finite element for portal

8/17/2019 Finite Element for portal

1/157

Untuk setiap batang

E = 2100

A = 35

4. Tentukan koordinat setiap joint dengan referensi sistem koordinat global

odal!joint 1 2 3 4 5" 0 141 4 2#2 # 424 2 5$5 $

t!,m2

,m2


2/157

*g1

0.'0'10' +0.'0'11 0 0 3$'.5 0

=

0.'0'10' 0.'0'10' 0 0

"

0 0

0 0 0.'0'10' +0.'0'11 +3$'.5 0

0 0 0.'0'10' 0.'0'10' 0 0

25-.#$1' 0 +25-.#$2 0 0.'0'10' 0.'0'10'

= 25-.#$1' 0 +25-.#$2 0

" +0.'0'11 0.'0'10'

+25-.#$2 0 25-.#$1' 0 0 0

+25-.#$2 0 25-.#$1' 0 0 0

U1 1 U2 2

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 U1

= 1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 1

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 U2

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 2

&atang 2

i = 1

j = 3

α = 0

) = 2#2.#4


3/157

= 0 0 0 0 1

+25-.#$4 0 25-.#$42 0 U3

0 0 0 0 3

&atang 3

i = 2

j = 3

α = 315

) = 200

*3

3$'.5 0 +3$'.5 0

= 0 0 0 0

+3$'.5 0 3$'.5 0

0 0 0 0

T3

0.'0'10' +0.'0'11 0 0

= 0.'0'10' 0.'0'10' 0 0

0 0 0.'0'10' +0.'0'11

0 0 0.'0'10' 0.'0'10'

*g3

0.'0'10' 0.'0'10' 0 0 3$'.5 0

= +0.'0'11 0.'0'10' 0 0

" 0 0

0 0 0 '0'10' 0 '0'10' +3$' 5 0


4/157

+3$'.5 0 3$'.5 0

0 0 0 0

T4

0.'0'10' 0.'0'10' 0 0

= +0.'0'11 0.'0'10' 0 0

0 0 0.'0'10' 0.'0'10'

0 0 +0.'0'11 0.'0'10'

*g4

0.'0'10' +0.'0'11 0 0 3$'.5 0

= 0.'0'10' 0.'0'10' 0 0 " 0 00 0 0.'0'10' +0.'0'11 +3$'.5 0

0 0 0.'0'10' 0.'0'10' 0 0

25-.#$1' 0 +25-.#$2 0 0.'0'10' 0.'0'10'

= 25-.#$1' 0 +25-.#$2 0

" +0.'0'11 0.'0'10'

+25-.#$2 0 25-.#$1' 0 0 0

+25-.#$2 0 25-.#$1' 0 0 0

U3 3 U4 4

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 U3

= 1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 3

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 U4

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 4


5/157

25-.#$42 0 +25-.#$4 0 1 0

= 0 0 0 0

" 0 1

+25-.#$4 0 25-.#$42 0 0 0

0 0 0 0 0 0

U3 3 U5 5

25-.#$42 0 +25-.#$4 0 U3

= 0 0 0 0 3

+25-.#$4 0 25-.#$42 0 U5

0 0 0 0 5

&atang $

i = 4

j = 5

α = 315

) = 200

*$

3$'.5 0 +3$'.5 0

= 0 0 0 0

+3$'.5 0 3$'.5 0

0 0 0 0

0 '0'10' +0 '0'11 0 0


6/157

i = 2

j = 4

α = 0

) = 2#2.#4

*'

25-.#$42 0 +25-.#$4 0

= 0 0 0 0

+25-.#$4 0 25-.#$42 0

0 0 0 0

T'

1 0 0 0

= 0 1 0 0

0 0 1 0

0 0 0 1

*g'

1 0 0 0 25-.#$42 0

= 0 1 0 0 " 0 00 0 1 0 +25-.#$4 0

0 0 0 1 0 0

25-.#$42 0 +25-.#$4 0 1 0

= 0 0 0 0

" 0 1

+25-.#$4 0 25-.#$42 0 0 0

0 0 0 0 0 0


7/157

U1 1 U2 2 U3 3 U4 4 U5

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 +25-.#$4 0

25-.#$42 0

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 0 0

0 0

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 +1#3.'5 1#3.'5 +25-.#$4 0

1#3.'5 +1#3.'5

25-.#$42 0

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 1#3.'5 +1#3.'5 0 0

+1#3.'5 1#3.'5

0 0

+25-.#$4 0 +1#3.'5 1#3.'5 25-.#$42 0 +1#3.'5 +1#3.'5 +25-.#$4

1#3.'5 +1#3.'5

1#3.'5 1#3.'5

25-.#$42 0

0 0 1#3.'5 +1#3.'5 0 0 +1#3.'5 +1#3.'5 0

+1#3.'5 1#3.'5


8/157

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 0 0 0 0 0

+1#3.'5 +1#3.'5 $2'.3$42 0 +1#3.'5 1#3.'5 +25-.#$4 0 0

+1#3.'5 +1#3.'5 0 3$'.5 1#3.'5 +1#3.'5 0 0 0

+25-.#$4 0 +1#3.'5 1#3.'5 ##'.22#5 0 +1#3.'5 +1#3.'5 +25-.#$4

0 0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5 +1#3.'5 0

0 0 +25-.#$4 0 +1#3.'5 +1#3.'5 $2'.3$42 0 +1#3.'5

0 0 0 0 +1#3.'5 +1#3.'5 0 3$'.5 1#3.'5

0 0 0 0 +25-.#$4 0 +1#3.'5 1#3.'5 443.$142

0 0 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5

'.

