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Unformatted text preview: A circular shaft AB consists of a 10in.long, %—in.~diameter
steel cylinder, in which a 5—in.long, gindiameter cavity has
been drilled from end B. The shaft is attached to ﬁxed sup
ports at both ends, and a 901b  ft torque is applied at its mid—
section (Fig. 3.28). Determine the torque exerted on )he shaft
by each of the supports. Fig. 3.23 Drawing the free—body diagram of the shaft and denoting
by TA and TB the torques exerted by the supports (Fig. 3.29a),
we obtain the equilibrium equation TA+TB=9Olbft Since this equation is not sufﬁcient to determine the two un
known torques T A and T B, the shaft is statically indeterminate. However, T A 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 b.y (1)1 and (#2, respectively,
the angles of twist of portions AC and CB, we write ¢=Q+Q=0 From the free—body diagram of a small portion of shaft in
cluding end A (Fig. 3.2927), we note that the internal torque T1
in AC is equal to TA; from the freebody diagram of a small
portion of shaft including end B (Fig. 3.29C), 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 DA EM
«b ca qt». 11G 12 G
Solving for TB, we have
L J
TB = Jim Substituting the numerical data
L1 = [Q = 5 in.
77(17—6in.)4 = 57.6 x 10—3 m“
7r[(17—61n.)4 — (155‘ my] = 42.6 x 1041114 I! NIH NIH J1
Jz
we obtain TB = 0.740 TA Substituting this expression into the original equilibrium equa—
tion, we write 1.740 TA = 90 lb  ft TA = 51.71b  ft TB = 38.31b  ft EXAMPLE 5—1 1 The solid steel shaft shown in Fig. 5—25a has a diameter of 20 m. If it is subjected to the two torques, determine the reactions at“ the .
fixed supports A and B. 500 NIn ‘ S t) i U T I 0 N
Equiiibrium. By inspection of the freebody diagram, Fig. 5~25b, it
is seen that the problem is statically indeterminate since there is only one available equation of equilibrium, whereas TA and T3 are
unknown. We require 2M, = 0; ~TB + 800 — 500 — TA = 0 (1) Cmnpatibiliry. Since the ends of the shaft are fixed, the angle of
twist of one end of the shaft with respect to the other must be zero.
Hence, the compatibility equation can be written as 924/3 = 0 This condition can be expressed in terms of the unknown torques by
using the load—displacement relationship, (1) = TL/J G. Here there are
three regions of the shaft where the internal torque is constant. On
the free—body diagrams in Fig. 5—25c we have shown these internal torques acting on segments of the shaft.* By the sign convention
established in Sec. 5.4, we have —TB(0.2 m) . (TA + 500 N . m)(1.5 m) + TA(0.3 m) _ 0
JG ' JG JG _ 1.8TA 7 0.2TB = —750 (2)
Solving Eqs. 1 and 2 yields
TA=—345N~m The negative sign indicates that TA acts in the opposite direction of
that shown in Fig. 5‘25b. A HS. *Altematively, using the other segments of the shaft, we can use internal loadings of
(TA — 300), (800‘ — TB), and (—TB + 300). @ 3.5 . 3.56 Two solid steel shaﬁs are ﬁtted with ﬂanges which are then connected by ﬁtted
bolts so that there is no relative rotation between the ﬂanges. Knowing that G = 77 , GPa, determine the maximum shearing stress in each shaﬁ when a 500 Nm torque is applied to ﬂange B. SOLUTION
Sha‘Ft‘ AB
T: TAB, Lu: dam c=iA= 0.015M
JAE) = 3&3: JE<0.OISY : 77.52 x104 .vaJ = 738143
‘5?” 61.3.3 _ gnxno“)(7a.52rnd’)
' 0.5 <93 . 4% = 14.102 ms 42c
Matching N‘i’w‘i’l‘ou at} 44: rﬁanf’es (p8: (PC : W
1—bit“? +ora'vc. On ‘F’pcmjcs T: TAqu'ED; goo N.M 500 = (“9.205 x103 4 H.103x103) 90 g) = 20.565xzo‘3 m4
Tm =. (10409103 )(20525 x/o") = 20 7. 27 Nm
T20 = (H. l02x103)(20$65 Ho's) = 290,.13 qu MWI'MUM shearins 54W”: in g .. 71°C _ awaken/5) _ e 4
Tu a.“  nflﬂoq — 3?.5‘7XI0 Pa. 37.; MP4. Mmimum Shealihﬁ Sires: iw CD = 12:9. : 31.47,“: Pa 31.7 MPa. '4
J2. lé‘tzﬁé. Ho"a 3.57 and 3.58 Two solid steel shaﬁs are ﬁtted with ﬂanges which are then connected
by bolts as shown. The bolts are slightly undersized and permit a l.5°rotation of one
ﬂange with respect to the other before the ﬂanges begin to rotate as a single unit.
Knowing that G = 77 GPa, determine the maximum shearing stress in each shaﬁ when
a 500 Nm torque T is applied to the ﬂange indicated. 3.57 The torque T is applied to ﬂange B. : OBLEM 3.57 SOLUTION Sh a¥+ A8  (17.529.0(78110")
‘ 0.6 LCD =O."Im) C=%_oi 20.0l9YnJ 9: Ii(OJJIB)? = mime H04 in"
Gm c 0" 144,296 to“)
720 : JD go; ~ (77% )( x Cpc _ 3
La, 0.6,  lililozx/o CI)c Cpeal‘ama. ro‘l‘wi’l'mn ‘Vod‘ ‘Fjan‘be 8 $8,: LSD '3 26‘ )3 XIX);s rag, Tara/ye +5; remole Jamaica: 71; ''(l0.QOb—ilo'SXQCJSNOz3 = 2€7J7 N°M
Torque T" +0 Cause “Ide m+a+.‘on_q9" .7: T” = 500  26, 7./7 = 232. 35 um
T” = 714' + 723
232.23 = (/0205><l03+ l‘lJoleos)?" q)": 7.576016er
TAB" : (10.205xlo’lCﬁ'.£~'..7é§” no") = Q'Iﬂs qu
Tm" = (H.loBHo‘Misus’xlo‘a) = 135. Io Nm MoochnUM/x Sl’lddli‘l'ﬂq $+P€S$ in : Tasc _§ 247.17 + 77.73 MomsL c
TAB J38 ‘ 7ﬁrszyloq  €3,8 X/O Pa, £3.8MPG. ad Maximum shearing siress in CD  72°C: USSJoMOle) 6 ‘_
f” J” t W = “75 v10 Pa HJS‘MPQ.‘ ...
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 Spring '10
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