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1337_Mechanics Homework Mechanics of Materials Solution

# 1337_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: PROB 11.40 The ship at A hasjust started to drill for oil on the ocean ﬂoor at a depth LEM "-49 of5000 s. The steel drill pipe has an outside diameter of8 in. and a mum wall thickness of 0.5 in. Knowing that the top of the drill pipe rotates through two complete revolutions before the drill bit at B starts to operate and using G = 11.2 x 106 psi, determine the maximum strain energt acquired bythe drill pipe. SOLUTION g) = (23(217‘) = 47: m4 L :- 5000 if 2 gowo‘ 'm if 1, LII-4. C;"'Co""t = 3.5 in. E(c.,"—C;") r [66. #05 .34" (2.: J: - L -G—Jep .125..-GI =._L___G ‘ cP“ £3 T" L U 263' (—52 ‘ﬂ U -_- . _(_\__l____.2‘x__|0_‘)(|c_§4:40é:)(4n31 — 2‘45 ,4 [0‘ in’ 11:. «an (23(60x103“) 11.41. The design speciﬁcations for the steel shaﬁ AB require that the shaﬁ acquire a strain energy of 300 hr“: as the 25-kip-in. torque is applied. Using G n 11.2 x 10‘ psi, determine (a) the largest inside diameter ofthe shnﬁ that can be used, (13) the eorrespondhig maximum shearing stress in the shaﬁ. PROBLEM 11.41 SOLUTION U 7 300 inn-J‘s T: 25‘; k1-P'I‘Vﬁ 225*”): gal-tn L '= 36 in. , T’L U 26 1 J ._. TZL. , (ZSXIDS) (3§)m__w= 3.3quinq ZG-Z'J' ‘ mmzxro‘nsoo) w I: tier-er] = \$50“ 0’9" 01;}: a]: _ .3342 2.54— 3%(334323 =- u.es7s7 n." 25 kip - in. F I a; :- .l.492, m. -‘ ,.. Tc. gZSon‘)(|.2s) _ 3 . _ . L 2 J = . 3.3'181 - q. 33"“) PSI - 9.33 k5: ‘ 1 . . 74 ...
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