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Overhanging beams

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  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data

    So I have a beam here, I'll post a Photoshop diagram of the problem at the bottom of this section, but its an overhanging beam simply supported at one end, and mid way down another, points A and B. I've changed up the values so I can do my work myself, I don't want someone blurting the answer to me. I need to learn rather than cruise through my class.

    The beam is carrying a continuous uniformly distributed load that represents it's own weight of 4kN/m and two vertical loads of 16kN and 10kN

    The setup looks like this

    6eQNQ.png

    I'm to create a shear force and bending moment diagram for this beam.

    2. Relevant equations

    So, my relevant equations are very simple, the sum of the moments = 0 because I'm in equilibrium, and so are the forces. I have these equations to work with;

    ∑M=0
    ∑F=0

    3. The attempt at a solution

    Well, to create my shear force diagram I decided to turn the UDL into a point load in the middle, considering I had a load of 4kN/m over 16 meters I ended up with 64kN in the middle of the beam at 8 meters. In hindsight I can see how this might not be right, considering it doesn't take into account support B but I don't know how to take it into account. It isn't at the end of the beam.

    Continuing on and hoping for the best I calculated my moments and did the following

    ∑MA=0
    0= 0+(16*4)+(64*8)-(MB*5)+(10*16)
    0= 64+512-(MB*5)+160

    MB*5=736

    ∴MB=736/5

    =147.2kN


    Ok, so I have now the reaction force at support B. I can now use a different calculation to get ahold of A.

    if ∑F=0
    then A= 64+512-147.2-160
    =268.8

    and from that I can set up a shear force diagram I'm not so sure is correct but I'll have to go with it.

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

    so now the bending moment diagram. Time to work out my bending moments.

    2. Relevant equations

    I'll take point A, and my equation is this

    MA=0

    3. The attempt at a solution

    so what I have in this beam are a number of forces to work out bending moments from.

    I have support A
    4 meters down the beam is force a at 16kN
    then the translated UDL is force b at 64kN, 8 meters away.
    support B exerting a force of 147.2kN upwards
    and force c exerting a force of 5kN 10m from A

    (Yes, I'm using capitals to represent supports and lowercase to represent force)

    so what I did is this

    MA=0
    a= (16*4)+(64*8)-(147.2*10)-(10*16)
    =-1056

    b=(64*4)-(147.2*6)-(10*12)
    =-747.2

    B=-(147.2*2)-(10*8)
    =-374.4

    c=-(10*6)
    =-60

    and I get a simple weird bending moment diagram that doesn't fit my situation, you can see it'd be kind of this shape √.

    Means I don't really get whats going on unfortunately, I need some help with this. What am I doing wrong?
     
  2. jcsd
  3. Jan 9, 2014 #2

    PhanthomJay

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    Hi, Hati, welcome to PF!
    When determining support reactions, you can always represent the uniformly distributed load as a resultant point load acting the cg. But when drawing shear and bending moment diagrams, you must keep the load as a distributed load, not as a point load at the cg.
    You should call the reaction force at B as RB or something else instead of MB , to avoid using the letter M which usually stands for moment, not force. Also, you must have mixed uo some numbers somewhere, because the distance from B to A is 10, not 5.
    you are summing forces, not moments, you need to be careful with your values.
    please google on 'Shear and bending moment diagrams' for help with drawing them
    you need to corect your numbers and throw that uniform load back in there and watch signs etc. and draw good FBD's of cut sections as well as the diagram.
     
  4. Jan 9, 2014 #3
    Thanks, nice to meet you to. Anyway, the thing that complicates matters is I know how to deal with a UDL but its a bit of a problem when it spans the entire beam, because its on both sides of support B. What do I do with it? I can deal with it if it only spanned between A and B, but it runs the full length of the beam here. Do I do anything different with it?

    Thanks, I'll amend that.

    Yeah, the confusion is because I changed the values so I could learn the method here and do the real question on my own. I got confused between the two, that 5 is meant to be 10. Shouldn't matter, really. As long as somehow I get the right idea.

    Already know how, the issue is I don't know if it's right but I'll have to go with it.

     
  5. Jan 9, 2014 #4

    PhanthomJay

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    No, when determining support reactions, just represent it (as you have done) as a single concentrated load of 64 kN acting 8 m from A. Then in determining the support reactions , you sum moments about A = 0 to solve for the support reaction at B, then sum moments about B = 0 to determine the support reaction at A, then check your results to be sure that the sum of the applied and support reaction forces in the y direction add to 0.
     
  6. Jan 9, 2014 #5
    Yes, thankyou. I've gotten as far as the shear force diagram, and it seems about right. I cannot turn the UDL into a point load for my bending moments which is giving me a headache. I have to do something different here.

    There's no way I know of to account for a UDL when I'm creating bending moments with the method above, which works fantastic for point loads. I heard somewhere that the area of shapes I have on each span of the beam I have on my shear force diagram gives me the right values to plot on my bending moment diagram, is this the case?

    I have a place in my shear force diagram, right after the 16kn load where the slope in the line crosses 0 and gives me two triangles. It crosses over at exactly 4 meters giving me one triangle, and where support B offers it's reaction force it crosses back over 0 and gives me that second triangle. Do I measure the areas of both triangles and add them?
     
  7. Jan 9, 2014 #6

    PhanthomJay

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    you can't turn the UDL into a point load when determining shear forces either. If you are getting triangles in your shear diagram, I guess you understand that correctly.
    Without seeing your diagram, I am not sure of the triangles of which you speak....nonetheless, when drawing the moment diagram, note the following, which comes from the calculus:
    -The area under the Shear Diagram between 2 points is the change in moment between those points....(with consideration given to positive and negative areas)....and
    -The slope of the Moment Diagram at any point is equal to the shear at that point.

    Also don't forget your friend; Free Body Diagrams of cut sections!
     
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