Beer F.P., Johnston E.R., DeWolf J.T., Mazurek D.F. Mechanics of Materials

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187

The value obtained is referred to as the maximum elastic torque,

since it is the largest torque for which the deformation remains

fully elastic. Recalling that for a solid circular shaft

c 5

, we

have

(3.29)

As the torque is further increased, a plastic region develops

in the shaft, around an elastic core of radius r

(Fig. 3.35c). In

the plastic region the stress is uniformly equal to t

, while in the

elastic core the stress varies linearly with r and may be expressed

t 5

(3.30)

As T is increased, the plastic region expands until, at the limit, the

deformation is fully plastic (Fig. 3.35d).

Equation (3.26) will be used to determine the value of the

torque T corresponding to a given radius r

of the elastic core.

Recalling that t is given by Eq. (3.30) for 0 # r # r

, and is equal

to t

for r

# r # c, we write

T 5 2p

rb dr 1 2p

T 5

1 2

(3.31)

or, in view of Eq. (3.29),

T 5

1 2

(3.32)

where T

is the maximum elastic torque. We note that, as r

approaches zero, the torque approaches the limiting value

(3.33)

This value of the torque, which corresponds to a fully plastic defor-

mation (Fig. 3.35d), is called the plastic torque of the shaft consid-

ered. We note that Eq. (3.33) is valid only for a solid circular shaft

made of an elastoplastic material.

Since the distribution of strain across the section remains linear

after the onset of yield, Eq. (3.2) remains valid and can be used to

express the radius r

of the elastic core in terms of the angle of twist

f. If f is large enough to cause a plastic deformation, the radius r

of the elastic core is obtained by making g equal to the yield strain

in Eq. (3.2) and solving for the corresponding value r

of the

distance r. We have

(3.34)

3.10 Circular Shafts Made of an Elastoplastic

Material

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188

Torsion

Let us denote by f

the angle of twist at the onset of yield, i.e., when

5 c. Making f 5 f

and r

5 c in Eq. (3.34), we have

c 5

(3.35)

Dividing (3.34) by (3.35), member by member, we obtain the follow-

ing relation:†



(3.36)

If we carry into Eq. (3.32) the expression obtained for r

yc, we

express the torque T as a function of the angle of twist f,

T 5

a1 2

(3.37)

where T

and f

represent, respectively, the torque and the angle

of twist at the onset of yield. Note that Eq. (3.37) may be used

only for values of f larger than f

. For f , f

, the relation

between T and f is linear and given by Eq. (3.16). Combining

both equations, we obtain the plot of T against f represented in

Fig. 3.39. We check that, as f increases indefinitely, T approaches

the limiting value T

corresponding to the case of a fully

developed plastic zone (Fig. 3.35d). While the value T

cannot

actually be reached, we note from Eq. (3.37) that it is rapidly

approached as f increases. For f 5 2f

, T is within about 3% of

, and for f 5 3f

within about 1%.

Since the plot of T against f that we have obtained for an ideal-

ized elastoplastic material (Fig. 3.36) differs greatly from the shearing-

stress-strain diagram of that material (Fig. 3.34), it is clear that the

shearing-stress-strain diagram of an actual material cannot be

obtained directly from a torsion test carried out on a solid circular

rod made of that material. However, a fairly accurate diagram may

be obtained from a torsion test if the specimen used incorporates a

portion consisting of a thin circular tube.‡ Indeed, we may assume

that the shearing stress will have a constant value t in that portion.

Equation (3.1) thus reduces to

T 5 rAt

where r denotes the average radius of the tube and A its cross-

sectional area. The shearing stress is thus proportional to the

torque, and successive values of t can be easily computed from

the corresponding values of T. On the other hand, the values

of the shearing strain g may be obtained from Eq. (3.2) and from

the values of f and L measured on the tubular portion of the

specimen.

†Equation (3.36) applies to any ductile material with a well-defined yield point, since its

derivation is independent of the shape of the stress-strain diagram beyond the yield

point.

‡In order to minimize the possibility of failure by buckling, the specimen should be made

so that the length of the tubular portion is no longer than its diameter.





Fig. 3.36 Load displacement relation for

elastoplastic material.

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189

EXAMPLE 3.08

A solid circular shaft, 1.2 m long and 50 mm in diameter, is subjected to

a 4.60 kN ? m torque at each end (Fig. 3.37). Assuming the shaft to be

made of an elastoplastic material with a yield strength in shear of 150 MPa

and a modulus of rigidity of 77 GPa, determine (a) the radius of the elastic

core, (b) the angle of twist of the shaft.

(a) Radius of Elastic Core. We first determine the torque T

the onset of yield. Using Eq. (3.28) with t

5 150 MPa, c 5 25 mm, and

p125 3 10

5 614 3 10

we write

1614 3 10

21150 3 10

Pa2

25 3 10

5 3.68 kN ? m

Solving Eq. (3.32) for (r

yc)

and substituting the values of T and T

, we

have

5 4 2

314.60 kN ? m

3.68 kN ? m

5 0.250

5 0.630

25 mm

5 15.8 mm

(b) Angle of Twist. We first determine the angle of twist f

the onset of yield from Eq. (3.16):

13.68 3 10

N ? m211.2 m2

614 3 10

77 3 10

5 93.4 3 10

rad

Solving Eq. (3.36) for f and substituting the values obtained for f

and

yc, we write

f 5

93.4 3 10

rad

0.630

5 148.3 3 10

rad

f 5 1148.3 3 10

rad2

360°

2p rad

5 8.50°

1.2 m

50 mm

4.60 kN

Fig. 3.37

*3.11 RESIDUAL STRESSES IN CIRCULAR SHAFTS

In the two preceding sections, we saw that a plastic region will develop

in a shaft subjected to a large enough torque, and that the shearing

stress t at any given point in the plastic region may be obtained from

the shearing-stress-strain diagram of Fig. 3.31. If the torque is

removed, the resulting reduction of stress and strain at the point

considered will take place along a straight line (Fig. 3.38). As you will

see further in this section, the final value of the stress will not, in

general, be zero. There will be a residual stress at most points, and

that stress may be either positive or negative. We note that, as was

the case for the normal stress, the shearing stress will keep decreasing

until it has reached a value equal to its maximum value at C minus

twice the yield strength of the material.

Consider again the idealized case of the elastoplastic material

characterized by the shearing-stress-strain diagram of Fig. 3.34.







Fig. 3.38 Unloading of shaft

with nonlinear stress-strain

diagram.

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190

Torsion

Assuming that the relation between t and g at any point of the shaft

remains linear as long as the stress does not decrease by more than 2t

we can use Eq. (3.16) to obtain the angle through which the shaft

untwists as the torque decreases back to zero. As a result, the unloading

of the shaft will be represented by a straight line on the T-f diagram

(Fig. 3.39). We note that the angle of twist does not return to zero after

the torque has been removed. Indeed, the loading and unloading of the

shaft result in a permanent deformation characterized by the angle

5 f 2 f9 (3.38)

where f corresponds to the loading phase and may be obtained from

T by solving Eq. (3.38), and where f9 corresponds to the unloading

phase and may be obtained from Eq. (3.16).

The residual stresses in an elastoplastic material are obtained by

applying the principle of superposition in a manner similar to that

described in Sec. 2.20 for an axial loading. We consider, on one hand, the

stresses due to the application of the given torque T and, on the other,

the stresses due to the equal and opposite torque which is applied to

unload the shaft. The first group of stresses reflects the elastoplastic

behavior of the material during the loading phase (Fig. 3.40a), and the

second group the linear behavior of the same material during the

unloading phase (Fig. 3.40b). Adding the two groups of stresses, we

obtain the distribution of the residual stresses in the shaft (Fig. 3.40c).





Fig. 3.39 Unloading of shaft

with elastoplastic material.





(a)(b)(c)









Fig. 3.40 Stress distributions for unloading of shaft with elastoplastic material.

We note from Fig. 3.40c that some residual stresses have the same

sense as the original stresses, while others have the opposite sense. This

was to be expected since, according to Eq. (3.1), the relation

t dA

5 0 (3.39)

must be verified after the torque has been removed.

EXAMPLE 3.09

For the shaft of Example 3.08 determine (a) the permanent twist, (b) the

distribution of residual stresses, after the 4.60 kN ? m torque has been

removed.

(a) Permanent Twist. We recall from Example 3.08 that the angle

of twist corresponding to the given torque is f 5 8.508. The angle f9

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191

through which the shaft untwists as the torque is removed is obtained

from Eq. (3.16). Substituting the given data,

T 5 4

60 3 10

N ? m

L 5

1.2

5 77 3 10

and the value J 5 614 3 10

obtained in the solution of Example 3.08,

we have

f¿ 5

14.60 3 10

N ? m211.2 m2

614 3 10

77 3 10

5 116.8 3 10

rad

f¿ 5 1116.8 3 10

rad2

360°

2p rad

5 6.69°

The permanent twist is therefore

5 f 2 f¿ 5 8.50° 2 6.69° 5 1.81°

(b) Residual Stresses. We recall from Example 3.08 that the

yield strength is t

5 150 MPa and that the radius of the elastic core

corresponding to the given torque is r

5 15.8 mm. The distribution of

the stresses in the loaded shaft is thus as shown in Fig. 3.41a.

The distribution of stresses due to the opposite 4.60 kN ? m torque

required to unload the shaft is linear and as shown in Fig. 3.41b. The

maximum stress in the distribution of the reverse stresses is obtained from

Eq. (3.9):

t¿

max

14.60 3 10

N ? m2125 3 10

614 3 10

5 18

3 MP

Superposing the two distributions of stresses, we obtain the residual

stresses shown in Fig. 3.41c. We check that, even though the reverse

stresses exceed the yield strength t

, the assumption of a linear distribu-

tion of these stresses is valid, since they do not exceed 2t

150

15.8 mm

25 mm

–187.3

31.6

–37.3

–118.4

(b)

(c)

(MPa)



(MPa)



(MPa)





(a)

Fig. 3.41

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192

SOLUTION

Geometric Properties

The geometric properties of the cross section are

1.5 in.

5 0.75 in. c

2.25 in.

5 1.125 in.

p1c

2 c

p311.125 in.2

2 10.75 in.2

45 2.02 in

a. Onset of Yield. For t

max

5 t

5 21 ksi, we find

121 ksi212.02 in

1.125 in.

5 37.7 kip ? in.

◀

Making r 5 c

and g 5 g

in Eq. (3.2) and solving for f, we obtain the

value of f

121 3 10

psi2160 in.2

11.125 in.2111.2 3 10

psi2

5 0.100 rad

5 5.738

◀

b. Fully Plastic Deformation. When the plastic zone reaches the inner

surface, the stresses are uniformly distributed as shown. Using Eq. (3.26),

we write

5 2pt

dr 5

2 c

21 ksi

231

1.125 in.

0.75 in.

5 44.1 kip ? in.

◀

When yield first occurs on the inner surface, the deformation is fully plastic;

we have from Eq. (3.2):

121 3 10

psi2160 in.2

10.75 in.2111.2 3 10

psi2

5 0.150 rad

5 8.598

◀

For larger angles of twist, the torque remains constant; the T-f diagram of

the shaft is as shown.

SAMPLE PROBLEM 3.7

Shaft AB is made of a mild steel that is assumed to be elastoplastic with

G 5 11.2 3 10

psi and t

5 21 ksi. A torque T is applied and gradually

increased in magnitude. Determine the magnitude of T and the correspond-

ing angle of twist (a) when yield first occurs, (b) when the deformation has

become fully plastic.

T´

2.25 in.

1.5 in.

60 in.

(ksi)

␶

␥

37.7 kip

in.

21 ksi

5.73

1.125 in.

0.75 in.

␾

␶

44.1 kip

in.

21 ksi

8.59

␾

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193

SOLUTION

Referring to Sample Prob. 3.7, we recall that when the plastic zone first

reached the inner surface, the applied torque was T

5 44.1 kip ? in. and

the corresponding angle of twist was f

5 8.598. These values are shown in

Fig. 1.

Elastic Unloading. We unload the shaft by applying a 44.1 kip ? in.

torque in the sense shown in Fig. 2. During this unloading, the behavior of

the material is linear. Recalling from Sample Prob. 3.7 the values found for

, c

, and J, we obtain the following stresses and angle of twist:

max

44.1 kip ? in.

1.125 in.

2.02 in

5 24.56 ksi

min

5 t

max

5 124.56 ksi2

0.75 in.

1.125 in.

5 16.37 ksi

f¿ 5

144.1 3 10

psi2160 in.2

12.02 in

2111.2 3 10

psi2

5 0.1170 rad 5 6.70°

Residual Stresses and Permanent Twist. The results of the loading

(Fig. 1) and the unloading (Fig. 2) are superposed (Fig. 3) to obtain the

residual stresses and the permanent angle of twist f

SAMPLE PROBLEM 3.8

For the shaft of Sample Prob. 3.7, determine the residual stresses and the

permanent angle of twist after the torque T

5 44.1 kip ? in. has been

removed.

(1)

(2)

(3)

⫽ 44.1 kip · in.

44.1 kip · in.

16.37 ksi

6.70⬚

1.89⬚

24.56 ksi

3.56 ksi

4.63 ksi

21 ksi

8.59⬚

⫽ 44.1 kip · in.

44.1 kip · in.

␾

⫽

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PROBLEMS

194

3.92 A 30-mm diameter solid rod is made of an elastoplastic material with

5 3.5 MPa. Knowing that the elastic core of the rod is 25 mm

in diameter, determine the magnitude of the applied torque T.

3.93 The solid circular shaft shown is made of a steel that is assumed

to be elastoplastic with t

5 21 ksi. Determine the magnitude

T of the applied torques when the plastic zone is (a) 0.8 in. deep,

(b) 1.2 in. deep.

3.94 The solid circular shaft shown is made of a steel that is assumed

to be elastoplastic with t

5 145 MPa. Determine the magnitude

T of the applied torque when the plastic zone is (a) 16 mm deep,

(b) 24 mm deep.

3.95 The solid shaft shown is made of a mild steel that is assumed to

be elastoplastic with G 5 11.2 3 10

psi and t

5 21 ksi. Deter-

mine the maximum shearing stress and the radius of the elastic

core caused by the application of a torque of magnitude (a) T 5

100 kip ? in., (b) T 5 140 kip ? in.

3.96 It is observed that a straightened paper clip can be twisted through

several revolutions by the application of a torque of approximately

60 mN ? m. Knowing that the diameter of the wire in the paper

clip is 0.9 mm, determine the approximate value of the yield stress

of the steel.

3.97 The solid shaft shown is made of a mild steel that is assumed to

be elastoplastic with t

5 145 MPa. Determine the radius of the

elastic core caused by the application of a torque equal to 1.1 T

where T

is the magnitude of the torque at the onset of yield.

3.98 For the solid circular shaft of Prob. 3.95, determine the angle of

twist caused by the application of a torque of magnitude (a) T 5

80 kip ? in., (b) T 5 130 kip ? in.

3.99 The solid shaft shown is made of a mild steel that is assumed to

be elastoplastic with G 5 77.2 GPa and t

5 145 MPa. Determine

the angle of twist caused by the application of a torque of magni-

tude (a) T 5 600 N ? m, (b) T 5 1000 N ? m.

c ⫽ 1.5 in.

Fig. P3.93

3 in.

4 ft

Fig. P3.95

30 mm

1.2 m

Fig. P3.97

1.2 m

15 mm

Fig. P3.99

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195

Problems

3.100 A 3-ft-long solid shaft has a diameter of 2.5 in. and is made of a

mild steel that is assumed to be elastoplastic with t

5 21 ksi and

G 5 11.2 3 10

psi. Determine the torque required to twist the

shaft through an angle of (a) 2.58, (b) 58.

3.101 For the solid shaft of Prob. 3.99, determine (a) the magnitude of

the torque T required to twist the shaft through an angle of 158,

(b) the radius of the corresponding elastic core.

3.102 The shaft AB is made of a material that is elastoplastic with t

5 12 ksi

and G 5 4.5 3 10

psi. For the loading shown, determine (a) the

radius of the elastic core of the shaft, (b) the angle of twist at end B.

3.103 A 1.25-in.-diameter solid circular shaft is made of a material that is

assumed to be elastoplastic with t

5 18 ksi and G 5 11.2 3 10

psi.

