40314987-EXAMPLE-Cantilever-Method.doc

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Example No. 1 Example No. 1

The steel jointed building frame show in the figure carry the lateral load as shown. The steel jointed building frame show in the figure carry the lateral load as shown. Analyse

Analyse the the frame frame and and draw draw the the bending bending moment moment diagram diagram using using the the cantilever cantilever method.

method.

Solution Solution

Step # 1 : Locate point of inflection Step # 1 : Locate point of inflection

A

A   ! ! ""

Step #  : Locate the centre of gravity Step #  : Locate the centre of gravity

$ $ mm A A A A A A A A A A 5 5 .. 3 3 4 4 )) 5 5 .. 7 7 (( )) 5 5 .. 4 4 (( )) 2 2 (( )) 0 0 ((         - Cantilever Method - Cantilever Method Page Page 11 1.% m 1.% m .& m .& m 

..& & mm ..% % mm ''..& & mm 1& () 1& () 1& () 1& () * $ * $ 1.% m 1.% m .& m .& m 

..& & mm ..% % mm ''..& & mm 1& () 1& () 1& () 1& () * $ '.% m * $ '.% m

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Step # ' : +etermine the a,ial force in each column

Since any column stress- - is proportional to its distance from the neutral a,is- )A-we can relate the column stress by proportional triangles.

5 . 1 1 5 . 3 P P  , P1 $ &./0 $ &./,&.''$&.'%2() 1 2 5 . 3 P P  , P2 $ &.30 $ &.'() 4 3 5 . 3 P P  , P3 $ 1.1'0 $ &./%()

alculate the value of 405 using e6uation of e6uilibrium

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 $ &.'' () 0 01 0 0' 1& () 0 01 0 0'

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Step #  : +etermine Shear forces of each part of frame and determine member end moments

At Point “a”

At Point “c” - Cantilever Method Page 3 ₌_>_{8 $ &} ;0 8 ? A@ $ & ?b $ 0 Vb = 0.833 kN #

₇_a _{8 $ &} ₇_a _{8 $ & 9down}

7a $ ?b91 7a$&.2%91.111 7a $ 9&.''91 7a$&.''()m9column Ma = 0.833 kNm (beam) ? b 0 $ &.'' () 1& ()  b  h ₇_b _{8 $ &} h 9&.2% $ &.''91 Hh = 1.111 kN # ₌_*_{8 $ &} 1& < b ; h $ & b $ 1& ; h Hb = 8.88 kN # 7_a ?_b  b d 01 $&.'%2() ?_d ? b $ &.'' ()  b $ ./ ()  i ₌_>_{8 $ &} V!"0.3$"0.833=0 V! = 1.1kN # ₇_d _{8 $ &} Hi(0.$)"0.833(%.%) "0.3$(1.%)=0 Hi = 3.0&kN # ₌_*_{8 $ &} ./;'.&/;d $& H! = .$kN # 7 c ? d _d ₇_d _{8 $ &} 7c$1.1/91.%()m M' = 1.&88kNm (beam) ₇_i _{8 $ &} 7c$'.&/9&.2% M' = %.3%1kNm ('olumn) 7_c 01  i

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At Point “e” At Point “g” ? f 0' $ &./% () _f  ( 0 $&.'() ? f ?_d $ 1.1/ () _d $ %.2/ () f _j ₌_>_{8 $ &} V"1.1"0.%38=0 V = 1.&%8kN # ₇_f _{8 $ &} H(0.$)"1.1(%.$) "0.%38(1.)=0 H  = &.8&kN # ₌_*_{8 $ &} ;f8%.2/;. $& H = 0.* kN # 7 e ? f  f ₇_f _{8 $ &} 7e$1.91.% Me = %.1&%kNm (beam) ₇_j _{8 $ &} 7e$.9&.2% Me = 3.*3kNm ('olumn) 7 e 0  j ₌_*_{8 $ &} ($f Hk = 0.*kN # ₇_f _{8 $ &} 7g$1.91.% M+ = %.1&%kNm (beam) ₇_f _{8 $ &} 7g$&.2%9&./3 7g$&.2()m9column 7 g ?_f  f

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Step # % : +raw bending moment diagram 9@eam A!" 1.()m .1()m &.''()m &.''()m 1.()m .1()m - Cantilever Method Page 5 1.% m .& m .& m .% m '.& m 1.% m .& m .& m .% m '.& m

