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Using this calculator you can visualise the shear force, bending moment and deflection of a simply supported beam when a trapezoidal load is applied spanning distance 'a' to 'b' from the fixed support.

Calculations

Applied force is negative (-) in the downwards direction.
Type 1 loading condition
Cantilever beam with Trapezoidal load

Free body diagram

Type 2 loading condition
Cantilever beam with Trapezoidal load

Free body diagram


Inputs

Geometry and Loading


Length of beam, L
:8.00m​


Distance from fixed support to minimum load end, a
:3.00m​


Distance from fixed support to maximum load end, b
:7.00m​


Magnitude of minimum load, F1
:-3.00kN/m​


Magnitude of maximum load, F2
:-5.00kN/m​
Beam Properties


Elastic Modulus, E
:200.00GPa​


Second Moment of Inertia, I
:142,000,000.00mm4​

Outputs

Geometry and Loading


Span of Loading, d
:4.00m​


Distance, c
:1.00m​


Distance, e
:0.00m​
Maximum forces and deflection


Max Shear
:16.00kN​


Max Moment
:-82.67kN m​


Max Deflection
:-48.27mm​


Distance, f
:5.17m​


Resultant force of trapezoidal load, w
:-16.00kN​

Output Diagrams

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



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
i) Type 1 loading condition:

V(x)=−M<x−0>−1+V<x−0>0+F1<x−a>1+(F2−F1)2d<x−a>2−F2<x−b>1−(F2−F1)2d<x−b>2V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F_1<x-a>^{1} + \frac{(F_2-F_1)}{2d}<x-a>^{2} - F_2<x-b>^{1} - \frac{(F_2-F_1)}{2d}<x-b>^{2}
ii) Type 2 loading condition:

V(x)=F1<x−c>1+(F2−F1)2d<x−c>2−F2<x−e>1−(F2−F1)2d<x−e>2V(x) = F_1<x-c>^{1} + \frac{(F_2-F_1)}{2d}<x-c>^{2} - F_2<x-e>^{1} - \frac{(F_2-F_1)}{2d}<x-e>^{2}

  1. Bending Moment
i) Type 1 loading condition:

M(x)=−M<x−0>0+V<x−0>1+F12<x−a>2+(F2−F1)6d<x−a>3−F22<x−b>2−(F2−F1)6d<x−b>3M(x) = -M<x-0>^0 + V<x-0>^{1} + \frac{F_1}{2}<x-a>^{2} + \frac{(F_2-F_1)}{6d}<x-a>^{3} - \frac{F_2}{2}<x-b>^{2} - \frac{(F_2-F_1)}{6d}<x-b>^{3}
ii) Type 2 loading condition:

M(x)=F12<x−c>2+(F2−F1)6d<x−c>3−F22<x−e>2−(F2−F1)6d<x−e>3M(x) = \frac{F_1}{2}<x-c>^{2} + \frac{(F_2-F_1)}{6d}<x-c>^{3} - \frac{F_2}{2}<x-e>^{2} - \frac{(F_2-F_1)}{6d}<x-e>^{3}

  1. Deflection
i) Type 1 loading condition:

Y(x)=1EI[−M2<x−0>2+V6<x−0>3+F124<x−a>4+(F2−F1)120d<x−a>5−F224<x−b>4−(F2−F1)120d<x−b>5]Y(x) = \frac{1}{EI}[-\frac{M}{2}<x-0>^2 + \frac{V}{6}<x-0>^{3} + \frac{F_1}{24}<x-a>^{4} + \frac{(F_2-F_1)}{120d}<x-a>^{5} - \frac{F_2}{24}<x-b>^{4} - \frac{(F_2-F_1)}{120d}<x-b>^{5}]
ii) Type 2 loading condition:

Y(x)=1EI[F124<x−c>4+(F2−F1)120d<x−c>5−F224<x−e>4−(F2−F1)120d<x−e>5+C1x+C2]where...C1=−[F1(L−c)46+(F2−F1)(L−c)424d−F2(L−e)36−(F2−F1)(L−e)424d]C2=−[F1(L−c)424+(F2−F1)(L−c)5120d−F2(L−e)424−(F2−F1)(L−e)5120d+C1L]Y(x) =\frac{1}{EI}[\frac{F_1}{24}<x-c>^{4} + \frac{(F_2-F_1)}{120d}<x-c>^{5} - \frac{F_2}{24}<x-e>^{4} - \frac{(F_2-F_1)}{120d}<x-e>^{5} + C_1x + C_2] \\ \text{where...} \\ C_1 = - [ \frac{F_1(L-c)^4}{6} + \frac{(F_2-F_1)(L-c)^4}{24d} - \frac{F_2(L-e)^3}{6} - \frac{(F_2-F_1)(L-e)^4}{24d}] \\ C_2 = - [ \frac{F_1(L-c)^4}{24} + \frac{(F_2-F_1)(L-c)^5}{120d} - \frac{F_2(L-e)^4}{24} - \frac{(F_2-F_1)(L-e)^5}{120d} + C_1L]
Want to know how to derive the equations? Keep reading!

Derivation

Step 1: Find the beam support reactions by taking moments at each end.
Type 1 loading condition
Cantilever beam with Trapezoidal load

Free body diagram

Type 2 loading condition
Cantilever beam with Trapezoidal load

Free body diagram


Distance formulas for Type 1

Distance formulas for Type 2

Beam support reactions (same for Type 1 and Type 2):

ΣFy=0V=−wwhere: w=12(F1+F2)d\Sigma F_y = 0 \\ V = -w\\\text{where: } w = \frac{1}{2}(F_1+F_2)d

ΣM0=0M=−w×fwhere: w=12(F1+F2)d\Sigma M_0 = 0 \\ M = -w\times f\\\text{where: } w = \frac{1}{2}(F_1+F_2)d
Step 2: Find the shear force

and bending moment

equations by using the table of Macaulay's Singularity Functions on the homepage. Because the load is not applied up to an end of the beam, there are a few extra steps to consider:
  1. Change the FBD so that the distributed load extends all the way to an end of the beam. Make sure the resultant directly gives the original FBD.
  2. Apply Macaulay's Theorem as normal to get the
    
    and
    
    equations considering all loads along the beam for Type 1 while only considering loads on the right side of section A-A for Type 2
  3. Find slope,
    
    given by:

m=y2−y1x2−x1=F2−F1b−a=F2−F1dm = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{F_2-F_1}{b-a} = \dfrac{F_2-F_1}{d}
Type 1 loading condition
Free body diagram adjusted for Macaulay's Theorem

Type 2 loading condition
Free body diagram adjusted for Macaulay's Theorem (Only consider up to Section A-A)

❗Note:


Type 1 loading condition:

V(x)=−M<x−0>−1+V<x−0>0+F1<x−a>1+(F2−F1)2d<x−a>2−F2<x−b>1−(F2−F1)2d<x−b>2V(x) = -M<x-0>^{-1} + V<x-0>^{0} + F_1<x-a>^{1} + \frac{(F_2-F_1)}{2d}<x-a>^{2} - F_2<x-b>^{1} - \frac{(F_2-F_1)}{2d}<x-b>^{2}

M(x)=−M<x−0>0+V<x−0>1+F12<x−a>2+(F2−F1)6d<x−a>3−F22<x−b>2−(F2−F1)6d<x−b>3M(x) = -M<x-0>^0 + V<x-0>^{1} + \frac{F_1}{2}<x-a>^{2} + \frac{(F_2-F_1)}{6d}<x-a>^{3} - \frac{F_2}{2}<x-b>^{2} - \frac{(F_2-F_1)}{6d}<x-b>^{3}
Type 2 loading condition:

V(x)=F1<x−c>1+(F2−F1)2d<x−c>2−F2<x−e>1−(F2−F1)2d<x−e>2V(x) = F_1<x-c>^{1} + \frac{(F_2-F_1)}{2d}<x-c>^{2} - F_2<x-e>^{1} - \frac{(F_2-F_1)}{2d}<x-e>^{2}

