sa_Lecture9_momentdistribution

sa_Lecture9_momentdistribution - !" # ! %') $ &(...

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Unformatted text preview: !" # ! %') $ &( " * ) + * " " + .) *" #! $ , " + " ) /" " 0 " 1 0) 1 "" " ") 2"" " 3 / 4 "* "* " ) M AB 1 = M BA 2 / 4 5 * ) 5 ) 0 * " " I M BA = 4 E tan θ L I = 4Eθ (for small values of θ ) L / 4 "* 5 )) 5 * 5 I (for small values of θ ) L "" * " 6 ) M BA = 3Eθ / ) + ") " 7) ) / 4 " "* " ) M AB = M BA 6EId =2 L "" " 5 / 4 " "* ) "" M BA 3EId 6EId = 2= 2 2L L 1! * 7 ) % 8 * 7 1! * 1 # $ 727 #! $ ,) θ % 8 7 7 ) 7 ) *1 " 0 1! * , ! FEM AB (being anti - clockwise) =− % 8 7 8 WL − 65.0 × 3.0 × 3.0 = = −48.75 kNm 12 12 FEM BA (being clockwise) clockwise) BA = +48.75 kNm 78 FEM BC (being anti - clockwise) =− WL − 32.0 × 3.6 × 3.6 = = −34.56 kNm 12 12 FEM CB (being clockwise) = +34.56 kNm 1! * 9 : 1 4 0 3 0 )3 % 8 72 " .) ; )> <= ? . 7 ; )> 0 ) <= ? 3 5 7 1% & 3 <)% 0 ; )>1' @ ; <)% 0 ) <= ? <)?A % & 3 1! * 7 7 " % 8 " θ) , 7 727 ) 7 B ") 1% & 3 <)% 0 ) 72 0 ) #, $ 1! * 3 A?C CC < % 8 7A' 7 A' ) )C )@ 5000 × 10 4 K AB = = 16.67 ×103 3.0 ×103 5000 ×10 4 K BC = = 13.89 ×103 10 3 3.6 × 10 16.67 × 103 = 0.545 DFAB = 3 3 16.67 × 10 + 13.89 × 10 13.89 × 103 DFBC = = 0.455 3 3 16.67 ×10 + 13.89 × 10 1! * % 8 " ) 0.545 × −14.19 = −7.73 kNm being added to the end BA 0.455 × −14.19 = −6.46 kNm being added to the end BC 1! * , 2 7 5 ") % 8 carry - over process 0.5 × −7.73 = −3.865 kNm transferred to the end A 0.5 × −6.46 = −3.23 kNm transferred to the end C 1! * " D ) " 7 72 4G ) H = % 8 " " ") 1?)@ 1<% ( 1 ( ' )C 2 ')' 0 ) % '3 9 EF 7 1! * 1 7! ( 8 ) B 7 =C A ?C < 8 B7 A ?C < B ! ?C @C A ?C < 1! * 2) * * ) ! . ( 8 " ", ! 72 " 1! * 72 5 " ) ) ( 8 7A) " 3 8500 ×10 4 3 K AB = × = 14.17 × 103 4 4.5 ×103 4 6500 × 10 4 K BC = = 17.33 × 103 3.75 × 103 3 3 5500 ×10 4 K CD = × = 11.00 × 103 4 4 3.75 ×103 1! * , DFBA = DFBC DFCB DFBC 3 K AB 4 ( 8 , 8 3 K AB + K BC 4 = 1 − 0.45 = 0.55 14.17 × 103 = 0.45 = 3 3 17.73 × 10 + 14.17 × 10 K BC 17.73 × 103 = = 0.612 = 3 3 3 17.73 × 10 + 11.00 × 10 K BC + K CD 4 = 1 − 0.612 = 0.388 1! * 3 72 * 3PL 3 ×120 × 4.5 = 101.25 kNm = 16 16 2 PL 2 × 70 × 3.75 = = −58.33 kNm FEM BC = − 9 9 FEM CB = +58.33 kNm kNm FEM BA = + FEM CD = − WL 220 × 3.75 = = −103.13 kNm 8 8 ( 8 8 8 M DE = − M DC = −50 × 1.2 = −60 kNm 1! * 4 , 2 )I ) " ( 8 ) ) 1! * ( 8 ; <)= % A; C #= ' C)%$1% 0 ) ';?)'1%' 'A <)= 3 0 3 72 ) Distribution BA = (+ 101.25 − 58.33)× 0.45 = 19.31 Distribution BC = (+ 101.25 − 58.33)× 0.55 = 23.61 Distribution CB = −14.8 × 0.612 = 9.06 Distribution CD = −14.8 × 0.388 = 5.74 1! * B 1 5 ( 8 ), ) 1! * ,7 ") ( 8 / 8 ) 1 " ) + 5 ) + * 1 * ) ) / " 5 1 ) ) ! !B ! ) 8 " *1 ( J " ,) ! ...
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This note was uploaded on 02/03/2011 for the course CE 573 taught by Professor Various during the Spring '11 term at American University of Science & Tech.

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