3.117 After the solid shaft of Prob. 3.116 has been loaded and unloaded as described in that problem, a torque T1 of sense opposite to the original torque T is applied to the shaft. Assuming no change in the value of fY, determine the angle of twist f1 for which yield is initiated in this second loading and compare it with the angle fY for which the shaft started to yield in the original loading.
3.118 The hollow shaft shown is made of a steel that is assumed to be elastoplastic with tY 5 145 MPa and G 5 77.2 GPa. The magni- tude T of the torques is slowly increased until the plastic zone first reaches the inner surface of the shaft; the torques are then removed.
Determine the magnitude and location of the maximum residual shearing stress in the rod.
3.119 In Prob. 3.118, determine the permanent angle of twist of the rod.
3.120 A torque T applied to a solid rod made of an elastoplastic material is increased until the rod is fully plastic and then removed.
(a) Show that the distribution of residual shearing stresses is as represented in the figure. (b) Determine the magnitude of the torque due to the stresses acting on the portion of the rod located within a circle of radius c0.
5 m
25 mm 60 mm
T
T'
Fig. P3.118
Y Y
c c0
1
3
Fig. P3.120
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Torsion Consider a small cubic element located at a corner of the cross section of a square bar in torsion and select coordinate axes parallel to the edges of the element (Fig. 3.43a). Since the face of the ele- ment perpendicular to the y axis is part of the free surface of the bar, all stresses on this face must be zero. Referring to Fig. 3.43b, we writetyx5 0 tyz5 0 (3.40) For the same reason, all stresses on the face of the element perpen- dicular to the z axis must be zero, and we write
tzx5 0 tzy5 0 (3.41) It follows from the first of Eqs. (3.40) and the first of Eqs. (3.41) that
txy5 0 txz5 0 (3.42) Thus, both components of the shearing stress on the face of the element perpendicular to the axis of the bar are zero. We conclude that there is no shearing stress at the corners of the cross section of the bar.
By twisting a rubber model of a square bar, one easily verifies that no deformations—and, thus, no stresses—occur along the edges of the bar, while the largest deformations—and, thus, the largest stresses—occur along the center line of each of the faces of the bar (Fig. 3.44).
y
x
zy xy
xz
yz
yx
zx
(a)
(b) z
z x
y
Fig. 3.43 Corner element.
max
max T T'
Fig. 3.44 Deformation of square bar.
†See S. P. Timoshenko and J. N. Goodier, Theory of Elasticity, 3d ed., McGraw-Hill, New York, 1969, sec. 109.
L a b
max
T T'
Fig. 3.45 Shaft with rectangular cross section.
The determination of the stresses in noncircular members sub- jected to a torsional loading is beyond the scope of this text. How- ever, results obtained from the mathematical theory of elasticity for straight bars with a uniform rectangular cross section will be indi- cated here for convenience.† Denoting by L the length of the bar, by a and b, respectively, the wider and narrower side of its cross section, and by T the magnitude of the torques applied to the bar (Fig. 3.45), we find that the maximum shearing stress occurs along the center line of the wider face of the bar and is equal to
tmax5 T
c1ab2 (3.43)
The angle of twist, on the other hand, may be expressed as f5 TL
c2ab3G (3.44)
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The coefficients c1 and c2 depend only upon the ratio ayb and are given in Table 3.1 for a number of values of that ratio. Note that Eqs. (3.43) and (3.44) are valid only within the elastic range.
We note from Table 3.1 that for ayb $ 5, the coefficients c1 and c2 are equal. It may be shown that for such values of ayb, we have
c15c25131120.630bya2 (for ayb % 5 only) (3.45) The distribution of shearing stresses in a noncircular member may be visualized more easily by using the membrane analogy. A homo- geneous elastic membrane attached to a fixed frame and subjected to a uniform pressure on one of its sides happens to constitute an analog of the bar in torsion, i.e., the determination of the deformation of the membrane depends upon the solution of the same partial differential equation as the determination of the shearing stresses in the bar.† More specifically, if Q is a point of the cross section of the bar and Q9 the corresponding point of the membrane (Fig. 3.46), the shearing stress t at Q will have the same direction as the horizontal tangent to the membrane at Q9, and its magnitude will be proportional to the maxi- mum slope of the membrane at Q9.‡ Furthermore, the applied torque will be proportional to the volume between the membrane and the plane of the fixed frame. In the case of the membrane of Fig. 3.46, which is attached to a rectangular frame, the steepest slope occurs at the midpoint N9 of the larger side of the frame. Thus, we verify that the maximum shearing stress in a bar of rectangular cross section will occur at the midpoint N of the larger side of that section.
The membrane analogy may be used just as effectively to visu- alize the shearing stresses in any straight bar of uniform, noncircular cross section. In particular, let us consider several thin-walled mem- bers with the cross sections shown in Fig. 3.47, which are subjected
TABLE 3.1. Coefficients for Rectangular Bars in Torsion
a/b c1 c2
1.0 0.208 0.1406
1.2 0.219 0.1661
1.5 0.231 0.1958
2.0 0.246 0.229
2.5 0.258 0.249
3.0 0.267 0.263
4.0 0.282 0.281
5.0 0.291 0.291
10.0 0.312 0.312
` 0.333 0.333
†See ibid. Sec. 107.
‡This is the slope measured in a direction perpendicular to the horizontal tangent at Q9.
N'
Rectangular frame Tangent of max. slope
Membrane Horizontal
tangent
N
Q b
a
a Q'
b
T
Fig. 3.46 Application of membrane analogy to shaft with rectangular cross section.
a b a
a b
b
Fig. 3.47 Various thin-walled members.
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Torsion to the same torque. Using the membrane analogy to help us visualize the shearing stresses, we note that, since the same torque is applied to each member, the same volume will be located under each mem- brane, and the maximum slope will be about the same in each case.Thus, for a thin-walled member of uniform thickness and arbitrary shape, the maximum shearing stress is the same as for a rectangular bar with a very large value of ayb and may be determined from Eq.
(3.43) with c1 5 0.333.†