Unformatted text preview: T“?! A—.\I 3.33 The ship at A has just started to drill for oil on the ocean ﬂoor at a depth of
PROBLEM 3‘33 5000 R. Knowing that the top of the 8in.diameter steel drill pipe (G 2 ”.2 x 10°psi) rotates through two complete revolutions before the drill bit at B starts to
operate, determine the maximum shearing stress caused in the pipe by torsion. SOLUTION WEIr TAP ._.I_§._.GJr:_Gc
7" J TLL*—EL g) = 2m mum .~ muses mt, c. = +9.1 . 40;"
L z. 5000+} = $0000 .3‘ 2/: (li.2¥lO‘2(l?.5‘6é§(ll.o) £0000
: 9,332.6)![03 FSII = 7.38 RSI. PROBLEM 3.34 334 Determine the largest allowable diameter of a 3mlong steel rod (G = 77 GPa)
if the rod is to be twisted through 30° without exceeding a shearing stress of 80 MPa. SOLUTION
't=3oxm‘Pa =eqv£ any}.
in L J "69’ 5.q535‘l63 M ‘= 5:953 mm
6" 2C. "' ”.5” MM 3.35 The torques shown are exerted on pulleys A and B. Knowing that the shafts are solid and made of aluminum (G = 77 GPa), determine the angle of twist between (a)
A and B, (b) A and C. A
l SOLUTION (at 17.3 = 300 NmJ LA. =03”, wag: =0.0l$m
J}. = %(o.ous)" = 7?.522 x10" m ‘* =M_ (3003(031 _ 4
G'J  WW  47.095 :40 rad @M : 253. ‘
03) Tie '7 300+‘1‘00 : 700 N‘mJ L“: 0.75m J Cu='k0l _. 0,013 m
I,“ = g (0.023)”: 939.5732103 mt  M.— im— _ .5
(Par Germ. ‘ (77xlo¢)(ng_573km.1) ~ 15.5uxlo um! (Pk = ‘Pae + 493:. = Jamar/0'3 m! = 3.42" ...
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