# Homework Help: Statics, forces in each memeber of a truss

1. Apr 10, 2013

### jonjacson

1. The problem statement, all variables and given/known data

I´ll show the problem using a picture:

2. Relevant equations

Sum Forces = 0

Sum Moments = 0

3. The attempt at a solution

Well, I´m a bit confused because I need to take care of the weight of the own structure.

1.- First attempt to solve to problem, since the structure is symmetric I think that the reactions at A and C are equal. And at C there is not any reaction on the horizontal axis X.

So the total weight of the structure is = 7 members * 400 lb = 2800 lb ( I don´t multiply by the gravity because the answers in the book are not given in Newtons, they are given in lb too).

Reaction at A Ra= 2800/2= 1400 lb

Now I try to use the method of joints at joint A, and I see three forces acting on it: Ra, the compression force AE, and the tension force AB.

If I´m right you don´t need to consider the weights mg of the bars, because that external force is acting at the bars, not at the joints, I´m not sure about this.

IF I only consider these three forces I get these equations:

Y axis----> Ra - AE sin(60) = 0; AE= 1400/ sin(60) = 1616,58 lb

But the book gives me AE= 2000√3 lb C = 3464 lb C

Well using that result I went backwards to guess the value of Ra and it looks like it´s 3000 lb.

What I´m doing wrong?

Even if I try to consider the weights of the two bars AE and AB I don´t get the correct result.

I used the moment around A to check if I was wrong:

-the moment arm of the bar AE is equal to 10 feet(since the cg is at the middle point) * sin (60)= 5 feet.

-the horizontal bar AB has a moment arm of 10, since it is in the middle too

-bar EB has a moment arm of 15 feet

...

-the bar DC has a moment arm of 35 feet

-Finally reaction force Rc has a moment arm of 40 feet so:

Ma= -mg (5+10+15+20+25+30+35) + Rc*40= 0

Rc = 140 * 400 /40= 1400 lb which confirms the previous result.

So I believe that I am not considering correctly other forces acting at joint A.

2. Apr 10, 2013

### voko

I do not think you can find forces in the members directly. You can find forces at joints, then the average force in a member is apparently the average of the forces at its ends.

3. Apr 10, 2013

### jonjacson

Maybe I don´t understand what the "average force" is. Do you know a formula?

I know which are the external forces acting on the structure, basically the weight of every beam, and the two reactions at A and C.

The internal forces are tension or compression for every beam, and an equal and opposite force at the joints.

If I consider the weight of the beam 400 acting at the joint A, there are only 1000 lb acting upwards from reaction force Ra, so AE sin(60) = 1000, so AE=1154 lb, assuming the force from joint E to joint A is equal and opposite to the force of A acting on E, I see that 2 * 1154 = 2308 lb are acting on the bar, if we add the 400 lb finally I find 2708lb acting on the beam AE, but that is not correct.

I don´t know if the average force acting on the beam is the sum of the forces acting on that beam or another forumla.

4. Apr 10, 2013

### voko

As I said, I think that you should take a beam and find the reaction forces at its ends. Then the average of these two forces should be the average compression/tension force in the beam.

Now what you need to do is find the forces at the joints properly.

5. Apr 10, 2013

### PhanthomJay

Your original answer for the force in AE is incorrect. And, unless you copied it down incorrectly, the book answer is also incorrect. You happened to stumble upon the correct answer for AE somewhere in post #3.

Because you have forces applied in between joints, this is not a pure truss, and members will be subject to shear and bending as well as axial forces.. Neglecting the shear and bending stresses, you may distribute 1/2 the weight of each member to the adjacent joints, and then solve for the truss member tension or compression forces. The end reactions are still 1400 pounds, but when you look at each joint, you must include the weight forces acting at that joint.

6. Apr 10, 2013

### jonjacson

Thanks Voko, I have tried another time and I found the mistakes.

Yes I made some mistakes,I found AE correctly at post three by chance because I thought that the 400 lb weight of the bar AE should be applied to joint A, but that is not true, casually the sum of W/2 two times, due to bar AE and bar AB is 400.

Now I think everything is clear, and my results agree with the book, I didn´t see properly the number it was 2000/√3 not 2000 * √3, so the result in the book is correct, it was my fault I´m sorry.

I show the calculations:

At joint A there are 4 forces acting, 200+200 downwards due to half the weight of the two bars connected to that joint, the 1400 force upwards and forces AE and AB.

Y---> 1400 - 400 - AE sin(60) = 0 ---------> AE= 1154 lb C

X--> AE cos(60) = AB=577.35 lb

At joint E I have three times W/2 so 600 lb pointing downwards.

Y---> AE sin(60) - 600 -EB sin (60)=0 ---------> EB= 461 lb

X ---> AE cos(60) + EB cos(60) - ED= 0 ----> ED= 808 lb

At joint B I have W/2 four times, so 800 lb, and EB should be equal to BD due to the symmetry of the structure so:

Y---> EB sin(60)* 2 = 800----> EB=461lb and this confirms previous results.

BD=461 lb

Other forces are equal to their symmetric ones, so it´s not necessary to show them.

Thank you very much Voko and PhantomJay!!!