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8.5-6 - Prahlsm 3.5-5 A cylindrical tan subjected to...

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Unformatted text preview: Prahlsm 3.5-5 A cylindrical tan}: subjected to internal pressure pl is simultaneously compressed by an axial force F = 1’2 kN {see figure}. The cylinder has diametert-rt.r = 101) mm and wall thickness t = 4 mm. F F Calculate the maximum allowable internal pressure pmu based upon an allowable shear stress in the wall of the tank of 60 MPa. Bnlutlnn 3.5-5 Cylindrical tank with compressive force ll3"1 —- .— F = 72 W ,2 p = internal pressure “‘— —" d= lflflrmn I=4mm rm=fiDMPa Cmcmmmmsmsss {TENSlCIIN} 1 pr p(5fl mm} = — = — = 12.5 U1 1 4 nun p Units: 0'1 = MP3 P = W3 UUT—flF—HANE SHEAR STRESSES [I1 Case 2: rm = Y = 6.25;}; til} MPa = 6.25;; Solving, p2 = 9.60 MPa Lcaonaomu smsss {mm} _pr F pr F U2 —_———_— 2: A 2’ 2” C35331Tm=fl=3.125p—23.643N[Pa _ s 25 _ rams N 2 ' p gfijg mm)t4 mm: as MPa = 3.125 p — 23.643 MPa 5 l ' , = mamas = 5.25;: — 57.295 MPa “”13 P3 Units: 0-: = MP3 p = mg CASE 2, UUT—OF—FLANE sass]: srsEss commas pm“ = 9.150 MPa 1— BIAXML stasss IN—PLANE sass]: smass {Cass l} (Tl—”'2 Tm: 2 =3.125p+23.643h-'IFa 61:! MPa = 3.12519 + 23.643MPa Solving, p1 = 10.03 MPa ...
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