This homework is due on Friday 10/26 at 4 pm. It is all from Chapter 6, and some of the questions are modified, so please read carefully.
Problems: 6-17, 6-30, 6-33, 6-41, 6-47, 6-78, 6-83, 6-103
Hint on 6-17: consider doing the problem as stated in 6-16 (i.e., determine the force in each as a function of P), then work with the maximum allowable loads. Also, you can do this one with either the method of joints or method of sections, whichever you prefer.
Hint on 6-78: remember this?
14 responses so far ↓
1 LittilLazz // Oct 25, 2007 at 2:19 pm
I was wondering if anyone had a little more advice than Professor Berger on the first one, 6-17. I’m having problems with too many variables
2 tpadden // Oct 25, 2007 at 4:31 pm
I am also having the same problems on 6-17
3 lew3r // Oct 25, 2007 at 4:32 pm
Hey—for 17
I started with the whole diagram, using Fx, Fy, and Me to solve for Ex, Ey, and Ax
I then used method of joints first at E, then C, then B, then D
I got AD, AB, BC, and CD = sqrt(5)/6
I got BD = (1/3)P
CE = Psqrt(13)/3
DE=0
I have no clue how to tell if something is in tension or compression. BUT, I do know that when I took the answer in the back of the book and used it for P in the eqtn for CE, I got 5….for what it’s worth
4 Natalie // Oct 25, 2007 at 4:53 pm
I did the same thing for #17, but in order to figure out tension and compression I made all of the forces positive in my equilibrium equations to begin with. Then I solved for the unknowns, and if it came out positive the force was in tension, and negative means compression. They mention this in the book on p.271.
For number 103, I solved for the tension in BC, but I’m not sure how to find the force acting along AC..does anyone have any suggestions?
5 katie // Oct 25, 2007 at 4:55 pm
if you solve them all as tension (going in the same direction), the negative ones will be compression.
6 katie // Oct 25, 2007 at 4:57 pm
sorryy my computer didnt refresh and show natalies comment.
7 berger // Oct 25, 2007 at 5:28 pm
This all sounds right on 6-17…now how about 103?
8 hambut // Oct 25, 2007 at 7:34 pm
wait how do you solve for the three unknowns with the whole diagram when you only have 3 equations and 4 unknowns? did yall solve them all in terms of P?
9 tpadden // Oct 25, 2007 at 8:16 pm
for 6-17 I am still confused as to when you plug in the values they give for max tension and max compression loads
10 cgo8a // Oct 25, 2007 at 8:37 pm
For 103 I thought AC was a 2 force member then I solved the equilibrium equations in x and y using the force in AC (unknown) the tension in CD (unknown) and the tension in BC from the first part. so there are 2 equations and 2 unknowns this seems to work
11 lew3r // Oct 25, 2007 at 8:38 pm
for 17
Solve everything in terms of P
I mis-typed the answer to AD, AB, BC, and CD ……they should equal Psqrt(5)/6 (i forgot the P)
Then, since those four are in compression (thanks Natalie), set Psqrt(5)/6 = 3 kn
solve for P
For BD set (1/3)P=5 and so forth
12 rsa8q // Oct 26, 2007 at 12:56 am
for 17, if you solve for the support rxns and then make a FBD at E, you can solve for EC in terms of P and know that EC is in tension. If you then substitute 5 kN for EC, you get the correct answer for P. I’m wondering if this is valid, since it doesnt take any of the other members into account. Can you assume that when this one member experiences its maximum force, P must be maximum as well?
13 berger // Oct 26, 2007 at 6:36 am
For 6-16, strictly speaking you want to solve for ALL the member forces in terms of P (you’ll get answer like 0.372P and 0.333P for some of the forces). Then, you plug in 5 kN for those in tension, and 3 kN for those in compression, and see what that tells you about the allowable values for P.
I wouldn’t go and simply assume that one member is a max or min; rather you should solve for each one.
But here’s a few more hints…can you use symmetry to really accelerate your progress on writing the member forces in terms of P?
14 berger // Oct 26, 2007 at 6:38 am
In 6-103, AC is indeed a 2-force member.
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