Sample Problem 3.4
No particular stressstrain 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 _{Y}. We know from Chap. 2 that, for all practical 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
where G is the modulus of rigidity or shear modulus of the material. Multiplying both members of Eq. (3.4) by G, we write
or, making use of Eq. (3.5),
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 ρ 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 c_{1} and outer radius c_{2}. From Eq. (3.6), we find that, in the latter case,
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:
Substituting for from (3.6) into (3.1), we write
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
or, solving for _{max},
Substituting for _{max} from (3.9) into (3.6), we express the shearing stress at any distance ρ from the axis of the shaft as
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 . In the case of a hollow circular shaft of inner radius c_{1} and outer radius c_{2}, the polar moment of inertia is
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 ρ in meters, and J in m^{4}; we check that the resulting shearing stress will be expressed in N/m^{2}, that is, pascals (Pa). If U.S. customary units are used, T should be expressed in lb · in., c or ρ in inches, and J in in^{4}, with the resulting shearing stress expressed in psi.
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?
The largest torque T that can be applied to the shaft is the torque for which _{max} = 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
Recalling that the polar moment of inertia J of the cross section is given by Eq. (3.11), where and , we write
Substituting for J and _{max} into (3.12), and letting c = c_{2} = 0.03 m, we have
The minimum value of the shearing stress occurs on the inner surface of the shaft. It is obtained from Eq. (3.7), which expresses that _{min} and _{max} are respectively proportional to c_{1} and c_{2}:
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 freebody 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 (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 element will be the shearing stresses defined by formula (3.9), namely _{max} = Tc/J. 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 45° 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 freebody 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 _{max} = Tc/J. The magnitude of the corresponding shearing forces is thus _{max} A_{0}, where A_{0} 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
The corresponding stress is obtained by dividing the force F by the area A of face DC. Observing that , we write
A similar analysis of the element of Fig. 3.18b shows that the stress on the face BE is σ = −_{max}. We conclude that the stresses exerted on the faces of an element c at 45° to the axis of the shaft (Fig. 3.19) are normal stresses equal to ±_{max}. Thus, while the element a in Fig. 3.19 is in pure shear, the element c in the same figure is subjected 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, Tc/J.^{†}
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 longitudinal 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 45° angle with the longitudinal axis of the specimen (Photo 3.2b).
SAMPLE
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 shearing stress in these shafts is 65 MPa.
SAMPLE
SAMPLE
Denoting by T_{AB} the torque in shaft AB, we pass a section through shaft AB and, for the free body shown, we write
We now pass a section through shaft BC and, for the free body shown, we have
For this hollow shaft we have
On the outer surface, we have
We write that the stresses are proportional to the distance from the axis of the shaft.
We note that in both of these shafts the magnitude of the torque is T = 6 kN · m and _{all} = 65 MPa. Denoting by c the radius of the shafts, we write
SAMPLE
SAMPLE
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 8in. outer diameter.
SAMPLE
SAMPLE
For the shaft as designed and this solid shaft to have the same weight and length, their crosssectional areas must be equal.
Since _{all} = 12 ksi, we write
For equal weight, the crosssectional areas again must be equal. We determine the inside diameter of the shaft by writing
For c_{5} = 3.317 in. and c_{4} = 4 in.,
With _{all} = 12 ksi and c_{4} = 4 in.,
SAMPLE
3.1  (a) Determine the maximum shearing stress caused by a 4.6kN · m torque T in the 76mmdiameter 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 24mm 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) Determine the maximum shearing stress caused by the same torque T in a solid cylindrical shaft of the same crosssectional area.

3.3  Knowing that d = 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 = 0.9 in., determine the maximum shearing stress caused by a torque of magnitude T = 9 kip · in. 
3.5  A torque T = 3 kN · m is applied to the solid bronze cylinder shown. Determine (a) the maximum shearing stress, (b) the shearing stress at point D, which lies on a 15mmradius circle drawn on the end of the cylinder, (c) the percent of the torque carried by the portion of the cylinder within the 15mm radius.