M(x)=F12<x−c>2+(F2−F1)6d<x−c>3−F22<x−e>2−(F2−F1)6d<x−e>3M(x) = \frac{F_1}{2}<x-c>^{2} + \frac{(F_2-F_1)}{6d}<x-c>^{3} - \frac{F_2}{2}<x-e>^{2} - \frac{(F_2-F_1)}{6d}<x-e>^{3}
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)=1EI∫M(x)dx\theta(x) = \frac{1}{EI}\int M(x) \hspace{0.1cm} dx
Type 1 loading condition:

θ(x)=1EI[−M<x−0>1+V2<x−0>2+F16<x−a>3+(F2−F1)24d<x−a>4−F26<x−b>3−(F2−F1)24d<x−b>4+C1]\\ \theta(x) = \frac{1}{EI}[-{M}<x-0>^1 + \frac{V}{2}<x-0>^{2} + \frac{F_1}{6}<x-a>^{3} + \frac{(F_2-F_1)}{24d}<x-a>^{4} - \frac{F_2}{6}<x-b>^{3} - \frac{(F_2-F_1)}{24d}<x-b>^{4} + C_1]
Type 2 loading condition:

θ(x)=1EI[F16<x−c>3+(F2−F1)24d<x−c>4−F26<x−e>3−(F2−F1)24d<x−e>4+C1]\theta(x) = \frac{1}{EI}[\frac{F_1}{6}<x-c>^{3} + \frac{(F_2-F_1)}{24d}<x-c>^{4} - \frac{F_2}{6}<x-e>^{3} - \frac{(F_2-F_1)}{24d}<x-e>^{4} + C_1]
  1. Integrate the Slope Equation to find the Deflection Equation.

Y(x)=1EI∫θ(x)dxY(x) = \frac{1}{EI}\int \theta(x) \hspace{0.1cm} dx
Type 1 loading condition:

Y(x)=1EI[−M2<x−0>2+V6<x−0>3+F124<x−a>4+(F2−F1)120d<x−a>5−F224<x−b>4−(F2−F1)120d<x−b>5+C1x+C2]Y(x) = \frac{1}{EI}[-\frac{M}{2}<x-0>^2 + \frac{V}{6}<x-0>^{3} + \frac{F_1}{24}<x-a>^{4} + \frac{(F_2-F_1)}{120d}<x-a>^{5} - \frac{F_2}{24}<x-b>^{4} - \frac{(F_2-F_1)}{120d}<x-b>^{5} + C_1x + C_2]
Type 2 loading condition:

Y(x)=1EI[F124<x−c>4+(F2−F1)120d<x−c>5−F224<x−e>4−(F2−F1)120d<x−e>5+C1x+C2]Y(x) =\frac{1}{EI}[\frac{F_1}{24}<x-c>^{4} + \frac{(F_2-F_1)}{120d}<x-c>^{5} - \frac{F_2}{24}<x-e>^{4} - \frac{(F_2-F_1)}{120d}<x-e>^{5} + C_1x + C_2]
  1. Apply the Boundary Conditions to find the constants
    
    and
    
    
Type 1 loading condition:

BC 1: @ x=0, θ(x)=00=1EI[−M<0−0>1+V2<0−0>2+F16<0−a>3+(F2−F1)24d<0−a>4−F26<0−b>3−(F2−F1)24d<0−b>4+C1]0=1EI[0+0+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_1}{6}<0-a>^{3} + \frac{(F_2-F_1)}{24d}<0-a>^{4} - \frac{F_2}{6}<0-b>^{3} - \frac{(F_2-F_1)}{24d}<0-b>^{4} + C_1] \\0=\frac{1}{EI}[ 0 + 0 + 0 + 0 + 0 + 0 +C_{1}] \\ C_{1}= 0

BC 2: @ x=0, Y(x)=00=1EI[−M2<0−0>2+V6<0−0>3+F124<0−a>4+(F2−F1)120d<0−a>5−F224<0−b>4−(F2−F1)120d<0−b>5+C2]0=1EI[0+0+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_1}{24}<0-a>^{4} + \frac{(F_2-F_1)}{120d}<0-a>^{5} - \frac{F_2}{24}<0-b>^{4} - \frac{(F_2-F_1)}{120d}<0-b>^{5} + C_2] \\0=\frac{1}{EI}[ 0 + 0 + 0 + 0 + 0 + 0 +C_{2}] \\ C_{2}= 0
Type 2 loading condition:

