# Homework Help: Unevenly distributed loads

1. May 28, 2006

### TSN79

In static problems, an evenly distributed load can be made into a point-force by multiplying the load by its length, and so this point-force will now act in the center of the distributed load. This is cool, but I can't find out how to do the same thing with loads that are unevenly distributed, those who are often shown as triangles, often with no force on one end and some force on the other.

This page gives me a formula for determining the value of this point-force, but it does not tell me at which distance from one of the ends it acts!

http://images.google.no/imgres?imgu...ributed+load&start=20&svnum=10&hl=no&lr=&sa=N

If anyone can help me out I'll appreciate it!

2. May 28, 2006

### vsage

Is there a particular shape you're looking at for a distributed load? If you assume the x axis is horizontal (like the page you linked to does), the x value of the centroid of the shape will give the x value of the point the total force acts on. There are bound to be loads of websites with the values of the centroids of different shapes.

3. May 28, 2006

### TSN79

Look at this page then:

http://physics.uwstout.edu/StatStr/statics/Stests/beams1/sol312.htm

If I am to find the torque the load exerts around point A, how would I know the length of the arm? I see in the example that it has been found to be 6.67 ft, but I don't get where that comes from...help!

Last edited by a moderator: Apr 22, 2017
4. May 29, 2006

### vsage

The length of the arm is definitely not 6.67 feet. The x value of the center of gravity of that distributed load is, however, 6.67 feet from the point A. For a right triangle, the center of gravity (centroid) is always 1/3 the base away from the upright side.

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