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Unformatted text preview: r ' T t M2 '— SoLu'neNS 5—50. The hydrofoil boat has an A36 steel propeller
shaft that is 100 ft long. It is connected to an inline diesel
engine that delivers a maximum power of 2500 hp and
causes the shaft to rotate at 1700 rpm. If the Outer
diameter of the shaft is 8 in. and the wall thickness is g in.,
determine the maximum shear stress developed in the
shaft. Also, what is the “wind up,” or angle of twist in the shaft at full power? Internal Torque: rev (217 rad) 1 mm = 56.677de/S = 17 '—.—
a) 00 mm rev 60 s 550 ft  lb/s P=2500hp( lhp ) = 1' 375 000 ftlb/s ng 1375000
to = 7723.7 Ib'ft Maximum Shear Stress: Applying torsion Formula. Tc
Tmax=7 7723.7(12)(4) = ————— = 2.83 k  . Ans.
g (44 — 3.6254) . 5‘ , Angle of Twist: ¢ = IE = W
JG ’r'(4“ — 3.6254)11.0(106) = 0.07725 rad = 4.43° Ans. ~5—53. The 20mmdiameter A36 steel shaft is subjected to
the torques shown. Determine the angle of twist of the end B. Internal Torque: As shown on FBD. Angle of Mist:
TL 4’3 = 75 = [800(0.8) + (—60.0)(o.6) + (—90.0)(0.2)] 80Nm = —0.1002 rad = 5.74° Am I : ‘BONW
7§c 12D? ‘0» in
T54: 90N' m
gourd aaﬂ'm ‘ ,
39".»? ' V « 80M *5—72. The 80min diameter shaft is made of.6osiT6
aluminum alloy and subjected to the torsional loading
shown. Determine the angle of twist at end A. Equilibrium: Referring to the free  body diagram of segment AB shown in Fig. a, 2M, = 0; —TAB — 2003) = 0 TAB = 2(103)Nm And the free  body diagram of segment BC, Fig. b,
EMx = 0; —TBC — 10(103)x — 2(103) = 0 » THC = —[10(103)x + 2(103)]Nm Angle of Twist: The polar moment of inertia of the shaft, is = 125(0042) = 1.28(10‘6)1r m4.We have TiLi TABLAB [an TBC dx
4“ 1.0. JG. 0 16.1 = 2(103)(0.6) + /°~6m—[10(103)x +2(103)]dx
1.28(10'5)7r(26)(109) o 1.28(10‘6)1r(26)(109) 0.6m
0 _ 1 3 3
128(10"")w(26)(1o9) {1200 + [500 >19 + 2(10 )x] 40.04017 rad = 230° Torsional Members Angle of Twist I The ﬁgure below shows four members all constructed of the same. material. The modulus of rigidity {
(G) is the same for all members. A torque of the same magnitude and direction is applied to each
member. Locations shown are at the midway point of the member. Rank these situations, from greatest to least, on the basis of the angle of twist ((p) at each point
relative to the left end of the member. \ r=1" r=1" Greatest 1 C/ 2 A _3 E 4 Least , Or, the angle of twist ((p) is the same for every beam. Please carefully explain your reasoning. A} ¢:':’\:— , ﬂ? w 4
TC» y a(\)(, VG! W ¢ : r, — ‘3 ‘6 3G E(1Y‘Q— '35 1G 2.
TL T U}
C, ¢o : :' I L —. E
N 36 UYG 0'1 «(A
[a 9 '56 v \T 4G 1 E, :6
i (ll How Sure are you of your ranking? (Circle one) Basically Guessed Sure Very Sure
). 1 2 3 4 5 6 7 , 8 9 10 ...
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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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