
#1
Dec2710, 04:39 PM


#2
Dec2710, 11:09 PM

Sci Advisor
HW Helper
P: 2,110

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
Dec2810, 12:41 AM

P: 692

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
Dec2810, 05:55 AM

P: 5

Statics  Distributed Load Over Beam
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. 



#5
Dec2810, 03:21 PM

P: 5





#6
Dec2810, 03:42 PM

P: 692

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
Dec2810, 07:29 PM

P: 5

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
Dec3010, 04:25 AM

P: 692

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.




#9
Dec3010, 04:33 PM

P: 5

What about this distribution over the same beam:
As far as I can tell the centroid is on the right half of the beam but the point of zero shear is in the left half of the beam. 



#10
Dec3010, 11:12 PM

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P: 5,965

Yes, that may be true, but this case is unlike the first. Without getting into the calculus behind it, when you distribute a load over a short length of the beam, the more will the distributed load behave as a concentrated load with the same resultant magnitude. Equivalent concentated loads produce greater moments than they would if distributed (compare, for example, the simple beam max moment with a concentated load at the center, which is PL/4, compared to PL/8 if that same load was uniformly distributed across the beam). So yes, there are cases where the max moment occurs left (or right) of the cg of the loads.



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