The problem involves finding the bending stress of a beam. The beam is 48 inches long, supported on both ends, and has a distributed load of 20 lb/in.

I know you have to use My/I to find the bending stress, but how do you find M? I thought since it's a distributed load, then it would just be 20lb/in (distributed load) x 48 in (length of beam) x 24 in. (distance from edge to midpoint), but the solutions uses Wx/2(l-x) where x = l/2. Why do they divide Wx by 2? Isn't that the equation for a triangle?

Thanks.

# NCEES 2001 #117

Started by
MetsFan
, Apr 11 2012 02:51 PM

4 replies to this topic

### #1

Posted 11 April 2012 - 02:51 PM

### #2

Posted 11 April 2012 - 03:14 PM

For a uniformly distributed load on a simply supported beam, maximum Bending Moment occurs at the center of the beam where the shear force line crosses the centerline. The reactions on the supported ends are WL/2 each. The shear force diagram has two triangles with the diagonals passing thru the center of the beam. Bending moment is the area of the shear force diagram (triangle in this case) on either side of the center of the beam. Hence the area of the triange is 1/2x WL2 x L/2 which gives WL/8 as the Bending Moment.

### #3

Posted 11 April 2012 - 03:31 PM

Use the beam tables to get the equation for the maximum moment.

### #4

Posted 11 April 2012 - 03:53 PM

Thanks guys, that makes sense. I skipped most of the statics/dynamics chapters so I guess it's coming back to haunt me.

hoosier2009, I think you mean 1/2 x WL/2 x L/2, which gives WL^2/8, right?

hoosier2009, I think you mean 1/2 x WL/2 x L/2, which gives WL^2/8, right?

### #5

Posted 11 April 2012 - 04:01 PM

Yeah. I meant WL^2/8.

#### 0 user(s) are reading this topic

0 members, 0 guests, 0 anonymous users