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Beam Analysis Calculator for cantilever beam with point moment's banner
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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.

Calculations

Applied moment is positive (+) in the clockwise direction.

Inputs

Geometry and Loading
  1. Length of beam,
    
    L
    :8.00 m
  2. Distance from wall to Moment,
    
    a
    :4.00 m
  3. Magnitude of Moment,
    
    M0
    :-5.00 kN m
Beam Properties
  1. Elastic Modulus,
    
    E
    :200.00 GPa
  2. Second Moment of Inertia,
    
    I
    :1.42e-4 m^4
    
Cantilever beam with Point Moment

Free body diagram


Outputs

Note, self-weight loading is excluded.
  1. 
    
    Max Shear
    :0.00 kN
    
  2. 
    
    Max Moment
    :5.00 kN m
    
  3. 
    
    Max Deflection
    :4.23 mm
    
  1. 
    
    V
    :0.00 kN
    
  2. 
    
    M
    :-5.00 kN m
    
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Explanation

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+M0<xa>1=V<x0>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + M_0<x-a>^{-1} \\= V<x-0>^{0}
  1. Bending Moment

M(x)=M<x0>0+V<x0>1+M0<xa>0M(x) = -M<x-0>^{0} + V<x-0>^{1} + M_0<x-a>^{0}
  1. Deflection

Y(x)=1EI[M2<x0>2+V6<x0>3+M02<x0>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}]
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=0\Sigma F_{y} = 0 \\ V = 0

ΣM@x=0=0M=M0\Sigma M_{@x=0} = 0 \\ M = M_0
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 moment at

.
❗Note:



V(x)=M<x0>1+V<x0>0+M0<xa>1=V<x0>0V(x) = -M<x-0>^{-1} + V<x-0>^{0} + M_0<x-a>^{-1} \\= V<x-0>^{0}

M(x)=M<x0>0+V<x0>1+M0<xa>0M(x) = -M<x-0>^{0} + V<x-0>^{1} + M_0<x-a>^{0}
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+M0<xa>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}]
  1. Integrate the Slope Equation to find the Deflection Equation.

Y(x)=1EIθ(x)dxY(x)=1EI[M2<x0>2+V6<x0>3+M02<xa>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}]
  1. Apply the Boundary Conditions to find the constants
    
    and
    
    

BC 1: @ x=0, θ(x)=00=1EI[M<00>1+V2<00>2+M0<0a>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}= 0

BC 2: @ x=0, Y(x)=00=1EI[M2<00>2+V6<00>3+M02<0a>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} = 0
  1. Both constants are zero so you final Deflection equation becomes:

Y(x)=1EI[M2<x0>2+V6<x0>3+M02<x0>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}]
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!