Simple truss question I can't seem to get right

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SUMMARY

The discussion centers on calculating reaction forces in a truss problem involving a pin and roller support. The user correctly applies static equilibrium equations but overlooks the vertical component of the load at point C, which contributes additional moments about point A. The calculated reaction forces are R_Ay = -4829 N and R_Dx = 3235 N, leading to an incorrect value for member BD, which should be 12.89 kN in compression. The oversight in accounting for the vertical load component results in the user being unable to proceed with the solution.

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Homework Statement


p6v4LnW.png


bRYWUNP.jpg


ignore all the other writing!

The Attempt at a Solution


I can't tell where I'm making a mistake. Probably a massive oversight somewhere but I don't know where!

First find the reaction forces at the pin and roller by treating everything as a rigid body. (I am now aware this isn't necessary since I can start at joint D since one of the equations has 1 unknown, but this method is still bugging me)

sum F_x = 0
R_Dx + R_Ax - 5000sin(15) = 0 (equation 1)

sum F_y = 0
-R_Ay - 5000cos(15) = 0
so R_Ay = -4829 N

sum Moments about point A = 0
(2*R_Dx) - (5*5000sin(15)) = 0
so R_Dx = 3235 N

so plugging R_Dx into equation 1 gives
R_Ax = -1941 Nstarting at joint D

sum F_x = 0

R_Dx + BD = 0
so BD = -R_Dx = -3235 N

but the answer says it is 12.89 kN compression!
I can't continue because everything else will be wrong.

What am I screwing up on?
 
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nothingsus said:

Homework Statement


p6v4LnW.png


bRYWUNP.jpg


ignore all the other writing!

The Attempt at a Solution


I can't tell where I'm making a mistake. Probably a massive oversight somewhere but I don't know where!

First find the reaction forces at the pin and roller by treating everything as a rigid body. (I am now aware this isn't necessary since I can start at joint D since one of the equations has 1 unknown, but this method is still bugging me)

sum F_x = 0
R_Dx + R_Ax - 5000sin(15) = 0 (equation 1)

sum F_y = 0
-R_Ay - 5000cos(15) = 0
so R_Ay = -4829 N

sum Moments about point A = 0
(2*R_Dx) - (5*5000sin(15)) = 0
so R_Dx = 3235 N

so plugging R_Dx into equation 1 gives
R_Ax = -1941 Nstarting at joint D

sum F_x = 0

R_Dx + BD = 0
so BD = -R_Dx = -3235 N

but the answer says it is 12.89 kN compression!
I can't continue because everything else will be wrong.

What am I screwing up on?
It appears you have accounted for the moments about A due to the horizontal component of the load at C.

However, additional moment at A will be created due to the vertical component of the load at C, since this component is offset horizontally from the line A-D.
 
SteamKing said:
It appears you have accounted for the moments about A due to the horizontal component of the load at C.

However, additional moment at A will be created due to the vertical component of the load at C, since this component is offset horizontally from the line A-D.

Dang, can't believe I missed that.

Thanks for the reply!
 

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