Simple truss question I can't seem to get right

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nothingsus
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Homework Statement


Truss diagram with joints A, B, C and D, 2 m vertical support at left, 4 m horizontal member, 3 m vertical member, and a 5 kN load at C angled 15 degrees.


Handwritten free-body diagram of a truss with joints A, B, C and D, showing labelled forces R_Ay, R_Dx, R_Ax, AD, AB, BD, BC and AC, a 15 degree angle at C and a 5000 N load, plus handwritten force component equations.


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?
 

Discussion

nothingsus said:

Homework Statement


Truss diagram of problem 4/9: pin support at A and roller support at D on a left wall, members AB, AC, AD, BC and BD, with 2 m, 4 m and 3 m dimensions, and a 5 kN load at C acting at 15 degrees from vertical.


Handwritten free-body diagram of a truss with joints A, B, C and D, showing reaction labels RAy and RDx, member labels AB, AD, BD, BC and AC, and force equations Fy = -5000 cos15 and Fx = -5000 sin15 with a 15 degree angle at C.


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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