# Stress problem

1. Oct 9, 2005

### snowygrouch

Hi,
Got to the last question in my mechanics problems list and just cant get it!

You have a brass tube with steel bar inside.

The brass tube is 400.12mm long; longer than the steel bar which is 400mm long.

You have to calculate the stress in both objects when a compression load of 60kN is applied.

I thought the first thing to do would be calculate how much force it required to shrink the brass tube down to the same length as the steel bar. Which is easy: F=stressxarea so F=16.488kN.

So now both materials are the same length so from this point on any change in length must be identical for both objects.

E is given as 100kN/mm2 for Brass and 200kN/mm2 for steel.

The area (cross sectional) of the brass tube is 550mm2 and 616mm2 for the steel bar.

At this point it all goes wrong! (I know what the answers are as theyre given but I cant get the right numbers....:grumpy: ).

I imagine its probably related to the fact the the change in lengths must be identical but how you apply that thinking has just irritated me.

Any ideas much appreciated!

Thanks

Calum

2. Oct 9, 2005

### lightgrav

Your 43.51 kN will compress *both* objects some distance,
their Forces now add (like in FBD's). Each one's Force is EA ...
usually you'd have to worry about the DeltaL/L ... but not here.
just add them together, since they're the same length.

3. Oct 9, 2005

### Pyrrhus

What did you do for the other part?

the resultant force from the stresses (sp?) of the brass + the resultant force from the stresses of the steel will be equal to the rest of the applied load. And remember the compatibility equation $\delta_{s} = \delta_{b}$ for the rest of the deformation.

Last edited: Oct 9, 2005
4. Oct 9, 2005

### snowygrouch

Not yet..

Hi,
Many thanks for the help so far.
I think my problem is that all the lectures were on single body stress,
Im just not sure how to apply the formulas to a parrallell system.
In other words if ive got 60kn and 2 parallell shafts do i divide the 60kn by 2 or??
This is only the second lecture so there are some fundamentals Im lacking.

Calum

5. Oct 9, 2005

### Pyrrhus

Are you supposed to find the displacement? if its the displacement, this is a 2 axial load problem, and can't be solved by the conventional formula for 1 axial load $\delta = \frac{PL}{EA}$, but by the more general $\delta = \sum_{i=1}^{n} \frac{N_{i}L_{i}}{E_{i}A_{i}}$

6. Oct 10, 2005

### snowygrouch

Hi,
No its not the displacement but the internal stress in each individual member.
All my lectures up to now have been on single bodies which is fairly easy; Im just not sure how to apply the equations Stress=force/area, E=stress/strain and Strain=change in length/length to multiple bodies in parallell.

In other words do you divide the total force by the number of objects or do you have to use a ratio of the surface areas etc.

Calum

7. Oct 10, 2005

### vanesch

Staff Emeritus
You ADD the forces. The individual forces are function of the displacement.

If x is the (common) displacement, then F1(x) is the force coming from the brass tube and F2(x) is the force coming from the steel bar. Both can be calculated individually from "single body compression".
The total force excerted is now of course F(x) = F1(x) + F2(x). You know the total force, you don't know x...