# Beam - Bending Moment

1. Mar 9, 2014

### Antex

1. The problem statement, all variables and given/known data

Problem: For the beam according to the image, solve graphical and analytical:

1.) Reactions in the supports,
2.) Bending moment diagram,
3.) Maximum bending moment

Given variables:
F = 14 kN
q = 40 N/cm
a = 100 cm
b = 55 cm
l = 180 cm

2. Relevant equations
This is a Beam with both overhanging supports, with a UDL and a point load on the right overhang. And I cannot find information for solving this.

Sorry for my terminology, I am doing physics in another language.

I am having a lot of trouble with the solving of this problem. What really confuses me is the UDL (I hope this is the correct terminology) that isnt over the entire beam, but hangs off on the left side.

I can't find any problems in my book that solve this kind of problem, and the professor didn't go into detail.

3. The attempt at a solution

This is a Beam with both overhanging supports, with a UDL and a point load on the right overhang.

My attempt at this was trying to figure out the force F$_{q}$, with: F$_{q}$ = q*l => 40 * 280 = 11200 N/cm, 112 N/m, 0,112 kN/m

And assuming that the above is correct, which I do not know, I can go ahead and find F$_{A}$ and F$_{B}$

Last edited: Mar 9, 2014
2. Mar 9, 2014

### haruspex

The numbers are right, but not the units. If you multiply N/cm by cm what do you get?

3. Mar 9, 2014

Just N?

4. Mar 9, 2014

### haruspex

Yes.

5. Mar 10, 2014

### Antex

Could you explain me the difference if there was no overhang at A, and the UDL ended at A? Is there a different process in calculation in that case?

And also: 11200 N is then 112 kN? or would it be N/m and then 1.12 kN?

6. Mar 10, 2014

### BvU

1. The difference is in the position where Fq "grabs". Otherwise same calculation.
2. No: 11200 N = 11200 N / (1000 N/kN)
or if you are word-oriented: eleven thousand 200 = eleven thousand plus (200 * thousand/thousand) = (eleven plus 200/thousand ) * thousand​

7. Mar 11, 2014

### PhanthomJay

When determining the support reactions, the UDL can be represented by a single concentrated load of 11,200 N acting at the center of gravity of UDL, that is, at (a + l)/2 from the left end of the beam.