10.1 INTRODUCTION

Many rotating shafts and axles used for power transmission experience torsional vibration, particularly when the prime mover is a reciprocating engine. The shafts used in high‐speed machinery, especially those carrying heavy wheels, are subjected to dynamic torsional forces and vibration. A solid or hollow cylindrical rod of circular section undergoes torsional displacement or twisting such that each transverse section remains in its own plane when a torsional moment is applied. In this case the cross‐sections of the rod do not experience any motion parallel to the axis of the rod. However, if the cross‐section of the rod is not circular, the effect of a twist will be more involved. In this case the twist will be accompanied by a warping of normal cross‐sections. The torsional vibrations of uniform and nonuniform rods with circular cross‐section and rods with noncircular section are considered in this chapter. For noncircular sections, the equations of motion are derived using both the Saint‐Venant and the Timoshenko–Gere theories. The methods of determining the torsional rigidity of noncircular rods are presented using the Prandtl stress function and the Prandtl membrane analogy.

10.2 ELEMENTARY THEORY: EQUATION OF MOTION

10.2.1 Equilibrium Approach

Consider an element of a nonuniform circular shaft between two cross‐sections at x and , as shown in Fig. 10.1(a). Let denote the torque induced in the shaft at x and time t and the torque induced in the shaft at and at the same time t. If the angular displacement of the cross‐section at x is denoted as , the angular displacement of the cross‐section at can be represented as . Let the external torque acting on the shaft per unit length be denoted . The inertia torque acting on the element of the shaft is given by , where is the mass polar moment of inertia of the shaft per unit length. Noting that and , Newton's second law of motion can be applied to the element of the shaft to obtain the equation of motion as

(10.1)

From the strength of materials, the relationship between the torque in the shaft and the angular displacement is given by [1]

(10.2)

where G is the shear modulus and is the polar moment of inertia of the cross‐section of the shaft. Using Eq. (10.2), the equation of motion, Eq. (10.1), can be expressed as

(10.3)

10.2.2 Variational Approach

The equation of motion of a nonuniform shaft, using the variational approach, has been derived in Section 4.11.1. In this section the variational approach is used to derive the equation of motion and the boundary conditions for a nonuniform shaft with torsional springs (with stiffnesses and ) and masses (with mass moments of inertia and ) attached at each end as shown in Fig. 10.1.

The cross‐sections of the shaft are assumed to remain plane before and after angular deformation. Since the cross‐section of the shaft at x undergoes an angular displacement about the center of twist, the shape of the cross‐section does not change. The cross‐section simply rotates about the x axis. A typical point P rotates around the x axis by a small angle as shown in Fig. 10.2. The displacements of point P parallel to the y and z axes are given by the projections of the displacement PP′ on oy and oz:

(10.4)

Since is small, we can write

(10.5)

so that

(10.6)

Thus, the displacement components of the shaft parallel to the three coordinate axes can be expressed as

(10.7)

The strains in the shaft are assumed to be

(10.8)

and the corresponding stresses are given by

(10.9)

The strain energy of the shaft and the torsional springs is given by

(10.10)

where . The kinetic energy of the shaft can be expressed as

(10.11)

The work done by the external torque can be represented as

(10.12)

The application of the generalized Hamilton's principle yields

or

(10.13)

The variations in Eq. (10.13) can be evaluated using integration by parts to obtain

(10.14)

(10.15)

(10.16)

(10.17)

Note that integration by parts with respect to time, along with the fact that at and , has been used in deriving Eqs. (10.16) and (10.17). By using Eqs. (10.14)– (10.17) in Eq. (10.13), we obtain

(10.18)

By setting the two expressions under the braces in each term of Eq. (10.18) equal to zero, we obtain the equation of motion for the torsional vibration of the shaft as

(10.19)

where is the mass moment of inertia of the shaft per unit length, and the boundary conditions as

(10.20)

Each of the equations in (10.20) can be satisfied in two ways but will be satisfied only one way for any specific end conditions of the shaft. The boundary conditions implied by Eqs. (10.20) are as follows. At , either is specified (so that ) or

(10.21)

At , either is specified (so that ) or

(10.22)

In the present case, the second conditions stated in each of Eqs. (10.21) and (10.22) are valid.