1 = 1

1 = 1

2 = 0

2 = +'3 = 0

3 = 0

4 = 0

4 = 0

5 = 0

5 = 5

&entukla/ 6ektor beban luar berdasarkan applied forces dan kondisi tumpuan


9/157

0 0 0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5

0 0 0 +25-.#$4 0 +1#3.'5 +1#3.'5 $2'.3$42

0 0 0 0 0 +1#3.'5 +1#3.'5 0

0 0 0 0 0 +25-.#$4 0 +1#3.'5

5 0 0 0 0 0 0 1#3.'5

10.

atriks 1

U2 2 U3 3 U4 4 U5

0

=

$2'.3$42 0 +1#3.'5 1#3.'5 +25-.#$4 0 0

+' 0 3$'.5 1#3.'5 +1#3.'5 0 0 0

0 +1#3.'5 1#3.'5 ##'.22#5 0 +1#3.'5 +1#3.'5 +25-.#$4

0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5 +1#3.'5 0

0 +25-.#$4 0 +1#3.'5 +1#3.'5 $2'.3$42 0 +1#3.'5

0 0 0 +1#3.'5 +1#3.'5 0 3$'.5 1#3.'50 0 0 +25-.#$4 0 +1#3.'5 1#3.'5 443.$142

1 +1#3.'5 +1#3.'5 +25-.#$4 0 0 0 0

1 +1#3.'5 +1#3.'5 0 0 0 0 0

5 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5

11.

)akukan rearrangement pada matri keseimbangan struktur agar siap ditentukan solusin7a

Tentukan solusi matri keseimbangan struktur 7ang tela/ di+ rearrangement untuk memperole

Menghitung Displacement


10/157

4 = +0.02$

U5 = 0.02'

Menghitung Reaksi Tumpuan

1

=

+1#3.'5 +1#3.'5 +25-.#$4 0 0 0 0

1 +1#3.'5 +1#3.'5 0 0 0 0 0

5 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5

1 = +#E+015

1 = 5.25

5 = 1.'5

aka 8ispla,ement

U1 = 0

1 = 0

U2 = 0.024

2 = +0.052

U3 = 0.020


11/157

+'.424$2

0

&atang 2

1

=

1 0 0 0 25-.#$42 0

1 0 1 0 0"

0 0

3 0 0 1 0 +25-.#$4 0

3 0 0 0 1 0 0

=

25-.#$42 0 +25-.#$4 0 0

0 0 0 0"

0

+25-.#$4 0 25-.#$42 0 0.020203

0 0 0 0 +0.045-#

=

+5.25

05.25

0

&atang 3

2 0.'0'10' +0.'0'11 0 0 1#3.'5 +1#3.'5

2 0 '0'10' 0 '0'10' 0 0 +1#3 '5 1#3 '5


12/157

+25-.#$2 +25-.#$2 25-.#$1' 25-.#$1' 0.010101

0 0 0 0 +0.02$3$

=

+2.4'4#'0

2.4'4#'4

0

&atang 5

3

=

1 0 0 0 25-.#$42 0

3 0 1 0 0"

0 0

5 0 0 1 0 +25-.#$4 0

5 0 0 0 1 0 0

=

25-.#$42 0 +25-.#$4 0 0.020203

0 0 0 0 " +0.045-#+25-.#$4 0 25-.#$42 0 0.02$-3'

0 0 0 0 0

=

+1.'5

0

1.'5

0


13/157

4 0 0 1 0 +25-.#$4 0

4 0 0 0 1 0 0

=

25-.#$42 0 +25-.#$4 0 0.0235'0 0 0 0

" +0.05214

+25-.#$4 0 25-.#$42 0 0.010101

0 0 0 0 +0.02$3$

=

3.5

0

+3.5

0


14/157


15/157

+3$'.5 0 0.'0'10' 0.'0'10' 0 0

0 0"

+0.'0'11 0.'0'10' 0 0

3$'.5 0 0 0 0.'0'10' 0.'0'10'0 0 0 0 +0.'0'11 0.'0'10'

0 0

0 0

0.'0'10' 0.'0'10'

+0.'0'11 0.'0'10'


16/157

+3$'.5 0 0.'0'10' +0.'0'11 0 0

0 0

"

0.'0'10' 0.'0'10' 0 0

3$' 5 0 0 0 0 '0'10' +0 '0'11


17/157

+3$'.5 0 0.'0'10' 0.'0'10' 0 0

0 0"

+0.'0'11 0.'0'10' 0 0

3$'.5 0 0 0 0.'0'10' 0.'0'10'

0 0 0 0 +0.'0'11 0.'0'10'

0 0

0 0

0.'0'10' 0.'0'10'

+0.'0'11 0.'0'10'


18/157

0 0

0 0

1 0

0 1


19/157


20/157

5

U1

1

U2

2

0

U3

0


21/157


22/157

+1#3.'5 0 0 3 3

0 +1#3.'5 1#3.'5 U4 U4

3$'.5 1#3.'5 +1#3.'5 4 4

1#3.'5 443.$142 +1#3.'5 U5 U5+1#3.'5 +1#3.'5 1#3.'5 5 0

U1 1 5

+1#3.'5 +1#3.'5 0 U2 U2

+1#3.'5 +1#3.'5 0 2 2

+25-.#$4 0 0 U3 U3

0 0 0 3 3

0 0 1#3.'5 U4 U4

0 0 +1#3.'5 4 40 0 +1#3.'5 U5 U5

443.$142 1#3.'5 0 U1 U1

1#3.'5 1#3.'5 0 1 1

0 0 1#3.'5 5 5

displacement setiap nodal dan reaksi tumpuan


23/157

0.024

" +0.052

0.020

+0.04$

0.010

+0.02$

0.02'


