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Internal reactions on bent bar

  1. Sep 27, 2012 #1
    Calculate the internal reactions for the member shown at the sections indicated;
    picture attached with FBD's

    I have the V & F forces correct but my moments are off. Can you point me in the right direction here?

    V M-M = 2k
    F M-M =-0.5k

    V N-N = -2k
    F N-N = 0.5k

    M M-M = -10k (2) + 0.5k (2) + 2k -0.5k + M=0
    M= -18.5 k/ft

    answer claims to be 21k/ft

    M N-N = -2k (3) + 0.5k (3) - 2k +0.5k +M=0
    M= - 6 k/ft

    answer claims to be + 6k/ft

    Thank You
    -Mike
     

    Attached Files:

  2. jcsd
  3. Sep 27, 2012 #2

    PhanthomJay

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    Your FBD's need some work in identifying force directions.
    You have labeled the sections MM and NN , but then reverse them in your calcs.
    I dont know what you mean about Ay = 10 K by inspection.
    You should indicate that the base at A is a fixed connection.
    The unit of moments or torques is ft*kips, not kips/ft.
    For equilibrium, the sum of forces in the x direction, and the sum of forces in the y direction, and the sum of moments about any point, inetrnally or externally, are each equal to zero.

    You are not calculating moments correctly. The moment of a force is the magnitude of the force times the perpendicular distance from its line of action to the point about which you are taking moments. If clockwise moments are considered plus, then counterclockwise moments are negative.
    If you look at section MM (which you call NN in error), the 2 k force produces a cw moment about MM of (2)(3) = 6 ft-k. That means that the momnet at MM must be - 6 ft-k (CCW) in order for the moments to sum to 0 for equilibrium. The 0.5 K force produces no moment about MM, because there is no 'perpendicular distance' moment arm. And dont add up forces in ths equation....which you did...when you are summing moments. Move on to the next section 'NN'...............
     
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