How do I analyze a body with a 14 kip force at a point?

AI Thread Summary
To analyze a body subjected to a 14 kip force, it is essential to apply the equilibrium equations: ΣFx = 0, ΣFy = 0, and ΣMz = 0. When detaching the members at point B, both Bx and By forces must be included alongside the 14 kip force. The discussion highlights the need to account for the correct distribution of forces across the members, particularly focusing on how the 14 kip force affects member AB. It is noted that there are six unknowns and six equations to solve for, emphasizing the importance of correctly identifying which forces to include in each equation. Ultimately, the approach to solving for Bx and By is confirmed as appropriate.
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Homework Statement


Screen Shot 2019-02-03 at 4.23.19 PM.png


Homework Equations


Using dimensional plane in the vertical y and horizontal x.
##\Sigma##Fx = 0
##\Sigma##Fy = 0
##\Sigma##Mz = 0

The Attempt at a Solution


I have detached the 2 members at B and analyze them both separately. The 14 kip force is throwing me off though. When I detach the body do I still include a Bx and By force at B, on top of the 14 kip force?
 

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fayan77 said:

Homework Statement


View attachment 238247

Homework Equations


Using dimensional plane in the vertical y and horizontal x.
##\Sigma##Fx = 0
##\Sigma##Fy = 0
##\Sigma##Mz = 0

The Attempt at a Solution


I have detached the 2 members at B and analyze them both separately. The 14 kip force is throwing me off though. When I detach the body do I still include a Bx and By force at B, on top of the 14 kip force?
In separating them, yes, you must include equal and opposite sets of Bx, By. For the X force, you just have to include it with one of them. If you need to answer for the values of Bx and By, it matters which you include it in. Looks like it is to be considered as applied to AB. But which you include it in does not affect any other reaction forces.
 
But then I'm still left with 3 equations 4 unknowns. member AB x positive to the right y positive up moment positive counter clockwise.
##\Sigma##Fx=0: Ax+14-Bx = 0
##\Sigma##Fy=0: Ay-By = 0
##\Sigma##MA=0: -14(12) + Bx(12) - By(9) = 0

same with other member
 
fayan77 said:
same with other member
Yes, but how many unknowns do you have altogether?
 
Ax
Ay
Bx
By
Dx
Dy
6 unknowns
 
fayan77 said:
Ax
Ay
Bx
By
Dx
Dy
6 unknowns
And six equations altogether, right?
 
the only ones I could combine are the ones at connection B. However, that 14 Kip force only act on member AB correct?
If so then from member BCD I use
##\Sigma##MD = 0: 10(5)-By(10)-Bx(12) = 0
50 = 10By+12Bx from member BCD
168 = 12Bx-9By from member AB
solving for Bx and By and solve for the rest
 
fayan77 said:
the only ones I could combine are the ones at connection B. However, that 14 Kip force only act on member AB correct?
If so then from member BCD I use
##\Sigma##MD = 0: 10(5)-By(10)-Bx(12) = 0
50 = 10By+12Bx from member BCD
168 = 12Bx-9By from member AB
solving for Bx and By and solve for the rest
Looks right.
 
Thank you
 
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