Bending Moment diagram

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  • #1
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i have the following problem and i have worked out the resultant forces that say A=397.8kN while B is 367.4 kN and the force of the UDL is 748.2kN. i have worked out the shear force diagram but am lost as to how to draw the bending moment for this problem any help would be appreicated. thank you
 
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  • #2
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Since this is homework, you should show some attempt. But let me ask you to look at your paper and answer this question? Are the free body diagram, the shear diagram, and the bending moment (to be) diagram lined up exactly vertically and to scale? Then where (physically) does your text tell you to start with the bending moment diagram? Then come back with some attempt please. perhaps you could post a picture of what you have so far.
 
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  • #3
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i have them lined up with one on top of the other but i have no idea about how to find the bending moments, so far i have tried to split the beam into two parts and try to solve for x but i get figures that cannot be correct.

i am merely asking how you calculate bending moments not to solve the following
 
  • #4
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i have them lined up with one on top of the other but i have no idea about how to find the bending moments, so far i have tried to split the beam into two parts and try to solve for x but i get figures that cannot be correct.

i am merely asking how you calculate bending moments not to solve the following
Bending moment is the summation of force X distance to the left of where you are at the moment on the BMD. Does that help you get started?
 
  • #5
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it does but i have to ask when i finish taking moments at either C or B i have two equations do i solve these simitanly or do i just use the equations to find the bending moment of at a cetain point eg equation C i say x=1,2,3,4 etc and calculate the bending moment for that point. or do i just find out x?
 
  • #6
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Well, you didn't give me any measurements, so I have to guess. Here's what I'm gonna guess. The distance from C to A is 1.7 m. The distance from A to B is 4.3 m. OK?

So, go to a point on the BMD that is 1.7 m from the left. In other words, you are right over top of point A. Look to your left.

The only force to your left is the downward pointing 17 kN. That force is 1.7 m distant. The bending moment is 17 kN X 1.7 m = 28.9 kNm (I'm not gonna mess with significant digits since I don't know how many there really are). That bending moment is also negative - why?

If you were to take a piece of thin plastic (or something similar) and support it like this and put similar loads on it, you would see that the strip is bent down at this point. That is, the very top fiber is in tension. That corresponds to a negative bending moment.

OK, now you try a point maybe 1 m to the right of point A. What do you get?
 
  • #7
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would it be 17*0.7=11.9 kNm?

are you saying that i should treat the two seperate parts of the beam as individual so i look first at the normal beam then the udl? so hence there is no unknow value of x distance?
 
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  • #8
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would it be 17*0.7=11.9 kNm?

are you saying that i should treat the two seperate parts of the beam as individual so i look first at the normal beam then the udl? so hence there is no unknow value of x distance?
No, don't worry about the beam. Think only of forces and distances to your left. So, if you are 1 m to the right of point A and you look to the left, you see an applied load of 17 kN that is a distance 2.7 m. So, that part of the bending moment is 17 kN X 2.7 m = 123.9 kNm CCW. Now there are 2 more forces to your left (forget your right - it doesn't exist yet). What are they and how far away are they and are they CW or CCW?
 
  • #10
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so is this what you mean?
http://img167.imageshack.us/img167/8245/problem2uy7.th.jpg [Broken]

or do i have to take the udl into account? ie +42.63 kN
Oops. Sorry about the 17 kN X 2.7 m calculation. Brain fog this am.

So, yes, so far, so good. You picked up the reaction force. Now add in only the part of the UDL that is to your left. Treat it as though it were a separate UDL from all the rest.

I notice you're treating CCW as positive and CW as negative. Just keep being consistent for now, but, when you have the whole BMD completed, we should discuss that a minute.

Keep going and watch to see if you notice any correspondence between the BMD and the shear diagram. BTW, if nobody has told you, UDLs always produce parabolic shaped BMD.
 
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  • #12
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so would my final BMD look like this:
updated: http://img518.imageshack.us/img518/4980/problem3up7.th.jpg [Broken]

i believe that at the supports we would not have any bending moments as they will be 0, however, i am unsure what would happen when the beam mets the udl
OK, you're almost there. Just a couple points. BTW, I haven't checked your calculations - they look about right but my wife needs me for a couple hours, so I'll check later.

Flip your BMD upside down. What you are calling negative is positive and vice-versa. The trick is to look at the actual deflection of the beam and see if the topmost fiber is in tension (Negative BM) or compression (positive BM).

Zero BM is called an inflection point and that is where the topmost fiber changes from tension to compression. You have to solve for this point by setting the BM summation = 0. It will not be over top the reaction forces.

The reason I stressed neatness and drawing to scale and lining up the diagrams is that the maximum BM will be where the shear diagram goes through zero. If you have several zero shears, you will have several maximum (positive or negative) BMs.

So, do that and, if you want me to check everything, post the beam diagram, the shear diagram, and the BMD and I'll go over it just after lunch.

Thanks for hanging in there.
 
