ANGLE OF TWIST IN THE ELASTIC RANGE 159

153SAMPLE PROBLEM 3.2

3.5 ANGLE OF TWIST IN THE ELASTIC RANGE 159

In this section, a relation will be derived between the angle of twist f of a circular shaft and the torque T exerted on the shaft. The entire shaft will be assumed to remain elastic. Considering first the case of a shaft of length L and of uniform cross section of radius c subjected to a torque T at its free end (Fig. 3.20), we recall from Sec. 3.3 that the angle of twist f and the maximum shearing strain g_max are related as follows:

g_max5 cf

L (3.3)

But, in the elastic range, the yield stress is not exceeded anywhere in the shaft, Hooke’s law applies, and we have g_max 5 t_maxyG or, recalling Eq. (3.9),

g_max5 t_max G 5 Tc

JG (3.15)

Equating the right-hand members of Eqs. (3.3) and (3.15), and solv- ing for f, we write

f5 TL

JG (3.16)

where f is expressed in radians. The relation obtained shows that, within the elastic range, the angle of twist f is proportional to the torque T applied to the shaft. This is in accordance with the experi- mental evidence cited at the beginning of Sec. 3.3.

Equation (3.16) provides us with a convenient method for determining the modulus of rigidity of a given material. A specimen of the material, in the form of a cylindrical rod of known diameter and length, is placed in a torsion testing machine (Photo 3.3). Torques of increasing magnitude T are applied to the specimen, and the corresponding values of the angle of twist f in a length L of the specimen are recorded. As long as the yield stress of the material is not exceeded, the points obtained by plotting f against T will fall on a straight line. The slope of this line represents the quantity JGyL, from which the modulus of rigidity G may be computed.

3.5 Angle of Twist in the Elastic Range

Photo 3.3 Torsion testing machine.

L

c T max

Fig. 3.20 Angle of twist f. bee80288_ch03_140-219.indd Page 159 9/21/10 3:05:30 PM user-f499

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160

EXAMPLE 3.02 What torque should be applied to the end of the shaft of Example 3.01 to produce a twist of 28? Use the value G 5 77 GPa for the modulus of rigidity of steel.

Solving Eq. (3.16) for T, we write T5JG

Lf Substituting the given values

G577310⁹ Pa

        

^{L51.5 m}

f52°a2p rad

360° b534.9310²³ rad

and recalling from Example 3.01 that, for the given cross section, J 5 1.021 3 10²⁶ m⁴

we have T5JG

L f511.021310²⁶ m⁴2 177310⁹ Pa2

1.5 m 134.9310²³ rad2 T51.829310³ N?m51.829 kN?m

EXAMPLE 3.03 What angle of twist will create a shearing stress of 70 MPa on the inner surface of the hollow steel shaft of Examples 3.01 and 3.02?

The method of attack for solving this problem that first comes to mind is to use Eq. (3.10) to find the torque T corresponding to the given value of t, and Eq. (3.16) to determine the angle of twist f corresponding to the value of T just found.

A more direct solution, however, may be used. From Hooke’s law, we first compute the shearing strain on the inner surface of the shaft:

g_min5 t_min

G 5 70310⁶ Pa

77310⁹ Pa5909310²⁶

Recalling Eq. (3.2), which was obtained by expressing the length of arc AA9 in Fig. 3.13c in terms of both g and f, we have

f5Lg_min c1

5 1500 mm

20 mm 1909310²⁶2568.2310²³ rad To obtain the angle of twist in degrees, we write

f5168.2310²³ rad2a 360°

2p radb53.91°

Formula (3.16) for the angle of twist can be used only if the shaft is homogeneous (constant G), has a uniform cross section, and is loaded only at its ends. If the shaft is subjected to torques at loca- tions other than its ends, or if it consists of several portions with various cross sections and possibly of different materials, we must divide it into component parts that satisfy individually the required

