Torsion Examples

# Torsion Examples - A circular shaft AB consists of a...

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Unformatted text preview: A circular shaft AB consists of a 10-in.-long, %—in.~diameter steel cylinder, in which a 5—in.-long, gin-diameter 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=9Olb-ft 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 free-body 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 N-In ‘ S t) i U T I 0 N Equiiibrium. By inspection of the free-body 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 N-m 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 N-m 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(7-8110") ‘ 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'SXQCJSNO-z3 = 2€7J7 N°M Torque T" +0 Cause “Ide m+a+.‘on_q9" .7: T” = 500 - 26, 7./7 = 232. 35 u-m T” = 714' + 723 232.23 = (/0-205><l03+ l‘lJoleos)?" q)": 7.576016er TAB" : (10.205xlo’lCﬁ'.£~'..-7é§” no") = Q'Iﬂs qu Tm" = (H.loBHo‘Misus’xlo‘a) = 135. Io N-m MoochnUM/x Sl’lddli‘l'ﬂq \$+P€S\$ in : Tasc _§ 247.17 + 77.73 Moms-L 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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Torsion Examples - A circular shaft AB consists of a...

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