# File Tutorial8withAnswers.pdf - Tutorial 8 w Problem 5-4...

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Unformatted text preview: Tutorial 8 w Problem 5-4 ' The copper pipe has an outer diameter of 40 mm and an inner diameter of 37 m. If it is tightly secured to the wall at A and three torques areapplied to it as shovtm, determine the absolute maximum shear stress developed in the pipe. %*2_ Problem S-ll The shaft consists of three concentric tubes, each made from the same material and having the inner and outer radii shown. If a torque of T = 800 Nm'is applied to the rigid disk ﬁxed to its end, determine the maximum shear stress in-the shaft. E—é "Solution 5-11 J = g ( (0.038)‘ - (0.032)‘ J + 3% (0.03504 — (0326)“) + .2’5 ( (0.025)“ — (0.020314) J = 2.545(10-6)m4 T . rm = J = MW 038) =11.9 MPa Ans - J 1545(10-6) - Problem 5-12 The solid shaft is ﬁxed to the support at C and subjected to the torsional loadings shown. Determine the shear stress at points A and B and sketch the shear stress On volume elements located at these points. Solution 5-12 ___________,.__.___ TB : gig: 800(0.02) __ 6'79M J ———————§(O.0354) — . Pa Ans‘ . 3m.» 7;=‘°””“ 73:67pm 2? .+ \$41!": {A = E: 500(0.035)_ 7 J mac-0354) - 42 MP3 Ans ﬁ=\$oa~m Problem 5—19 The steel shaft is subjected to the torsional loading shown. Determine the absolute maximum shear stress in the shaft and sketch the shear—stress distribution along a radial line where it is maximum. “xvi Solution 5-19 __,,,___.....-—-—-~—-— quus is 3.5 W . m ;, imam, two mm of the shaft shoukﬁ be; mnsidamd sins: J is diffan _ T6 = 5(103 new} — tam: an!!! 7 gmmr ‘22: -8.5(i0’)(0.36} nun- law 7 £40.06)“ = 49.? MP3 = 749.? MPa = 25.1 MP; Ans 46-93 48331-Mechanics of Solid Student Guide, ‘I'uteria! and AssignmeuterbleIgs PROBLEMS ON SMLE TORSION 14-1 (a ) For the hollowalminium shaft (G = 27 GPa) shown, if the applied torque To is 3 kN—m, determine - (i) The shear stress at the inner edge (inner circle ) of the shaft (ii) The maximum shear stress in the shaft (iii) The angle of twist (b ) (1) Determine the torque To causing an angle of twist of 2°. (ii) Determine the angle of twist if the same torque is applied to a solid shaft of the same length and cross-sectional area. 40 mm \w- . .3- e ‘Tovslm Thao‘ré.‘ _..___.____- §__'___’_T__.£§__= '5 a "d"— "' ,7 "H K 3 -_-.' o‘.oL\--+%q read. - Vii thxxowés’szwgbh _ rmiw" 3 S.%_~A\D“‘ :- 9007‘? 'W‘ ‘ ' R 3; PM Ok SDUA ska 4:, " I ‘ 4 “1..— _— :1.—L'..'_':_ _ - s- R, -..—. Born—mm QS= —-—-— = \‘1'SQ'ZMDT'TE‘ a _ TL ><\o Kzusmo . i; Thus) ><\-'l lama 7 ‘35:“— :thnu ‘ L 48331-Mechanics of Solid Student Guide, Tutorial and Assignmt Problem 14-3 The solid spindle AB of diameter d3 = 66 mm passes through the sleeve and is welded to the sleeve at C. The sleeve has an Outer diameter of 80 mm and a Solid Spindle Hollow Sleeve Weld Fan-am I 30 ‘J Section 3-21 69 Th: 'A-36 swel pests an: "drilled" at census! angu— f 3 in speed ink) the soil using the rotary ha an inn=r diamem 225 mm. delermine: Eh: the past with respect to end B whnn the post reaches the depth indicated. Due to soil hicﬁou,assum= 1h: torque along thc past Van's: Encarly as shown. and a communist] ion-qua orEGkN-mamatthcbit. I}; - 30—%(15)(3) ~—- 0 2M =‘o; 13 = 102.5 m -m ‘2: 1' Mum 1 '1 j 2Mz =0. 102.5~E(Sz)(z)—T=0 T: (102.5 4.53) kN - m I? 3242 UT ‘4; I335 MW“. “3 JG JG = '10:.5(103)(4) + [a (102.5—2.522)(103}dz gaunzs)‘»(0.1)4)(75)(109) a §{(0.1125)4- (0.1)4}(75)(109) ,. .k a = 0.0980 rad Ans 235 ...
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## This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.

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File Tutorial8withAnswers.pdf - Tutorial 8 w Problem 5-4...

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