24/157

+25-.#$4 0 0

0 0"

0

25-.#$42 0 0.020

0 0 +0.04$

+1#3.'5 1#3.'5 0.024

1#3 '5 1#3 '5 0 052


25/157

+25-.#$4 0 0.020

0 0"

+0.04$

25-.#$42 0 0.02'

0 0 0.000


26/157

25-.#$42 0 0.010

0 0 +0.02$


27/157


28/157


29/157


30/157


31/157


32/157


33/157


34/157


35/157


36/157


37/157


38/157


39/157


40/157


41/157

*g1

0.'0'10' +0.'0'1 0 0 3$'.5 0 +3$'.5

= 0.'0'10' 0.'0'11 0 0

" 0 0 0

0 0 0.'0'11 +0.'0'1 +3$'.5 0 3$'.50 0 0.'0'11 0.'0'11 0 0 0

0.'0'10' 0.'0'11 0 0

+0.'0'11 0.'0'11 0 0

0 0 0.'0'11 0.'0'11

0 0 +0.'0'1 0.'0'11

25-.#$1' 0 +25-.#$ 0 0.'0'11 0.'0'11 0

= 25-.#$1' 0 +25-.#$ 0

" +0.'0'1 0.'0'11 0

+25-.#$2 0 25-.#$2 0 0 0 0.'0'11

+25-.#$2 0 25-.#$2 0 0 0 +0.'0'1

U1 1 U2 2

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 U1

= 1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 1

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 U2

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 2

&atang 2


42/157

0 0 1 +2E+01$

0 0 2E+01$ 1

25-.#$42 0 +25-.#$ 0 1 +2E+01$ 0=

+$E+014 0 $E+014 0"

2E+01$ 1 0

+25-.#$4 0 25-.#$4 0 0 0 1

$.4E+014 0 +$E+014 0 0 0 2E+01$

U1 1 U3 3

25-.#$42 +$E+014 +25-.#$ $E+014 U1

= +$E+014 2E+02- $E+014 +2E+02- 1+25-.#$4 $E+014 25-.#$4 +$E+014 U3

$.4E+014 +2E+02- +$E+014 2E+02- 3

&atang 3

i = 2

j = 3

α = 315

) = 200

k3

3$'.5 0 +3$'.5 0

= 0 0 0 0


43/157

U2 2 U3 3

1#3.'5 +1#3.'5 +1#3.'5 1#3.'5 U2

= +1#3.'5 1#3.'5 1#3.'5 +1#3.'5 2

+1#3.'5 1#3.'5 1#3.'5 +1#3.'5 U3

1#3.'5 +1#3.'5 +1#3.'5 1#3.'5 3

&atang 4

i = 3

j = 4

α = 45

) = 200

k4

3$'.5 0 +3$'.5 0

= 0 0 0 0

+3$'.5 0 3$'.5 0

0 0 0 0


44/157

= 1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 3

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 U4

+1#3.'5 +1#3.'5 1#3.'5 1#3.'5 4

&atang 5

i = 3

j = 5

α = 3$0) = 2#2.#4

k5

25-.#$42 0 +25-.#$ 0

= 0 0 0 0

+25-.#$4 0 25-.#$4 0

0 0 0 0

t5

1 +2E+01$ 0 0

= 2.5E+01$ 1 0 0

0 0 1 +2E+01$

0 0 2E+01$ 1


45/157

i = 4

j = 5

α = 315

) = 200

k$

3$'.5 0 +3$'.5 0

= 0 0 0 0

+3$'.5 0 3$'.5 0

0 0 0 0

t$

0.'0'10' +0.'0'1 0 0

= 0.'0'10' 0.'0'11 0 0

0 0 0.'0'11 +0.'0'1

0 0 0.'0'11 0.'0'11

*g$

0.'0'10' 0.'0'11 0 0 3$'.5 0 +3$'.5

= +0.'0'11 0.'0'11 0 0

" 0 0 0

0 0 0.'0'11 0.'0'11 +3$'.5 0 3$'.5

0 0 +0.'0'1 0.'0'11 0 0 0

0.'0'10' +0.'0'1 0 0


46/157

k' = 0 0 0 0

+25-.#$4 0 25-.#$4 0

0 0 0 0

t'

1 +2E+01$ 0 0

= 2.5E+01$ 1 0 0

0 0 1 +2E+01$

0 0 2E+01$ 1

*g'

1 2E+01$ 0 0 25-.#$4 0 +25-.#$

= +2E+01$ 1 0 0 " 0 0 00 0 1 2E+01$ +25-.#$ 0 25-.#$4

0 0 +2E+01$ 1 0 0 0

1 +2E+01$ 0 0

2.5E+01$ 1 0 0

0 0 1 +2E+01$

0 0 2E+01$ 1

25-.#$42 0 +25-.#$ 0 1 +2E+01$ 0

= +$E+014 0 $E+014 0

" 2E+01$ 1 0

+25-.#$4 0 25-.#$4 0 0 0 1

$.4E+014 0 +$E+014 0 0 0 2E+01$


47/157

&atang 5 &atang $

U3 3 U5 5 U4 4 U5 5

25-.#$4 +$E+014 +25-.#$4 $E+014 U3 1#3.'5 +1#3.'5 +1#3.'5 1#3.'5

+$E+014 2E+02- $.4E+014 +2E+02- 3 +1#3.'5 1#3.'5 1#3.'5 +1#3.'5+25-.#$ $E+014 25-.#$42 +$E+014 U5 +1#3.'5 1#3.'5 1#3.'5 +1#3.'5

