No particular stress-strain relationship has been assumed so far in our discussion of circular shafts in torsion. Let us now consider the case when the torque T is such that all shearing stresses in the shaft remain below the yield strength tY. We know from Chap. 2 that, for all practi- cal purposes, this means that the stresses in the shaft will remain below the proportional limit and below the elastic limit as well. Thus, Hooke’s law will apply and there will be no permanent deformation.
Recalling Hooke’s law for shearing stress and strain from Sec.
2.14, we write
t 5 Gg (3.5)
where G is the modulus of rigidity or shear modulus of the material.
Multiplying both members of Eq. (3.4) by G, we write Gg5r
c Ggmax or, making use of Eq. (3.5),
t5r
ctmax (3.6)
The equation obtained shows that, as long as the yield strength (or proportional limit) is not exceeded in any part of a circular shaft, the shearing stress in the shaft varies linearly with the distance r from the axis of the shaft. Figure 3.14a shows the stress distribution in a solid circular shaft of radius c, and Fig. 3.14b in a hollow circular shaft of inner radius c1 and outer radius c2. From Eq. (3.6), we find that, in the latter case,
tmin 5 c1
c2
tmax (3.7)
We now recall from Sec. 3.2 that the sum of the moments of the elementary forces exerted on any cross section of the shaft must be equal to the magnitude T of the torque exerted on the shaft:
er(tdA) 5 T (3.1)
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149
Substituting for t from (3.6) into (3.1), we write T5 ert dA5 tmax
c er2 dA
But the integral in the last member represents the polar moment of inertia J of the cross section with respect to its center O. We have therefore
T 5tmax J
c (3.8)
or, solving for tmax,
tmax5 Tc
J (3.9)
Substituting for tmax from (3.9) into (3.6), we express the shearing stress at any distance r from the axis of the shaft as
t5 Tr
J (3.10)
Equations (3.9) and (3.10) are known as the elastic torsion formulas.
We recall from statics that the polar moment of inertia of a circle of radius c is J512pc4. In the case of a hollow circular shaft of inner radius c1 and outer radius c2, the polar moment of inertia is
J5 12pc242 12pc14 512p1c242 c412 (3.11) We note that, if SI metric units are used in Eq. (3.9) or (3.10), T will be expressed in N ? m, c or r in meters, and J in m4; we check that the resulting shearing stress will be expressed in N/m2, that is, pascals (Pa). If U.S. customary units are used, T should be expressed in lb ? in., c or r in inches, and J in in4, with the resulting shearing stress expressed in psi.
3.4 Stresses in the Elastic Range
max
max
min
(a) (b)
c
O O c1 c2
Fig. 3.14 Distribution of shearing stresses.
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150
EXAMPLE 3.01 A hollow cylindrical steel shaft is 1.5 m long and has inner and outer diameters respectively equal to 40 and 60 mm (Fig. 3.15). (a) What is the largest torque that can be applied to the shaft if the shearing stress is not to exceed 120 MPa? (b) What is the corresponding minimum value of the shearing stress in the shaft?
(a) Largest Permissible Torque. The largest torque T that can be applied to the shaft is the torque for which tmax 5 120 MPa. Since this value is less than the yield strength for steel, we can use Eq. (3.9).
Solving this equation for T, we have T5Jtmax
c (3.12)
Recalling that the polar moment of inertia J of the cross section is given by Eq. (3.11), where c1512140 mm250.02 m and c2512160 mm2 50.03 m, we write
J512p 1c422c412512p10.03420.024251.02131026 m4
Substituting for J and tmax into (3.12), and letting c 5 c2 5 0.03 m, we have
T5 Jtmax
c 5 11.02131026 m4211203106 Pa2
0.03 m 54.08 kN?m
(b) Minimum Shearing Stress. The minimum value of the shear- ing stress occurs on the inner surface of the shaft. It is obtained from Eq.
(3.7), which expresses that tmin and tmax are respectively proportional to c1 and c2:
tmin5c1
c2tmax50.02 m
0.03 m 1120 MPa2 580 MPa
1.5 m
40 mm 60 mm T
Fig. 3.15
The torsion formulas (3.9) and (3.10) were derived for a shaft of uniform circular cross section subjected to torques at its ends.
However, they can also be used for a shaft of variable cross section or for a shaft subjected to torques at locations other than its ends (Fig. 3.16a). The distribution of shearing stresses in a given cross section S of the shaft is obtained from Eq. (3.9), where J denotes the polar moment of inertia of that section, and where T represents the internal torque in that section. The value of T is obtained by drawing the free-body diagram of the portion of shaft located on one side of the section (Fig. 3.16b) and writing that the sum of the torques applied to that portion, including the internal torque T, is zero (see Sample Prob. 3.1).