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161

conditions for the application of formula (3.16). In the case of the shaft AB shown in Fig. 3.21, for example, four different parts should be considered: AC, CD, DE, and EB. The total angle of twist of the shaft, i.e., the angle through which end A rotates with respect to end B, is obtained by adding algebraically the angles of twist of each component part. Denoting, respectively, by Ti, Li, Ji, and Gi the inter- nal torque, length, cross-sectional polar moment of inertia, and mod- ulus of rigidity corresponding to part i, the total angle of twist of the shaft is expressed as

f5 a_i Ti Li

Ji Gi (3.17)

The internal torque Ti in any given part of the shaft is obtained by passing a section through that part and drawing the free-body dia- gram of the portion of shaft located on one side of the section. This procedure, which has already been explained in Sec. 3.4 and illus- trated in Fig. 3.16, is applied in Sample Prob. 3.3.

In the case of a shaft with a variable circular cross section, as shown in Fig. 3.22, formula (3.16) may be applied to a disk of thick- ness dx. The angle by which one face of the disk rotates with respect to the other is thus

df 5 T dx JG

3.5 Angle of Twist in the Elastic Range TC

TD

T_A TB

A C

B

E D

Fig. 3.21 Multiple sections and multiple torques.

x

A dx

B

L T'

T

Fig. 3.22 Shaft with variable cross section.

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162

^Torsion where J is a function of x, which may be determined. Integrating in x from 0 to L, we obtain the total angle of twist of the shaft:

f5

#

^L

0

T dx

JG (3.18)

The shaft shown in Fig. 3.20, which was used to derive formula (3.16), and the shaft of Fig. 3.15, which was discussed in Examples 3.02 and 3.03, both had one end attached to a fixed support. In each case, therefore, the angle of twist f of the shaft was equal to the angle of rotation of its free end. When both ends of a shaft rotate, however, the angle of twist of the shaft is equal to the angle through which one end of the shaft rotates with respect to the other. Con- sider, for instance, the assembly shown in Fig. 3.23a, consisting of two elastic shafts AD and BE, each of length L, radius c, and modu- lus of rigidity G, which are attached to gears meshed at C. If a torque T is applied at E (Fig. 3.23b), both shafts will be twisted.

Since the end D of shaft AD is fixed, the angle of twist of AD is measured by the angle of rotation f_A of end A. On the other hand, since both ends of shaft BE rotate, the angle of twist of BE is equal to the difference between the angles of rotation f_B and f_E, i.e., the angle of twist is equal to the angle through which end E rotates with respect to end B. Denoting this relative angle of rotation by f_EyB, we write

f_EyB5f_E2f_B5 TL JG

(a) C

B L

rB

A ^rA

E Fixed support D

(b)

C'' T

E

B

C Fixed end

B L

A D

A

C'

E

Fig. 3.23 Gear assembly.

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EXAMPLE 3.04 For the assembly of Fig. 3.23, knowing that r_A 5 2r_B, determine the angle

of rotation of end E of shaft BE when the torque T is applied at E.

We first determine the torque T_AD exerted on shaft AD. Observing that equal and opposite forces F and F9 are applied on the two gears at C (Fig. 3.24), and recalling that r_A 5 2r_B, we conclude that the torque exerted on shaft AD is twice as large as the torque exerted on shaft BE;

thus, T_AD 5 2T.

Since the end D of shaft AD is fixed, the angle of rotation f_A of gear A is equal to the angle of twist of the shaft and is obtained by writing

f_A5TADL JG 52TL

JG

Observing that the arcs CC9 and CC0 in Fig. 3.26b must be equal, we write rAf_A 5 rBf_B and obtain

f_B51r_Ayr_B2f_A5 2f_A We have, therefore,

f_B52f_A5 4TL JG

Considering now shaft BE, we recall that the angle of twist of the shaft is equal to the angle f_EyB through which end E rotates with respect to end B. We have

f_EyB5T_BEL JG 5 TL

JG The angle of rotation of end E is obtained by writing

f_E5f_B1f_EyB 5 4TL

JG 1TL