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I am reading Mechanics of Materials by Beer and Johnston.The author points out if the torque in the shaft is removed the resulting reduction of stress and strain at a point considered will take place along a straight line .As was the case for normal stresses the shearing stress will keep on decreasing until it has reached a value equal to its maximum value at C minus twice the yield strength of the material.

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Can anyone throw some insight into it?I am not aware regarding normal stresses as well.Thanks.

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  • $\begingroup$ Essentially a duplicate of engineering.stackexchange.com/questions/18866/… $\endgroup$ – Solar Mike Jan 13 '18 at 8:47
  • $\begingroup$ @solarmike not at all.why the factor twice is there? $\endgroup$ – user471651 Jan 13 '18 at 9:30
  • $\begingroup$ Timoshenko, Strength of Materials, part II, chapter IX, section 75 should answer your question. $\endgroup$ – Robin Jan 15 '18 at 8:40
  • $\begingroup$ @Robin I don't have that book.can you please cite that section as an answer $\endgroup$ – user471651 Jan 15 '18 at 11:35
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When loaded above the yield point ($Y$ in your sketch), ductile material will start to deform plastically. When you stop at some point before the material fails and reduce the load again, the material will relax.

Citing the wikipedia article on plasticity: "Plasticity in a crystal of pure metal is primarily caused by two modes of deformation in the crystal lattice, slip and twinning. Slip is a shear deformation which moves the atoms through many interatomic distances relative to their initial positions. Twinning is the plastic deformation which takes place along two planes due to a set of forces applied to a given metal piece."

Therefore, the relaxation will be elastic. However, since the internal structure has been changed somewhat during plastic deformation, residual stresses will remain inside your specimen. There also is some lasting deformation. When loaded into the opposite direction, the yield limit is reduced. The length of the elastic line is the same for the pre-deformed specimen as for a new specimen (which is $2\tau_Y$).

The underlying principle is the same for torsion and for pure tension. The major difference is that when twisted, only an outer layer of the material will undergo plastic deformation when loaded above the yield point, while the tension probe will undergo plastic deformation in the whole cross section. This is because strain and stress increase linearly with radius in a torsion probe (see here for an example), but are independant on the location in a tension probe.

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