# Statics - Distributed Load Over Beam

1. ### sael

5
Hello,

I am attempting a problem in an older statics text book an image of which is attached.

No matter what I try, I get a solution of between 0.52 and 0.53 metres and was hoping someone could verify if the text book answer of 0.48m is right or wrong.

Thanks Heaps,

Scott.

2. ### nvn

2,124
sael: There is a homework template with three parts, which should have appeared in your browser. The second part is "list relevant equations." And the third part is "show your work." Would you be able to show your work? And then someone might check your math.

3. ### pongo38

707
There is scope for a difference here that might explain your question. For example, in the first section, 0<x<100 mm, how did you decide the value on the graph in the range 0.15<y<0.23 ?

4. ### sael

5
As I don't know the equation of the curve ( although it looks like some function of ln ) I assumed a linear increase from 0.15 to 0.23.

The total force for the segment was calculated by taking the average of 0.15 and 0.23 and multiplying it by 0.1, i.e. 0.1*(0.15+0.23)/2.

I then took the moment for section 1 about point A by applying the force through the centroid of the area of the approximated segment. Initially I used an exact centroid but found that it was really close to just applying the force down the centre of the element. I used both methods in a spreadsheet and they didn't appear to make any significant difference to the answer.

I did all nine other segments this way, summed the moments which equals the reaction at B. The reaction at A was found by summing the forces of all segments and then subtracting the reaction at B.

I then calculated the shear forces at 100mm intervals along the beam by starting with reaction A at x=0, subtracted the force for section 1 to get the shear force at x=0.1m, subtracted the force for section 2 from the shear force at x=0.1m to get the shear force for x=0.2m etc.

To find the point of maximum bending moment I looked for the shear going from +ve to -ve.

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6. ### pongo38

707
I can't see anything wrong with what you've done and assume that the distance given in the answer is the distance from B not A. The centroid of the loading is clearly in the right half of the beam, and so the answer given cannot be correct as it stands.

7. ### sael

5
Thanks heaps for that. I can see how in this case that the distribution being greater on the right half of the beam should push the centroid and point of zero shear to the right half of the beam as well.

What I am having trouble with now is whether, for any distribution, the centroid being in the right half of the beam would indicate the zero shear point is also in the right half of the beam.

If for example the distribution was initially constant across the entire beam, the centroid and zero shear point would coincide in the centre of the beam. If some of the distribution is taken from the very left hand side and moved closer to the centre but still resides left of centre the centroid would shift to the right but the zero shear point would shift to the left ( I think ). Maybe I am wrong?

Thanks again :-)

8. ### pongo38

707
The centroid is not necessarily the same point as the point of zero shear because the centroid is based on a balance of moments, whereas the point of zero shear is based on a balance of forces. However, because the centroid of the loads is to the right of centre, so that is where the point of maximum moment will be. Therefore the point of zero shear is to the right of centre.

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