10.3 FREE VIBRATION OF UNIFORM SHAFTS

For a uniform shaft, Eq. (10.19) reduces to

(10.23)

By setting , we obtain the free vibration equation

(10.24)

where

(10.25)

It can be observed that Eqs. (10.23)–(10.25) are similar to the equations derived in the cases of transverse vibration of a string and longitudinal vibration of a bar. For a uniform shaft, and Eq. (10.25) takes the form

(10.26)

By assuming the solution as

(10.27)

Eq. (10.24) can be written as two separate equations:

(10.28)

(10.29)

The solutions of Eqs. (10.28) and (10.29) can be expressed as

(10.30)

(10.31)

where A, B, C, and D are constants. If denotes the nth frequency of vibration and the corresponding mode shape, the general free vibration solution of Eq. (10.24) is given by

(10.32)

The constraints and can be evaluated from the initial conditions, and the constraints and can be determined (not the absolute values, only their relative values) from the boundary conditions of the shaft. The initial conditions are usually stated in terms of the initial angular displacement and angular velocity distributions of the shaft.

10.3.1 Natural Frequencies of a Shaft with Both Ends Fixed

For a uniform circular shaft of length l fixed at both ends, the boundary conditions are given by

(10.33)

(10.34)

The free vibration solution is given by Eq. (10.27):

(10.35)

Equations (10.33) and (10.35) yield

(10.36)

and the solution can be expressed as

(10.37)

where C′ and D′ are new constants. The use of Eq. (10.34) in (10.37) gives the frequency equation

(10.38)

The natural frequencies of vibration are given by the roots of Eq. (10.38) as

or

(10.39)

The mode shape corresponding to the natural frequency can be expressed as

(10.40)

The free vibration solution of the fixed–fixed shaft is given by a linear combination of its normal modes:

(10.41)

10.3.2 Natural Frequencies of a Shaft with Both Ends Free

Since the torque, , is zero at a free end, the boundary conditions of a free–free shaft are given by

(10.42)

(10.43)

In view of Eq. (10.27), Eqs. (10.42) and (10.43) can be expressed as

(10.44)

(10.45)

Eq. (10.30) gives

(10.46)

(10.47)

Eqs. (10.44) and (10.47) yield

(10.48)

and Eqs. (10.45) and (10.47) result in

(10.49)

The roots of Eq. (10.49) are given by

(10.50)

The nth normal mode is given by

(10.51)

The free vibration solution of the shaft can be expressed as (see Eq. [10.32])

(10.52)

where the constants and can be determined from the initial conditions of the shaft.

10.3.3 Natural Frequencies of a Shaft Fixed at One End and Attached to a Torsional Spring at the Other

For a uniform circular shaft fixed at and attached to a torsional spring of stiffness at , as shown in Fig. 10.3, the boundary conditions are given by

(10.53)

(10.54)

The free vibration solution of a shaft is given by Eq. (10.27):

(10.55)

The use of the boundary condition of Eq. (10.53) in Eq. (10.55) gives

(10.56)

and the solution can be expressed as

(10.57)

The use of the boundary condition of Eq. (10.54) in Eq. (10.57) yields the frequency equation

(10.58)

Using Eq. (10.26), Eq. (10.58) can be rewritten as

(10.59)

where

(10.60)

The roots of the frequency equation (10.60) give the natural frequencies of vibration of the shaft as

(10.61)

and the corresponding mode shapes as

(10.62)

Finally, the free vibration solution of the shaft can be expressed as

(10.63)

Several possible boundary conditions for the torsional vibration of a uniform shaft are given in Table 10.1, along with the corresponding frequency equations and the mode shapes.