Up to this point, our analysis of stresses in a shaft has been limited to shearing stresses. This is due to the fact that the element we had selected was oriented in such a way that its faces were either parallel or perpendicular to the axis of the shaft (Fig. 3.5). We know from earlier discussions (Secs. 1.11 and 1.12) that normal stresses, shearing stresses, or a combination of both may be found under the same loading condition, depending upon the orientation of the element that has been chosen. Consider the two elements a and b located on the surface of a circular shaft subjected to torsion
B
(a)
(b)
TC TE
TB
TA
E B
S C
A E S
TB T
TE
Fig. 3.16 Shaft with variable cross section.
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151 (Fig. 3.17). Since the faces of element a are respectively parallel and
perpendicular to the axis of the shaft, the only stresses on the ele- ment will be the shearing stresses defined by formula (3.9), namely tmax5 TcyJ. On the other hand, the faces of element b, which form arbitrary angles with the axis of the shaft, will be subjected to a combination of normal and shearing stresses.
Let us consider the stresses and resulting forces on faces that are at 458 to the axis of the shaft. In order to determine the stresses on the faces of this element, we consider the two triangular elements shown in Fig. 3.18 and draw their free-body diagrams. In the case of the element of Fig. 3.18a, we know that the stresses exerted on the faces BC and BD are the shearing stresses tmax 5 TcyJ. The magnitude of the corresponding shearing forces is thus tmax A0, where A0 denotes the area of the face. Observing that the components along DC of the two shearing forces are equal and opposite, we conclude that the force F exerted on DC must be perpendicular to that face.
It is a tensile force, and its magnitude is
F 521tmaxA02cos 45°5 tmaxA012 (3.13)
The corresponding stress is obtained by dividing the force F by the area A of face DC. Observing that A5A012, we write
s5 F
A 5 tmaxA012
A012 5tmax (3.14)
A similar analysis of the element of Fig. 3.18b shows that the stress on the face BE is s52tmax. We conclude that the stresses exerted on the faces of an element c at 458 to the axis of the shaft (Fig. 3.19) are normal stresses equal to 6tmax. Thus, while the element a in Fig. 3.19 is in pure shear, the element c in the same figure is sub- jected to a tensile stress on two of its faces, and to a compressive stress on the other two. We also note that all the stresses involved have the same magnitude, TcyJ.†
As you learned in Sec. 2.3, ductile materials generally fail in shear. Therefore, when subjected to torsion, a specimen J made of a ductile material breaks along a plane perpendicular to its longitu- dinal axis (Photo 3.2a). On the other hand, brittle materials are weaker in tension than in shear. Thus, when subjected to torsion, a specimen made of a brittle material tends to break along surfaces that are perpendicular to the direction in which tension is maximum, i.e., along surfaces forming a 458 angle with the longitudinal axis of the specimen (Photo 3.2b).
3.4 Stresses in the Elastic Range
†Stresses on elements of arbitrary orientation, such as element b of Fig. 3.18, will be dis- cussed in Chap. 7.
Photo 3.2 Shear failure of shaft subject to torque.
(a) Ductile failure T'
T
(b) Brittle failure T'
T a
max T T'
b
Fig. 3.17 Circular shaft with elements at different orientations.
(a) (b)
C C
B B
D E
maxA0
maxA0
maxA0
maxA0
45⬚ 45⬚
F F'
Fig. 3.18 Forces on faces at 458 to shaft axis.
⫽TcJ
max 45⬚⫽⫾TcJ a
T T'
c
Fig. 3.19 Shaft with elements with only shear stresses or normal stresses.
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152
SAMPLE PROBLEM 3.1
Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid and of diameter d. For the loading shown, determine (a) the maximum and minimum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shear- ing stress in these shafts is 65 MPa.
0.9 m
d
A
B
TC
TD 0.7 m
0.5 m 120 mm
d
C D
TA 6 kN ã m 14 kN ã m
26 kN ã m 6 kN ã m TB
SOLUTION
Equations of Statics. Denoting by TAB the torque in shaft AB, we pass a section through shaft AB and, for the free body shown, we write
©Mx50: 16 kN?m22TAB50 TAB56 kN?m
We now pass a section through shaft BC and, for the free body shown, we have
©Mx50: 16 kN?m21114 kN?m22TBC50 TBC520 kN?m
a. Shaft BC. For this hollow shaft we have J5p
21c422c412 5p
23 10.06024210.045244 513.9231026 m4 Maximum Shearing Stress. On the outer surface, we have
tmax5t25TBCc2
J 5120 kN?m2 10.060 m2
13.9231026 m4 tmax586.2 MPa b
Minimum Shearing Stress. We write that the stresses are propor- tional to the distance from the axis of the shaft.
