Axial loading of a composite beam

[ski0331]

Homework Statement

Homework Equations

In my attempt at a solution

The Attempt at a Solution

 

[ski0331]

Basically I'm asking if I did this correctly and have the correct answer and if I don't, what I did wrong and where I went wrong in my thinking.

 

[Chestermiller]

I confirm the result in the first part, but I don't follow what you did in the 2nd part. It seems to me for the 2nd part, all you need to do is add a compressive force to the entire rod that compresses it 13.04 mm: $$\frac{FL_1}{A_1E_1}+\frac{FL_2}{A_2E_2}=-0.01304$$

 

[ski0331]

Where did you get the 13.04 mm? And I thought I did add them together f=-ax and f2=-ax+4000N

 

[Chestermiller]

Where did you get the 13.04 mm? And I thought I did add them together f=-ax and f2=-ax+4000N

I'm just doing it as 2 superimposed problems. In the first part, you showed that the growth of the bar would be 14.04 mm if the bar were totally unconstrained. Since the original gap was 1 mm, that would mean that you would have to compress the bar 13.04 mm to get it to fit into the available space.

Assuming purely elastic behavior, this gives a compressive stress of 106 MPa in the Brass and a compressive stress of 131 MPa in the steel. But the yield stress of the Brass is only 78 MPa. How do you think that would affect the answer?

 

[ski0331]

I'm just doing it as 2 superimposed problems. In the first part, you showed that the growth of the bar would be 14.04 mm if the bar were totally unconstrained. Since the original gap was 1 mm, that would mean that you would have to compress the bar 13.04 mm to get it to fit into the available space.

Assuming purely elastic behavior, this gives a compressive stress of 106 MPa in the Brass and a compressive stress of 131 MPa in the steel. But the yield stress of the Brass is only 78 MPa. How do you think that would affect the answer?

I'm not sure. would it? I've struggled with this subject (axial loads). But a guess would be that it would bend the brass?

 

[Chestermiller]

I'm not sure. would it? I've struggled with this subject (axial loads). But a guess would be that it would bend the brass?

When you complete the solution of the purely elastic problem and have confirmed my answer, we can talk about what to do because the compressive yield stress of the brass is reached.

Your solution for the elastic problem is incorrect because you included the 4000 with the steel rather than with the brass.

 

[ski0331]

When you complete the solution of the purely elastic problem and have confirmed my answer, we can talk about what to do because the compressive yield stress of the brass is reached.

Your solution for the elastic problem is incorrect because you included the 4000 with the steel rather than with the brass.

ahhh ok I see where I did that! That makes sense, but I make my first cut to determine what Ax is for my free body diagram before the 4Kn. Like I said axial loads have been giving me fits to solve.

 

[Chestermiller]

ahhh ok I see where I did that! That makes sense, but I make my first cut to determine what Ax is for my free body diagram before the 4Kn. Like I said axial loads have been giving me fits to solve.

OK. Let's see what you get now. It seemed to me you were doing pretty well up to this point.

 

[ski0331]

OK. Let's see what you get now. It seemed to me you were doing pretty well up to this point.

 

[Chestermiller]

View attachment 223003

No. You need to be careful with your arithmetic: $$-\frac{(A_x+4000)(10)}{(101\times 10^9)(100\times 10^{-6})}-\frac{(A_x)(10)}{(200\times 10^9)(50\times 10^{-6})}+0.018=0.001$$ where $A_x$ is a positive compressive force. What do you get for $A_x$? What do you get for F from my equation in post #3?

 

[ski0331]

No. You need to be careful with your arithmetic: $$-\frac{(A_x+4000)(10)}{(101\times 10^9)(100\times 10^{-6})}-\frac{(A_x)(10)}{(200\times 10^9)(50\times 10^{-6})}+0.018=0.001$$ where $A_x$ is a positive compressive force. What do you get for $A_x$? What do you get for F from my equation in post #3?

A(x) I got is equal to 6552.24, F=-6552.44

 

[Chestermiller]

A(x) I got is equal to 6552.24, F=-6552.44

I confirm. I treated F as tension and got a negative number, while you treated it as compression and got a positive number. So this is the compressive force on the steel rod, and the compressive force on the brass rod is 6552+4000=10552 N. What do these solutions to the elastic problem (neglecting the yield stress) give you for the stresses in the brass and the steel?

 

[ski0331]

I confirm. I treated F as tension and got a negative number, while you treated it as compression and got a positive number. So this is the compressive force on the steel rod, and the compressive force on the brass rod is 6552+4000=10552 N. What do these solutions to the elastic problem (neglecting the yield stress) give you for the stresses in the brass and the steel?

 

[Chestermiller]

View attachment 223007

 

[ski0331]

View attachment 223008

ok you lost me on this chart. why are there 4p? 2 ax?

 

[Chestermiller]

ok you lost me on this chart. why are there 4p? 2 ax?

The forces on the left are the reaction forces from the constraint. The 2 P forces on the right are the applied forces. The Ax on the right is the compressive force applied on the brass section by the steel section.

You are aware that tension and compression are bi-directional, arising from the application of the Cauchy stress relationship to determine the forces acting on a surface? You remember from freshman physics the tension in a rope pointing to the right (for the force the material on the right applies to the material on the left), and to the left (for the force the material on the left applies to the material on the right), correct?

 

[ski0331]

The forces on the left are the reaction forces from the constraint. The 2 P forces on the right are the applied forces. The Ax on the right is the compressive force applied on the brass section by the steel section.

You are aware that tension and compression are bi-directional, arising from the application of the Cauchy stress relationship to determine the forces acting on a surface? You remember from freshman physics the tension in a rope pointing to the right (for the force the material on the right applies to the material on the left), and to the left (for the force the material on the left applies to the material on the right), correct?

yes but Ax is the reaction forces of the wall. In my FBD Ax pushes back on 2p and the Normal force.

 

[Chestermiller]

yes but Ax is the reaction forces of the wall. In my FBD Ax pushes back on 2p and the Normal force.

Taking each section as a free body, the forces on the two ends of each section must be equal in magnitude and opposite in direction. Otherwise, the sections could not each be in force equilibrium.

 

[Chestermiller]

An overall force balance on the composite beam tells you that the reaction force on the left boundary has to be $A_x+2P$.