BC 1: @ x=L, θ(x)=00=1EI[F16<L−c>3+(F2−F1)24d<L−c>4−F26<L−e>3−(F2−F1)24d<L−e>4+C1]C1=−[F1(L−c)46+(F2−F1)(L−c)424d−F2(L−e)36−(F2−F1)(L−e)424d]\text{BC 1: @ x=L, $\theta$(x)=0} \\ 0 =\frac{1}{EI}[\frac{F_1}{6}<L-c>^{3} + \frac{(F_2-F_1)}{24d}<L-c>^{4} - \frac{F_2}{6}<L-e>^{3} - \frac{(F_2-F_1)}{24d}<L-e>^{4} + C_1] \\ C_1 = - [ \frac{F_1(L-c)^4}{6} + \frac{(F_2-F_1)(L-c)^4}{24d} - \frac{F_2(L-e)^3}{6} - \frac{(F_2-F_1)(L-e)^4}{24d}]

BC 2: @ x=L, Y(x)=00=1EI[F124<L−c>4+(F2−F1)120d<L−c>5−F224<L−e>4−(F2−F1)120d<L−e>5+C1(L)+C2]C2=−[F1(L−c)424+(F2−F1)(L−c)5120d−F2(L−e)424−(F2−F1)(L−e)5120d+C1L]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[\frac{F_1}{24}<L-c>^{4} + \frac{(F_2-F_1)}{120d}<L-c>^{5} - \frac{F_2}{24}<L-e>^{4} - \frac{(F_2-F_1)}{120d}<L-e>^{5} + C_1(L) + C_2] \\ C_2 = - [ \frac{F_1(L-c)^4}{24} + \frac{(F_2-F_1)(L-c)^5}{120d} - \frac{F_2(L-e)^4}{24} - \frac{(F_2-F_1)(L-e)^5}{120d} + C_1L]
  1. So you final Deflection equations are:
Type 1 loading condition:

Y(x)=1EI[−M2<x−0>2+V6<x−0>3+F124<x−a>4+(F2−F1)120d<x−a>5−F224<x−b>4−(F2−F1)120d<x−b>5]Y(x) = \frac{1}{EI}[-\frac{M}{2}<x-0>^2 + \frac{V}{6}<x-0>^{3} + \frac{F_1}{24}<x-a>^{4} + \frac{(F_2-F_1)}{120d}<x-a>^{5} - \frac{F_2}{24}<x-b>^{4} - \frac{(F_2-F_1)}{120d}<x-b>^{5}]
Type 2 loading condition:

Y(x)=1EI[F124<x−a>4+(F2−F1)120d<x−a>5−F224<x−b>4−(F2−F1)120d<x−b>5+C1x+C2]where...C1=−[F1(L−c)46+(F2−F1)(L−c)424d−F2(L−e)36−(F2−F1)(L−e)424d]C2=−[F1(L−c)424+(F2−F1)(L−c)5120d−F2(L−e)424−(F2−F1)(L−e)5120d+C1L]Y(x) =\frac{1}{EI}[\frac{F_1}{24}<x-a>^{4} + \frac{(F_2-F_1)}{120d}<x-a>^{5} - \frac{F_2}{24}<x-b>^{4} - \frac{(F_2-F_1)}{120d}<x-b>^{5} + C_1x + C_2] \\ \text{where...} \\ C_1 = - [ \frac{F_1(L-c)^4}{6} + \frac{(F_2-F_1)(L-c)^4}{24d} - \frac{F_2(L-e)^3}{6} - \frac{(F_2-F_1)(L-e)^4}{24d}] \\ C_2 = - [ \frac{F_1(L-c)^4}{24} + \frac{(F_2-F_1)(L-c)^5}{120d} - \frac{F_2(L-e)^4}{24} - \frac{(F_2-F_1)(L-e)^5}{120d} + C_1L]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a cantilever beam with a trapezoidal load!