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Maximum weight a bar and cables can hold before breaking

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A weight W is supported by bar AO and cables AB, AC, and AD. Point A lies in the xy plane, if the bar and cables have the strengths listed below, determine the largest value W may have.

    Hint given: A FBD of joint A is recommendedto being this problem. Bar AO should be assumed to carry load along its length. Unit vectors point from A to B and from A to C are needed to describe the direction of cable tensions. Since cables are required to be in tension, it should be clear from the figure that AO will be in compression. It is usually not obvious which structural member first reaches its limit load. Loads carried by each member should be obtained in terms of W. Individual failure criteria can then be applied to each member to find W inducing failure in that member. The largest weight that may be supported is the minimum of this set of failure loads.
    Screen Shot 2015-09-22 at 2.08.53 PM.png
    2. Relevant equations
    Unit vector equation
    ΣF=0, ΣFx=0, ΣFy=0,
    Basic trig

    3. The attempt at a solution
    First I drew out the FBD which was basically exactly the same as the picture given so I didn't find it necessary to post it. Then I found the point coordinates to be:
    A (15,48,0) in
    B (15,0,36) in
    C (0,48,36) in
    O (0,0,0) in

    Like the hint suggests, I found the unit vector of AB and AC to be (0,-4/5,3/5) and (-5/13,0,12/3) respectively.
    Then I multiplied those by the strengths given making the, equal to (0,-1568,1176) for AB and (-538.5,0,1292) for AC. The problem statement says the force of AO is already in it's direction so AO would be (29400,94080,0).

    From here I don't really know how to go about "Loads carried by each member should be obtained in terms of W." I just need a good push in the right direction.

    Thank you so much for any help, I really appreciate it!
     
    Last edited: Sep 22, 2015
  2. jcsd
  3. Sep 22, 2015 #2

    SteamKing

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    You might want to check the coordinates for point A.

    What you have to do in order to check the strength of each member is to calculate the loads in OA, AB, and AC using the equations of equilibrium, assuming that the load is W applied at point D. These loads will be in terms of W. Once you have calculated the loads, you then compare them with the allowable strengths listed in the table to find out what the maximum value of W can be without overloading any part of this structure.
     
  4. Sep 22, 2015 #3
    My bad, on my actual work I have written out my A coordinate is (15,48,0) I just had a typo. So when I'm finding the components of all the cables, D is the only one that will have the vertical W force? How does that make every other cable in terms of W?
     
  5. Sep 22, 2015 #4

    SteamKing

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    Because W is the only load applied to this structure. All of the loads in the cables and the bar OA are produced by this one load; hence, it stands to reason that these loads will also be proportional to the magnitude of W.

    Like I said in my previous post, you'll have to solve for the loads in each member using the equations of static equilibrium. The cables will be in tension, because they don't work if they're not, but the bar OA will probably be in compression.
     
  6. Sep 22, 2015 #5
    Let me see if I got this right, I found the unit vectors and I know the tension/compression of all of them because of the table given.

    AB = (0, -1568, 1176)
    AC = (-538.5, 0, 1292)
    OA = (29400, 94080, 0)

    So, from theses I can find
    ΣFx = 0 - 538.5 + 29400 = 28861.5
    ΣFy = -1568 + 0 + 94080 = 92512
    ΣFz = 1176 + 1292 + 0 = 2468

    I know w is in the z-direction, but I'm not really sure where to apply it from here.

    EDIT: Do I find the magnitudes of AB, AC and OA? I attempted that and tried to put in the lowest number for the answer and that didn't work. :frown:
     
    Last edited: Sep 22, 2015
  7. Sep 22, 2015 #6

    SteamKing

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    No, these are not the forces created by the weight W. You are misinterpreting the maximum allowable forces from the given table as the actual forces in the structure.
    For some reason, you are resisting my repeated suggestions that you find the forces in the structure due to the applied weight W, whose maximum magnitude is what you are supposed to be finding. You must use the equations of static equilibrium in order to do this. This is not a simple plug-and-chug problem, which you assume it to be.
     
  8. Sep 22, 2015 #7
    ...I don't know what you are referring to when you say equations of static equilibrium. I get they all have W acting on them but we don't know W so I don't understand how I
    "find the forces in the structure due to the applied weight W". How are the loads of the structures not the magnitudes. I don't get it.
     
  9. Sep 22, 2015 #8

    SteamKing

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    The equations of static equilibrium are:

    ΣM = 0, ΣFx = 0, ΣFy = 0, ΣFz = 0

    Thus, there should be no net force and no net moment acting at point A.

    If you are not comfortable with using W for the load on the structure, then assume for the purposes of this calculation that W = 1 lb or 100 lbs or 1000 lbs, whichever value is more convenient. The forces in the structure you obtain from your calculations can then be scaled by this assumed value to find out their proportion to the applied load W. Then, you may use the values from the table to find out what the actual maximum magnitude of W is such that the structure is not overloaded.
     
