Timoshenko, Strength of materials example solution

[akmkeng]

Hi everyone, I am new here, so please be kind if I make mistakes on this forum, I will try to learn fast.

I need help in explaining the solution of one of the examples from Timoshenko's Strength of Materials book. The solution is provided in the book, but I did not get it, no matter at which angle I looked at it.
Can somebody please bother to explain it to me? I don't understand why the stresses' balance in elongation formula is as it is in the solution.
See the example with the solution in the attached picture.

 

[PhanthomJay]

Hi everyone, I am new here, so please be kind if I make mistakes on this forum, I will try to learn fast.

I need help in explaining the solution of one of the examples from Timoshenko's Strength of Materials book. The solution is provided in the book, but I did not get it, no matter at which angle I looked at it.
Can somebody please bother to explain it to me? I don't understand why the stresses' balance in elongation formula is as it is in the solution.
See the example with the solution in the attached picture.

i am not sure what you mean by the stresses being balanced. Beyond that comment, the deformation of the member is the sum of the deformations of each
of the three parts. If you draw a free body diagram of the top part, the only force acting in that section is Q, so it’s deflection is QL1/AE. Similarly the deflection of the bottom section is the same, so that’s where the first term come from, that is , 2QL1/AE. Now draw a free body diagram of the middle and top section , cut thru the middle section, and the force in the middle section is Q- P (do you see why?). That gives the second term for the deflection of the middle piece. Then add ‘em up.

Another way to do this is use the superposition of forces principle. First forget about P and look at deflection under Q only. That’s Q(2L1 +L2)/AE. Then look at the deflection under P only. The middle section only deforms, -PL2/AE, and the top and bottom section go along for the ride. Now add them up to get same result. Make sense?

 

[akmkeng]

i am not sure what you mean by the stresses being balanced. Beyond that comment, the deformation of the member is the sum of the deformations of each
of the three parts. If you draw a free body diagram of the top part, the only force acting in that section is Q, so it’s deflection is QL1/AE. Similarly the deflection of the bottom section is the same, so that’s where the first term come from, that is , 2QL1/AE. Now draw a free body diagram of the middle and top section , cut thru the middle section, and the force in the middle section is Q- P (do you see why?). That gives the second term for the deflection of the middle piece. Then add ‘em up.

Another way to do this is use the superposition of forces principle. First forget about P and look at deflection under Q only. That’s Q(2L1 +L2)/AE. Then look at the deflection under P only. The middle section only deforms, -PL2/AE, and the top and bottom section go along for the ride. Now add them up to get same result. Make sense?

English is not my first language, so sorry for probably unclear use of words. In the language I learned physics in, we use the term of balancing forces, stresses etc, a lot.

I don't really get it still.
1) Why do we only account for deflection due to 1 P in the middle section, when there are clearly 2 equal P forces of opposite direction acting on the middle section.
2) Also, if we say that top and bottom sections go along with the middle section due to its shrinking due to P, then why can't we look at it the other way, saying, the elongation on top and bottom sections is due to Q+P(I see that directions of Q and P are opposite, but the act on the opposite ends of the same section, thus sort of stretching it, isn't it?)? Because clearly P is acting on the border between middle and top/bottom sections.

 

[Tom.G]

Does it help if you consider P generated as an external clamp attached to the beam/post. This external clamp then compresses the l₂ section; perhaps even before Q is applied?

Cheers,
Tom

 

[PhanthomJay]

When a system is in equilibrium , you are correct that external forces (not stresses) are balanced, that is to say, they add to 0. Here, you have 4 external forces on the system, so
you have Q + P - Q - P = 0.
But this problem asks about deflection, and you are not drawing your free body diagrams correctly. A knowledge of free body diagrams is essential. When you draw a free body diagram of the top piece , you cut the top section away from the system to identify the forces acting on it, and since Q is acting at the top down on it, then the force at the ‘cut’ section must be Q acting up on it, for equilibrium balance. So the internal force in the top section is Q, not Q + P. And when you cut the middle and top section away from the system, since Q + P acts down, then the internal force at the midsection cut must be - (Q+ P) , for equilibrium.