Determinate Structure Cantilever-Method.doc | Applied And Interdisciplinary Physics

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FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275 Example No. 1 The steel jointed building frame show in the figure carry the lateral load as shown. Analyse the frame and draw the bending moment diagram using the cantilever method. 10 kN 1.5 m 10 kN
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  FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275 Example No. 1 The steel jointed building frame show in the figure carry the lateral load as shown. Analyse the frame and draw the bending moment diagram using the cantilever method. Solution Step # 1 : Locate point of inflection A  ! Step #  : Locate the centre of gravity $ m A A A A A 5.34)5.7()5.4()2()0( =+++ - Cantilever Method Page   1 1.% m.& m.& m.% m'.& m1& ()1& ()* $   1.% m.& m.& m.% m'.& m1& ()1& ()* $ '.% m  FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275 Step # ' : +etermine the a,ial force in each column Since any column stress- σ - is proportional to its distance from the neutral a,is- )A-we can relate the column stress by proportional triangles. 5.115.3  P  P  = , P1  $ &./0 $ &./,&.''$&.'%2() 125.3  P  P  = , P2   $ &.30 $ &.'() 435.3  P  P  = , P3  $ 1.1'0 $ &./%()alculate the value of 405 using e6uation of e6uilibrium Σ 7 h   8 $ &1& 9&.2% 8 01 9 ; 0 9.% ; 0' 92.% $ &1& 9&.2% 8 9&./09 < 9&.309.% < 91.1'092.% $ &0 $ &.'' () - Cantilever Method Page   2  00100'1& ()00100'  FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275 Step #  : +etermine Shear forces of each part of frame and determine member end moments  At Point “a”  At Point “c”  - Cantilever Method Page   3 Σ = > ↑ 8 $ &;0 8 ?  A@  $ &? b  $ 0  V b  = 0.833 kN # Σ 7 a   8 $ & Σ 7 a   8 $ & 9down7 a  $ ? b 91 7a$&.2%91.111 7 a  $ 9&.''91 7a$&.''()m9column M a  = 0.833 kNm (beam) ? b 0 $ &.'' () 1& () b  h Σ 7 b   8 $ & h  9&.2% $ &.''91 H h  = 1.111 kN # Σ = * → 8 $ &1& <  b  ;  h  $ & b  $ 1& ;  h   H b  = 8.88 kN # 7 a ? b  b  d 01 $&.'%2() ? d ? b  $ &.'' () b  $ ./ ()  i Σ = > ↑ 8 $ & V! 0.3$ 0.833=0V !  = 1.1kN # Σ 7 d   8 $ & Hi(0.$) 0.833(%.%) 0.3$(1.%)=0H i  = 3.0&kN # Σ = * → 8 $ &./;'.&/;d$& H !  = .$kN # 7 c ? d  d Σ 7 d   8 $ &7c$1.1/91.%()m M '  = 1.&88kNm (beam) Σ 7 i   8 $ &7c$'.&/9&.2% M '  = %.3%1kNm ('olumn) 7 c 01  i  FAKULTI KEJURUTERAAN AWAM Faculty of Civil Engineering Tel:(+603) 5543 6189 Fax: (+603) 5543 5275  At Point “e”  At Point “g”  - Cantilever Method Page   4 ? f  0' $ &./% ()  f   ( 0 $&.'() ? f  ? d  $ 1.1/ () d  $ %.2/ ()  f    j Σ = > ↑ 8 $ & V 1.1 0.%38=0V    = 1.&%8kN # Σ 7 f    8 $ & H(0.$) 1.1(%.$) 0.%38(1.)=0H    = &.8&kN # Σ = * → 8 $ &;f8%.2/;.$& H    = 0.* kN # 7 e ? f   f  Σ 7 f    8 $ &7e$1.91.% M e  = %.1&%kNm (beam) Σ 7  j   8 $ &7e$.9&.2% M e  = 3.*3kNm ('olumn) 7 e 0  j Σ = * → 8 $ &($f  H k  = 0.*kN # Σ 7 f    8 $ &7g$1.91.% M +  = %.1&%kNm (beam) Σ 7 f    8 $ &7g$&.2%9&./37g$&.2()m9column7 g ? f   f 
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