Bending Moment and Force calculations two beams connected

1. May 8, 2012

olski1

Hello,

I need to find the bending moment and forces at the mid point cross-sections of two separate beams connected together. One is a horizontal beam connected to the top of a vertical beam that sort of look like this:

H========
H
H
H
H

There is a UDL of 10kn/m on the top member, with the bottom of the vertical beam is fixed to the ground.

I need to find the bending moment and Forces at the midpoint cross-section of both beams separately. How would I go about doing this?

So far in my course we have only dealt with single beams that are simply supported.

My first attempt at this question was to treat them as two separate beams and firstly find the reaction bending moment at the joining of the two beams using the udl. Then I would take that moment as a acting on the end of the vertical beam and work my way from there. Does that moment act at the very tip of the upright beam or half the width down of the horizontal beam?

But honestly I am quite confused with this one..... any help on the method (not answer) would be greatly appreciated :)

Last edited: May 8, 2012
2. May 8, 2012

PhanthomJay

I assume you have also dealt with beams fixed at one support?
you can do it this way, but don't forget the reaction force at that joint as well.
..and force...
For the purposes of this problem, you can use the beam centerlines for the analysis.
You should note that it it is not necessary to separate the beams when solving for the forces in the vertical beam. You can take a free body diagram at the mid point of the horiz beam to get it's internal moment and force at that point, and you can take a FBD at the midpoint of the vertical beam to get its internal force and moment

3. May 9, 2012

olski1

For the top member which is 1m long I found:

ƩFy=0=-10kN + RAY => RAY=10KN

ƩMo=5<x>2 + MA => MA= -5<x>2

when x = 1 => MA=-5kNm

At the mid point section,

F=-5kN and M=-1.25KN

For the Vertical Memeber which is also 1 m long.

ƩFx=0=-10kN + RBx => RBx=10KN

ƩMo=-MA + Mo => Mo= -5kNm

hence at x=0.5, M=-5knm and F=-10kN.

Just to check I ran this through F.E.A with a beam with cross section 80mm high and 40mm wide, which is the dimensions given in the problem. But I am getting slightly different results.

For example, the top member moment is 1.133kNm and the Force is 4.733kN compared to 1.25kNm and 5kN.

Why are they different? where did I go wrong?

4. May 9, 2012

PhanthomJay

You didn't go wrong, your calculation was excellent. When you do the computer analysis, the solution is dependent on the end connections.....for example, whether the top beam frames into the vertical beam onto its top, or onto its side, will yield slightly different results, depending on how you input the member properties and lengths into the computer, and how you modeled the joint. Heck, the solution is only accurate to 1 significant figure anyway.