# Statics, Truss Anyalysis

Hello,

I'm having trouble with a truss analysis problem. We are supposed to use the method of joints & the method of sections. We have to find all the external reaction forces and internal member forces.

The method of joints is not problem, and I THINK I'm doing the method of sections correctly, but I can't get all the forces in all the joints into equilibrium.

Here is how the problem is given:
http://img290.imageshack.us/img290/7383/staticsproblemgiven5qv.png [Broken]

Here is my attempt
http://img374.imageshack.us/img374/2958/staticsmyattempt9vf.png [Broken]

And HERE is my procedure:

I started by taking a moment about the lower left hand corner (with CCW being positive).

(-40*15)+(-25*20)+(x*30) = 1100/30 = 36.67 kips

Then I sum the 'Y' forces (up being positive):
36.67-25-x=0 X = 11.67 kips

By inspection the horizontal reaction at the lower left hand corner is 40 kips to the left.

Now I started method of joints on the top left joint.

The horizontal component of the diagonal is 40 kips to the left, since it is a 1:2 slope, the vertical component is 20 kips up (the diagonal member is in compression)

The vertical member must counter act the 20 kips from the diagonal, so that member is 20 kips down (tension)

Now I skip over to the bottom right joint, and by similar process I get a vertical force of 36.67 kips down and 73.34 kips to the right (for the diagonal member). The horizontal member is 73.34 kips in tension to counter act the diagonal.

Now I used method of section, I cut my section just to the left of the 25 kip vertical load and took a moment about the lower right joint. Since the diagonal member & horizontal members intersect with my moment joint, they don't have a moment (correct?). The vertical member of the truss is a zero force member, so the only remaining member has a 25 kip force UP and a 50 kip force to the right (the member is in compression).

The problem is, there is no way to zero out the rest of the truss now using method of joints. Either I messed up the method of sections or there is a flaw in the problem.....I've been stuck on this problem for 3 days - HELP!!

:yuck:

Eric

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Pyrrhus
Homework Helper
Well the first i noticed is that your external reactions are wrong, look the sum of Fy is not zero!.

Y reacations

36.67 - 25 - X = 0

11.67 - X = 0
+X = 0 + X

11.67 = X

Therefore

36.67-25-11.67 = 0

Pyrrhus
Homework Helper
The vector representation of the Force with magnitude 36.67 is wrong then, it should be pointing up. There could be also more mistakes of this nature.

Pyrrhus
Homework Helper
Ok i see your problem besides the other one i pointed out!. The slope is 1 on 1, if you notice the opposite side of the triangle is 10 feet and so is the adjacent side of the triangle, therefore there's a hypotenuse of quadratic root of 2, or an angle with the horizontal of 45 degrees.

Finally...

I finally figured out the answer, I was all messed up.:rofl:

I get it now! Thanks for the help.