# Homework Help: Need help with a Statically Indeterminate System!

1. Jul 8, 2009

### TA1068

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

Three steel bars are pin-connected to a rigid member K. Determine the force developed in each bar. Determine the load carried be each of the tension members and the elongation of each member

http://img3.imageshack.us/img3/6850/problemdiagram.jpg [Broken]

Known:
$$A_A_B$$ = 0.10 in^2
$$E_A_B$$ = 30E6 psi
$$A_C_D$$ = 0.20 in^2
$$E_C_D$$ = 15E6 psi
$$A_F_H$$ = 0.30 in^2
$$E_F_H$$ = 10E6 psi

2. Relevant equations

$$\delta$$ = (PL) / (AE)

3. The attempt at a solution

10$$P_C_D$$ + 20$$P_F_H$$ = 15(15000)
or
$$P_F_H$$ = 7500 - (1/3)$$P_C_D$$

The equation $$\delta$$ = (PL) / (AE) yields:
$$P_A_B$$ = 150,000$$\delta_A_B$$
$$P_C_D$$ = 200,000$$\delta_C_D$$
$$P_F_H$$ = 300,000$$\delta_F_H$$

And the sum of forces in the Y direction gives:
$$P_A_B$$ + $$P_C_D$$ + $$P_F_H$$ = 15000

This is where I'm stuck. If any point along K was fixed it would be easy; K is rigid, so then the distance from the fixed point can be turned into a ratio to find the other $$\delta$$ values. I think all 3 points (B, D, and H) are pulled downward, but I'm not sure what there relation is to each other. Any clues?

Last edited by a moderator: May 4, 2017
2. Jul 8, 2009

### TA1068

Yikes, I think my TEX tags are all messed up. Not sure how to fix it, let me know if you have any questions!

3. Jul 8, 2009

### TA1068

Now if it was something like this:
http://img194.imageshack.us/img194/8167/fixedpoint.jpg [Broken]
I would say sigma_CD = 10(theta) and sigma_FH = 30(theta) and all would be good.

But since it's like this:
http://img199.imageshack.us/img199/9023/nonfixedpoint.jpg [Broken]
I have another variable in there with x. Now sigma_AB = (x)(theta), sigma_CD = (10+x)(theta), and sigma_FH = (30+x)(theta)

Hmm...

Last edited by a moderator: May 4, 2017