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 contain ed .ithin the outer core of the shaft, which has an inner radius of c/2 .md outer radius c. :e>LUTION 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) 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 - bs tituting 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 re r core of the shaft is highly effective in resisting torque, which the use of tubular shafts as an efficient means for transmitting ""qUe, and thereby saving material. 5.2 THE TORS ION FORMULA 1 8 9 (a) (b) Fig. 5-10
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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. ,