﻿
M(x)=−M<x−0>0+V<x−0>1+M0<x−a>0M(x) = -M<x-0>^{0} + V<x-0>^{1} + M_0<x-a>^{0} M(x)=−M<x−0>0+V<x−0>1+M0​<x−a>0
Deflection 
﻿
Y(x)=1EI[−M2<x−0>2+V6<x−0>3+M02<x−0>2]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2} + \frac{V}{6}<x-0>^{3} + \frac{M_0}{2}<x-0>^{2}]Y(x)=EI1​[−2M​<x−0>2+6V​<x−0>3+2M0​​<x−0>2]
Want to know how to derive the equations? Keep reading!
DerivationStep 1: Find the beam support reactions by using the equilibrium equations.
﻿
Free body diagram
﻿
ΣFy=0V=0\Sigma F_{y} = 0 \\ V = 0ΣFy​=0V=0
﻿
ΣM@x=0=0M=M0\Sigma M_{@x=0} = 0 \\  M = M_0ΣM@x=0​=0M=M0​
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 three terms in the ﻿V(x)
 and ﻿M(x)
 equations since there is ﻿V
 and ﻿M
 reaction at ﻿a=0
 and ﻿M0​
 applied moment at ﻿a
. 
❗Note: ﻿ifn<0→⟨x−a⟩n=0
﻿
﻿
V(x)=−M<x−0>−1+V<x−0>0+M0<x−a>−1=V<x−0>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + M_0<x-a>^{-1}  \\= V<x-0>^{0}V(x)=−M<x−0>−1+V<x−0>0+M0​<x−a>−1=V<x−0>0
﻿
M(x)=−M<x−0>0+V<x−0>1+M0<x−a>0M(x) = -M<x-0>^{0} + V<x-0>^{1} + M_0<x-a>^{0} M(x)=−M<x−0>0+V<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. 
﻿
θ(x)=1EI∫M(x)dxθ(x)=1EI[−M<x−0>1+V2<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} [-M<x-0>^{1} + \frac{V}{2}<x-0>^{2} + M_0<x-a>^{1} + C_{1}]θ(x)=EI1​∫M(x)dxθ(x)=EI1​[−M<x−0>1+2V​<x−0>2+M0​<x−a>1+C1​]
Integrate the Slope Equation to find the Deflection Equation.
﻿
Y(x)=1EI∫θ(x)dxY(x)=1EI[−M2<x−0>2+V6<x−0>3+M02<x−a>2C1x+C2]Y(x) = \frac{1}{EI}\int \theta(x) \hspace{0.1cm} dx  \\ Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2} + \frac{V}{6}<x-0>^{3} +\frac{M_0}{2}<x-a>^{2}C_{1}x +\hspace{0.1cm} C_{2}]Y(x)=EI1​∫θ(x)dxY(x)=EI1​[−2M​<x−0>2+6V​<x−0>3+2M0​​<x−a>2C1​x+C2​]
Apply the Boundary Conditions to find the constants ﻿C1​
 and ﻿C2​
﻿
﻿
BC 1: @ x=0, θ(x)=00=1EI[−M<0−0>1+V2<0−0>2+M0<0−a>1+C1]0=1EI[0+0+0+C1]C1=0\text{BC 1: @ x=0, $\theta$(x)=0} \\ 0 = \frac{1}{EI} [-M<0-0>^{1} + \frac{V}{2}<0-0>^{2} + M_0<0-a>^{1} + C_{1}] \\0=\frac{1}{EI}[ 0 + 0 + 0 +C_{1}] \\ C_{1}= 0BC 1: @ x=0, θ(x)=00=EI1​[−M<0−0>1+2V​<0−0>2+M0​<0−a>1+C1​]0=EI1​[0+0+0+C1​]C1​=0
﻿
BC 2: @ x=0, Y(x)=00=1EI[−M2<0−0>2+V6<0−0>3+M02<0−a>2+C2]0=1EI[0+0+0+C2]C2=0\text{BC 2: @ x=0, Y(x)=0} \\ 0=\frac{1}{EI}[ -\frac{M}{2}<0-0>^{2} + \frac{V}{6}<0-0>^{3} + \frac{M_0}{2}<0-a>^{2} + C_{2}] \\ 0=\frac{1}{EI}[ 0 + 0 +0 +C_{2}] \\ C_{2} = 0BC 2: @ x=0, Y(x)=00=EI1​[−2M​<0−0>2+6V​<0−0>3+2M0​​<0−a>2+C2​]0=EI1​[0+0+0+C2​]C2​=0
Both constants are zero so you final Deflection equation becomes:
﻿
Y(x)=1EI[−M2<x−0>2+V6<x−0>3+M02<x−0>2]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2} + \frac{V}{6}<x-0>^{3} + \frac{M_0}{2}<x-0>^{2}]Y(x)=EI1​[−2M​<x−0>2+6V​<x−0>3+2M0​​<x−0>2]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a cantilever beam with a point moment!
﻿