Beam Analysis Calculator for cantilever beam with UDL

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Using this calculator you can visualise the shear force, bending moment and deflection of a cantilever beam when a UDL of length 'c' is applied at a distance 'a' from the wall support.
CalculatorApplied force is negative (-) in the downwards direction.
InputsGeometry and Loading
Length of beam,﻿L
:10.00m​
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Distance from wall to start of UDL,﻿a
:4.00m​
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Distance from wall to end of UDL,﻿b
:8.00m​
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Magnitude of UDL,﻿F
:-5.00kN/m​
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Beam Properties
Elastic Modulus,﻿E
:200.00GPa​
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Second moment of Inertia, ﻿I
:142,000,000.00mm4​
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Cantilever beam with UDL
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Free body diagram
Outputs﻿﻿Max Shear
:20.00kN​
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﻿﻿Max Deflection
:-103.286mm​
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﻿﻿Max Moment
:-120.00kN m​
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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)=−M<x−0>−1+V<x−0>0+F<x−a>1−F<x−b>1=V<x−0>0+F<x−a>1−F<x−b>1V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} \\=V<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} V(x)=−M<x−0>−1+V<x−0>0+F<x−a>1−F<x−b>1=V<x−0>0+F<x−a>1−F<x−b>1
Bending Moment 
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M(x)=−M<x−0>0+V<x−0>1+F2<x−a>2−F2<x−b>2M(x) = -M<x-0>^{0} + V<x-0>^{1} + \frac{F}{2}<x-a>^{2} - \frac{F}{2}<x-b>^{2}M(x)=−M<x−0>0+V<x−0>1+2F​<x−a>2−2F​<x−b>2
Deflection 
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Y(x)=1EI[−M2<x−0>2+V6<x−0>3+F24<x−a>4−F24<x−b>4]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2}+\frac{V}{6}<x-0>^{3} + \frac{F}{24}<x-a>^{4} - \\\frac{F}{24}<x-b>^{4}]Y(x)=EI1​[−2M​<x−0>2+6V​<x−0>3+24F​<x−a>4−24F​<x−b>4]
Want to know how to derive the equations? Keep reading!
DerivationStep 1: Find the beam support reactions by using the equilibrium equations.
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Cantilever beam with UDL
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Free body diagram
Equations for d & W
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ΣFy=0V=−w\Sigma F_y = 0 \\ V = -wΣFy​=0V=−w
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ΣM0=0M=−wd\Sigma M_0 = 0 \\M = -wdΣM0​=0M=−wd
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.
Apply Macaulay's Theorem as normal to get the ﻿V(x)
 and ﻿M(x)
 equations considering all forces and moments along the beam
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Free body diagram adjusted for Macaulay's Theorem
❗Note: ﻿ifn<0→⟨x−a⟩n=0
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V(x)=−M<x−0>−1+V<x−0>0+F<x−a>1−F<x−b>1=V<x−0>0+F<x−a>1−F<x−b>1V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} \\=V<x-0>^{0} + F<x-a>^{1} - F<x-b>^{1} V(x)=−M<x−0>−1+V<x−0>0+F<x−a>1−F<x−b>1=V<x−0>0+F<x−a>1−F<x−b>1
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M(x)=−M<x−0>0+V<x−0>1+F2<x−a>2−F2<x−b>2M(x) = -M<x-0>^{0} + V<x-0>^{1} + \frac{F}{2}<x-a>^{2} - \frac{F}{2}<x-b>^{2}M(x)=−M<x−0>0+V<x−0>1+2F​<x−a>2−2F​<x−b>2
To find deflection there are a few more steps to consider. We will be doing the Double Integration method to get the Deflection Equation of the beam. These steps are outlined below.
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[−M<x−0>1+V2<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} [-M<x-0>^{1} + \frac{V}{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​[−M<x−0>1+2V​<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[−M2<x−0>2+V6<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{M}{2}<x-0>^{2} + \frac{V}{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​[−2M​<x−0>2+6V​<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, θ(x)=00=1EI[−M<0−0>1+V2<0−0>2+F6<0−a>3+F6<x−a>3+C1]0=1EI[0+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} + \frac{F}{6}<0-a>^{3} + \frac{F}{6}<x-a>^{3} +\hspace{0.1cm} C_{1}] \\0=\frac{1}{EI}[ 0 + 0 + 0 + 0 +C_{1}] \\ C_{1}= 0BC 1: @ x=0, θ(x)=00=EI1​[−M<0−0>1+2V​<0−0>2+6F​<0−a>3+6F​<x−a>3+C1​]0=EI1​[0+0+0+0+C1​]C1​=0
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BC 2: @ x=0, Y(x)=00=1EI[−M2<0−0>2+V6<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 2: @ x=0, Y(x)=0} \\ 0=\frac{1}{EI}[ -\frac{M}{2}<0-0>^{2} + \frac{V}{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 2: @ x=0, Y(x)=00=EI1​[−2M​<0−0>2+6V​<0−0>3+24F​<0−a>4−24F​<0−b>4+C1​(0)+C2​]0=EI1​[0+0+0+0+C2​]C2​=0
Both constants are zero so you final Deflection equation becomes:
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Y(x)=1EI[−M2<x−0>2+V6<x−0>3+F24<x−a>4−F24<x−b>4]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2}+\frac{V}{6}<x-0>^{3} + \frac{F}{24}<x-a>^{4} - \\\frac{F}{24}<x-b>^{4}]Y(x)=EI1​[−2M​<x−0>2+6V​<x−0>3+24F​<x−a>4−24F​<x−b>4]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a cantilever beam with a UDL!