1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    member 1-4

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


    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

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted