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CalcTree
Using this calculator you can visualise the shear force, bending moment and deflection of a cantilever beam when a point load is applied at a distance 'a' from the wall support.

Calculations

Applied force is negative (-) in the downwards direction.

Inputs

Geometry and Loading


Length of beam, L
:10.00m


Distance between wall and point load, a
:5.00m


Magnitude of force applied, F
:5.00kN
Beam Properties


Elastic Modulus, E
:200.00GPa


Second Moment of Inertia
:142,000,000.00mm4
Cantilever Beam with Point load

Free body diagram


Outputs



Max Shear
:-5.00kN


Max Moment
:25.00kN m


Max Deflection
:18.339mm

Output Diagrams

Can’t display the image because of an internal error. Our team is looking at the issue.



Beam Analysis Equations

Using Macaulay's Theorem and the Double Integration Method, we can create the equations for shear force, bending moment and deflection as follows:
  1. Shear Force

V(x)=M<x0>1+V<x0>0+F<xa>0=V<x0>0+F<xa>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F<x-a>^{0}\\=V<x-0>^{0} + F<x-a>^{0}
  1. Bending Moment

M(x)=M<x0>0+V<x0>1+F<xa>1M(x) = -M<x-0>^{0} + V<x-0>^{1} + F<x-a>^{1}
  1. Deflection

Y(x)=1EI[M2<x0>2+V6<x0>3+F6<xa>3]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2} + \frac{V}{6}<x-0>^{3} + \frac{F}{6}<x-a>^{3}]
Want to know how to derive the equations? Keep reading!

Derivation

Step 1: Find the beam support reactions by using the equilibrium equations.
Free body diagram


ΣFy=0V=F\Sigma F_{y} = 0 \\ V = -F

ΣM@x=0=0M=F×a\Sigma M_{@x=0} = 0 \\ M = -F\times a
Step 2: Find the shear force

and bending moment

equations by using the table of Macaulay's Singularity Functions on the homepage. There will be three terms in the

and

equations since there is

and

reaction at

and

applied force at

.
❗Note:



V(x)=M<x0>1+V<x0>0+F<xa>0=V<x0>0+F<xa>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F<x-a>^{0}\\=V<x-0>^{0} + F<x-a>^{0}

M(x)=M<x0>0+V<x0>1+F<xa>1M(x) = -M<x-0>^{0} + V<x-0>^{1} + F<x-a>^{1}
Step 3: Perform the Double Integration Method to find the deflection equation.
  1. Integrate the Bending moment equations once to get the Slope Equation.

θ(x)=1EIM(x)dxθ(x)=1EI[M<x0>1+V2<x0>2+F2<xa>2+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}{2}<x-a>^{2} +\hspace{0.1cm} C_{1}]
  1. Integrate the Slope Equation to find the Deflection Equation.

Y(x)=1EIθ(x)dxY(x)=1EI[M2<x0>2+V6<x0>3+F6<xa>3+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}{6}<x-a>^{3} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]
  1. Apply the Boundary Conditions to find the constants
    
    and
    
    

BC 1: @ x=0, θ(x)=00=1EI[M<00>1+V2<00>2+F2<0a>2+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} + \frac{F}{2}<0-a>^{2} +\hspace{0.1cm} C_{1}] \\0=\frac{1}{EI}[ 0 + 0 + 0 +C_{1}] \\ C_{1}= 0

BC 2: @ x=0, Y(x)=00=1EI[M2<00>2+V6<00>3+F6<0a>3+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}{6}<0-a>^{3} +\hspace{0.1cm} C_{1}(0) +\hspace{0.1cm} C_{2}] \\ 0=\frac{1}{EI}[ 0 + 0 + 0 + 0 +C_{2}] \\ C_{2} = 0
  1. Both constants are zero so you final Deflection equation becomes:

Y(x)=1EI[M2<x0>2+V6<x0>3+F6<xa>3]Y(x) =\frac{1}{EI}[ -\frac{M}{2}<x-0>^{2} + \frac{V}{6}<x-0>^{3} + \frac{F}{6}<x-a>^{3}]
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 load!