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  • #13
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here is what i have got so far, but i do not know what the distance on the BMD when the second point is supposed to hit 0, for example i know that when above point A we have a lift while at d we have the greatest bending moment. i know that at c it intepets the x axis but i dont know at what distance on the beam from the left side its 0.

if you can check the following diagram and tell me how i can find out the point c (on BMD) distance from the point where we have the 17 kN load acting thanks.

http://img223.imageshack.us/img223/6454/problem4ku6.th.jpg [Broken]
 
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  • #14
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I think that I solved your problem. Look at the picture above:
 

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  • #15
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here is what i have got so far, but i do not know what the distance on the BMD when the second point is supposed to hit 0, for example i know that when above point A we have a lift while at d we have the greatest bending moment. i know that at c it intepets the x axis but i dont know at what distance on the beam from the left side its 0.

if you can check the following diagram and tell me how i can find out the point c (on BMD) distance from the point where we have the 17 kN load acting thanks.

http://img223.imageshack.us/img223/6454/problem4ku6.th.jpg [Broken]
OK, that looks pretty good, but there are a few things yet. I'm a little confused by your nomenclature for the various points, so I'm gonna use labels instead.

You seem to have the reaction forces correct although on the beam diagram at Rb, you have 36 +/-.38 . What is that? The reaction force is 367 kN and you seem to know that since you use it correctly in the shear diagram.

The shear diagram is correct except that you should use negative signs below the zero line. It would also be good to solve for the zero crossing since it's easy here.

The BMD is reversed positive/negative. Flip it upside down. You seem to have solved for the maximum BM at the halfway point between A and B. Your calculation looks OK but that is not the maximum. The maximum is exactly where the shear crosses through zero.

You're doing good. Just a few small points.
 
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  • #17
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OK, that looks pretty good, but there are a few things yet. I'm a little confused by your nomenclature for the various points, so I'm gonna use labels instead.

You seem to have the reaction forces correct although on the beam diagram at Rb, you have 36 +/-.38 . What is that? The reaction force is 367 kN and you seem to know that since you use it correctly in the shear diagram.

The shear diagram is correct except that you should use negative signs below the zero line. It would also be good to solve for the zero crossing since it's easy here.

The BMD is reversed positive/negative. Flip it upside down. You seem to have solved for the maximum BM at the halfway point between A and B. Your calculation looks OK but that is not the maximum. The maximum is exactly where the shear crosses through zero.

You're doing good. Just a few small points.
o thats supposed to be 367.38kN my bad!

ok so on the SFD is the 0 value supposed to be 2.15m? (the middle of A to B?) if it is then wouldnt the maximum bending moment be half way of A to B +/- the 17 kN?

the other point i was curious as to knowing is at what point would i hit 0 on the BMD so i can show a parabolic curve for the UDL, as i dont believe we can just drop from 28.9 to zero like a SFD.

by the way TVP45 i would like to thank you for all your help, really helped me into understanding how to draw bending moment diagrams!
 
  • #18
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On the SFD, you're going in a straight line from +381 kN at reaction A to -367 kN at reaction B, so the zero crossing would be (381/(381+367))*4.30 m = 2.19 m. See if you get that also.

And, you're right that you wouldn't just drop in a straight line from the -28.9 kNm to 0 on the BMD. The parabola starts at the -28.9 kNm cause that's where the UDL starts. Zero crossing can be found by setting the summations = to 0. I always just sort of guess the parabola shape unless it's something criticial; then I plot maybe a dozen points.

The important thing to remember is that the maximum BM is wherever the SFD goes through zero.

You're doing great. Keep at it.
 
  • #19
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im sorry but what do you mean by setting the summations = to 0 is that just taking moments at a point and saying it is zero? also wouldnt the UDL have to cross 0 on the SFD at 2.15 as that is 4.3/2?
 
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  • #20
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im sorry but what do you mean by setting the summations = to 0 is that just taking moments at a point and saying it is zero?
Exactly that. Just calculate moments from left to right and say =0 (or from right to left). Sorry on bad english.
 
  • #21
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im sorry but what do you mean by setting the summations = to 0 is that just taking moments at a point and saying it is zero? also wouldnt the UDL have to cross 0 on the SFD at 2.15 as that is 4.3/2?
Look at the SFD. From A to B, you have a left side that is 381 kN. On the right side, you have -367 kN. They are joined by a straight line. They are similar triangles. The sum of the bases is 4.30 m. The base of the left hand, by similar triangles, is (4.30 m) X (381 kN/748 kN) = 2.19 m. 748 kN is just 381 kN - (-367 kN).

About the summation. Go down to the BMD. You can see by examination that the zero crossing will be a little to the right of reaction A point. Let's call that distance s meters. Move to s and calculate the bending moments. You have (without units)
-17 X (1.70 + s ) = -28.9 - 17s
398 X s = 398s
-((174 ) X s) X (s/2) = -87s^2
Sum them to zero
-87s^2 + 381s - 28.9 = 0
s = 0.077 m and 4.30 m
You already knew 4.30 m, so the one you want is 0.077 m to the right of reaction A.
 

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