For an 8-ft length of the shaft, determine the maximum shearing

stress and the angle of twist caused by a 7.5-kip ? in. torque.

3.104 An 18-mm-diameter solid circular shaft is made of a material that

is assumed to be elastoplastic with t

5 145 MPa and G 5 77 GPa.

For an 1.2-m length of the shaft, determine the maximum shearing

stress and the angle of twist caused by a 200 N ? m-torque.

3.105 A solid circular rod is made of a material that is assumed to be

elastoplastic. Denoting by T

and f

, respectively, the torque and

the angle of twist at the onset of yield, determine the angle of

twist if the torque is increased to (a) T 5 1.1 T

, (b) T 5 1.25 T

3.106 The hollow shaft shown is made of steel that is assumed to be elas-

toplastic with t

5 145 MPa and G 5 77.2 GPa. Determine the

magnitude T of the torque and the corresponding angle of twist

(a) at the onset of yield, (b) when the plastic zone is 10 mm deep.

6.4 ft

 2560 lb · in.

in.

Fig. P3.102

5 m

25 mm

60 mm

Fig. P3.106

2.5 in.

3 in.

5 in.

Fig. P3.108

3.107 For the shaft of Prob. 3.106, determine (a) angle of twist at which

the section first becomes fully plastic, (b) the corresponding magni-

tude T of the applied torque. Sketch the T-f curve for the shaft.

3.108 A steel rod is machined to the shape shown to form a tapered solid

shaft to which torques of magnitude T 5 75 kip ? in. are applied.

Assuming the steel to be elastoplastic with t

5 21 ksi and G 5 11.2

3 10

psi, determine (a) the radius of the elastic core in portion AB

of the shaft, (b) the length of portion CD that remains fully elastic.

3.109 If the torques applied to the tapered shaft of Prob. 3.108 are slowly

increased, determine (a) the magnitude T of the largest torques

that can be applied to the shaft, (b) the length of the portion CD

that remains fully elastic.

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196

Torsion

3.110 A hollow shaft of outer and inner diameters respectively equal to

0.6 in. and 0.2 in. is fabricated from an aluminum alloy for which the

stress-strain diagram is given in the diagram shown. Determine the

torque required to twist a 9-in. length of the shaft through 108.

3.111 Using the stress-strain diagram shown, determine (a) the torque

that causes a maximum shearing stress of 15 ksi in a 0.8-in.-diameter

solid rod, (b) the corresponding angle of twist in a 20-in. length of

the rod.

3.112 A 50-mm-diameter cylinder is made of a brass for which the stress-

strain diagram is as shown. Knowing that the angle of twist is 58 in

a 725-mm length, determine by approximate means the magnitude

T of torque applied to the shaft.

3.113 Three points on the nonlinear stress-strain diagram used in Prob.

3.112 are (0, 0), (0.0015, 55 MPa), and (0.003, 80MPa). By fitting

the polynomial T 5 A 1 Bg 1 Cg

through these points, the fol-

lowing approximate relation has been obtained.

T 5 46.7 3 10

g 2 6.67 3 10

Solve Prob. 3.112 using this relation, Eq. (3.2), and Eq. (3.26).

3.114 The solid circular drill rod AB is made of a steel that is assumed to

be elastoplastic with t

5 22 ksi and G 5 11.2 3 10

psi. Knowing

that a torque T 5 75 kip ? in. is applied to the rod and then removed,

determine the maximum residual shearing stress in the rod.

0.002 0.004 0.006 0.008 0.010

 (ksi)



Fig. P3.110 and P3.111

100

0.001 0.002 0.003

 (MPa)



725 mm

⫽

50 mm

Fig. P3.112

3.115 In Prob. 3.114, determine the permanent angle of twist of the rod.

3.116 The solid shaft shown is made of a steel that is assumed to be elas-

toplastic with t

5 145 MPa and G 5 77.2 GPa. The torque is

increased in magnitude until the shaft has been twisted through 68;

the torque is then removed. Determine (a) the magnitude and loca-

tion of the maximum residual shearing stress, (b) the permanent

angle of twist.

1.2 in.

35 ft

Fig. P3.114

16 mm

0.6 m

Fig. P3.116

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