End conditions of shaft	Boundary conditions	Frequency equation	Mode shape (normal function)	Natural frequencies
1. Fixed–free
2. Free–free
3. Fixed–fixed
4. Fixed–disk
5. Fixed–torsional spring
6. Free–disk
7. Free–torsional spring
8. Disk–disk

10.4 FREE VIBRATION RESPONSE DUE TO INITIAL CONDITIONS: MODAL ANALYSIS

The angular displacement of a shaft in torsional vibration can be expressed in terms of normal modes using the expansion theorem, as

(10.64)

where is the ith generalized coordinate. Substituting Eq. (10.64) into Eq. (10.24), we obtain

(10.65)

where and . Multiplication of Eq. (10.65) by and integration from 0 to l yields

(10.66)

In view of the orthogonality relationships, Eqs. (E10.3.8) and (E10.3.10), Eq. (10.66) reduces to

or

(10.67)

Equation (10.67) yields

(10.68)

The solution of Eq. (10.68) is given by

(10.69)

where and denote the initial values of the generalized coordinate and the generalized velocity , respectively.

Initial Conditions

If the initial conditions of the shaft are given by

(10.70)

(10.71)

Eq. (10.64) gives

(10.72)

(10.73)

By multiplying Eqs. (10.72) and (10.73) by and integrating from 0 to l, we obtain

(10.74)

(10.75)

in view of the orthogonality of normal modes (Eq. [E10.3.8]). Using the initial values of and , Eqs. (10.74) and (10.75), the free vibration response of the shaft can be determined from Eqs. (10.69) and (10.64):

(10.76)

10.5 FORCED VIBRATION OF A UNIFORM SHAFT: MODAL ANALYSIS

The equation of motion of a uniform shaft subjected to distributed external torque, , is given by Eq. (10.23):

(10.77)

The solution of Eq. (10.77) using modal analysis is expressed as

(10.78)

where is the nth normalized normal mode and is the nth generalized coordinate. The normal modes are determined by solving the eigenvalue problem

(10.79)

by applying the boundary conditions of the shaft. By substituting Eq. (10.78) into (10.77), we obtain

(10.80)

where

(10.81)

Using Eq. (10.79), Eq. (10.80) can be rewritten as

(10.82)

Multiplication of Eq. (10.82) by and integration from 0 to l result in

(10.83)

In view of the orthogonality relationships, Eq. (E10.3.8), Eq. (10.83) reduces to

(10.84)

where the normal modes are assumed to satisfy the normalization condition

(10.85)

and , called the generalized force in nth mode, is given by

(10.86)

The complete solution of Eq. (10.84) can be expressed as

(10.87)

where the constants and can be determined from the initial conditions of the shaft. Thus, the forced vibration response of the shaft (i.e. the solution of Eq. [10.77]), is given by

(10.88)

The steady‐state response of the shaft, without considering the effect of initial conditions, can be obtained from Eq. (10.88), as

(10.89)

Note that if the shaft is unrestrained (free at both ends), the rigid‐body displacement, , is to be added to the solution given by Eq. (10.89). If denotes the torque applied to the shaft, the rigid‐body motion of the shaft, , can be determined from the relation

(10.90)

where denotes the mass moment of inertia of the shaft and indicates the acceleration of rigid‐body motion.

10.6 TORSIONAL VIBRATION OF NONCIRCULAR SHAFTS: SAINT‐VENANT'S THEORY

For a shaft or bar of noncircular cross‐section subjected to torsion, the cross‐sections do not simply rotate with respect to one another as in the case of a circular shaft, but they are deformed, too. The originally plane cross‐sections of the shaft do not remain plane but warp out of their own planes after twisting, as shown in Fig. 10.7. Thus, the points in the cross‐section undergo an axial displacement. A function , known as the warping function, is used to denote the axial displacement as

(10.91)

where denotes the rate of twist along the shaft, assumed to be a constant. The other components of displacement in the shaft are given by

(10.92)

(10.93)

The strains are given by

(10.94)

The corresponding stresses can be determined as

(10.95)

The strain energy of the shaft is given by

(10.96)

Defining the torsional rigidity (C) of its noncircular section of the shaft as

(10.97)

the strain energy of the shaft can be expressed as

(10.98)

