Statically Indeterminate Systems

[skaboy607]

Homework Statement

A composite shaft consists of a brass bar 50mm in diameter and 200mm long, to each end of which are concetrically friction-welded steel rods of 20mm and 100mm long. During a tensile test to check the welds on the composite bar, at a particular stage the overall extension is measured as 0.15mm. What are the axial stresses in the bar given that Ebrass=120Gpa and Esteel=208 Gpa.

Homework Equations

Equations of equilibrium
Equations to describe the geometry of deformation
Stress-strain relationships.

The Attempt at a Solution

For the bar, the equations of equilibrium is that the total force is the same as force exerted on steel bar and brass bar.

The overall change in length must the sum of the change in rod 1 (steel), rod 2(brass) and rod 3(steel).

Stress strain equations are: E=Stress/Strain for each rod (material)

Ive got this far but there seem to always be too many unknowns in the problem to solve for stress in each rod?

Many thanks for your help.

 

[PhanthomJay]

Since the force in each part of the rod ,and the area of each part of the rod, is the same, then the stress in each part will be the same.

 

[skaboy607]

How can the area in each part of the rod be the same? They have different diameters, so surely they will have different cross sectional areas.

 

[PhanthomJay]

I assumed they were the same diameters because you did not list them. What are they? If known, you can proceed in the same manner, noting that the force in each part of the rod is the same, and the stress in each part is just P/A. You need first to solve for P. Note that your equation \sigma = \epsilon E can be rewritten as \Delta = PL/AE.

 

[skaboy607]

If we are talking about the same thing, then the two diameters are given above, 50mm and 20mm. The problem is I am not given the force that is loaded in tension on the bar?

 

[PhanthomJay]

If we are talking about the same thing, then the two diameters are given above, 50mm and 20mm. The problem is I am not given the force that is loaded in tension on the bar?

Is this the way the problem statement should read

A composite shaft consists of a brass bar 50mm in diameter and 200mm long, to each end of which are concetrically friction-welded steel rods of 20mm in diameter[/color] and 100mm long...?

After confirming this, then compute the extension of each part of the rod in terms of P, add em up, and set them = to the given total extension, and solve for P. Then solve for the stress in each part.

 

[skaboy607]

Yes sorry, it should have read 'diameter'.

ok, I got as far getting what I think are formula's for doing this but there are two many unknowns which is what I can't get my head around. The fomulae for each individual part that I got was F/A=E*(extension/length). If I rearrange for extension, I still have an unknown which is the force?

Am I understanding this right or being a bit stupid?

Thanks for your help.

 

[srvs]

Your formula is correct. It's the same one PhantomJay listed. The point now is to rearrange for displacement, fill them into your compatibility equation and then solve this together with your equilibrium equation. Two equations, two unknowns.

 

[skaboy607]

ok...I have done some more but come across some more confusion. From the question, I read that the bar consists of three parts. One center piece made material 1 and then two out pieces of the same diameter, length and material 2. From this I got that my deformation eq would be: total change in length=change in length 1 + change in length 2 + change in length 3. So after substitution of my stress-strain eq's, I would have total change in length=(F/A=E*(extension/length))'Piece1'+(F/A=E*(extension/length))'Piece 2'+(F/A=E*(extension/length))'Piece 3'.

Now if this is correct, I don't know what to do next! Unknowns are F and extension for each of the pieces. My equilibrium eq is F=F1=F2=F3 but how do I use this to solve?

 

[PhanthomJay]

Since F1=F2=F3=F, then the sum of the deltas for each piece, F(L1)/A1E1 + F(L2)/A2E2+ F(L3)/A3E3, is equal to 0.15 mm. Solve for F.