Statically indeterminate problem

[Queren Suriano]

Homework Statement

Calculate the stress in every segment of the system. AD and BC are laton and AB is steel.

Homework Equations

geometry compatibility-δ(A/D) +δ(A/B) - δ(C/B) = 0

F(D) - F(C) - 50 =0

The Attempt at a Solution

I supossed that DA and BC are in compression and AB is in tension. But I don't get the answer of the book, so I would like you help to understand what I am analyzing wrong.

The solution is this but I don't understand it...

 

[Chestermiller]

Is there something omitted from the figure that accounts for the fact that Fc and Fd are different? From the figure, it looks like they should be equal if the system is in static equilibrium.

Chet

 

[Queren Suriano]

Is there something omitted from the figure that accounts for the fact that Fc and Fd are different? From the figure, it looks like they should be equal if the system is in static equilibrium.

Chet

Do you refer to the figure of segment BC??

 

[Chestermiller]

Oh. I wasn't able to see them until I enlarged the figure.

Be back shortly.

Chet

 

[Chestermiller]

OK. I didn't look at what the answer key did, but I did look at what you did.

You just had some sign issues. You can't arbitrarily assume that certain members are in compression and certain others are in tension. You need to let the math do the work for you. I let the δ's represent the increases in length of the three members. If they are in compression, then the signs will come out negative. So,

$$δ_{AD}+δ_{AB}+δ_{CB}=0$$
$$-\frac{F_DL_{AD}}{E_1A_1}-\frac{(F_D-150)L_{AB}}{E_2A_2}-\frac{F_CL_{CB}}{E_1A_1}=0$$

Chet

 

[Queren Suriano]

OK. I didn't look at what the answer key did, but I did look at what you did.

You just had some sign issues. You can't arbitrarily assume that certain members are in compression and certain others are in tension. You need to let the math do the work for you. I let the δ's represent the increases in length of the three members. If they are in compression, then the signs will come out negative. So,

$$δ_{AD}+δ_{AB}+δ_{CB}=0$$
$$-\frac{F_DL_{AD}}{E_1A_1}-\frac{(F_D-150)L_{AB}}{E_2A_2}-\frac{F_CL_{CB}}{E_1A_1}=0$$

Chet

OK. I didn't look at what the answer key did, but I did look at what you did.

You just had some sign issues. You can't arbitrarily assume that certain members are in compression and certain others are in tension. You need to let the math do the work for you. I let the δ's represent the increases in length of the three members. If they are in compression, then the signs will come out negative. So,

$$δ_{AD}+δ_{AB}+δ_{CB}=0$$
$$-\frac{F_DL_{AD}}{E_1A_1}-\frac{(F_D-150)L_{AB}}{E_2A_2}-\frac{F_CL_{CB}}{E_1A_1}=0$$

Chet

 

[Queren Suriano]

But what do you write all the terms negative?Anyway if I multiplicate (-1) it turns positive; I solved assumed all terms positive and I get the correct answers, and they give possitive...So does this mean that the 3 segments are in tension?

 

[Queren Suriano]

Thanks for your help

 

[Chestermiller]

No. The way the figure is drawn, Fc and Fd are both compressional forces. So the two end sections are in compression. What does that tell you about the center section? Also, look at how the forces are applied to the ends of the center section.

Chet

 

[Queren Suriano]

No. The way the figure is drawn, Fc and Fd are both compressional forces. So the two end sections are in compression. What does that tell you about the center section? Also, look at how the forces are applied to the ends of the center section.

Chet

Ok, but in the center section If I substitute the force F(D) I obtain -42.11 kN. So the center section would be in compression, but in the answer is in tension...

 

[Chestermiller]

Ok, but in the center section If I substitute the force F(D) I obtain -42.11 kN. So the center section would be in compression, but in the answer is in tension...

Throughout this analysis, compressive stresses have consistently been treated as positive (see the figure). So, -42.11 kN means that the center member is in tension.

Chet