3.6^3.6. STATICALLY INDETERMINATE SHAFTS^163^175^,,^3592^3864%
3.6
STATICALLY INDETERMINATE SHAFTS

You saw in Sec. 3.4 that, in order to determine the stresses in a shaft, it was necessary to first calculate the internal torques in the various parts of the shaft. These torques were obtained from statics by drawing the free-body diagram of the portion of shaft located on one side of a given section and writing that the sum of the torques exerted on that portion was zero.

There are situations, however, where the internal torques cannot be determined from statics alone. In fact, in such cases the external torques themselves, i.e., the torques exerted on the shaft by the supports and connections, cannot be determined from the free-body diagram of the entire shaft. The equilibrium equations must be complemented by relations involving the deformations of the shaft and obtained by considering the geometry of the problem. Because statics is not sufficient to determine the external and internal torques, the shafts are said to be statically indeterminate. The following example, as well as Sample Prob. 3.5, will show how to analyze statically indeterminate shafts.

Page 164
EXAMPLE 3.05

A circular shaft AB consists of a 10-in.-long, -diameter steel cylinder, in which a 5-in.-long, -diameter cavity has been drilled from end B. The shaft is attached to fixed supports at both ends, and a 90 lb · ft torque is applied at its midsection (Fig. 3.25). Determine the torque exerted on the shaft by each of the supports.

Fig. 3.25

Drawing the free-body diagram of the shaft and denoting by TA and TB the torques exerted by the supports (Fig. 3.26a), we obtain the equilibrium equation

Since this equation is not sufficient to determine the two unknown torques TA and TB, the shaft is statically indeterminate.

Fig. 3.26

However, TA and TB can be determined if we observe that the total angle of twist of shaft AB must be zero, since both of its ends are restrained. Denoting by ϕ1 and ϕ2, respectively, the angles of twist of portions AC and CB, we write

From the free-body diagram of a small portion of shaft including end A (Fig. 3.26b), we note that the internal torque T1 in AC is equal to TA; from the free-body diagram of a small portion of shaft including end B (Fig. 3.26c), we note that the internal torque T2 in CB is equal to TB. Recalling Eq. (3.16) and observing that portions AC and CB of the shaft are twisted in opposite senses, we write

Solving for TB, we have

Substituting the numerical data gives

we obtain

Substituting this expression into the original equilibrium equation, we write

Page 165

SAMPLE

SAMPLE PROBLEM 3.3

The horizontal shaft AD is attached to a fixed base at D and is subjected to the torques shown. A 44-mm-diameter hole has been drilled into portion CD of the shaft. Knowing that the entire shaft is made of steel for which G = 77 GPa, determine the angle of twist at end A.

SAMPLE

SAMPLE

SOLUTION

Since the shaft consists of three portions AB, BC, and CD, each of uniform cross section and each with a constant internal torque, Eq. (3.17) may be used.

Statics.

Passing a section through the shaft between A and B and using the free body shown, we find

Passing now a section between B and C, we have

Since no torque is applied at C,

Polar Moments of Inertia
Angle of Twist.

Using Eq. (3.17) and recalling that G = 77 GPa for the entire shaft, we have

SAMPLE

Page 166

SAMPLE

SAMPLE PROBLEM 3.4

Two solid steel shafts are connected by the gears shown. Knowing that for each shaft G = 11.2 × 106 psi and that the allowable shearing stress is 8 ksi, determine (a) the largest torque T0 that may be applied to end A of shaft AB, (b) the corresponding angle through which end A of shaft AB rotates.

SAMPLE

SAMPLE

SOLUTION
Statics.

Denoting by F the magnitude of the tangential force between gear teeth, we have

(1)
Kinematics.

Noting that the peripheral motions of the gears are equal, we write

(2)
a. Torque T0
Shaft AB.

With TAB = T0 and c = 0.375 in., together with a maximum permissible shearing stress of 8000 psi, we write

Shaft CD.

From (1) we have TCD = 2.8T0. With c = 0.5 in. and all = 8000 psi, we write

Maximum Permissible Torque.

We choose the smaller value obtained for T0

b. Angle of Rotation at End A.

We first compute the angle of twist for each shaft.

Shaft AB.

For TAB = T0 = 561 lb · in., we have

Shaft CD.

TCD = 2.8T0 = 2.8(561 lb · in.)

Since end D of shaft CD is fixed, we have ϕC = ϕC/D = 2.95°. Using (2), we find the angle of rotation of gear B to be

For end A of shaft AB, we have

SAMPLE

Page 167

SAMPLE

SAMPLE PROBLEM 3.5

A steel shaft and an aluminum tube are connected to a fixed support and to a rigid disk as shown in the cross section. Knowing that the initial stresses are zero, determine the maximum torque T0 that can be applied to the disk if the allowable stresses are 120 MPa in the steel shaft and 70 MPa in the aluminum tube. Use G = 77 GPa for steel and G = 27 GPa for aluminum.

SAMPLE

SAMPLE

SOLUTION
Statics.

Free Body of Disk. Denoting by T1 the torque exerted by the tube on the disk and by T2 the torque exerted by the shaft, we find

(1)
Deformations.

Since both the tube and the shaft are connected to the rigid disk, we have

(2)
Shearing Stresses.

We assume that the requirement alum ≤ 70 MPa is critical. For the aluminum tube, we have

Using Eq. (2), we compute the corresponding value T2 and then find the maximum shearing stress in the steel shaft.

We note that the allowable steel stress of 120 MPa is exceeded; our assumption was wrong. Thus the maximum torque T0 will be obtained by making steel = 120 MPa. We first determine the torque T2.

From Eq. (2), we have

Using Eq. (1), we obtain the maximum permissible torque

SAMPLE

Page 168
PROBLEMS
3.31 (a) For the solid steel shaft shown (G = 77 GPa), determine the angle of twist at A. (b) Solve part a, assuming that the steel shaft is hollow with a 30-mm-outer diameter and a 20-mm-inner diameter.
Fig. P3.31
3.32 For the aluminum shaft shown (G = 27 GPa), determine (a) the torque T that causes an angle of twist of 4°, (b) the angle of twist caused by the same torque T in a solid cylindrical shaft of the same length and cross-sectional area.
Fig. P3.32
3.33 Determine the largest allowable diameter of a 10-ft-long steel rod (G = 11.2 × 106 psi) if the rod is to be twisted through 30° without exceeding a shearing stress of 12 ksi.
3.34 While an oil well is being drilled at a depth of 6000 ft, it is observed that the top of the 8-in.-diameter steel drill pipe rotates though two complete revolutions before the drilling bit starts to rotate. Using G = 11.2 × 106 psi, determine the maximum shearing stress in the pipe caused by torsion.
3.35 The electric motor exerts a 500 N · m-torque on the aluminum shaft ABCD when it is rotating at a constant speed. Knowing that G = 27 GPa and that the torques exerted on pulleys B and C are as shown, determine the angle of twist between (a) B and C, (b) B and D.
Fig. P3.35
 
3.36
The torques shown are exerted on pulleys B, C, and D. Knowing that the entire shaft is made of aluminum (G = 27 GPa), determine the angle of twist between (a) C and B, (b) D and B.
Fig. P3.36
3.37 The aluminum rod BC (G = 26 GPa) is bonded to the brass rod AB (G = 39 GPa). Knowing that each rod is solid and has a diameter of 12 mm, determine the angle of twist (a) at B, (b) at C.
Fig. P3.37
3.38 The aluminum rod AB (G = 27 GPa) is bonded to the brass rod BD (G = 39 GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at A.
Fig. P3.38
3.39 The solid spindle AB has a diameter ds = 1.5 in. and is made of a steel with G = 11.2 × 106 psi and all = 12 ksi, while sleeve CD is made of a brass with G = 5.6 × 106 psi and all = 7 ksi. Determine the largest angle through which end A can be rotated.
Fig. P3.39 and P3.40
3.40 The solid spindle AB has a diameter ds = 1.75 in. and is made of a steel with G = 11.2 × 106 psi and all = 12 ksi, while sleeve CD is made of a brass with G = 5.6 × 106 psi and all = 7 ksi. Determine (a) the largest torque T that can be applied at A if the given allowable stresses are not to be exceeded and if the angle of twist of sleeve CD is not to exceed 0.375°, (b) the corresponding angle through which end A rotates.
 
3.41
Two shafts, each of diameter, are connected by the gears shown. Knowing that G = 11.2 × 106 psi and that the shaft at F is fixed, determine the angle through which end A rotates when a 1.2 kip · in. torque is applied at A.
Fig. P3.41
3.42 Two solid shafts are connected by gears as shown. Knowing that G = 77.2 GPa for each shaft, determine the angle through which end A rotates when TA = 1200 N · m.
Fig. P3.42
 
3.43
A coder F, used to record in digital form the rotation of shaft A, is connected to the shaft by means of the gear train shown, which consists of four gears and three solid steel shafts each of diameter d. Two of the gears have a radius r and the other two a radius nr. If the rotation of the coder F is prevented, determine in terms of T, l, G, J, and n the angle through which end A rotates.
Fig. P3.43
3.44 For the gear train described in Prob. 3.43, determine the angle through which end A rotates when T = 5 lb · in., l = 2.4 in., d = in., G = 11.2 × 106 psi, and n = 2.
3.45 The design of the gear-and-shaft system shown requires that steel shafts of the same diameter be used for both AB and CD. It is further required that max ≤ 60 MPa and that the angle ϕD through which end D of shaft CD rotates not exceed 1.5°. Knowing that G = 77 GPa, determine the required diameter of the shafts.
Fig. P3.45
 
3.46
The electric motor exerts a torque of 800 N · m on the steel shaft ABCD when it is rotating at a constant speed. Design specifications require that the diameter of the shaft be uniform from A to D and that the angle of twist between A and D not exceed 1.5°. Knowing that max ≤ 60 MPa and G = 77 GPa, determine the minimum diameter shaft that can be used.
Fig. P3.46
3.47 The design specifications of a 2-m-long solid circular transmission shaft require that the angle of twist of the shaft not exceed 3° when a torque of 9 kN · m is applied. Determine the required diameter of the shaft, knowing that the shaft is made of (a) a steel with an allowable shearing stress of 90 MPa and a modulus of rigidity of 77 GPa, (b) a bronze with an allowable shearing stress of 35 MPa and a modulus of rigidity of 42 GPa.
3.48 A hole is punched at A in a plastic sheet by applying a 600-N force P to end D of lever CD, which is rigidly attached to the solid cylindrical shaft BC. Design specifications require that the displacement of D should not exceed 15 mm from the time the punch first touches the plastic sheet to the time it actually penetrates it. Determine the required diameter of shaft BC if the shaft is made of a steel with G = 77 GPa and all = 80 MPa.
Fig. P3.48
 
3.49
The design specifications for the gear-and-shaft system shown require that the same diameter be used for both shafts and that the angle through which pulley A will rotate when subjected to a 2-kip · in. torque TA while pulley D is held fixed will not exceed 7.5°. Determine the required diameter of the shafts if both shafts are made of a steel with G = 11.2 × 106 psi and all = 12 ksi.
Fig. P3.49
3.50 Solve Prob. 3.49, assuming that both shafts are made of a brass with G = 5.6 × 106 psi and all = 8 ksi.
3.51 A torque of magnitude T = 4 kN · m is applied at end A of the composite shaft shown. Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine (a) the maximum shearing stress in the steel core, (b) the maximum shearing stress in the aluminum jacket, (c) the angle of twist at A.
Fig. P3.51 and P3.52
3.52 The composite shaft shown is to be twisted by applying a torque T at end A. Knowing that the modulus of rigidity is 77 GPa for the steel and 27 GPa for the aluminum, determine the largest angle through which end A can be rotated if the following allowable stresses are not to be exceeded: steel = 60 MPa and aluminum = 45 MPa.
 
3.53
The solid cylinders AB and BC are bonded together at B and are attached to fixed supports at A and C. Knowing that the modulus of rigidity is 3.7 × 106 psi for aluminum and 5.6 × 106 psi for brass, determine the maximum shearing stress (a) in cylinder AB, (b) in cylinder BC.
Fig. P3.53
3.54 Solve Prob. 3.53, assuming that cylinder AB is made of steel, for which G = 11.2 × 106 psi.
3.55 and  3.56Two solid steel shafts are fitted with flanges that are then connected by bolts as shown. The bolts are slightly undersized and permit a 1.5° rotation of one flange with respect to the other before the flanges begin to rotate as a single unit. Knowing that G = 11.2 × 106 psi, determine the maximum shearing stress in each shaft when a torque of T of magnitude 420 kip · ft is applied to the flange indicated.
  • 3.55

    The torque T is applied to flange B.

  • 3.56

    The torque T is applied to flange C.

Fig. P3.55 and P3.56
 
3.57
Ends A and D of the two solid steel shafts AB and CD are fixed, while ends B and C are connected to gears as shown. Knowing that a 4-kN · m torque T is applied to gear B, determine the maximum shearing stress (a) in shaft AB, (b) in shaft CD.
Fig. P3.57 and P3.58
3.58 Ends A and D of the two solid steel shafts AB and CD are fixed, while ends B and C are connected to gears as shown. Knowing that the allowable shearing stress is 50 MPa in each shaft, determine the largest torque T that can be applied to gear B.
3.59 The steel jacket CD has been attached to the 40-mm-diameter steel shaft AE by means of rigid flanges welded to the jacket and to the rod. The outer diameter of the jacket is 80 mm and its wall thickness is 4 mm. If 500 N · m-torques are applied as shown, determine the maximum shearing stress in the jacket.
Fig. P3.59
3.60 A solid shaft and a hollow shaft are made of the same material and are of the same weight and length. Denoting by n the ratio c1/c2, show that the ratio Ts/Th of the torque Ts in the solid shaft to the torque Th in the hollow shaft is (a) if the maximum shearing stress is the same in each shaft, (b) (1 − n2)/(1 + n2) if the angle of twist is the same for each shaft.
3.61 A torque T is applied as shown to a solid tapered shaft AB. Show by integration that the angle of twist at A is
Fig. P3.61
 
3.62
The mass moment of inertia of a gear is to be determined experimentally by using a torsional pendulum consisting of a 6-ft steel wire. Knowing that G = 11.2 × 106 psi, determine the diameter of the wire for which the torsional spring constant will be 4.27 lb · ft/rad.
Fig. P3.62
3.63 An annular plate of thickness t and modulus G is used to connect shaft AB of radius r1 to tube CD of radius r2. Knowing that a torque T is applied to end A of shaft AB and that end D of tube CD is fixed, (a) determine the magnitude and location of the maximum shearing stress in the annular plate, (b) show that the angle through which end B of the shaft rotates with respect to end C of the tube is
Fig. P3.63