C!.

~

ar

}c

i :

lai

· ~~ "-

The solid shaft of radius c is subjected to a torque T, Fig. 5-10a.

Determine the fraction of T that is resisted by the material contained

.ithin the outer core of the shaft, which has an inner radius of c/2

.md outer radius c.

---~

:e>LUTION "~

~ e stress in the shaft varies linearly, such that T = (p/ c)T max' Eq. 5-3.

-=t:lerefore, the torque dT' on the ring (area) located within the outer

:me, Fig. 5-106, is

dT' = p(TdA) = p(pjc)Tmax(27Tpdp)

~or the entire outer core area the torque is

c

1 ~ 27TTmaxJ p3dp T ~ c c/2

27TT max .!_ p4 1 c

= C 4 c/2

- that

T' ~ 157T TITmaxC3 (1)

This torque T' can be expressed in terms of the applied torque T

_- first using the torsion formula to determine the maximum stress in

...:.e shaft. We have

Tc Tc ~ ~ = - 4

T max ~ 1 (7T /2)c

2T T max 7TC3

- bstituting this into Eq. 1 yields

15 T' =-T

16 Ans.

OTE: Here, approximately 94% of the torque is resisted by the outer

:me, and the remaining 6% (or fr,) of Tis resisted by the inner core of

_e shaft, p = 0 to p = c /2. As a result, the material located at the

rer core of the shaft is highly effective in resisting torque, which

~ Lifi es the use of tubular shafts as an efficient means for transmitting

""qUe, and thereby saving material.

5.2 THE TORSION FORMULA 1 8 9

(a)

(b)

Fig. 5-10

~

190 C HAPTER 5 TORSION

EXAMPLE

42.5 kip·in.

(a)

* 18.9ksi

(c)

Fig. 5- 11

The shaft shown in Fig. 5-1la is supported by two bearings and is

subjected to three torques. Determine the shear stress developed at

points A and B, located at section a-a of the shaft, Fig. 5-1lc.

42.5 kip·in.

(b)

SOLUTION

Internal Torque. Since the bearing reactions do not offer

resistance to shaft rotation, the applied torques satisfy momenr

equilibrium about the shaft's axis.

The internal torque at section a-a will be determined from the

free-body diagram of the left segment, Fig. 5-1lb. We have

'i,Mx = 0; 42.5 kip· in. - 30 kip· in. - T = 0 T = 12.5 kip· in.

Section Property. The polar moment of inertia for the shaft is

7r J = - (0.75 in . )4 = 0.497 in.4

2

Shear Stress. Since point A is at p = c = 0.75 in. ,

Tc

J

( 12.5 kip· in.) ( 0.75 in.) -=-------"'----------'----'-c:----'-- = 18.9 ksi

( 0.497 in.4)

Likewise for point B , at p = 0.15 in., we hllve

Tp ( 12.5 kip· in.) (0.15 in.) T B = - = -'---=----:____:__,--------'--- = 3. 77 ksi

J (0.497in.4 )

Ans.

Ans.

NOTE: The directions of these stresses on each element at A and B .

Fig. 5-11c, are established from the direction of the resultant internal

torque T , shown in Fig. 5-1lb. Note carefully how the shear stress act

on the planes of each of these elements.

-::e pipe shown in Fig. 5-12a has an inner diameter of 80 mm and an

_.er diameter of 100 mm. If its end is tightened against the support at

.. using a torque wrench at B, determine the shear stress developed in

....:e material at the inner and outer walls along the central portion of

· -e pipe when the 80-N forces_are applied to the wrench. "- ~4.

:::>LUTION '<~

:ternal Torque. A section is taken at an intermediate location C

_{mg the pipe's axis, Fig. 5-12b. The only unknown at the section is

· e internal torque T. We require

'i.My = 0; SON (0.3m ) +SON (0.2 m)- T = 0

T = 40N·m

Section Property. The polar moment of inertia for the pipe's

:.ass-sectional area is

1= ~ [(0.05m) 4 -(0.04m) 4] = 5.796(10- 6 )m4

Shear Stress. For any point lying on the outside surface of the pipe,

o = c0

= 0.05 m, we have

Teo 40 N. m(0.05 m) = 0.345 MPa Ans. 70 =]= 5.796(10- 6 )m4

--\.nd for any point located on the inside surface, p = c; = 0.04 m,

•' 0 that

Tc; 40N·m(0.04m) = 0.276MPa Ans. T; = J = 5.796 ( 10-6 )m4

NOTE: To show how these stresses act at representative points D

and E on the cross-section, we will first view the cross section

from the front of segment CA of the pipe, Fig. 5-12a. On this

section, Fig. 5-12c, the resultant internal torque is equal but

opposite to that shown in Fig. 5-12b. The shear stresses at D and

E contribute to this torque and therefore act on the shaded faces

of the elements in the directions shown. As a consequence, notice

how the shear-stress components act on the other three faces.

