{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

895_Mechanics Homework Mechanics of Materials Solution

895_Mechanics Homework Mechanics of Materials Solution -...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 7.160 The compressed-air tankAB has an inside-diameter of 450 mm and a uniform PROBLEM 7.160 wall thickness of 6 mm. Knowing that the gage prwsurc in the tank is 1.2 MPa, dammine the maximum nonnaisu‘ess and the maximum in-plane shearing stress at - intsa andbonthctopofthetank SOLUTION c,= 225 m , cz=22§+ts e 231m Tor-su'on‘: J = ¥(ng'clq) 7 4413.“! )1”)6 mm“! 3 qq‘_qx[O-‘ W14 T= ('Svto‘)(.soonc53) = 2500 N-m -3 f -.- 19. = (25°°‘(23"‘° :- 1.2?2'HO‘ Pa. = uzezMPa I 4%}: “0“ Transverse shear; _. ’1‘.’="O' a} Poifl'l’s at and lo. Bending: I-‘- "5: '~' 223.‘IS¥10_" M.i J C." 23iM’O-s m, Pol'n't a, PG;V\“ l) = (sxto‘ )(750sto'5 )= 3750 N-m M= (snowman-owe" ) - 7500 N-m M -3 6= c: :53750)(25mo 2: 3,8814% I anaemic" T041.) s+nsses (Mpa) '-' -'- - 7.75 MPG. Téa‘l'afi 5+r~esses (MP4\ Lon3;+oa;na1> 5,: 225+ 3.88 =~ 26.38 6, = 21-5 + 7.75 = 30.2; Cimumremfld 5] : 45 . g; = 45" Shaun To = 1.732 4.3 —- 1.291 Sufi “Side-63% 35.69 MR: R= (fig—@th; = W m 6M=Ji(6}+63)- szézs MPq R = (45$)H1; “—' 7.437 MP4 6' =' €M+R= 45.1MPO.‘ mm 61,“: 6;¢+R= 45.! MPa -‘ Z’Mm(in-r)m) = R = 51.40 MPCL “ T~m(;._rj.")= ’R -‘ 7-‘4‘7 MP4. ‘I‘ f’" '1: ...
View Full Document

{[ snackBarMessage ]}