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193_Mechanics Homework Mechanics of Materials Solution

# 193_Mechanics Homework Mechanics of Materials Solution -...

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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 8-in.-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 WEI-r- TAP ._.I_§._.GJr:_Gc 7" J TLL*—EL g) = 2m mum -.~ muses mt, c. = +9.1 .- 4-0;" L z. 5000+} = \$0000 .3‘ 2/: (li.2¥lO‘2(l?.5‘6é§(l-l.o) £0000 : 9,332.6)![03 FSII = 7.38 RSI. PROBLEM 3.34 3-34 Determine the largest allowable diameter of a 3-m-long 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 N-mJ 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 : 2-5-3. ‘ 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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