  10. Sep 22, 2015 #9
    Alright, but I did take the sum of all the components and you told me I wasn't listening to you. I'm assume I take the sum without adding the strengths.

    That would be
    ΣFx = 0 - 5/13 - 15
    ΣFy = -4/5 + 0 - 48
    ΣFz = 3/5 + 12/3 + 0

    Making ΣFM = ( -200/13, -244/5, 23/5)

    Is this along the line of what you are talking about? I am having trouble seeing where W is going to be applied here, I know it scales the the applied load, but is that found by the all the components?
     
  11. Sep 22, 2015 #10

    SteamKing

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    I have no idea what these calculations represent. They certainly are not the components of the forces in the cables AB and AC or the bar OA.

    You should start by calculating the unit vectors for cables AB and AC and the bar OA.

    The unit vector for AB looks OK, but you should check the unit vector for AC. You also should calculate the unit vector for OA. Remember to check the lengths of the unit vectors to see if each equals 1 exactly.
     
  12. Sep 22, 2015 #11
    The unit vector for OA is (-0.2983, -0.9545, 0)
    AB = (0, -4/5, 3/5)
    AC = (-5/13, 0, 12/13)

    I still got the same thing for AC, AC = (0-15, 48-48, 36-0) = (-15, 0, 36) and the magnitude of this is 39. I reduced the fraction and got (-5/13, 0, 12/13) for the unit vector
     
    Last edited: Sep 22, 2015
  13. Sep 22, 2015 #12

    SteamKing

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    You still need to calculate a unit vector for the bar OA.
     
  14. Sep 22, 2015 #13
    I did... (-0.2983, -0.9545, 0)
     
  15. Sep 22, 2015 #14

    SteamKing

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    Now, the forces in the cables, call them FAB and FBC, along with the force in the bar, FOA, are all unknown at this point.

    By multiplying the magnitude of each force by the corresponding unit vector for the member, you can obtain the components of each force. These components are then used to write the equations of equilibrium for this structure. You can write a single force equation each for the x-, y-, and z- directions. You can also write one moment equation.
     
  16. Sep 22, 2015 #15
    Are the magnitudes the same as the strengths given?
    OA = 1960(-0.2983, -0.9545, 0) = (-584.668, -1870.82, 0) (1960, because AO is in compression)
    AB = 1960(0, -4/5, 3/5) = (0, -1568, 1176)
    AC = 1400(-5/13, 0, 12/13) = (-538.462, 0, 1292.31)

    From these
    ΣFx = -584.668 + 0 + -538.462 = -1123.13
    ΣFy = -0.9545 - 1568 + 0 = -1568.9545
    ΣFz = 0 + 1176 + 1292.31 = 2468.31

    ΣFm = (-1123.13, -1567.9545, 2468.31)
     
  17. Sep 22, 2015 #16

    SteamKing

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    No, they are not. You must find the forces in the members before you can use the loads from the tables.

    The forces in the members will be some factor times W for each member.
     
  18. Sep 22, 2015 #17
    I know that the magnitudes of the unit vectors are going to be 1 so it must be from the original coordinates.

    AB = (0, -48, 36) has a magnitude of 60
    AC = (-15, 0, 36) has a magnitude of 39
    OA = (-15, -48, 0) has a magnitude of 50.289

    but this seems a little redundant.. I don't know what else the magnitudes could be
     
  19. Sep 22, 2015 #18

    SteamKing

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    We need to find the magnitudes of the forces, not the lengths of the members.

    You must use the equations of static equilibrium to find the magnitudes of these forces.
     
  20. Sep 22, 2015 #19
    OA = (-0.2983, -0.9545, 0)
    AB = (0, -4/5, 3/5)
    AC = (-5/13, 0, 12/13)

    ΣFx = -0.2983 + 0 - 5/13, Magx = 0.4867
    ΣFy = -0.9545 - 4/5 + 0, Magy = 1.2454
    ΣFz = 0 + 3/5 + 12/13, Magz = 1.1009
     
    Last edited: Sep 22, 2015
  21. Sep 22, 2015 #20

    SteamKing

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    You don't seem to be grasping the concept here.

    The unit vectors for each member are as follows:
    OA: (-0.2983, -0.9545, 0.0000)
    AB: ( 0.0000, -0.8000, 0.6000)
    AC: (-0.3846, 0.0000, 0.9231)

    Multiply each member's unit vector by the corresponding magnitude of the unknown force:
    FOA = FOA * (-0.2983, -0.9545, 0.0000)
    FAB = FAB * ( 0.0000, -0.8000, 0.6000)
    FAC = FAC * (-0.3846, 0.0000, 0.9231)

    The vector for the weight W is simply W = W * (0, 0, -1)

    If you want to write the equations of equilibrium, then you use the force vector components defined above.

    For example:

    ∑ Fx = -0.2983 * FOA - 0.3846 * FAC = 0

    I'll leave it to you to write the equations for ∑ Fy and ∑ Fz
     
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