Beam Analysis using Macaulay's Theorem

Verified by the CalcTree engineering team on July 23, 2024
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On this page you will find a library of calculators that determine the shear force, bending moment and deflection of a single-span beam with different types of applied loading. The calculators are for a simply supported beam or a cantilever beam. 
Keep reading this page to discover the theory behind the calculators!
CalculatorsClick on one of these calculators below for a specific loading condition on a simply supported beam!
Simply Supported Beams - Individual Load Calculators
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Beam Analysis Calculator for simply supported beam with point load
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Beam Analysis Calculator for simply supported beam with point moment
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Beam Analysis Calculator for simply supported beam with UDL
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Beam Analysis Calculator for simply supported beam with triangular load
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Beam Analysis Calculator for simply supported beam with trapezoidal load
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Click on one of these calculators below for a specific loading condition on a cantilever beam!
Cantilever Beams - Individual Load Calculators
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Beam Analysis Calculator for cantilever beam with point load
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Beam Analysis Calculator for cantilever beam with point moment
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Beam Analysis Calculator for cantilever beam with UDL 
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Beam Analysis Calculator for cantilever beam with triangular load
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Beam Analysis Calculator for cantilever beam with trapezoidal load
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Click on this calculator that holds up to 10 loads on either a simply supported beam or a cantilever beam!
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Beam Analysis Calculator for beam with multiple loads
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Explore this other calculator which provides beam analysis for a beam with multiple spans! Beam Analysis using Anastruct
ExplanationFinding SFD and BMD Equations: Macaulay's TheoremMacaulay's theorem enables us to describe the applied load, shear force diagram  (SFD) and bending moment diagram (BMD) for the full length of the beam with single equations. 
Macaulay's theorem is a singularity function, which means the function has discontinuities at singular points. Macaulay's theorem says:
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For n≥0⟨x−a⟩n={0when x < a(x−a)nwhen x≥a\text{For }n\geq0\hspace{1cm}\langle x - a\rangle^n =\begin{cases}0 \hspace{1.58cm} \text{when x < a}\\(x - a)^n \hspace{0.5cm} \text{when x$\geq$a} \end{cases}For n≥0⟨x−a⟩n={0when x < a(x−a)nwhen x≥a​
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For n<0⟨x−a⟩n=0\text{For }n < 0  \hspace{1cm}\langle x-a\rangle^n = 0For n<0⟨x−a⟩n=0
Where:
﻿﻿x
 is the position along the beam
﻿﻿﻿a
 is the position of the load (or start of the load) on the beam
﻿﻿n
 is an integer based on the load applied, with the following categories:
         If ﻿n=−2→
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         If ﻿n=−1→
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         If ﻿n=0→
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         If ﻿n=1→
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It is a Moment
It is a Point Load
It is a Uniformly Distributed Load
It is a Triangular Load / Uniformly Varying Load
﻿﻿⟨⟩
 angle brackets is the notation used for singularity functions. The angle brackets for ﻿⟨x−a⟩
 imply that if ﻿x<a
 then the function goes to zero as there is no effect on the beam due to the load, and if ﻿x≥a
 then there is an effect on the beam so the function is treated as a continuous function
The applied load, shear force and bending moments can be described by singularity functions as tabulated below. You will notice how the applied load equation is integrated once to get the shear force equation and then integrated again to get the bending moment equation. Note the integration rules are slightly different for singularity functions!
Table of Macaulay's Singularity Functions
Loading
Load 
Shear   
Moment 
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﻿﻿w(x)=M0​⟨x−a⟩−2
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﻿﻿V(x)=M0​⟨x−a⟩−1
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﻿﻿M(x)=M0​⟨x−a⟩0
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﻿﻿w(x)=P⟨x−a⟩−1
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﻿﻿V(x)=P⟨x−a⟩0
