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Unformatted text preview: l‘i’WE‘l l3 6 SDLUWNS 5—62. The two shafts are made of A36 steel. Each has a
diameter of 1 in., and they are supported by bearings at A,
B, and C, which allow free rotation. If the supportat D is
ﬁxed, determine the angle of twist of end A when the
torques are applied to the assembly as shown. Internal Torque: As shown on FBD. Angle of Twist:
_ T_L_
“55 — 2 JG
= W [—60.0(12)(30) + 20.0(12)(10)]
2 . . —0.01778 rad = 0.01778 rad 6
¢F = 2455 = 2 (0.01778) = 0.02667 rad _ TGF LGF
4W — ‘76— —40(12)(10) .
g (0.54)(11.0)(106) = —0.004445 rad = 0.004445 rad 4M = 4’1: + ¢A/F
= 0.02667 + 0.004445 = 0.03111 rad = 178" 5—70. The shafts are made of A—36 steel and each has a
diameter of 80 mm. Determine the angle of twist of gear D. Equilibrium: Referring to the freebody diagram of shaft CDE shown in Fig. (1,
2M, = 0; 10(103) — 2(103) — F(O.2) = 0 F = 40(103) N Internal Loading: Referring to the free  body diagram of gear B, Fig. b, 2M, = 0; —TAB  40(103)(0.15) = 0 ~ TAB = —6(103) Nm
Referring to the free  body diagram of gear D, Fig. c, 2M,r = 0; 1000’) — 2(103) — TCD = 0 TCD = 8(103) Nm Angle of Twist: The polar moment of inertia of the shafts are
= $0.04“) = 1.28(10'6)1rm4.We have ¢ = TAB LAB = 3 JG“ 1.28(10”6)7r(75)(109)' = —0.01194 rad = 0.01194 rad Using the gear ratio, ’3 150
= — = — = _
¢C ¢B<rc) 01194(200). 0 008952 rad Also, TCD LCD = 8(103)(06) = ..__.____.— = 0.01592 fad
4W0 16,. 1.28(10‘6)vr(75)(109) Thus,
(#0 = ¢c + d’D/C
(110 = 0.008952 + 0.01592
= 0.02487 rad = 1.42° 5—78. The A36 steel shaft has a diameter of 60 mm and is ﬁxed at its ends A and B. Ifit is subjected to the torques shown,
determine the absolute maximum shear stress in the shaft. Referring to the FBD of the shaft shown in Fig. a,
I . M N.“
LavtwaWﬂ EM, = 0; TA + TB — 506‘— 200~= 0 (1) ¢A¢ + dc? * (‘93 >9! CONWMLAT‘I E TM, 0’3 + TW + Tm 1 (5 L1)
Nﬁ"sum~Ac; ﬁajgnt ashes wktn
“ “ A L? 115%,}, 931%:14 :‘t “BOON‘M
(y'\’ ‘I MON 1 500 N.“ 09
T . 1T1:
30*“ “w W0“ 4’; $17, ‘¢: ‘95 1T; 500w“,M mwuomwt: M47565? Nwv TB — Tom) 4‘ z 325
'3, : $451143 NM
;.Tg=Tt=4maNM
To? =T,, ‘300 N»: M4,?) NMJSOO NM :35} N .M
Tu — Tk — TOOM—v‘nSJ NM Tm 4.3 N.“ 0.03“
[TM 7' C :— : N/M1
3’ £0.03M) M ’v _ =v CL—UD‘K MEL 5—79. The steel shaft is made from two segments: AC has a
diameter of 0.5 in, and CB has a diameter of 1 in. If it is
ﬁxed at its ends A and B and subjected to a torque of
determine the maximum shear stress in the shaft. Gst = 10.8(103) ksi. Equilibrium:
bTA+T,3—500=0 ~ (1) Compatibility condition: 1‘ m. ¢D/A = (I’D/B sow1t T, .
TA(5) + 731(3) = T302) g(0.254)G g (0.54)G g (0.5% 1408 TA = 192 T3 (2) Solving Eqs. (1) and (2) yields TA=601bft T3=4401bft TAG = :Q = = 29.3 ksi (max) Ans'
§(0.254) J
T_C 440(12)(0.5)
J 1’03 —— = ———' = 26.9 ksi % (0.54) ...
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This note was uploaded on 02/09/2011 for the course CIVIL ENGI 215 taught by Professor Dr.pollock during the Fall '10 term at Washington State University .
 Fall '10
 Dr.Pollock

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