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222_Mechanics Homework Mechanics of Materials Solution

# 222_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 3.73 While a steel shaﬁ of the cross section shown rotates at 120 rpm, 3 stroboscopic measurement indicates that the angle of twist is 2° in a 4-m length. Using G = 77 GPa, detennine the power being transmitted. SOLUTION O 30mm75m’“ Twisi' angle CP = 2° = 39-?0’7W0’3 Paul ____L o. = i A. = omsm, cﬁﬁ, = 0.0375... 3': 1}(c,f-c."\ : —E(o.os7\$"- 0.015“) J: 30253::0" um") L=Urm cp . 15 _ GJQ _ (17w:o*)§3.oaczuo“)gamw “(5‘) GI T ’ L ' 4 PROBLEM 3.73 T = 2.033%:0’ N-m ‘Fr. IZOnpw. = T”?- —, 2H2 P= (217F3T = 21r(2)(.2.0339wo‘)= 25.6 up" u a 25.6 kw .- - 3.74 Determine the required thickness of the 50-min tubular shaft of Example 3.07, PROBLEM 3.74 . . _ . . _ if It 18 to transmlt the same power whale rotatmg at a ﬁ-equency of: ’10 H z. SOLUTION From Exampie. 3.07 P e: [00 kw T.» = GOMPo. = 60 XID‘ Pa. Given 5? = 40 Hz 3 :- szngﬁgmmﬂ :- 397. gq N . M 111140) '7” 21‘ If. it u = To ._ c ‘i _ Ca. 1(C2 ' C, ) r “3:4 E‘cﬁq_c'9) W 2TC3, (2\(3‘]7_3Q)C0‘01§} _ 285.03! ”IO... «9 _ a _ 4‘ C. - CZ "" Tr? ~— 0-025 W— _ c,I : 23.1! Mo's m t =.'_C,_ -"C. = MN Ho‘s». ‘- 1:24 Mm ...
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