3.6  (a) Determine the torque that can be applied to a solid shaft of 20mm 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 crosssectional area and with an inner diameter equal to half of its outer diameter. 
The solid spindle AB has a diameter d_{s} = 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. Determine 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 d_{s} 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.

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 shearing stress occurs, (b) the magnitude of that stress.

3.12  Knowing that an 8mmdiameter 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. 
Under normal operating conditions, the electric motor exerts a 12kip · in. torque at E. Knowing that each shaft is solid, determine the maximum shearing stress in (a) shaft BC, (b) shaft CD, (c) shaft DE.


3.14  Solve Prob. 3.13, assuming that a 1in.diameter hole has been drilled into each shaft. 
3.15  The allowable shearing stress is 15 ksi in the 1.5in.diameter steel rod AB and 8 ksi in the 1.8in.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 = 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 = 1250 N · m is applied at A, determine the required diameter of (a) rod AB, (b) rod BC.

3.18  The solid rod BC has a diameter of 30 mm and is made of an aluminum 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. 
The solid rod AB has a diameter d_{AB} = 60 mm. The pipe CD has an outer diameter of 90 mm and a wall thickness of 6 mm. Knowing 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 d_{AB} = 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 = 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.

3.22  A torque of magnitude T = 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 magnitude T_{F} = 1200 lb · in. at F. Knowing that r_{D} = 8 in., r_{G} = 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.

3.24  Under normal operating conditions a motor exerts a torque of magnitude T_{F} at F. The shafts are made of a steel for which the allowable shearing stress is 12 ksi and have diameters d_{CDE} = 0.900 in. and d_{FGH} = 0.800 in. Knowing that r_{D} = 6.5 in. and r_{G} = 4.5 in., determine the largest allowable value of T_{F}. 
The two solid shafts are connected by gears as shown and are made of a steel for which the allowable shearing stress is 8500 psi. Knowing that a torque of magnitude T_{C} = 5 kip · in. is applied at C and that the assembly is in equilibrium, determine the required diameter of (a) shaft BC, (b) shaft EF.


3.26  The two solid shafts are connected by gears as shown and are made of a steel for which the allowable shearing stress is 7000 psi. Knowing the diameters of the two shafts are, respectively, d_{BC} = 1.6 in. and d_{EF} = 1.25 in., determine the largest torque T_{C} that can be applied at C. 
3.27  A torque of magnitude T = 100 N · m is applied to shaft AB of the gear train shown. Knowing that the diameters of the three solid shafts are, respectively, d_{AB} = 21 mm, d_{CD} = 30 mm, and d_{EF} = 40 mm, determine the maximum shearing stress in (a) shaft AB, (b) shaft CD, (c) shaft EF.

3.28  A torque of magnitude T = 120 N · m is applied to shaft AB of the gear train shown. Knowing that the allowable shearing stress is 75 MPa in each of the three solid shafts, determine the required diameter of (a) shaft AB, (b) shaft CD, (c) shaft EF. 
3.29  (a) For a given allowable shearing stress, determine the ratio T/w of the maximum allowable torque T and the weight per unit length w for the hollow shaft shown. (b) Denoting by (T/w)_{0} the value of this ratio for a solid shaft of the same radius c_{2}, express the ratio T/w for the hollow shaft in terms of (T/w)_{0} and c_{1}/c_{2}.

3.30  While the exact distribution of the shearing stresses in a hollow cylindrical shaft is as shown in Fig. P3.30a, an approximate value can be obtained for _{max} by assuming that the stresses are uniformly distributed over the area A of the cross section, as shown in Fig. P3.30b, and then further assuming that all of the elementary shearing forces act at a distance from O equal to the mean radius of the cross section. This approximate value _{0} = T/Ar_{m}, where T is the applied torque. Determine the ratio _{max}/_{0} of the true value of the maximum shearing stress and its approximate value _{0} for values of c_{1}/c_{2} respectively equal to 1.00, 0.95, 0.75, 0.50 and 0.