$E+014 +2E+02- +$E+014 2E+02- 5 1#3.'5 +1#3.'5 +1#3.'5 1#3.'5

&atang 'U2 2 U4 4

25-.#$4 +$E+014 +25-.#$4 $E+014 U2

+$E+014 2E+02- $.4E+014 +2E+02- 2

+25-.#$ $E+014 25-.#$42 +$E+014 U4

$E+014 +2E+02- +$E+014 2E+02- 4

U1 1 U2 2 U3 3 U4 4 U5 5

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 +25-.#$ $E+014

25-.#$4 +$E+014

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 $E+014 +2E+02-


48/157

$.4E+014 +2E+02- +1#3.'5 +1#3.'5 1#3.'5 1#3.'5 1#3.'5 +1#3.'5

+1#3.'5 1#3.'5

+$E+014 2E+02-

+25-.#$ $E+014 +1#3.'5 1#3.'5 25-.#$4 +$E+014

1#3.'5 +1#3.'5

$E+014 +2E+02- 1#3.'5 +1#3.'5 +$E+014 2E+02-

+1#3.'5 1#3.'5

atriks

443.$14 1#3.'5 +1#3.'5 +1#3.'5 +25-.#$ $E+014 0 0 0 0

1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 $E+014 +2E+02- 0 0 0 0

+1#3.'5 +1#3.'5 $2'.3$42 +$E+014 +1#3.'5 1#3.'5 +25-.#$ $E+014 0 0

+1#3.'5 +1#3.'5 +$E+014 3$'.5 1#3.'5 +1#3.'5 $E+014 +2E+02- 0 0

+25-.#$ $E+014 +1#3.'5 1#3.'5 ##'.22# 0 +1#3.'5 +1#3.'5 +25-.#$ $E+014

$E+014 +2E+02- 1#3.'5 +1#3.'5 +$E+014 3$'.5 +1#3.'5 +1#3.'5 $E+014 +2E+02-

0 0 +25-.#$4 $E+014 +1#3.'5 +1#3.'5 $2'.3$4 0 +1#3.'5 1#3.'5

0 0 $.4E+014 +2E+02- +1#3.'5 +1#3.'5 0 3$'.5 1#3.'5 +1#3.'5

0 0 0 0 +25-.#$ $E+014 +1#3.'5 1#3.'5 443.$14 +1#3.'5


49/157

3 = 3

U4 = U4

4 = 4

U5 = U5

5 = 0


50/157

#.

untuk memperole/ displa,ement setiap nodal dan reaksi tumpuan

+1

0

=

$2'.3$42 +3E+014 +1#3.'5 1#3.'5 +25-.#$ 0 0

+' 0 3$'.5 1#3.'5 +1#3.'5 0 0 0

0 +1#3.'5 1#3.'5 ##'.22# 0 +1#3.'5 +1#3.'5 +25-.#$

0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5 +1#3.'5 0 "

Tentukan solusi matri keseimbangan struktur 7ang tela/ di+ rearrangement

Menghitung Displacement


51/157

1

=

+1#3.'5 +1#3.'5 +25-.#$ 0 0 0 0

1 +1#3.'5 +1#3.'5 0 0 0 0 0 "

5 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5

1 = +#E+015

1 = 5.25

5 = 1.'5

aka 8ispla,ement

U1 = 0.00

1 = 0.00

U2 = 0.02

2 = +0.05

U3 = 0.02

3 = +0.05

U4 = 0.01

4 = +0.03

U5 = 0.03


52/157

= 0 Tekan

+'.424$2

0

&atang 2

f1

=

1 +2E+01$ 0 0 25-.#$4 +$E+014 +25-.#$

g1 2.5E+01$ 1 0 0"

+$E+014 2E+02- $E+014

f3 0 0 1 +2E+01$ +25-.#$ $E+014 25-.#$4


53/157

&atang 3

f2

=

0.'0'10' +0.'0'1 0 0 1#3.'5 +1#3.'5 +1#3.'5

g2 0.'0'10' 0.'0'11 0 0"

+1#3.'5 1#3.'5 1#3.'5


54/157

&atang 4

f3 0.'0'10' 0.'0'11 0 0 1#3.'5 1#3.'5 +1#3.'5


55/157


56/157


57/157

&atang '


58/157


59/157


60/157

0

0"

0

0

0

0

0.'0'11

0.'0'11


61/157

0

0

+2E+01$

1


62/157


63/157

0


64/157

00

"0

0


65/157

0

0 "0

0

0

0

+2E+01$

1


66/157

U4

4

U5

5

U1

1


67/157

4

U5

5


68/157


69/157

U2

2

U3

3

0.02

0 05


70/157

+0.05

0.02

+0.05

0.01

+0.03

0.03


71/157

$E+014

+2E+02-"

+$E+014

2E+02-


72/157

1#3.'5

+1#3.'5"

+1#3 '5


73/157

+1#3.'5

+1#3 '5


74/157

$E+014


75/157


76/157


77/157

$. &entuk matri keseimbangan struktur

U1 1 U2 2 U3 3 U4 4 U5 5


78/157

1

=

443.$14 1#3.'5 +1#3.'5 +1#3.'5 +25-.#$4 0 0 0 0 0 U1

1 1#3.'5 1#3.'5 +1#3.'5 +1#3.'5 0 0 0 0 0 0 1

0 +1#3.'5 +1#3.'5 $2'.3$4 +3E+014 +1#3.'5 1#3.'5 +25-.#$4 0 0 0 U2

+' +1#3.'5 +1#3.'5 0 3$'.5 1#3.'5 +1#3.'5 0 0 0 0 20 +25-.#$4 0 +1#3.'5 1#3.'5 ##'.22# 0 +1#3.'5 +1#3.'5 +25-.#$4 0 U3

0 0 0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5 +1#3.'5 0 0 3

0 0 0 +25-.#$4 0 +1#3.'5 +1#3.'5 $2'.3$4 0 +1#3.'5 1#3.'5 U4

0 0 0 0 0 +1#3.'5 +1#3.'5 0 3$'.5 1#3.'5 +1#3.'5 4

0 0 0 0 0 +25-.#$4 0 +1#3.'5 1#3.'5 443.$14 +1#3.'5 U5

5 0 0 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5 1#3.'5 5

'.