tmin tmax 5c1
c2 86.2 MPatmin 5 45 mm60 mm tmin564.7 MPa b
b. Shafts AB and CD. We note that in both of these shafts the mag- nitude of the torque is T 5 6 kN ? m and tall 5 65 MPa. Denoting by c the radius of the shafts, we write
t5Tc
J 65 MPa5 16 kNp?m2c
2 c4 c3558.831026 m3 c538.931023 m
d52c52138.9 mm2 d577.8 mm b
A TAB
x TA 6 kN ã m
TB
A
B TBC
xx TA 6 kN ã m
14 kN ã m
c1 45 mm c2 60 mm
2 1
A
B 6 kN ã m
6 kN ã m
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153 SAMPLE PROBLEM 3.2
The preliminary design of a large shaft connecting a motor to a generator calls for the use of a hollow shaft with inner and outer diameters of 4 in.
and 6 in., respectively. Knowing that the allowable shearing stress is 12 ksi, determine the maximum torque that can be transmitted (a) by the shaft as designed, (b) by a solid shaft of the same weight, (c) by a hollow shaft of the same weight and of 8-in. outer diameter.
8 ft
T'
T
6 in.
4 in.
SOLUTION
a. Hollow Shaft as Designed. For the hollow shaft we have J5p
21c422c412 5p
23 13 in.24212 in.244 5102.1 in4 Using Eq. (3.9), we write
tmax5 Tc2
J 12 ksi5 102.1 inT13 in.24 T5408 kip?in. b
b. Solid Shaft of Equal Weight. For the shaft as designed and this solid shaft to have the same weight and length, their cross-sectional areas must be equal.
A1a25A1b2
p3 13 in.222 12 in.224 5pc23 c352.24 in.
Since tall 5 12 ksi, we write tmax5 Tc3
J 12 ksi5 pT12.24 in.2
212.24 in.24
T5211 kip?in. b
c. Hollow Shaft of 8-in. Diameter. For equal weight, the cross- sectional areas again must be equal. We determine the inside diameter of the shaft by writing
A1a25A1c2
p3 13 in.22212 in.2245p3 14 in.222c254 c553.317 in.
For c5 5 3.317 in. and c4 5 4 in., J5p
23 14 in.242 13.317 in.2445212 in4 With tall 5 12 ksi and c4 5 4 in.,
tmax5 Tc4
J 12 ksi5T212 in14 in.42 T5636 kip?in. b
c2 ⫽ 3 in.
c1 ⫽ 2 in.
T
c3
T
c4 ⫽ 4 in.
c5
T
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PROBLEMS
154
3.1 (a) Determine the maximum shearing stress caused by a 4.6-kN ? m torque T in the 76-mm-diameter solid aluminum shaft shown.
(b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same outer diameter and of 24-mm inner diameter.
3.2 (a) Determine the torque T that causes a maximum shearing stress of 45 MPa in the hollow cylindrical steel shaft shown. (b) Deter- mine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same cross-sectional area.
76 mm 1.2 m
T Fig. P3.1
3.3 Knowing that d 5 1.2 in., determine the torque T that causes a maximum shearing stress of 7.5 ksi in the hollow shaft shown.
3.4 Knowing that the internal diameter of the hollow shaft shown is d 5 0.9 in., determine the maximum shearing stress caused by a torque of magnitude T 5 9 kip ? in.
3.5 A torque T 5 3 kN ? m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shear- ing stress at point D, which lies on a 15-mm-radius circle drawn on the end of the cylinder, (c) the percent of the torque carried by the portion of the cylinder within the 15-mm radius.
2.4 m
30 mm
45 mm T
Fig. P3.2
d 1.6 in.
T
Fig. P3.3 and P3.4 60 mm
30 mm
200 mmD T 3 kN ã m
Fig. P3.5
3.6 (a) Determine the torque that can be applied to a solid shaft of 20-mm diameter without exceeding an allowable shearing stress of 80 MPa. (b) Solve part a, assuming that the solid shaft has been replaced by a hollow shaft of the same cross-sectional area and with an inner diameter equal to half of its outer diameter.
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155
Problems
3.7 The solid spindle AB has a diameter ds 5 1.5 in. and is made of a steel with an allowable shearing stress of 12 ksi, while sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Deter- mine the largest torque T that can be applied at A.
3.8 The solid spindle AB is made of a steel with an allowable shearing stress of 12 ksi, and sleeve CD is made of a brass with an allowable shearing stress of 7 ksi. Determine (a) the largest torque T that can be applied at A if the allowable shearing stress is not to be exceeded in sleeve CD, (b) the corresponding required value of the diameter ds of spindle AB.
3.9 The torques shown are exerted on pulleys A and B. Knowing that both shafts are solid, determine the maximum shearing stress in (a) in shaft AB, (b) in shaft BC.
4 in.
8 in.
ds
t 14in.
3 in.