Neglecting the inertia due to axial motion, the kinetic energy of the shaft can be written as in Eq. (10.11). The work done by the applied torque is given by Eq. (10.12). Hamilton's principle can be written as

(10.99)

The first integral of Eq. (10.99) can be evaluated as

(10.100)

The first integral term on the right‐hand side of Eq. (10.100) can be expressed as (see Eq. [10.97])

(10.101)

The second integral term on the right‐hand side of Eq. (10.100) is set equal to zero independently:

(10.102)

Integrating Eq. (10.102) by parts, we obtain

(10.103)

where is the bounding curve of the cross‐section and is the cosine of the angle between the normal to the bounding curve and the direction. Equation (10.103) yields the differential equation for the warping function as

(10.104)

and the boundary condition on as

(10.105)

Physically, Eq. (10.105) represents that the shear stress normal to the boundary must be zero at every point on the boundary of the cross‐section of the shaft. When Eq. (10.101) is combined with the second and third integrals of Eq. (10.99), it leads to the equation of motion as

(10.106)

and the boundary conditions on as

(10.107)

10.7 TORSIONAL VIBRATION OF NONCIRCULAR SHAFTS, INCLUDING AXIAL INERTIA

Love included the inertia due to the axial motion caused by the warping of the cross‐section in deriving the equation of motion of a shaft in torsional vibration [4, 10]. In this case, the kinetic energy of the shaft is given by

(10.108)

where

(10.109)

(10.110)

Note that denotes the axial inertia term. The variation associated with in Hamilton's principle leads to

(10.111)

Denoting

(10.112)

(10.113)

the integrals in Eq. (10.111) can be evaluated to obtain

(10.114)

The first, second, and third terms on the right‐hand side of Eq. (10.114) contribute to the equation of motion, Eq. (10.106), the boundary conditions on , Eq. (10.107), and to the differential equation for , Eq. (10.104), respectively. The new equations are given by

(10.115)

(10.116)

(10.117)

(10.118)

where

(10.119)

10.8 TORSIONAL VIBRATION OF NONCIRCULAR SHAFTS: THE TIMOSHENKO–GERE THEORY

In this theory also, the displacement components of a point in the cross‐section are assumed to be [4, 11, 17]

(10.120)

(10.121)

(10.122)

where is not assumed to be a constant. The components of strains can be obtained as

(10.123)

(10.124)

(10.125)

(10.126)

(10.127)

(10.128)

The components of stress are given by

(10.129)

that is,

(10.130)

(10.131)

(10.132)

(10.133)

(10.134)

Note that the effect of Poisson's ratio is neglected in Eqs. (10.130) and (10.131). The strain energy of the shaft can be determined as

(10.135)

where

(10.136)

(10.137)

The variation of the integral can be evaluated as

(10.138)

where

(10.139)

When the second term on the right‐hand side of Eq. (10.138) is integrated by parts, we obtain

(10.140)

The variation of the integral can be evaluated as indicated in Eqs. (10.100), (10.101) and (10.103). The expressions for the kinetic energy and the work done by the applied torque are given by Eqs. (10.108) and (10.12), respectively, and hence their variations can be evaluated as indicated earlier. The application of Hamilton's principle leads to the equation of motion for :

(10.141)

and the boundary conditions

(10.142)

(10.143)

The differential equation for the warping function becomes

(10.144)

with the boundary condition on given by

(10.145)

10.9 TORSIONAL RIGIDITY OF NONCIRCULAR SHAFTS

It is necessary to find the torsional rigidity C of the shaft in order to find the solution of the torsional vibration problem, Eq. (10.106). The torsional rigidity can be determined by solving the Laplace equation, Eq. (10.104):

(10.146)

subject to the boundary condition, Eq. (10.105):

(10.147)

which is equivalent to

(10.148)

