- #1
paul11273
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This is a HW question, and I posted it in the HW section, but no one there seems interested. Perhaps somewhere here would be willing to lend a hand?
See the problem in the attachment.
I basically don't know if I am on the right track.
I started with part a) by trying to find the force (tension) pulling on member AB. I did this by calculating the moment about point D.
Md = 6kN * .3m = 1.8kNm (ccw)
I took this torque and used it to compute the x component of the tension at AB.
Md=Force*distance
Fab=Md / distance
Fab= 1.8kNm * .320m = 5.625kN to the left
Using this tension, I tried to find the x component of the force at point C by using the moment created at E by tension AB.
Me = .460m * 5.625kN = 2.59kNm (ccw)
Fcx = Me / distance
Fcx = 2.59kNm / .300m = 8.63kN
Finally, I try to compute the force along CD using the x component I just found (or think I found). I do this using the right triangle created between D, C and the horizontal from D. This gives me .4 along the horz, and .16 along the vertical. That gives a hypotenuse of 80*sqrt(29). Also, the angle created by DC with the horz is 21.8 degrees. By using similar triangles, I find the force along CD to be
Fcd=(80*sqrt(29)/400) * Fcx = 9.29kN
This answer is wrong. The book has 10.72kN at 21.8 degrees. Atleast I found the angle correctly. I am close, but not correct.
Also, I don't feel like I followed any process that we were given for the analysis of structures. Ofcourse, the class examples were painfully simple, and I left class with a great feeling that this was a piece of cake. This problem is changing my mind. Maybe this is simple too, but I am just not seeing it.
Thanks in advance.
See the problem in the attachment.
I basically don't know if I am on the right track.
I started with part a) by trying to find the force (tension) pulling on member AB. I did this by calculating the moment about point D.
Md = 6kN * .3m = 1.8kNm (ccw)
I took this torque and used it to compute the x component of the tension at AB.
Md=Force*distance
Fab=Md / distance
Fab= 1.8kNm * .320m = 5.625kN to the left
Using this tension, I tried to find the x component of the force at point C by using the moment created at E by tension AB.
Me = .460m * 5.625kN = 2.59kNm (ccw)
Fcx = Me / distance
Fcx = 2.59kNm / .300m = 8.63kN
Finally, I try to compute the force along CD using the x component I just found (or think I found). I do this using the right triangle created between D, C and the horizontal from D. This gives me .4 along the horz, and .16 along the vertical. That gives a hypotenuse of 80*sqrt(29). Also, the angle created by DC with the horz is 21.8 degrees. By using similar triangles, I find the force along CD to be
Fcd=(80*sqrt(29)/400) * Fcx = 9.29kN
This answer is wrong. The book has 10.72kN at 21.8 degrees. Atleast I found the angle correctly. I am close, but not correct.
Also, I don't feel like I followed any process that we were given for the analysis of structures. Ofcourse, the class examples were painfully simple, and I left class with a great feeling that this was a piece of cake. This problem is changing my mind. Maybe this is simple too, but I am just not seeing it.
Thanks in advance.
