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Stiffness Matrix method

  1. May 15, 2014 #1
    This problem has alot of calculations from the beginning so i have skipped to the part i am stuck with and tried to include relevant information, apologies if i have missed anything. Assume all working is correct as i was given the answers.

    After carrying out the stiffness matrix method I am trying to construct a bending moment diagram using member equilibrium, see attached.

    The specific problem is member 1-5

    working

    member 1-4

    bending moment at 1 (given) 123.33Kn
    vertical force 57.44Kn
    length 2.5

    therefore

    123.33 - 57.44Kn*2.5 = -20.27 (correct)

    Member 4 -5
    -20.67 + 160KN(moment from pre-eliminated beam) = 139.63Kn (correct)

    however i can not get the moment of -123.98 at point 5

    attempt member 4-5
    vertical force relative to beam = 80 (sin(36.87)
    moment = 139.63Kn
    139.63Kn - 80 sin(36.87)*2.5 = 19.629 (incorrect)
     

    Attached Files:

  2. jcsd
  3. May 17, 2014 #2
    It's difficult to follow what you doing here but is there not a shear force at point 4 in addition to the 80?
     
  4. May 17, 2014 #3
    Yes I appreciate it is fairly difficult to follow. Essentially I have got the bending moments on one side using stiffness matrix method and the used these values and member equilibrium for the rest of the beam. There is a shear force yes, I have tried to account for that, that's why I have used 80sin(36.87) as this would give me the vertical component of the force (vertical relative to the member)
     
  5. May 17, 2014 #4
    The shear force is the internal shear in addition to the 80 external force.
     
  6. May 17, 2014 #5
    Would that have to be included in the bending moment diagram? If so, how is the force calculated? And would it be used in the same way as a vertical force e.g F x distance
     
  7. May 17, 2014 #6
    1) Yes.

    2) Same way you got the bending moment at point 4.

    3) Yes.
     
  8. May 18, 2014 #7
    I got the moment by using (UDL x L) x L/2 ?
     
  9. May 18, 2014 #8
    But you added another term to it as well.
     
  10. May 18, 2014 #9
    Yes I added the moment from point one. For point 5 this is the method I have tried. I don't follow what you are suggesting?
     
  11. May 18, 2014 #10
    OK, we are trying to get the moment at point 5, correct? So you need to take the moment at point 4 plus the shear at point 4 multiplied by L = 2.92m.

    The shear at point 4 is 80kN*sin(37.9°) plus the shear from member 4-1.
     
  12. May 18, 2014 #11
    Attempt

    139.63-((80sin(37.9))+57.44)*25 = -126.83

    This is pretty close to -123.98 but still off,

    Have i made an error?
     
  13. May 18, 2014 #12
    You probably made a rounding error.
     
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