Problem: Please refer to the explanation of the problem on page 110 in Beer, Johnston, & DeWolf's 3rd Edition of Mechanics of Materials. However, I have revised the problem so that you are asked to find the following:
Solution: The plate is firmly attached to the rod and tube, so that both elongate together and, when the force is removed, both shrink together. I won't copy the calculations that the authors do - you can read them on page 110.
Because the rod has a different cross sectional
area, a different yield strength, and a different elastic modulus (than
the tube), it requires a different value of the force to reach the yield
point. Your text computes the yield force for the rod to be 2.7 kips and
the deformation at yield to be 36 x 10-3 in. That's how he gets
the first of the following plots:
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This plot shows that, if the force were applied to the rod alone, we would be unable to increase the force beyond 2.7 kips since the rod would simply continue to stretch (being in the plastic region). |
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Your text also calculates the force required to reach the yield point for the tube and plots it. Compare this with the plot for the rod. The different slope means that, when the deformation has reached 36 x 10-3 in (yield for the rod), there's only 1.8 kips being applied to the tube, whereas 2.7 kips are applied to the rod. An increase in force beyond that is applied to the tube, not the rod, since it impossible to apply more force to the rod. By the time the deformation reaches 90 x 10-3 in, the rod has been in the plastic region for some time. |
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This is the overall
(total) load-deflection diagram. It was obtained by adding the loads at
each deflection. Clearly, 7.2 kips are required to reach 0.090 in (or larger)
of deflection. This is the answer to question (b).
To reach 0.065 in of deflection, we can use the slope of the second line:
This is the answer to question (a). |
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To obtain a deflection
of 0.1 in, both the rod and the tube must be plastic. After this,
as the load decreases from 7.2 kips, the rod and tube recover elastically
along lines parallel to their original load lines. The combination of these
two load lines is already drawn; it is the single line OA. The return path
is shown by the dashed line CD. The permanent set is the length of OD.
Therefore,
The permanent set is 42.4 x 10-3 in. This is the answer to part (c). |
d,e) The residual stresses can be found,
as the text points out, by superposing the stresses due to loading and
the reverse stresses under unloading. I think it's easier to use the plots
to do this. It's certainly easier to see it from the plots! From
the diagrams above, when the external load P has been returned to zero
at point D, the internal forces on the rod and tube have not been returned
to zero. You can see this by drawing a line vertically up from D for each
of the original plots for the rod and tube alone (see below).
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From 100 x 10-3 in to 42.4 x 10-3 in, the internal force in the rod decreases by
Therefore the residual force in the rod is 2.7 -4.32 = - 1.62 kips (compression). |
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From 100 x 10-3
in to 42.4 x 10-3 in, the internal force in the tube decreases
by
Therefore the residual force in the tube is 4.5 -2.88 = 1.62 kips (tension). |
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Notice that the
net force. according to Newton's First Law, must be zero (and it is).
The stresses, however, are different:
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