# Singularity Functions for Beam Bending

1. Jan 24, 2010

### tangodirt

There is a beam of width 10cm, and vertical reaction loads on each end (x1 = 0cm, x2 = 10cm). Starting from the left end of the beam, we have a vertical distributed load of 2,000 N/m spanning from 0cm to 5cm. Finally, we have a 1,000 N point load located 7.5cm from the left end of the beam.

Through statics, it can be said that the left most reaction load (x1 = 0cm) is of magnitude 325 N while the right most reaction load (x2 = 10cm) has a magnitude of 775 N.

My singularity function for this system is shown below:

$$V = 325<x - 0>^{0} - 2000<x - 0>^{1} + 2000<x - 0.05>^{1} - 1000<x - 0.075>^{0} + 775<x - 0.1>^{0}$$

Which, when plotted (my end goal here), works perfectly and as it should. My issue comes when I switch the shear (V) singularity function to a moment function by increasing the exponents by one (as I've been told).

Through integration of the shear singularity function, the moment equation then becomes:

$$M = 325<x - 0>^{1} - 2000<x - 0>^{2} + 2000<x - 0.05>^{2} - 1000<x - 0.075>^{1} + 775<x - 0.1>^{1}$$

Which doesn't work quite as well. The moment function falls completely apart, but from every source I've read so far, it shouldn't. Also, if I draw the moment equation by hand (through the "area under the curve" approach), it hardly matches the output of the moment singularity equation.

Any ideas?

Last edited: Jan 24, 2010
2. Jan 25, 2010

### nvn

tangodirt: But for n ≥ 0, w*integral[(<x - a>^n)*dx] = w*[1/(n+1)]*<x - a>^(n+1), not w*<x - a>^(n+1). Therefore, don't the second and third terms of your M equation contain a typographic mistake? See if this resolves the problem.