D C
A B
T Fig. P3.7 and P3.8
30 mm
46 mm
C A
B TA 300 N ã m
TB 400 N ã m
Fig. P3.9
3.10 In order to reduce the total mass of the assembly of Prob. 3.9, a new design is being considered in which the diameter of shaft BC will be smaller. Determine the smallest diameter of shaft BC for which the maximum value of the shearing stress in the assembly will not increase.
3.11 Knowing that each of the shafts AB, BC, and CD consists of a solid circular rod, determine (a) the shaft in which the maximum shear- ing stress occurs, (b) the magnitude of that stress.
D
dCD 21 mm
B
dBC 18 mmC 60 N ã m
48 N ã m
A dAB 15 mm 144 N ã m
Fig. P3.11 and P3.12
3.12 Knowing that an 8-mm-diameter hole has been drilled through each of the shafts AB, BC, and CD, determine (a) the shaft in which the maximum shearing stress occurs, (b) the magnitude of that stress.
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156 Torsion 3.13 Under normal operating conditions, the electric motor exerts a 12-kip ? in. torque at E. Knowing that each shaft is solid, deter- mine the maximum shearing stress in (a) shaft BC, (b) shaft CD, (c) shaft DE.
2.25 in.
2 in.
1.75 in.
1.50 in.
E
A
B
C
D 5 kip ã in.
4 kip ã in.
3 kip ã in.
Fig. P3.13
3.14 Solve Prob. 3.13, assuming that a 1-in.-diameter hole has been drilled into each shaft.
3.15 The allowable shearing stress is 15 ksi in the 1.5-in.-diameter steel rod AB and 8 ksi in the 1.8-in.-diameter brass rod BC. Neglecting the effect of stress concentrations, determine the largest torque that can be applied at A.
3.16 The allowable shearing stress is 15 ksi in the steel rod AB and 8 ksi in the brass rod BC. Knowing that a torque of magnitude T 5 10 kip ? in. is applied at A, determine the required diameter of (a) rod AB, (b) rod BC.
3.17 The allowable stress is 50 MPa in the brass rod AB and 25 MPa in the aluminum rod BC. Knowing that a torque of magnitude T 5 1250 N ? m is applied at A, determine the required diameter of (a) rod AB, (b) rod BC.
B
C
Brass T A
Steel
Fig. P3.15 and P3.16
Brass Aluminum
B C
A T
Fig. P3.17 and P3.18
3.18 The solid rod BC has a diameter of 30 mm and is made of an alu- minum for which the allowable shearing stress is 25 MPa. Rod AB is hollow and has an outer diameter of 25 mm; it is made of a brass for which the allowable shearing stress is 50 MPa. Determine (a) the largest inner diameter of rod AB for which the factor of safety is the same for each rod, (b) the largest torque that can be applied at A.
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157
Problems
3.19 The solid rod AB has a diameter dAB 5 60 mm. The pipe CD has an outer diameter of 90 mm and a wall thickness of 6 mm. Know- ing that both the rod and the pipe are made of steel for which the allowable shearing stress is 75 MPa, determine the largest torque T that can be applied at A.
3.20 The solid rod AB has a diameter dAB 5 60 mm and is made of a steel for which the allowable shearing stress is 85 MPa. The pipe CD, which has an outer diameter of 90 mm and a wall thickness of 6 mm, is made of an aluminum for which the allowable shearing stress is 54 MPa. Determine the largest torque T that can be applied at A.
3.21 A torque of magnitude T 5 1000 N ? m is applied at D as shown.
Knowing that the diameter of shaft AB is 56 mm and that the diameter of shaft CD is 42 mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD.
D
A
B 90 mm
dAB C
T
Fig. P3.19 and P3.20
A
100 mm 40 mm C
B D
T 1000 N ã m
Fig. P3.21 and P3.22
3.22 A torque of magnitude T 5 1000 N ? m is applied at D as shown.
Knowing that the allowable shearing stress is 60 MPa in each shaft, determine the required diameter of (a) shaft AB, (b) shaft CD.
3.23 Under normal operating conditions a motor exerts a torque of mag- nitude TF 5 1200 lb ? in. at F. Knowing that rD 5 8 in., rG 5 3 in., and the allowable shearing stress is 10.5 ksi in each shaft, determine the required diameter of (a) shaft CDE, (b) shaft FGH.
F
TE
H E
A
D B C
rG G
rD TF
Fig. P3.23 and P3.24
3.24 Under normal operating conditions a motor exerts a torque of mag- nitude TF at F. The shafts are made of a steel for which the allow- able shearing stress is 12 ksi and have diameters dCDE 5 0.900 in.
and dFGH 5 0.800 in. Knowing that rD 5 6.5 in. and rG 5 4.5 in., determine the largest allowable value of TF.
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