# Bending Moment, UDL and Forces

1. May 14, 2012

### mm391

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

Find the equations for the bending moment along AB and BC?
Reactions at A = -15kN
Reactions at B = 60kN

2. Relevant equations

F in the Y direction = 0
Moments with clockwise being positive = 0

3. The attempt at a solution

Moments about cross-section from the left before the point load (clockwise is +ve)= 15x-10(x/2)-M=0
M=10x

(clockwise is +ve) = 10x-15(x-1)-10((x-1)/2)-M=0 M=-10x-20

Moments about cross-section from RIGHT hand side = M+60(3-x)+10(3-x)((3-x)/2)=0
M= -5(x^2-18x+45)

As I have no answers I am not sure if these are correct or where to go from here?

Any help would be greatly appreciated.

Thanks

Matt

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Last edited: May 14, 2012
2. May 14, 2012

### PhanthomJay

Show your calcs for the reaction forces. They should both be positive upward and sum to 45.

3. May 14, 2012

### pongo38

In these problems you can check it yourself. Sum of vertical loads = sum of vertical reactions? Checks ok. BUT Sum of moments about any point must be zero. Emphasise any. Do you, for example, have a balance of moments about C? In checking this, make sure that, whatever sign convention you adopt, clockwise moments balance anti-clockwise moments. This should reveal that your reactions are in error. You could have (should have) checked this before attempting any bending moment equation. I suggest you start again by getting reactions right, and check for your self they are correct by checking moments balance about ANY point. Do this for several points to make really sure.

4. May 15, 2012

### mm391

My fault. Have just calculated the reaction forces again:

ƩMoments about A (clockwise +ve) = (15*1)+((10*3)/2)+Reactions at C*3)=0

∴ Reactions at C = 10kN

ƩForces in the Y direction = Reactions at A - 15 - 30 + Reactions at C = 0

Reactions at C = 10

∴ Reactions at A = 35kN

5. May 15, 2012

### PhanthomJay

you are incorrecty calculating the moment from the distributed load. When determining force reactions, the distributed load can be represented as a single resultant concentrated load acting at its center of gravity. The uniform load is 10 kN/m over the 3 m length of beam. So its total resultant load is ___?___ and its moment arm is ___?___
Your error carries over when doing it this way. Sum moments about C instead, then check your work to be sure ƩForces in y direction = 0.

6. May 15, 2012

### pongo38

Jay's advice good. Please check your answer by taking moments about any point you haven't used before. This is what engineers have to do in practical cases, because real life examples don't have an answer "in the back of the book". A good way to get reactions in a case like this is to take moments about A to get reaction at C. Then take moments about C to get reaction at A. Then check that vertical forces are balanced.

7. Jun 10, 2012

### mm391

So I find the Reactions to be;

Reactions at A = 25kN
Reactions at B = 20kN

Then the bending moment equation from 0<x<1 meter:

25x-5x^2

Now I am unsure how to carry on, do I make another cut before Reactions at B and take moments then after solving that make another cut and take moments from the left hand side. Sorry I know this may sound vague but I am not sure how to explain my method. I have tried it before but when I look at it none of the bending moments equaitons seem to incude the 15kN load. Should I have three bending moment with one from 0 to 1m, one from 1m to 1<x<3 and then find another eqations.

8. Jun 11, 2012

### PhanthomJay

Reactions are at A and C, not A and B.
Your moment equation between x = 0 and x= 1 is OK.
Between x = 1 and x = 3, there are several ways to find the moment. One way is to cut the beam to the right of B and use the left section as your FBD and sum moments about the cut. Note that since the distance from A to the cut is x, then for example the distance from B to the cut is (x - 1), etc.