Furthermore, since the top face of D and the inner face of E are

in stress-free regions taken from the pipe's outer and inner walls, no

shear stress can exist on these faces or on the other corresponding

faces of the elements.

5.2 T HE TORSION FORMULA 1 91

SON z

kY/y

~

~ · .·. ,..

(b) ""

X

D

E

(c)

Fig. 5-12

-_.. -_

ft;-

IE

·-; '-'-'~

ec:

nc..

!t~

12

ilue

ha

the

, or

solid steel shaft AB, shown in Fig. 5- 13, is to be used to transmit

: hp from the motor M to which it is attached. If the shaft rotates

:.~ w = 175 rpm and the steel has an allowable shear stress of

- ~o w = 14.5 ksi, determine the required diameter of the shaft to the 1.

=earest 8 m.

Fig. 5-13

SOLUTION

-:be torque on the shaft is determined from Eq. 5-10, that is, P = Tw.

:=:xpressing P in foot-pounds per second and w in radians/second,

·e have

Thus,

o = Tw;

(550ft·lb/s ) = 2750ft·lb/s

p = 5 hp 1 hp

w-175 rev

( 27T rad ) ( 1 min ) = 18.33 rad/ s

1 rev 60s mm

2750ft·lb/s = T(18.33rad/s )

T = 150.1 ft·lb

Applying Eq. 5-12 yields

J 7T c4 T -- ==-

c 2 C Tallow

c = (____3!_)113

= (2 ( 150.1 ft ·lb) ( 12 in.jft) )1/3

7TTallow 7T(14500lb/in2)

c = 0.429 in.

- ince 2c = 0.858 in., select a shaft having a diameter of

7 d = 8 in. = 0.875 in. Ans.

5.3 POWER TRANSMISSION 1 9 3

"

---=-=c

- ..,~ a< each section and

show the torsional ::-- -- - ;=e-e;::. -al Yolume elements

located at A. B. C.,_ ~ D_

300 ·m

\

(a)

PS-1

PS-2. Determine the internal torque at each section and

show the torsional stress on differential volume elements

located at A, B, C, and D.

(b)

PS-2

PS-3. The solid and hollow shafts are each subjecte:

the torque T. In each case, sketch the shear s:-c-_

distribution along the two radial lines.

---+

PS-3

PS-4. The motor delivers 10 hp to the shaft. If it rot at es~

1200 rpm, detemine the torque produced by the ~otor .

PS-4

s-1. The solid circular shaft is subjected to an internal

;-que ofT= 5 kN · m. Determine the shear stress developed

points A and B. Represent each state of stress on a

:ume element. .__;4

~

T

FS-1

~1. 'Th.e hollow circular shaft is subjected to an internal

:;.ue of T = 10 kN · m. Determine the shear stress

eloped at points A and B. Represent each state of stress

,... :1. volume element.

T= lOkN·m

60mm

FS-2-

5.3 POWER TRANSM ISSION 195

FS-3. The shaft is hollow from A to B and solid

from B to C. Determine the maximum shear stress developed

in the shaft. The shaft has an outer diameter of 80 mm, and

the thickness of the wall of the hollow segment is 10 mm.

· ·~ A ~ ~'PJ. 'f .~ J 4kN · m ~~ '

2kN·m

FS-3

FS-4. Determine the maximum shear stress developed in ~

the 40-mm-diameter shaft.

6kN

FS-4

I 196 CHAP TER 5 TORSION

F5-5. Determine the maximum shear stress developed in the shaft at section a-a.

a

600N·m

~m·Omm 0 4omW 1sooN·mc

Section a-a

600 N·m

F5-5

F5-6. Determine the shear stress developed at point A on

the surface of the shaft. Represent the state of stress on a

volume element at this point. The shaft has a radius of 40 mm.

F5-7. The solid 50-mm-diameter shaft is used to tra.L..S;:

the torques applied to the gears. Determine the abso!

maximum shear stress in the shaft.

F5-7

F5-8. The gear motor can develop 3 hp when it turn _

150 rev/ min. If the allowable shear stress for the shaft

Tal low = 12 ksi, determine the smallest diameter of the s ~

to the nearest kin. that can be used.

PROBLEMS

'--L The solid shaft of radius r is subjected to a torque T.

