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Homework Help: Analysis of Structures

  1. Nov 3, 2013 #1
    1. The problem statement, all variables and given/known data
    A structure is made of 3 members, AD, AC, and BD (as shown in the attached image). A and B are NOT directly connected. There is also a pulley attached to D with radius 0.15 m. 1 end of the cable is attached to a 500 N weight, the other on is fixed in a horizontal sense. A is on a frictionless pin support. C is on a roller support; its line of action is directed along the y-axis. There is also a 300 N force directed downwards in the center of AD. Determine the forces on member AD.

    2. Relevant equations
    Sum of forces in the X = 0
    Sum of forces in the Y = 0
    Sum of moments about a point = 0

    3. The attempt at a solution
    I took the reaction on the center of D, which is +500 N in both the X and Y, then applied this force on member BD, reversing its sense. B is the only other point, therefore the forces on point B will be equal to the forces on D, reversed in sense. Then I moved to member AC. At B there is 500 N down, 500 N left. I took the moment about A: -500(.5) + Cy(1.0), Cy = +250 N, therefore Ay = -250 N, and Ax = +500 N.

    At this point I got stuck because I just don't understand why the book gives the answer as "On AD: Ax = -300 N, Ay = 0 N, Dx = +300 N, and Dy = +300 N". So it became pretty clear I did something wrong. I've been moving the numbers around in my head but I don't get how the book got 300 N as a magnitude for any force of Dx or Dy. There's probably something simple I'm missing.

    Any help is appreciated.

    Attached Files:

    Last edited: Nov 3, 2013
  2. jcsd
  3. Nov 3, 2013 #2


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    The diagram attached to your post has members AB, BC, and BD. I see no member AD. Is your diagram correct?
  4. Nov 3, 2013 #3
    Member BD is 1 structure, and the L-shaped AD is another structure. A and B are not directly connected. Sorry for the confusion. I don't have a scanner so I couldn't just scan the picture.

    EDIT: I uploaded another image. Hopefully this one clears up any confusion.
    Last edited: Nov 3, 2013
  5. Nov 3, 2013 #4


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    But this force is applied on both BD and AD
    No, not correct; you are assuming the pulley load at D all goes into BD. Not true because you have shear forces in AD. This is not a truss.
    It is almost always best to find support reactions first before taking apart the frame. Look at the entire frame and sum moments about A. Solve for Cy, then Ax and Ay. Now look at AC in a FBD and solve for BD. BD is a 2-force member.
  6. Nov 4, 2013 #5
    Thank you, I managed to figure it out. You were right, I was treating it like a truss for some reason, and kept getting weird answers. I eventually just took apart the entire frame pin by pin, member by member. It helped me understand frames better, and I managed to get the correct answer eventually.
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