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

Calculations

Applied force is negative (-) in the downwards direction.
Type 1 loading condition
Simply Supported Beam with Triangular Load

Type 2 loading condition
Simply Supported Beam with Triangular Load

Free body diagram

Free body diagram


Inputs

Geometry and Loading


Length of beam, L
:7.00m​


Distance from left support to the zero load end, a
:6.00m​


Distance from left support to the maximum load end, b
:1.00m​


Magnitude of maximum load, F
:-12.00kN/m​
Beam Properties


Elastic Modulus, E
:200.00GPa​


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

Outputs

Geometry and Loading


Span of load, d
:5.00m​


Distance, c
:1.00m​


Distance, e
:6.00m​


Distance, f
:2.67m​
Maximum forces and deflection


Max Shear
:18.57kN​


Max Moment
:34.94kN m​


Max Deflection
:-5.98mm​


Distance, g
:4.33m​


Resultant force of triangular load, w
:-30.00kN​

Output Diagrams

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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
i) Type 1 loading condition:

V(x)=Ra<x−0>0+F2d<x−a>2−F<x−b>1−F2d<x−b>2\small{V(x) = R_a<x-0>^{0} + \frac{F}{2d}<x-a>^{2} - F<x-b>^{1} - \frac{F}{2d}<x-b>^{2}}
ii) Type 2 loading condition:

V(x)=Rb<x−0>0+F2d<x−c>2−F<x−e>1−F2d<x−e>2\small{V(x) = R_b<x-0>^{0} + \frac{F}{2d}<x-c>^{2} - F<x-e>^{1} - \frac{F}{2d}<x-e>^{2}}

  1. Bending Moment
i) Type 1 loading condition:

M(x)=Ra<x−0>1+F6d<x−a>3−F2<x−b>2−F6d<x−b>3\small{M(x) = R_a<x-0>^{1} + \frac{F}{6d}<x-a>^{3} - \frac{F}{2}<x-b>^{2} -\frac{F}{6d}<x-b>^{3}}
ii) Type 2 loading condition:

M(x)=Rb<x−0>1+F6d<x−c>3−F2<x−e>2−F6d<x−e>3\small{M(x) = R_b<x-0>^{1} + \frac{F}{6d}<x-c>^{3} - \frac{F}{2}<x-e>^{2} -\frac{F}{6d}<x-e>^{3} }

  1. Deflection
i) Type 1 loading condition:

Y(x)=1EI[Ra6<x−0>3+F120d<x−a>5−F24<x−b>4−F120d<x−b>4+C1x]whereC1=−[RaL26+F(L−a)5120Ld−F(L−b)424L−F(L−b)5120Ld]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{F}{120d}<x-a>^{5} - \frac{F}{24}<x-b>^{4} - \frac{F}{120d}<x-b>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_aL^2}{6} + \frac{F(L-a)^5}{120Ld} - \frac{F(L-b)^4}{24L} - \frac{F(L-b)^5}{120Ld} ]
ii) Type 2 loading condition:

Y(x)=1EI[Rb6<x−0>3+F120d<x−c>5−F24<x−e>4−F120d<x−e>4+C1x]whereC1=−[RbL26+F(L−c)5120Ld−F(L−e)424L−F(L−e)5120Ld]Y(x) =\frac{1}{EI}[ \frac{R_b}{6}<x-0>^{3} + \frac{F}{120d}<x-c>^{5} - \frac{F}{24}<x-e>^{4} - \frac{F}{120d}<x-e>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_bL^2}{6} + \frac{F(L-c)^5}{120Ld} - \frac{F(L-e)^4}{24L} - \frac{F(L-e)^5}{120Ld} ]
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
Simply Supported Beam with Triangular Load

Type 2 loading condition
Simply Supported Beam with Triangular Load

Free body diagram

Free body diagram


Distance formulas for Type 1

Distance formulas for Type 2

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

ΣM@x=0=0Rb×L=−w×fRb=−w×fLwhere: w=12F×d\Sigma M_{@x=0} = 0 \\ R_b\times L= -w\times f \\ R_b = \frac{-w\times f}{L}\\\text{where: } w = \frac{1}{2}F\times d