Since the solution of Eq. (10.146), for the warping function , subject to the boundary condition of Eq. (10.147) or (10.148) is relatively more difficult, we use an alternative procedure which leads to a differential equation similar to Eq. (10.146), and a boundary condition that is much simpler in form than Eq. (10.147) or (10.148). For this we express the stresses and in terms of a function , known as the Prandtl stress function, as [3, 7]

(10.149)

The stress field corresponding to Saint‐Venant's theory, Eq. (10.95), along with Eq. (10.149), satisfies the equilibrium equations:

(10.150)

By equating the corresponding expressions of and given by Eqs. (10.95) and (10.149), we obtain

(10.151)

(10.152)

Differentiating Eq. (10.151) with respect to z and Eq. (10.152) with respect to y and subtracting the resulting equations one from the other leads to the Poisson equation:

(10.153)

where

(10.154)

is assumed to be a constant. The condition to be satisfied by the stress function on the boundary can be derived by considering a small element of the rod at the boundary as shown in Fig. 10.8. The component of shear stress along the normal direction n can be expressed as

(10.155)

since the boundary is stress‐free. In Eq. (10.155), the direction cosines are given by

(10.156)

where t denotes the tangential direction. Using Eq. (10.149), the boundary condition, Eq. (10.155), can be written as

(10.157)

The rate of change of along the tangential direction at the boundary (t) can be expressed as

(10.158)

using Eqs. (10.156) and (10.157). Equation (10.158) indicates that the stress function is a constant on the boundary of the cross‐section of the rod. Since the magnitude of this constant does not affect the stress, which contains only derivatives of , we choose, for convenience,

(10.159)

to be the boundary condition.

Next we derive a relation between the unknown angle (angle of twist per unit length) and the torque acting on the rod. For this, consider the cross‐section of the twisted rod, as shown in Fig. 10.9. The moment about the x axis of all the forces acting on a small elemental area dA located at the point is given by

(10.160)

The resulting moment can be found by integrating the expression in Eq. (10.160) over the entire area of cross‐section of the bar as

(10.161)

Each term under the integral sign in Eq. (10.161) can be integrated by parts to obtain (see Fig. 10.9):

(10.162)

since at the points and . Similarly,

(10.163)

Thus, the torque on the cross‐section is given by

(10.164)

The function satisfies the linear differential (Poisson) equation given by Eq. (10.153) and depends linearly on , so that Eq. (10.164) produces an equation of the form , where J is called the torsional constant (J is the polar moment of inertia of the cross‐section for a circular section) and C is called the torsional rigidity. Thus, Eq. (10.164) can be used to find the torsional rigidity (C).

10.10 PRANDTL'S MEMBRANE ANALOGY

Prandtl observed that the differential equation for the stress function, Eq. (10.153), is of the same form as the equation that describes the deflection of a membrane or soap film under transverse pressure (see Eq. [13.1] without the right‐hand‐side inertia term). This analogy between the torsion and membrane problems has been used in determining the torsional rigidity of rods with noncircular cross‐sections experimentally [3, 4]. An actual experiment with a soap bubble would consist of an airtight box with a hole cut on one side (Fig. 10.11). The shape of the hole is the same as the cross‐section of the rod in torsion. First, a soap film is created over the hole. Then air under pressure (p) is pumped into the box. This causes the soap film to deflect transversely as shown in Fig. 10.11. If P denotes the uniform tension in the soap film, the small transverse deflection of the soap film (w) is governed by the equation (see Eq. [13.1] without the right‐hand‐side inertia term)

(10.165)

or

(10.166)

in the hole region (cross‐section) and

(10.167)

on the boundary of the hole (cross‐section). Note that the differential equation and the boundary condition, Eqs. (10.166) and (10.167), are of precisely the same form as for the stress function , namely, Eqs. (10.153) and (10.159):

(10.168)

in the interior, and

(10.169)

on the boundary. Thus, the soap bubble represents the surface of the stress function with

(10.170)

or

(10.171)

where denotes a proportionality constant:

(10.172)

The analogous quantities in the two cases are given in Table 10.2.

