# hw21-10sol - HW WU e SOLUTWNS*8—4 The tank of the air...

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Unformatted text preview: HW WU e SOLUTWNS *8—4.- The tank of the air compressor is subjected to an internal pressure of 90 psi. If the internal diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element. Hoop Stress for Cylindrical Vessels: Since é = 6% = 44 > 10, then thin wall 0&5" k"; analysis can be used. Applying Eq. 8—1 ‘ Dyna :36 -pr-90(11)-3960 '-3961<' Ans 0-1— I — 0.25 — p51— . 51 . Longitudinal Stress for Cylindrical Vessels: Applying Eq. 8—2 -ﬂv2ﬂ—1980 '-198k' A 02- 2t — 2(0-25) - p51" . 51 ns. 35500 ' Vgomm “6—H (Wm) GWEN: ?: V‘— “an at z in T hour 4’ ' i. FW’D: (rump M41 01m: Eh of armor. st‘muz EM,“ 2 T; = 15rd. (3(4):!)C ml.) = \OXOLG 7 Shaw, VFW“ - 2PM, =9! . \OS/O \$1M)”, 1 1 FMuf - . “W ’ A“... 1:». 0m ’ 54°!” Va“? 540 w 6' ; -/— :. __,_____, z 00 ‘ :1 O - “"7 MI WQOJ'L‘SRV “‘0 vs“ “I ‘05" Quawa ‘6 '1‘6 (re/«sum GWEN: T) : \00 w w——\.15;n (wk Ar: 05;“ YWWLD: 3W WW,“ £ o~ om U'\§‘w}:}~\ (1" “WT” M”) sowmu; 4351—46: N— wow-«<15 Nzwu LT) +5? Meir-t6: Now 0.15:1) — M = (6 L M: 2.63:4 N \oou; - 0’ — — - —. u . A M” A \15.\(05.~D 4 3 9‘” Um” ) y 3 L HEMSB‘” - o awry! 0 "’ 03;; 10.15'm 1 Mo mm. (0.15% q’ _, : ___Ir_____‘“_ 1 n ‘25 . MW I 0-0W’L’5'm" M f“ gﬁNTW~ 1 0km“, + WNW '~ 4’ ‘ _ \‘Klqrsi (Murrbhtmw . - \ '00 0 : \‘oucrv x-->+ 35. I - 0,55')‘ \Y'ché w [\$001.4 Mex : mwm - mu ‘ 313.92) a 0.11;; (ma, .4 M. x; MM Wm M; QM) X ' \£00+\‘62.°\ 35500 ' ‘ 0.3 = —- = —1067 ps1 = 1067 p51 (C) 0849. The joint is subjected to a force of P = 200 lb and F = 150 lb. Determine the state of stress at points A and B and sketch ‘the results on differential elements located at these points. The member has a rectangular cross-sectional area of width 0.75 in. and thickness 0.5 in. A ' +IT€FY1¢1 N—vloo‘ﬁ' :4) N124”)La . LZFL>¢‘-"'ksow “I”! V:\SOU3 +5§MJ=¢E M+200m015ﬂ -\[email protected]’.\ =51 . ' ' N: 500% A = 0.5(0.75) = 0.375 in2 ‘ _ - ' 41 QA = y’AA’ = 0.125(0.75)(0.25) = 0.0234375 in’; Q, = 0 _ ' 5‘ mu =V = 315 (0.75)(0.53) = 0.0078125 in‘ . ' m, Normal Stress: 1 15b» 00 . 0.375 + 0 - 533 p51 (T) 200 50(025) (TA: E: E Shear stress: _ 150(00234373) ._ 600 , > 7‘ " (0.0078125)(0.75) ‘ ps‘ 5 75:0 - I Ans. ...
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