ΣM@x=L=0Ra×L=−w×gRa=−w×gLwhere: w=12F×d\Sigma M_{@x=L} = 0 \\ R_a\times L= -w\times g \\ R_a = \frac{-w\times g}{L}\\\text{where: } w = \frac{1}{2}F\times 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 the right end of the beam, there are a few extra steps to consider:
  1. Change the FBD so that the load does a wrap-around the beam and stops at
    
    from the wall on the underside of the beam
  2. Cut a section 'A' just before the right support reaction
  1. Apply Macaulay's Theorem as normal to get the
    
    and
    
    equations considering only the forces present on the left side of the section cut
  2. Find slope,
    
    given by:

m=y2−y1x2−x1=F−0b−a=Fdm = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{F-0}{b-a} = \dfrac{F}{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

Type 1 loading condition:

V(x)=Ra<x−0>0+F2d<x−a>2−F<x−b>1−F2d<x−b>2\small{V(x) = R_a<x-0>^{0} + \frac{F}{2d}<x-a>^{2} - F<x-b>^{1} - \frac{F}{2d}<x-b>^{2}}

M(x)=Ra<x−0>1+F6d<x−a>3−F2<x−b>2−F6d<x−b>3\small{M(x) = R_a<x-0>^{1} + \frac{F}{6d}<x-a>^{3} - \frac{F}{2}<x-b>^{2} -\frac{F}{6d}<x-b>^{3}}
Type 2 loading condition:

V(x)=Rb<x−0>0+F2d<x−c>2−F<x−e>1−F2d<x−e>2\small{V(x) = R_b<x-0>^{0} + \frac{F}{2d}<x-c>^{2} - F<x-e>^{1} - \frac{F}{2d}<x-e>^{2}}

M(x)=Rb<x−0>1+F6d<x−c>3−F2<x−e>2−F6d<x−e>3\small{M(x) = R_b<x-0>^{1} + \frac{F}{6d}<x-c>^{3} - \frac{F}{2}<x-e>^{2} -\frac{F}{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
i) Type 1 loading condition:

θ(x)=1EI[Ra2<x−0>2+F24d<x−a>4−F6<x−b>3−F24d<x−b>4+C1]\\ \theta(x) = \frac{1}{EI} [\frac{R_a}{2}<x-0>^{2} + \frac{F}{24d}<x-a>^{4} - \frac{F}{6}<x-b>^{3} - \frac{F}{24d}<x-b>^{4} +\hspace{0.1cm} C_{1}]
ii) Type 2 loading condition:

θ(x)=1EI[Rb2<x−0>2+F24d<x−c>4−F6<x−e>3−F24d<x−e>4+C1]\theta(x) = \frac{1}{EI} [\frac{R_b}{2}<x-0>^{2} + \frac{F}{24d}<x-c>^{4} - \frac{F}{6}<x-e>^{3} - \frac{F}{24d}<x-e>^{4} +\hspace{0.1cm} 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
i) Type 1 loading condition:

Y(x)=1EI[Ra6<x−0>3+F120d<x−a>5−F24<x−b>4−F120d<x−b>5+C1x+C2]Y(x) =\frac{1}{EI}[\frac{R_a}{6}<x-0>^{3} + \frac{F}{120d}<x-a>^{5} - \frac{F}{24}<x-b>^{4} - \frac{F}{120d}<x-b>^{5} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]
ii) Type 2 loading condition:

Y(x)=1EI[Rb6<x−0>3+F120d<x−c>5−F24<x−e>4−F120d<x−e>5+C1x+C2]Y(x) =\frac{1}{EI}[\frac{R_b}{6}<x-0>^{3} + \frac{F}{120d}<x-c>^{5} - \frac{F}{24}<x-e>^{4} - \frac{F}{120d}<x-e>^{5} +\hspace{0.1cm} C_{1}x +\hspace{0.1cm} C_{2}]
  1. Apply the Boundary Conditions to find the constants
    
    and
    
    
i) Type 1 loading condition:

BC 1: @ x=0, Y(x)=00=1EI[Ra6<0−0>3+F120d<0−a>5−F24<0−b>4−F120d<0−b>5+C1(0)+C2]0=1EI[0+0+0+0+0+C2]C2=0\text{BC 1: @ x=0, Y(x)=0} \\ 0 =\frac{1}{EI}[\frac{R_a}{6}<0-0>^{3} + \frac{F}{120d}<0-a>^{5} - \frac{F}{24}<0-b>^{4} - \frac{F}{120d}<0-b>^{5} +\hspace{0.1cm} C_{1}(0) +\hspace{0.1cm} C_{2}] \\0=\frac{1}{EI}[ 0 + 0 + 0 + 0 + 0 +C_{2}] \\ C_{2}= 0

