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hw14_3.35_3.51(2)

# hw14_3.35_3.51(2) - wuwsmmm Problem 3.35 3.35 The electric...

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Unformatted text preview: wuwsmmm Problem 3.35 3.35 The electric motor exerts a 500 N - m torque on the alurm'num shaﬁ ABCD when it is rotating at a constant speed. Knowmg that G =.27 GPa and that the torques exerted on pulleys B and C are as shown, determme the angle of mm between (a) B and C, (b) B and D- 200 N-u. (CL) Angle ﬁ-twfst between 8233! . T8c=tzoo New) Lu: LZM Créd = 0-0mm} G— = 27X/Oa Po. 300N-m a; l£c"= strenuo" m -l- - (mum " _3 , .. CPB/c GJ' v-WWT)‘ ”24.57on mm! £98k; 1.38% «a 1% Us) An ﬂe o‘F +VES+ he'l‘weew B “J 13.. H‘ Ta, :500 N-mJ LCD-=03,“ C=%J=0.024mj G=27¥/0"Pq J; 316" = %Z(0-02+l"= 52.1.153Ylo'7 w,” ' C? = (500M033 ' ‘ (PB/p : @8/4" Goa/D = 29-357x/0—34- 3L7SO>¢[O‘% :: 56‘137y/0‘3W‘J _ Cpapa: 1 3.220 Q : 3]-?8OXIO'3 my .800N~m Problem 3.36 13.36 The torques shown are exerted on pulleys A, B, and C. Knowing that both shafts are solid and made of brass (G = 39 GPa), determine the angle of twist v _ 40 mm between (a) A and B, (b) A and C. 1200 N ‘ m (as A 19. °¥+w.-'s1‘ 5611mm ml 13 - 400mm / TAB 2 [+00 Him.) LAB: L2 M . . I . . C =ilfo": O;C>!Sm G: 373/0731 «35%[cqr 79.52 we“ m 63 (3? >‘t '0“ l (79. 52x 10“?) t _ = 0.154772 W! 9 90M; _— 5127" f) 4: (b5 Angie OP, \$013+ Between A 4AA) C , T3: = 800 N-~m J Lu: [.8 m) <:- 4:! =o.~ozo "H G = sex/0‘91 T “I - JM=§6 = \$101020) r 25:.327uo" m” _ (gowns) ‘ -R z ‘98]; (39¥107)(25l.327xlo'7) 0. mean) 2 (pm- TL - (L100)(I.Zl —___________ Fromm-"151 lﬂmnlﬂcﬂhdasﬂmxmbmdadhﬁhrulndmﬂchdm mmdgmlﬁ thhm' ofn‘pdm" iﬂJxlﬂ‘paiﬁzt 3: pi _ damn-mm mm] hwlinduﬂmilicyﬂuulc mm EM); =12.5—TA —TC :0 TA 12.5 Kip.in TAB = TA U as TA Tc THC = 7;, —12.5 12.5 Kip.in TB C ( 2) Geometric compatibility: A¢AC = 0 = A¢AB + A¢BC TL (3)A¢ = a Combining (2) and (3) we get, = TAx12 + (TA —12.5)x(18) 3.7x106 xzx1.54 /32 5.6x106x7rx2.04 /32 TA = 2.9841Kip.in = TAB, TBC = 2.9841 —12.5 = —9.516Ktp.in ”Meow “1.54/32 “2.04/32— ...
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