atriks 1



79/157

U2 2 U3 3 U4 4 U5 U1 1 5

0

=

$2'.3$4 +3E+014 +1#3.'5 1#3.'5 +25-.#$4 0 0 +1#3.'5 +1#3.'5 0 U2

+' 0 3$'.5 1#3.'5 +1#3.'5 0 0 0 +1#3.'5 +1#3.'5 0 20 +1#3.'5 1#3.'5 ##'.22# 0 +1#3.'5 +1#3.'5 +25-.#$4 +25-.#$4 0 0 U3

0 1#3.'5 +1#3.'5 0 3$'.5 +1#3.'5 +1#3.'5 0 0 0 0 3

0 +25-.#$4 0 +1#3.'5 +1#3.'5 $2'.3$4 0 +1#3.'5 0 0 1#3.'5 U4

0 0 0 +1#3.'5 +1#3.'5 0 3$'.5 1#3.'5 0 0 +1#3.'5 4

0 0 0 +25-.#$4 0 +1#3.'5 1#3.'5 443.$14 0 0 +1#3.'5 U5

1 +1#3.'5 +1#3.'5 +25-.#$4 0 0 0 0 443.$14 1#3.'5 0 U1

1 +1#3.'5 +1#3.'5 0 0 0 0 0 1#3.'5 1#3.'5 0 1

5 0 0 0 0 1#3.'5 +1#3.'5 +1#3.'5 0 0 1#3.'5 5


80/157

0

0

U2

2U3

3

U4

4

U5

0


81/157

U2

2U3

3

U4

4

U5

U1

1

5


82/157


83/157


84/157

Untuk setiap batang

E = 2100

A = -$

9 = 1152

: = 0.01 t!,m

aka

;2 = 3 t 2 = 300 t.,m;3 = 3 t 3 = +300 t.,m

1 Tentukan koordinat setiap joint dengan referensi sistem koordinat global

odal!joint 1 2 3 4

" 0 0 $00 $00

% 0 500 500 300

t!,m2

,m2

,m4

−

=

100000

0,ossin000

0sin,os000

α α

α α


85/157

f. ?itung matri kekakuan global batang

&atang 1

i = 1

j = 2

α = -0

) = 500

k1

403.2 0 0 +403.2 0 0

0 4#3#.4 241-.2 0 +4#3#.4 241-.2

= 0 241-.2 1$12.# 0 +241-.2 #0$.4

+403.2 0 0 403.2 0 0

0 +4#3#.4 +241-.2 0 4#3#.4 +241-.2

0 241-.2 #0$.4 0 +241-.2 1$12.#

t1

$E+01' 1 0 0 0 0

+1 $E+01' 0 0 0 0

= 0 0 1 0 0 0

0 0 0 $E+01' 1 0

0 0 0 +1 $E+01' 0

0 0 0 0 0 1

$E+01' +1 0 0 0 0 403.2 0

1 $E 01' 0 0 0 0 0 4#3# 4

100000

[ ] [ ] [ ] [ ]T k T k l T

g ⋅⋅=

k2

33$ 0 0 +33$ 0 0

0 4032 201$ 0 +4032 201$

0 201$ 1344 0 +201$ $'2


86/157

k2 =+33$ 0 0 33$ 0 0

0 +4032 +201$ 0 4032 +201$

0 201$ $'2 0 +201$ 1344

T2

1 +2E+01$ 0 0 0 0

2E+01$ 1 0 0 0 0

= 0 0 1 0 0 0

0 0 0 1 +2E+01$ 0

0 0 0 2E+01$ 1 0

0 0 0 0 0 1

*g2

1 2E+01$ 0 0 0 0 33$ 0

+2E+01$ 1 0 0 0 0 0 4032

= 0 0 1 0 0 0

0 201$

0 0 0 1 2E+01$ 0 +33$ 0

0 0 0 +2E+01$ 1 0 0 +4032

0 0 0 0 0 1 0 201$

33$ 1E+012 5E+013 +33$ +1E+012 5E+013 1 +2E+01$

+#E+014 4032 201$ #E+014 +4032 201$ 2E+01$ 1

= 0 201$ 1344 0 +201$ $'2

0 0

+33$ +1E+012 +5E+013 33$ 1E+012 +5E+013 0 0

#E+014 +4032 +201$ +#E+014 4032 +201$ 0 0

0 201$ $'2 0 +201$ 1344 0 0

U2 2 @2 U3 3 @3

33$ -E 013 5E 013 33$ -E 013 5E 013 U2

*g3

+2E+01$ 1 0 0 0 0 100# 0

+1 +2E+01$ 0 0 0 0 0 120-$

= 0 0 1 0 0 0

0 $04#


87/157

*g3 = 0 0 0 +2E+01$ 1 0 +100# 0

0 0 0 +1 +2E+01$ 0 0 +120-$

0 0 0 0 0 1 0 $04#

+2E+013 120-$ $04# 2E+013 +120-$ $04# +2E+01$ +1

+100# +2E+012 +1E+012 100# 2E+012 +1E+012 1 +2E+01$

= 0 $04# 4032 0 +$04# 201$

0 0

2E+013 +120-$ +$04# +2E+013 120-$ +$04# 0 0

100# 2E+012 1E+012 +100# +2E+012 1E+012 0 0

0 $04# 201$ 0 +$04# 4032 0 0

U3 3 @3 U4 4 @4

120-$ +2E+012 $04# +120-$ 2E+012 $04# U3

+2E+012 100# +1E+012 2E+012 +100# +1E+012 3

= $04# +1E+012 4032 +$04# 1E+012 201$ @3

+120-$ 2E+012 +$04# 120-$ +2E+012 +$04# U4

2E+012 +100# 1E+012 +2E+012 100# 1E+012 4

$04# +1E+012 201$ +$04# 1E+012 4032 @4

3

atri kekakuan seluru/ struktur dapat ditentukan dengan menggabungkan seluru/