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﻿﻿M(x)=P⟨x−a⟩1
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﻿﻿w(x)=w0​⟨x−a⟩0
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﻿﻿V(x)=w0​⟨x−a⟩1
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﻿﻿M(x)=2w0​​⟨x−a⟩2
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﻿﻿w(x)=m⟨x−a⟩1
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﻿﻿V(x)=2m​⟨x−a⟩2
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﻿﻿M(x)=6m​⟨x−a⟩3
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🗒️Integration rules for Singularity Functions
If ﻿n≥0
, the integration laws are followed as per normal:
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∫Fn(x)=Fn+1(x)n+1\int F_n(x) = \cfrac{F_{n+1}(x)}{n+1}∫Fn​(x)=n+1Fn+1​(x)​
If ﻿n<0
, the integration laws change and become:
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∫Fn(x)=Fn+1(x)\int F_n(x) = F_{n+1}(x)∫Fn​(x)=Fn+1​(x)
Now that we know the singularity functions for each type of force, we can form the ﻿V(x)
 and ﻿M(x)
 equations for our beam. To do this: 
Make a section cut just before the right end of the beam, hence ignoring any forces or reactions at the right beam support
Sweep from left to right of your beam's free body diagram and add a singularity function from the table of Macaulay's Singularity Functions each time an applied load or reaction is present
For example, for a simply supported beam with a point load there will be two terms in the equations since there is ﻿Ra​
 reaction force at ﻿a=0
 and ﻿F
 applied force at ﻿a
. Use the singularity functions in the second row of the table to form the ﻿V(x)
 and ﻿﻿M(x)
 equations:
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w(x)=Ra⟨x−0⟩−1+F⟨x−a⟩−1\small{w(x) = R_a\langle x-0\rangle^{-1}+F\langle x-a\rangle^{-1}}w(x)=Ra​⟨x−0⟩−1+F⟨x−a⟩−1
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V(x)=Ra⟨x−0⟩0+F⟨x−a⟩0\small{V(x) = R_a\langle x-0\rangle^{0} + F\langle x-a\rangle^{0}}V(x)=Ra​⟨x−0⟩0+F⟨x−a⟩0
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M(x)=Ra⟨x−0⟩1+F⟨x−a⟩1\small{M(x) = R_a\langle x-0\rangle ^{1} + F\langle x-a\rangle ^{1}}M(x)=Ra​⟨x−0⟩1+F⟨x−a⟩1
❗Sign convention: all terms in the example above are positive (+) even though a downwards load ﻿F
 is shown because when values are substituted in the equations, the negative (-) sign for ﻿F
 shall be included. 
Finding Deflection Equations: Double Integration MethodThere are three main ways to calculate the deflection equations, but the most convenient is the Double Integration Method. 
Using the bending moment equation you created from the Macaulay's theorem, integrating it once gives you the slope equation and integrating a second time gives you the deflection equation. The formulas are given by:
Slope Equation
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θ(x)=1EI∫M(x)dx\theta(x) = \dfrac{1}{EI}\int M(x) \hspace{0.1cm} dxθ(x)=EI1​∫M(x)dx
Deflection Equation
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Y(x)=1EI∫θ(x)dxY(x) = \dfrac{1}{EI} \int \theta(x) \hspace{0.1cm} dxY(x)=EI1​∫θ(x)dx
💡Learn about stiffness, ﻿EI
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A common mistake made when integrating is to forget the integration constants. Since there are two integrations there should be a the term ﻿C1​x+C2​
 at the end of your ﻿Y(x)
 equation. 
To find these constants you need to apply boundary conditions (BCs) which differ to each support condition of your beam. A cantilever beam has a fixed support on one end and a simply supported beam has fixed/roller supports at both ends. The boundary condition can be stated as follows:
Cantilever Beam
﻿﻿BC1​:x=0,θ(x)=0
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﻿﻿BC2​:x=0,Y(x)=0
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Simply Supported Beam
﻿﻿BC1​:x=0,Y(x)=0
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﻿﻿BC2​:x=L,Y(x)=0
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SummaryThis library of calculators allow you to create the shear force, bending moment and deflection diagrams of either a cantilever or simply supported beam with various loading conditions. Macaulay's theorem, which is the theory behind how the loaded beams are analysed, and how the ﻿V(x)
, ﻿M(x)
 and ﻿Y(x)
 equations are derived is also provided. 
Additional ResourcesRead more about Macaulay's theorem here!
Moment of Inertia Calculators
Beam Analysis using Anastruct
Beam Analysis using Macaulay's Theorem

Calculators

Simply Supported Beams - Individual Load Calculators

Cantilever Beams - Individual Load Calculators

Explanation

Finding SFD and BMD Equations: Macaulay's Theorem

Table of Macaulay's Singularity Functions

🗒️Integration rules for Singularity Functions

Finding Deflection Equations: Double Integration Method

💡Learn about stiffness, ﻿EI﻿

Summary

Additional Resources

💡Learn about stiffness,

$E I$
