Further insight into the plastic behavior of a shaft in torsion is obtained by considering the idealized case of a solid circular shaft made of an elastoplastic material. The shearing-stress-strain diagram of such a material is shown in Fig. 3.34. Using this diagram, we can proceed as indicated earlier and find the stress distribution across a section of the shaft for any value of the torque T.
As long as the shearing stress t does not exceed the yield strength tY, Hooke’s law applies, and the stress distribution across the section is linear (Fig. 3.35a), with tmax given by Eq. (3.9):
tmax5 Tc
J (3.9)
Y
Fig. 3.34 Elastoplastic stress- strain diagram.
O
(b)
max Y
c O
(d)
c Y
Fig. 3.35 Stress-strain diagrams for shaft made of elastoplastic material.
O
(a)
max Y
c O
(c)
c Y
Y
As the torque increases, tmax eventually reaches the value tY (Fig.
3.35b). Substituting this value into Eq. (3.9), and solving for the cor- responding value of T, we obtain the value TY of the torque at the onset of yield:
TY5 J
ctY (3.28)
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The value obtained is referred to as the maximum elastic torque, since it is the largest torque for which the deformation remains fully elastic. Recalling that for a solid circular shaft Jyc512pc3, we have
TY512pc3tY (3.29) As the torque is further increased, a plastic region develops in the shaft, around an elastic core of radius rY (Fig. 3.35c). In the plastic region the stress is uniformly equal to tY, while in the elastic core the stress varies linearly with r and may be expressed as
t5 tY
rYr (3.30)
As T is increased, the plastic region expands until, at the limit, the deformation is fully plastic (Fig. 3.35d).
Equation (3.26) will be used to determine the value of the torque T corresponding to a given radius rY of the elastic core.
Recalling that t is given by Eq. (3.30) for 0 # r# rY, and is equal to tY for rY #r # c, we write
T52p
#
rY0
r2 atY
rYrb dr12p
#
crY
r2tY dr 5 1
2pr3YtY1 2
3pc3tY22 3pr3YtY T5 2
3pc3tYa12 1 4
r3Y
c3b (3.31)
or, in view of Eq. (3.29), T5 4
3TYa12 1 4
r3Y
c3b (3.32)
where TY is the maximum elastic torque. We note that, as rY approaches zero, the torque approaches the limiting value
Tp5 4
3 TY (3.33)
This value of the torque, which corresponds to a fully plastic defor- mation (Fig. 3.35d), is called the plastic torque of the shaft consid- ered. We note that Eq. (3.33) is valid only for a solid circular shaft made of an elastoplastic material.
Since the distribution of strain across the section remains linear after the onset of yield, Eq. (3.2) remains valid and can be used to express the radius rY of the elastic core in terms of the angle of twist f. If f is large enough to cause a plastic deformation, the radius rY of the elastic core is obtained by making g equal to the yield strain gY in Eq. (3.2) and solving for the corresponding value rY of the distance r. We have
rY5 LgY
f (3.34)
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Torsion Let us denote by fY the angle of twist at the onset of yield, i.e., when rY 5 c. Making f 5 fY and rY5 c in Eq. (3.34), we havec5 LgY
fY (3.35)
Dividing (3.34) by (3.35), member by member, we obtain the follow- ing relation:†
r Y
c 5 fY
f (3.36)
If we carry into Eq. (3.32) the expression obtained for rYyc, we express the torque T as a function of the angle of twist f,
T5 4
3TYa121 4
f3Y
f3b (3.37)
where TY and fY represent, respectively, the torque and the angle of twist at the onset of yield. Note that Eq. (3.37) may be used only for values of f larger than fY. For f , fY, the relation between T and f is linear and given by Eq. (3.16). Combining both equations, we obtain the plot of T against f represented in Fig. 3.39. We check that, as f increases indefinitely, T approaches the limiting value Tp543TY corresponding to the case of a fully developed plastic zone (Fig. 3.35d). While the value Tp cannot actually be reached, we note from Eq. (3.37) that it is rapidly approached as f increases. For f 5 2fY, T is within about 3% of Tp, and for f 5 3fY within about 1%.
Since the plot of T against f that we have obtained for an ideal- ized elastoplastic material (Fig. 3.36) differs greatly from the shearing- stress-strain diagram of that material (Fig. 3.34), it is clear that the shearing-stress-strain diagram of an actual material cannot be obtained directly from a torsion test carried out on a solid circular rod made of that material. However, a fairly accurate diagram may be obtained from a torsion test if the specimen used incorporates a portion consisting of a thin circular tube.‡ Indeed, we may assume that the shearing stress will have a constant value t in that portion.
Equation (3.1) thus reduces to
T 5rAt
where r denotes the average radius of the tube and A its cross- sectional area. The shearing stress is thus proportional to the torque, and successive values of t can be easily computed from the corresponding values of T. On the other hand, the values of the shearing strain g may be obtained from Eq. (3.2) and from the values of f and L measured on the tubular portion of the specimen.
†Equation (3.36) applies to any ductile material with a well-defined yield point, since its derivation is independent of the shape of the stress-strain diagram beyond the yield point.
‡In order to minimize the possibility of failure by buckling, the specimen should be made so that the length of the tubular portion is no longer than its diameter.
0 Y
3 Y TY
Tp 4TY T
Y 2 Y 3
Fig. 3.36 Load displacement relation for elastoplastic material.
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