matrik kekakuan batang dalam koordinat global dengan memila/+mila/ sesuai dengan

perpinda/an global 7ang terjadi

U1 1 @1 U2 2 @2 U3 3 @3 U4 4

4#3#.4 +3E+013 +241-.2 +4#3#.4 3E+013 +241-.2

&entukla/ matri kekakuan seluru/ struktur overall stiffnes matrix

+3E+013 403.2 1E+013 3E+013 +403.2 1E+013 0 0 0 0 0

+241-.2 1E+013 1$12.# 241-.2 +1E+013 #0$.4 0 0 0 0 0

+4#3#.4 3E+013 241-.2 51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0 0


88/157

3E+013 +403.2 +1E+013 $E+013 4435.2 201$ +-E+013 +4032 201$ 0 0

+241-.2 1E+013 #0$.4 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0 0

0 0 0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# +120-$ 2E+0120 0 0 +-E+013 +4032 +201$ +1E+012 5040 +201$ 2E+012 +100#

0 0 0 5E+013 201$ $'2 $04# +201$ 53'$ +$04# 1E+012

0 0 0 0 0 0 +120-$ 2E+012 +$04# 120-$ +2E+012

0 0 0 0 0 0 2E+012 +100# 1E+012 +2E+012 100#

0 0 0 0 0 0 $04# +1E+012 201$ +$04# 1E+012

aka

4#3#.4 +3E+013 +241-.2 +4#3#.4 3E+013 +241-.2 0 0 0 0 0

+3E+013 403.2 1E+013 3E+013 +403.2 1E+013 0 0 0 0 0

+241-.2 1E+013 1$12.# 241-.2 +1E+013 #0$.4 0 0 0 0 0

+4#3#.4 3E+013 241-.2 51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0 0

3E+013 +403.2 +1E+013 $E+013 4435.2 201$ +-E+013 +4032 201$ 0 0

+241-.2 1E+013 #0$.4 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0 0

0 0 0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# +120-$ 2E+012

0 0 0 +-E+013 +4032 +201$ +1E+012 5040 +201$ 2E+012 +100#

0 0 0 5E+013 201$ $'2 $04# +201$ 53'$ +$04# 1E+012

0 0 0 0 0 0 +120-$ 2E+012 +$04# 120-$ +2E+012

0 0 0 0 0 0 2E+012 +100# 1E+012 +2E+012 100#

0 0 0 0 0 0 $04# +1E+012 201$ +$04# 1E+012

4

1 1

&entukla/ 6ektor beban luar berdasarkan applied forces dan kondisi tumpuan

$ &entuk matri keseimbangan struktur

[ ] [ ] [ ]U K P ⋅=


89/157

U1 1 @1 U2 2 @2 U3 3 @3

1

=

4#3#.4 +3E+013 +241-.2 +4#3#.4 3E+013 +241-.2 0 0 0

1 +3E+013 403.2 1E+013 3E+013 +403.2 1E+013 0 0 01 +241-.2 1E+013 1$12.# 241-.2 +1E+013 #0$.4 0 0 0

0 +4#3#.4 3E+013 241-.2 51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013

+3 3E+013 +403.2 +1E+013 $E+013 4435.2 201$ +-E+013 +4032 201$

300 +241-.2 1E+013 #0$.4 241-.2 201$ 2-5$.# +5E+013 +201$ $'2

0 0 0 0 +33$ +-E+013 +5E+013 12432 +1E+012 $04#

+3 0 0 0 +-E+013 +4032 +201$ +1E+012 5040 +201$

+300 0 0 0 5E+013 201$ $'2 $04# +201$ 53'$

4 0 0 0 0 0 0 +120-$ 2E+012 +$04#

4 0 0 0 0 0 0 2E+012 +100# 1E+012

0 0 0 0 0 0 0 $04# +1E+012 201$

'

0

=

51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0 +4#3#.4 3E+013

+3 $E+013 4435.2 201$ +-E+013 +4032 201$ 0 3E+013 +403.2

300 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0 +241-.2 1E+013

0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# $04# 0 0

+3 +-E+013 +4032 +201$ +1E+012 5040 +201$ +1E+012 0 0

+300 5E+013 201$ $'2 $04# +201$ 53'$ 201$ 0 0

0 0 0 0 $04# +1E+012 201$ 4032 0 0

1 4#3# 4 3E 013 241- 2 0 0 0 0 4#3# 4 3E 013


=

2

1

2

1

U

U

d c

ba

P

P

3 +0.0211-

@3 +0.32022

@4 +0.2$142


90/157

eng/itung Beaksi Tumpuan

1

=

+4#3#.4 3E+013 +241-.2 0 0 0 0 +0.0'1''