BC 2: @ x=L, Y(x)=00=1EI[Ra6<L−0>3+F120d<L−a>5−F24<L−b>4−F120d<L−b>5+C1(L)+C2]0=1EI[RaL36+F(L−a)5120d−F(L−b)424−F(L−b)5120d+C1L]C1=−[RaL26+F(L−a)5120dL−F(L−b)424L−F(L−b)5120dL]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[\frac{R_a}{6}<L-0>^{3} + \frac{F}{120d}<L-a>^{5} - \frac{F}{24}<L-b>^{4} - \frac{F}{120d}<L-b>^{5} +\hspace{0.1cm} C_{1}(L) +\hspace{0.1cm} C_{2}] \\ 0=\frac{1}{EI}[\frac{R_aL^3}{6} + \frac{F(L-a)^5}{120d} - \frac{F(L-b)^4}{24} - \frac{F(L-b)^5}{120d} +C_{1}L] \\ C_{1} = -[\frac{R_aL^2}{6} + \frac{F(L-a)^5}{120dL} - \frac{F(L-b)^4}{24L} - \frac{F(L-b)^5}{120dL}]
ii) Type 2 loading condition:

BC 2: @ x=L, Y(x)=00=1EI[Rb6<L−0>3+F120d<L−c>5−F24<L−e>4−F120d<L−e>5+C1(L)+C2]0=1EI[RbL36+F(L−c)5120d−F(L−e)424−F(L−e)5120d+C1L]C1=−[RbL26+F(L−c)5120dL−F(L−e)424L−F(L−e)5120dL]\text{BC 2: @ x=L, Y(x)=0} \\ 0 =\frac{1}{EI}[\frac{R_b}{6}<L-0>^{3} + \frac{F}{120d}<L-c>^{5} - \frac{F}{24}<L-e>^{4} - \frac{F}{120d}<L-e>^{5} +\hspace{0.1cm} C_{1}(L) +\hspace{0.1cm} C_{2}] \\ 0=\frac{1}{EI}[\frac{R_bL^3}{6} + \frac{F(L-c)^5}{120d} - \frac{F(L-e)^4}{24} - \frac{F(L-e)^5}{120d} +C_{1}L] \\ C_{1} = -[\frac{R_bL^2}{6} + \frac{F(L-c)^5}{120dL} - \frac{F(L-e)^4}{24L} - \frac{F(L-e)^5}{120dL}]
  1. So you final Deflection equations are:
i) Type 1 loading condition:

Y(x)=1EI[Ra6<x−0>3+F120d<x−a>5−F24<x−b>4−F120d<x−b>4+C1x]whereC1=−[RaL26+F(L−a)5120Ld−F(L−b)424L−F(L−b)5120Ld]Y(x) =\frac{1}{EI}[ \frac{R_a}{6}<x-0>^{3} + \frac{F}{120d}<x-a>^{5} - \frac{F}{24}<x-b>^{4} - \frac{F}{120d}<x-b>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_aL^2}{6} + \frac{F(L-a)^5}{120Ld} - \frac{F(L-b)^4}{24L} - \frac{F(L-b)^5}{120Ld} ]
ii) Type 2 loading condition:

Y(x)=1EI[Rb6<x−0>3+F120d<x−c>5−F24<x−e>4−F120d<x−e>4+C1x]whereC1=−[RbL26+F(L−c)5120Ld−F(L−e)424L−F(L−e)5120Ld]Y(x) =\frac{1}{EI}[ \frac{R_b}{6}<x-0>^{3} + \frac{F}{120d}<x-c>^{5} - \frac{F}{24}<x-e>^{4} - \frac{F}{120d}<x-e>^{4} + \hspace{0.1cm}C_1x] \\ \text{where} \hspace{0.3cm} C_1 =-[\frac{R_bL^2}{6} + \frac{F(L-c)^5}{120Ld} - \frac{F(L-e)^4}{24L} - \frac{F(L-e)^5}{120Ld} ]
You are now ready to plot the curves to determine the overall shear force, bending moment and deflection of a simply supported beam with a triangular load!