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Example No. %

The building frame in figure below is subjected to horiBontal forces of 1&& ()- & () and %& () at node +-  and L- respectively- and all columns has a dimension of %& mm , %& mm. Csing the cantilever methodD

i. establish the location of the centroidal a,is of the frame ii. calculate the a,ial forces for columns E- -- / and H

iii. draw the bending moment diagram for beam 245

2  4 5

E - / H

   

M N 6 7

.&m 3.&m .&m

1&& () & () %& () '. m '. m .& m

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Solution

i. !stablish the location of the centroidal a,is of the frame

= m A A A A A 8 4 ) 18 ( ) 10 ( ) 4 ( ) 0 (    

ii. calculate the a,ial forces for columns !E- =F- "G and L

=rom the distribution diagram shown calculate the value of 01- 0 and 0' in term of 0 . 1  P P  , 01 $ &.% 0 2 2 8 P P  , 0 $ &.% 0 1& '  P P  , 0' $ 1.%0 - Cantilever Method Page 7 0 01 0 0'  m

.&m .& m .& m .& m

.& m

 m

.& m .& m .&m

1&& () & () '. m 1.3 m 0 01 0 0' * =

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@y ta(ing 7oment at mid point of column !E- calculate the value of 0

₇_!E _{8 $ &}

0 91& 8 0' 91 < 01 9 < & 91.3 < 1&& 9. $ &

9&.%091& 8 91.%091 < 9&.%09 < 9&91.3 < 91&&9. $& .%0 8 .%0 < 0 < 1 < & $ &

0 $ 3.'% ()

Therefore the value of a,ial forces for columns !E- =F- "G and L are as followsD !E $ 0 $ 3.'% ()

=F $ 01 $ &.%0 $ 9&.%9 3.'% $ 1'.1 () "G $ 0 $ &.%0 $ 9&.%93.'% $ 3.3&/ () L $ 0' $ 1.%0 $ 91.%93.'% $ ''.& ()

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@y ta(ing 7oment at mid point of column A!- calculate the value of 0

₇_A! _{8 $ &}

0 91& 8 0' 91 < 01 9 < 1&& 91.3 $ &

9&.%091& 8 1.%091 < 9&.%09 < 91&&91.3 $& .%0 8 .%0 < 0 < 13& $ &

0 $ 3./%2 ()

Therefore the value of a,ial forces for columns A!- @=- " and + are as followsD A! $ 0 $ 3./%2 ()

@= $ 01 $ &.%0 $ 9&.%9 3./%2 $ '.2/ () " $ 0 $ &.%0 $ 9&.%9 3./%2 $ 1.2'/ () + $ 0' $ 1.%0 $ 91.%93./%2 $ .3/2 ()

+etermine shear force and moment for beam A@+

- Cantilever Method

Page 9  m

.& m .& m .& m .&m

1&& ()

1.3 m

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At Point A At Point B ₌_>_{8 $ &} 0 8 ? A@ $ & ? A@ $ ; 0 V2 = " *.$ kN # ₇_A _{8 $ &}

7 A ; ? A@9 $ &

7 A < 9;3./%29 $ & M2 = " 13.1& kNm # ₇_@ _{8 $ &} 7@ ; ?@9' $ & 7@ < 9;1&.'39' $ & M = " 31.308 kNm # 01 $ '.2/ () ₌_>_{8 $ &} ; ? A@ 8 ?@ 8 01 $ & ?@ = " 3./%2 ; '.2/ V4 = " 10.&3* kN # ?@ $ ; 1&.'3 () 7@ ?@ ? A@ $ ; 3./%2 () ?_A@ 0 $ 3./%2 () ? A@ $ ; 3./%2 () 7_A

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At Point C

@ending moment diagram for beam A@+

- Cantilever Method Page 11 ₇_ _{8 $ &} 7 ; ?+9 $ & 7 < 9;.3/29 $ & M4 = " 3&.$88 kNm # 0 $ 1.2'/ () ₌_>_{8 $ &} ; ?@ 8 ?+ ; 0 $ & ?+ = ; 1&.'3 8 1.2'/ V45 = " 8.*$ kN # ?+ $ ; .3/2 () 7 ?+ ?@ $ ; 1&.'3 ()

.&m 3.& m .& m

1'./1 ()m 1'./1 ()m '1.'& ()m '1.'& ()m '.2 ()m '.2 ()m