Soap bubble (membrane) problem	Torsion problem
	G
p
2 (volume under bubble)

The membrane analogy provides more than an experimental technique for the solution of torsion problem. It also serves as the basis for obtaining approximate analytical solutions for rods with narrow cross‐sections and open thin‐walled cross‐sections. Table 10.3 gives the values of the maximum shear stress and the angle of twist per unit length for some commonly encountered cross‐sectional shapes of rods.

Cross‐section	Angle of twist per unit length,	Maximum shear stress,
1. Solid circular shaft
2. Thick‐walled tube
3. Thin‐walled tube
4. Solid elliptic shaft
5. Hollow elliptic tube
6. Solid square shaft
7. Solid rectangular shaft	1.0 0.141 0.208  2.0 0.229 0.246  3.0 0.263 0.267  5.0 0.291 0.292 10.0 0.312 0.312 0.333 0.333
8. Hollow rectangular shaft
9. Solid equilateral triangular shaft
10. Thin‐walled tube

10.11 RECENT CONTRIBUTIONS

The torsional vibration of tapered rods with rectangular cross‐section, pre‐twisted uniform rods, and pre‐twisted tapered rods is presented by Rao [4]. In addition, several refined theories of torsional vibration of rods are also presented in Ref. [4].

REFERENCES

1. 1 W.F. Riley, L.D. Sturges, and D.H. Morris, Mechanics of Materials, 5th ed., Wiley, New York, 1999.
2. 2 S. Timoshenko, D.H. Young, and W. Weaver, Jr., Vibration Problems in Engineering, 4th ed., Wiley, New York, 1974.
3. 3 W.B. Bickford, Advanced Mechanics of Materials, Addison‐Wesley, Reading, MA, 1998.
4. 4 J.S. Rao, Advanced Theory of Vibration, Wiley, New York, 1992.
5. 5 S.K. Clark, Dynamics of Continuous Elements, Prentice‐Hall, Englewood Cliffs, NJ, 1972.
6. 6 S.S. Rao, Mechanical Vibrations, 6th ed., Pearson Education, Inc., Hoboken, NJ, 2017.
7. 7 J.H. Faupel and F.E. Fisher, Engineering Design, 2nd ed., Wiley‐Interscience, New York, 1981.
8. 8 M. Vet, Torsional vibration of beams having rectangular cross‐sections, Journal of the Acoustical Society of America, Vol. 34, p. 1570, 1962.
9. 9 J.M. Vance, Rotordynamics of Turbomachinery, Wiley, New York, 1988.
10. 10 A.E.H. Love, A Treatise on the Mathematical Theory of Elasticity, 4th ed., Dover, New York, 1944.
11. 11 J.M. Gere, Torsional vibrations of beams of thin walled open section, Journal of Applied Mechanics, Vol. 21, p. 381, 1954.
12. 12 H. Lo and M. Goulard, Torsion with warping restraint from Hamilton's principle, Proceedings of the 2nd Midwestern Conference on Solid Mechanics, 1955, p. 68.
13. 13 A.K. Darpe, K. Gupta, and A. Chawla, Coupled bending, longitudinal and torsional vibration of a cracked rotor, Journal of Sound and Vibration, Vol. 269, No. 1–2, pp. 33–60, 2004.
14. 14 I.Y. Shen, W. Guo, and Y.C. Pao, Torsional vibration control of a shaft through active constrained layer damping treatments, Journal of Vibration and Acoustics, Vol. 119, No. 4, pp. 504–511, 1997.
15. 15 L. Ran, R. Yacamini, and K.S. Smith, Torsional vibrations in electrical induction motor drives during start up, Journal of Vibration and Acoustics, Vol. 118, No. 2, pp. 242–251, 1996.
16. 16 C.P. Filipich and M.B. Rosales, Free flexural–torsional vibrations of a uniform spinning beam, Journal of Sound and Vibration, Vol. 141, No. 3, pp. 375–387, 1990.
17. 17 S.P. Timoshenko, Theory of bending, torsion and buckling of thin‐walled member of open cross‐section, Journal of the Franklin Institute, Vol. 239, pp. 201, 249, and 343, 1945.