1 3E+013 +403.2 1E+013 0 0 0 0 0.03#1

1 241-.2 +1E+013 #0$.4 0 0 0 0 0.1-253

4 0 0 0 +120-$ 2E+012 +$04# +$04# 0.2#102

4 0 0 0 2E+012 +100# 1E+012 1E+012 +0.0211-

+0.32022

+0.2$142

1

=

+11#.53$

1 +15.4

1 +1#.3$0- 0.0

4 11#.53$

4 21.3$0-

8ispla,ement

U1 = 0

1 = 0

@1 = 0

U2 = +0.0'1''

2 = 0.03#1

@2 = 0.1-253

U3 = 0.2#102

3 0 0211-

+241-.2 1E+013 #0$.4 241-.2 +1E+013 1$12.# 0.1-253

f1 +15.3$0-

1 11# 53$


91/157

=

g1 11#.53$

m1 +1#.3$0-

f2 15.3$0-g2 +11#.53$

m2 13$.#-'

&atang 2

f2

=

1 +2E+01$ 0 0 0 0 33$ -E+013

g2 2E+01$ 1 0 0 0 0 -E+013 4032

m2 0 0 1 0 0 0 5E+013 201$f3 0 0 0 1 +2E+01$ 0 +33$ +-E+013

g3 0 0 0 2E+01$ 1 0 +-E+013 +4032

m3 0 0 0 0 0 1 5E+013 201$

=

33$ +#E+014 0 +33$ #E+014 0 +0.0'1''

1E+012 4032 201$ +1E+012 +4032 201$ 0.03#1

5E+013 201$ 1344 +5E+013 +201$ $'2

0.1-253

+33$ #E+014 0 33$ +#E+014 0 0.2#102

+1E+012 +4032 +201$ 1E+012 4032 +201$ +0.0211-

5E+013 201$ $'2 +5E+013 +201$ 1344 +0.32022

=

+11#.53$

+1#.4

1$3.103

11#.53$

1# 3$0-

Untuk menentukan ga7a+ga7a dalam internal for,es lakukan seperti pada atakulia/

ekanika Beka7asa 9 99 999 dan 9

11 ambarkan ga7a ga7a luar external forces dan ga7a ga7a dalam internal


92/157

11

for,es pada setiap batang

ambarkan ga7a+ga7a luar external forces dan ga7a+ga7a dalam internal


93/157


94/157

0 +403.2 0 0 $E+01' 1 0 0 0 0

241- 2 0 4#3# 4 241- 2 1 $E 01' 0 0 0 0


95/157

0 +33$ 0 0 1 +2E+01$ 0 0 0 0

201$ 0 +4032 201$ 2E+01$ 1 0 0 0 0

1344 0 +201$ $'2

0 0 1 0 0 0

0 33$ 0 0 0 0 0 1 +2E+01$ 0

+201$ 0 4032 +201$ 0 0 0 2E+01$ 1 0

$'2 0 +201$ 1344 0 0 0 0 0 1

0 0 0 0

0 0 0 01 0 0 0

0 1 +2E+01$ 0

0 2E+01$ 1 0

0 0 0 1

0 +100# 0 0 +2E+01$ +1 0 0 0 0

$04# 0 +120-$ $04# 1 +2E+01$ 0 0 0 0

4032 0 +$04# 201$

0 0 1 0 0 0

0 100# 0 0 0 0 0 +2E+01$ +1 0


96/157

0 00# 0 0 0 0 0 0 $ 0

+$04# 0 120-$ +$04# 0 0 0 1 +2E+01$ 0

201$ 0 +$04# 4032 0 0 0 0 0 1

0 0 0 0

0 0 0 0

1 0 0 0

0 +2E+01$ +1 0

0 1 +2E+01$ 0

0 0 0 1

@4

U1 U1 1 @1 U2 2 @2

4#3# 4 3E 013 241- 2 4#3# 4 3E 013 241- 2 U1

0 1

0 @1

0 U2

0 2


97/157

0 @2

$04# U3

+1E+012 3

201$ @3

+$04# U4

1E+012 4

4032 @4

0

0

0

0

0

0

$04#

+1E+012

201$+$04#

1E+012

4032

U4 4 @4


98/157

0 0 0 U1

"

0

0 0 0 1 0

0 0 0 @1 0

0 0 0 U2 U2

0 0 0 2 2

0 0 0 @2 @2

+120-$ 2E+012 $04# U3 U3

2E+012 +100# +1E+012 3 3

+$04# 1E+012 201$ @3 @3

120-$ +2E+012 +$04# U4 0

+2E+012 100# 1E+012 4 0

+$04# 1E+012 4032 @4 @4

241-.2 0 0 U2+1E+013 0 0 2

#0$.4 0 0 @2

0 +120-$ 2E+012 U3

0 2E+012 +100# 3

0 +$04# 1E+012

@3

0 +$04# 1E+012 @4

241- 2 0 0 0


99/157


100/157

5E+013 +33$ +-E+013 5E+013 +0.0'1''

201$ +-E+013 +4032 201$ 0.03#1

1344 +5E+013 +201$ $'2

0.1-253

+5E+013 33$ -E+013 +5E+013 0.2#102

+201$ -E+013 4032 +201$ +0.0211-

$'2 +5E+013 +201$ 1344 +0.32022


101/157


102/157


103/157


104/157


105/157


106/157


107/157


108/157


109/157


110/157


111/157

Untuk setiap batang

E = 2100

A = -$9 = 1152

: = 0.01 t!,m

aka

;2 = 3 t 2 = 300 t.,m

3 3 3 300

t!,m2

,m2

,m4

d. ?itung matri kekakuan batang dalam koordinat lokal


112/157

e. ?itung matri transformasi batang

f. ?itung matri kekakuan global batang

[ ]