:::ermine the radius r' of the inner core of the shaft

: resists one-half of the applied torque (T/ 2). Solve

.:: problem two ways: (a) by using the torsion formula,

- ·- v finding the resultant of the shear-stress 'flistribution. . "<i)

~ -.:. The solid shaft of radius r is subjected to a torque T .

. :::ermine the radius r' of the inner core of the shaft

;;.::.: resists one-quarter of the applied torque (T/4). Solve

-:: problem two ways: (a) by using the torsion formula,

y finding the resultant of the shear-stress distribution.

r

T

Probs. 5-112

.3--3. The solid shaft is fixed to the support at C and

-ected to the torsional loadings shown. Determine the

- ~ar stress at points A and B and sketch the shear stress on

ume elements located at these points.

c

75mm

Prob.5-3

5.3 POWER TRANSMISSION 197

*5-4. The copper pipe has an outer diameter of 40 mm

and an inner diameter of 37 mm. If it is tightly secured to

the wall at A and three torques are applied to it as shown,

determine the absolute maximum shear stress developed

in the pipe.

( ~ » '---30N ·m

~ - 20N·m

SON·m

Prob. 5-4

5-5. The copper pipe has an outer diameter of 2.50 in.

and an inner diameter of 2.30 in. If it is tightly secured to •

the wall at C and three torques are applied to it as shown, ~ determine the shear stress developed at points A and B .

These points lie on the pipe's outer surface. Sketch the

shear stress on volume elements located at A and B.

B"" s~ (A/N )~>? ~

0 « '--- 450 lb·ft

/' :X ~ ~A ~ '-' 350 lb·ft

600 lb·ft

Prob. 5-5

198 CHAPTER 5 TORS ION

5-<i. The solid shaft has a diameter of 0.75 in. If it is subjected

to the torques shown, determine the maximum shear stress

developed in regions BC and DE of the shaft. The bearings at

A and Fallow free rotation of the shaft.

5-7. The solid shaft has a diameter of 0.75 in. If it is

subjected to the torques shown, determine the maximum

shear stress developed in regions CD and EF of the shaft.

The bearings at A and Fallow free rotation of the shaft.

Probs. 5-6/7

*5-8. The solid 30-mm-diameter shaft is used to transmit

the torques applied to the gears. Determine the absolute

maximum shear stress on the shaft.

300 N·m

Prob.5-8

5-9. The solid shaft is fixed to the support at C ~

subjected to the torsional loadings shown. Determine ·­

shear stress at points A and Bon the surface, and sketch -

shear stress on volume elements located at these points.

c

35~ · A ·· . ··· .

~~ ~ . ~.:~v ;o mm

35mm ~ (~ ~ J ~~V - 3oON·m

800 N·m

Prob.5-9

5-10. The coupling is used to connect the two s h ~

together. Assuming that the shear stress in the bolL

uniform, determine the number of bolts necessary to m

the maximum shear stress in the shaft equal to the she-­

stress in the bolts. Each bolt has a diameter d.

r

1 Prob. 5-10

rt at C .=.::;:

~termin.::...;;

td sketd: -­

;e poin

~-o shaf..,

~ bolts !3

r to mak::

the shec:-

The assembly consists of two sections of galvanized

cipe connected together using a reducing coupling at B.

.aller pipe has an outer diameter of 0.75 in. and an inner

ter of 0.68 in., whereas the larger pipe has an outer

ter of 1 in. and an inner diameter of 0.86 in. If the pipe is

_.-secured to the wall at C, determine the maximum shear

- developed in each section of the pipe when the couple

is applied to the handles of the wrench.

~~

151b

Prob.5-11

~ U. The 150-mm-diameter shaft is supported by a

th journal bearing at E and a smooth thrust bearing

F. Determine the maximum shear stress developed in .:.::..:h segment of the shaft.

5--13. If the tubular shaft is made from material having an

owable shear stress of Tallow = 85 MPa, determine the

-.=.:juired minimum wall thickess of the shaft to the nearest

,...,1limeter. Th ~ shaft has an outer diameter of 150 mm.

30kN·m

Probs. 5-12/13

5.3 POWER TRANSM ISSION 199

5-14. A steel tube having an outer diameter of 2.5 in. is

used to transmit 9 hp when turning at 27 rev /min. Determine

the inner diameter d of the tube to the nearest k in. if the

allowable shear stress is r allow= 10 ksi.

Prob.5-14

5-15. The solid shaft is made of material that has an

allowable shear stress of r allow = 10 MPa. Determine the

required diameter of the shaft to the nearest millimeter.

*5-16. The solid shaft has a diameter of 40 rum. Determine

the absolute maximum shear stress in the shaft and sketch

the shear-stress distribution along a radial line of the shaft

where the shear stress is maximum.

70N·m

Probs. 5-15/16