Maximum weight a bar and cables can hold before breaking

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SUMMARY

The discussion centers on determining the maximum weight (W) that a bar (AO) and cables (AB, AC, AD) can support without breaking. Participants emphasize the importance of using static equilibrium equations (ΣF=0, ΣFx=0, ΣFy=0, ΣFz=0) to calculate the loads in each member in terms of W. The unit vectors for cables AB and AC, as well as bar OA, are critical for these calculations. The maximum allowable weight is found by comparing the calculated loads with the strengths provided in the discussion.

PREREQUISITES
  • Understanding of static equilibrium equations (ΣF=0, ΣFx=0, ΣFy=0, ΣFz=0)
  • Knowledge of unit vector calculations for structural members
  • Familiarity with tension and compression forces in structural analysis
  • Basic trigonometry for resolving forces in three dimensions
NEXT STEPS
  • Calculate the unit vectors for all structural members involved in the problem.
  • Apply static equilibrium equations to find the forces in cables AB, AC, and bar AO.
  • Compare calculated forces with allowable strengths to determine maximum weight W.
  • Explore failure criteria for structural members to ensure safety and compliance.
USEFUL FOR

Engineering students, structural analysts, and professionals involved in load-bearing design and analysis will benefit from this discussion.

  • #31
Alison A. said:
So I am finding all the forces from the moments first then applying those to the force equations?
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
 
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  • #32
SteamKing said:
No, you use the one moment equation I indicated,

ΣMz = FAB(12) - FAC(18.46) = 0 to find FAB in terms of FAC

Then, you can substitute FAB back into the force equations to find the other forces in terms of W.
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W
 
  • #33
Alison A. said:
Alright I found FAB to be (4615/5539)W and FAC to be (3000/5539)W

And from there force equations I found FOA to be -0.6983W
 
  • #34
WOOOOOOOOOOOOOO:oldsurprised: I got the right answer. Wow that was a lot of work, I think I've gone through about 10 pages of paper. :bow:Thank you so much for sticking with me even when it seemed like I couldn't grasp the most simple concepts . Could you help me find the last part to my other problem you've been answering? That is my last problem... then I'm finally done.
 

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