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229_Mechanics Homework Mechanics of Materials Solution

# 229_Mechanics Homework Mechanics of Materials Solution -...

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Unformatted text preview: 3.84 A I.5-in.-diameter steel shaft of length 4 it will be used to transmit 60 hp PR BLEM 3.84 . . . 0 between a motor and a pump. Knowmg that G = l 1 .2 X 10 6 psr, determine the lowest speed of rotation at which the shearing stress will not exceed 8500 psi and the angle SOLUTION of twist will not exceed 2°. c=-',:a.- 0.755..J L= 4H = #3:". Torque based CM de‘muh Sheaw'na 5+V~¢SS Jilwu‘i ”if: 8500 psi 7: = id; = “\$712 .: T =§fcsz=-;£(o;7533(8\$oo)reicastO‘JL-in Tongue 5:»st on +vis'i angle [Pf-m"; so =‘ 2° 1' 3‘1_<io7 “IO-3 Mud 2' Ii: .1 T :é‘JBBL- TC~G?_ MW g) (“I ’- 7-1— — (21m) = 4.048 x lo‘ .tb-iu Smaliew +quue jovewns T = 4.043 "I03 4%- ha 1’ = TIT-FT who». PI: éO LP 2 (GOXQQOO) t 394. .103 M4,, /S 01. ~13- : ———3°’—”1'———-— : 15.57 Hz r s39 rpm .- ZﬁT “(4.0% H03) 3.85 A l .6-m-long tubular shaft of 42~mm outer diameter d, having the cross section shown is to be made of a steel for which 1;“ = 75 MPa and G = 77 GPa. Knowing that the angle of twist of the shaft must not exceed 4° when the shaﬁ is subjected to a torque of 900 Nm, determine the largest inner diameter d2 which can be speciﬁed in the design. :12 SOLUTION (1.: lack: 0.021 m L= u; m Rascal on stress Jim-i Z = 75 MP4 = 75:10‘ PA J: T3! 1., (q00)(0.011 i =' 252*10-‘7 Mn] L 75m:- Eucoi on ulna): O‘P +Wf\$+ pin-‘1' 9 "' 4° 7 69.313 *(0-3 Mid ‘1 le; EM...” 8 -9 g) G-J ’ J 6.9 (77xf0?)(6q‘8‘3K[o-3) - 297.8 *IO M Laurie.» Van/PVC how J govern-5 J: 267.88 mo" IMJ’t J = %(CI‘" (2:) -+ 4' _ C2 .—. a. 0.01! _ W = 23.9% xto " " IQ-Li‘f MM 6’2 1" 2C2- " 2"."7 MM 4 ...
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