PROBLEMS

1. 10.1 A shaft with a uniform circular cross‐section of diameter d and length l carries a heavy disk of mass moment of inertia at the center. Find the first three natural frequencies and the corresponding modes of the shaft in torsional vibration. Assume that the shaft is fixed at both the ends.
2. 10.2 A shaft with a uniform circular cross‐section of diameter d and length l carries a heavy disk of mass moment of inertia at the center. If both ends of the shaft are fixed, determine the free vibration response of the system when the disk is given an initial angular displacement of and a zero initial angular velocity.
3. 10.3 A uniform shaft supported at and rotating at an angular velocity is suddenly stopped at the end . If the end is free and the cross‐section of the shaft is tubular with inner and outer radii and , respectively, find the subsequent time variation of the angular displacement of the shaft.
4. 10.4 A uniform shaft of length l is fixed at and free at . Find the forced vibration response of the shaft if a torque is applied at the free end. Assume the initial conditions of the shaft to be zero.
5. 10.5 Find the first three natural frequencies of torsional vibration of a shaft of length 1 m and diameter 20 mm for the following end conditions:
   1. Both ends are fixed.
   2. One end is fixed and the other end is free.
   3. Both ends are free.

   Material of the shaft: steel with and .

6. 10.6 Solve Problem 10.5 by assuming the material of the shaft to be aluminum with and .
7. 10.7 Consider two shafts each of length l with thin‐walled tubular sections, one in the form of a circle and the other in the form of a square, as shown in Fig. 10.13.

   

   Assuming the same wall thickness of the tubes and the same total area of the region occupied by the material (material area), compare the fundamental natural frequencies of the torsional vibration of the shafts. Assume the shaft to be fixed at both the ends.

8. 10.8 Solve Problem 10.7 by assuming the tube wall thickness and the interior cavity areas of the tubes to be the same.
9. 10.9 Determine the velocity of propagation of torsional waves in the drive shaft of an automobile for the following data:
   1. Cross‐section: circular with diameter 100 mm; material: steel with and .
   2. Cross‐section: hollow with inner diameter 80 mm and outer diameter 120 mm; material: aluminum with and .
10. 10.10 Find the first three natural frequencies of torsional vibration of a shaft fixed at and a disk of mass moment of inertia attached at . Shaft: uniform circular cross‐section of diameter 20 mm and length 1 m; material of shaft: steel with and .
11. 10.11 Find the fundamental natural frequency of torsional vibration of the shaft described in Problem 10.10 using a single‐degree‐of‐freedom model.
12. 10.12 Find the free torsional vibration response of a uniform shaft of length l subjected to an initial angular displacement and an initial angular velocity using modal analysis. Assume the shaft to be fixed at and free at .
13. 10.13 Find the free torsional vibration response of a uniform shaft of length l subjected to an initial angular displacement and an initial angular velocity using modal analysis. Assume the shaft to be fixed at and free at .
14. 10.14 Derive the frequency equation for the torsional vibration of a uniform shaft with a torsional spring of stiffness attached to each end.
15. 10.15 Find the steady‐state response of a shaft fixed at both ends when subjected to a torque at using modal analysis.
16. 10.16 A shaft with a uniform circular cross‐section of diameter d and length l is fixed at and carries a heavy disk of mass moment of inertia at the other end . Derive the frequency equation for the torsional vibration of the shaft.
17. 10.17 Consider two shafts each of length l with thin‐walled tubular sections, one in the form of a circle and the other in the form of a square, as shown in Fig. 10.13. Compare the angles of twist per unit length of the two shafts for unit torque.
18. 10.18 Consider two shafts each of length l with thin‐walled tubular sections, one in the form of a circle and the other in the form of a square, as shown in Fig. 10.13. What is the difference in the fundamental natural frequencies of vibration of the two shafts?
19. 10.19 A uniform circular shaft of polar area moment of inertia and polar mass moment of inertia and length l is fixed at one end and free at the other end. Find the free vibration response of the shaft.