Beam Example 2

Beam Example 2 - Problem 1 The tapered beam shown below has...

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Unformatted text preview: Problem 1 The tapered beam shown below has a rectangular cross-section whose depth varies linearly in x: w ‘ghw U . x: .Yv g) Reqmwd: Iticuéizc {hr beam with 1M; {fili’ilitti'lia liiuil’i}? antiwar” section properties db inflow: « ,1] s W l e i / firmwmrrwm WWWW “Mi Few—«w 1_ 9:2 wwwm~w~avl+~ ~~~~ ~~-~ 142 WWW»; Use the finite element method to determine an approximate solution for the displacement and rotation at x = L. Assume that E =107 psi, b = 0.1 in, h] = 0.5 in, h2 = 1.0 in and L =20 in. Assume that the depth of each element is equal to the average value of the depth in the tapered beam at the center of each element. .L x}; xix} HEOUMDAM Cmbrrmps (AG-u": 9'20 FHL= FU—LiMZ‘: 0 F743 :""320 _':> FEWs -1 1 . 51/0 PSL . Ar e/cmm 7z Q3 fly" c/M‘”ZZ® H F L: /0 :14. r Ar-IOI'M— l was __./ A: 0. 0875/4 Z 0-625" A = 0A 0425!“ m f=o_005583.'u7 Jr I” I:o.ooZDS$/,:‘ H 0-1.» 3 Efiag'f'SK/oz/és Eff: 625 x10 [55 z 2 El :55 .‘BSX/Oz/és I'n E1: = 20.3‘1x163/AS/L4 I 3 ' [343710 8?.5 o o [—8¥.5 o o 0 0.67m) 3.350 i o —~o.c,3\oo 3.350 0 3.350 22.3.: I o —3.;:§3_ ILL; —8¥.s o o 34.5 o o 0 "6.6400 —3.350' c a .6700 “3.350 0 3.350 1/1? '1 o ~335a 22.33 CL) . 3 l Dd“: ,0 62,5 0 o | 4.2.5 o o o 0.277, [.220 l 0 ~52qu 1.27.9 _O___ “1.22; 8435 _o :szzo' 7.068 —(,2.5 o o igzg- o o o —-o.2‘f‘fl 4.2201 0 0.2%” 4.225 0 1-210 7.068‘ 0 4.220 8.86 l I I ' [11¢ “ L4.L_I_K_.z_._ o [<21’J IC'Z‘L‘tK'FI Kn. z 1 O : Kzu I K17. 5K1 CfIFEFj 3 I I [D 87.5 No C) “8?.5 o 0 l0 0 o 0 0.6700 3.355 ‘ o -047“ 3.350 0 o o _o £350 _z_z.33 __o —3__._3_5o ILH‘ lo 0 o~~ ~37; o o [(50 o 0 P625 0 o o —o.(a7-oo #3350 O 0.714, 4430 f o 40.”!me ' _0 k3 350 UPI 0 “2430 309? 0 'I220 #068 o o o “62.5 o o [62.5 o o“ o o o ‘ O «9.2%; 4.220! 0 a 29w ‘/ 220 o o o 0 L220 4.06% 0 4.22.0 3.136 1 'O 1:“ 8 Fv'n “a = M‘ 15, o 91 8 “3 v; E _——————-——————_'— ff ,FINH‘L. I 5TEP~ DLSTmBu'rED OP (FRAM Ea. 7~. Iota) ELEM WT 57.0 m x )PZ‘ —P‘3 t F2141 -= ~235Mb lb LE]: 0 ——2,ooolb -33’3ao1b{m 17mm, L. EQUA-T (cs/Ki ‘ Dc] L71: ETD] fix—rm 7 (a) 146 Luv/2:10pm. O/rZSP/Iucmgafi aft 1’3qu 9—? ) Llosz fiuafl‘dus fl,” Hz? 4/: “071 A a V5! -_ So we W ‘fltm “ma/«1,17 14/ 15f 93, [as 0.7/97 -Z..ISO «0.297, 1.220 ’U’L o —2.130 30-‘1’7- -I:'zzo 7.068 9;, = 0 -o.zI-ILH "Lzzo 0.294, 4.20.0 V; P O I. no 9. 068 4.230 88¢ 93 .3 1); = 5?.8? P ;../up*!o 93:5.1/ Frag/u, No ...
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This note was uploaded on 05/19/2010 for the course MAE 472 taught by Professor Peters during the Spring '08 term at N.C. State.

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Beam Example 2 - Problem 1 The tapered beam shown below has...

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