−

−−−

−

−

−

−

=

L EI

L

EI L EI

L

EI

L

E

L

EI

L

EI

L

EI

L AE

L AE

L EI

L

EI L EI

L

EI

L

EI

L

EI

L

EI

L

EI

L AE

L AE

l k

4$2$

$12$12

2$4$

$12$12

22

2323

22

2323

00

00

0000

00

00

0000

[ ]

−

−

=

100000

0,ossin000

0sin,os000

000100

0000,ossin

0000sin,os

α α

α α

α α

α α

T

[ ] [ ] [ ] [ ]T k T k l T

g ⋅⋅=

$E+01' +1 0 0 0 0

1 $E+01' 0 0 0 0

0 0 1 0 0 0


113/157

*g1 =

0 0 1 0 0 0

0 0 0 $E+01' +1 0

0 0 0 1 $E+01' 0

0 0 0 0 0 1

403.2 0 0 +403.2 0 0

0 4#3#.4 241-.2 0 +4#3#.4 241-.2

0 241-.2 1$12.# 0 +241-.2 #0$.4

+403.2 0 0 403.2 0 0

0 +4#3#.4 +241-.2 0 4#3#.4 +241-.2

0 241-.2 #0$.4 0 +241-.2 1$12.#

$E+01' 1 0 0 0 0

+1 $E+01' 0 0 0 0

0 0 1 0 0 0

0 0 0 $E+01' 1 0

0 0 0 +1 $E+01' 0

0 0 0 0 0 1

2E+014 +4#3#.4 +241-.2 +2E+014 4#3#.4 +241-.2

403.2 3E+013 1E+013 +403.2 +3E+013 1E+013

i = 2

j 3


114/157

j = 3

α = 3$0

) = $00

k2

33$ 0 0 +33$ 0 0

0 4032 201$ 0 +4032 201$

= 0 201$ 1344 0 +201$ $'2

+33$ 0 0 33$ 0 0

0 +4032 +201$ 0 4032 +201$

0 201$ $'2 0 +201$ 1344

T2

1 +2E+01$ 0 0 0 0

2E+01$ 1 0 0 0 0

= 0 0 1 0 0 0

0 0 0 1 +2E+01$ 0

0 0 0 2E+01$ 1 0

0 0 0 0 0 1

33$ 1E 012 5E 013 33$ 1E 012 5E 013


115/157

33$ 1E+012 5E+013 +33$ +1E+012 5E+013

+#E+014 4032 201$ #E+014 +4032 201$

= 0 201$ 1344 0 +201$ $'2

+33$ +1E+012 +5E+013 33$ 1E+012 +5E+013

#E+014 +4032 +201$ +#E+014 4032 +201$

0 201$ $'2 0 +201$ 1344

1 +2E+01$ 0 0 0 0

2E+01$ 1 0 0 0 0

0 0 1 0 0 0

0 0 0 1 +2E+01$ 0

0 0 0 2E+01$ 1 0

0 0 0 0 0 1

U2 2 @2 U3 3 @3

33$ -E+013 5E+013 +33$ +-E+013 5E+013 U2

-E+013 4032 201$ +-E+013 +4032 201$ 2

= 5E+013 201$ 1344 +5E+013 +201$ $'2 @2

+33$ +-E+013 +5E+013 33$ -E+013 +5E+013 U3

+-E+013 +4032 +201$ -E+013 4032 +201$ 3

5E+013 201$ $'2 +5E+013 +201$ 1344 @3

+2E+01$ 1 0 0 0 0

+1 +2E+01$ 0 0 0 0

0 0 1 0 0 0


116/157

*g3 =

0 0 1 0 0 0

0 0 0 +2E+01$ 1 0

0 0 0 +1 +2E+01$ 0

0 0 0 0 0 1

100# 0 0 +100# 0 0

0 120-$ $04# 0 +120-$ $04#

0 $04# 4032 0 +$04# 201$

+100# 0 0 100# 0 0

0 +120-$ +$04# 0 120-$ +$04#

0 $04# 201$ 0 +$04# 4032

+2E+01$ +1 0 0 0 0

1 +2E+01$ 0 0 0 0

0 0 1 0 0 0

0 0 0 +2E+01$ +1 0

+120-$ 2E+012 +$04# 120-$ +2E+012 +$04# U4

2E+012 +100# 1E+012 +2E+012 100# 1E+012 4

$04# +1E+012 201$ +$04# 1E+012 4032 @4


117/157

$04# 1E 012 201$ $04# 1E 012 4032 @4

+2E+012 100# +1E+012 2E+012

5E+013 201$ $'2 +5E+013 +201$ 1344

$04# +1E+012 4032 +$04#


118/157

$04# 1E 012 4032 $04#

+120-$ 2E+012 +$04# 120-$

2E+012 +100# 1E+012 +2E+012

$04# +1E+012 201$ +$04#

adi matri kekakuan tersebut adala/

U1 1 @1 U2 2 @2 U3 3 @3 U4

4#3#.4 +3E+013 +241-.2 +4#3#.4 3E+013 +241-.2 0 0 0 0

+3E+013 403.2 1E+013 3E+013 +403.2 1E+013 0 0 0 0

+241-.2 1E+013 1$12.# 241-.2 +1E+013 #0$.4 0 0 0 0

+4#3#.4 3E+013 241-.2 51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0

3E+013 +403.2 +1E+013 $E+013 4435.2 201$ +-E+013 +4032 201$ 0

+241-.2 1E+013 #0$.4 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0

0 0 0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# +120-$

0 0 0 +-E+013 +4032 +201$ +1E+012 5040 +201$ 2E+012

0 0 0 5E+013 201$ $'2 $04# +201$ 53'$ +$04#

0 0 0 0 0 0 +120-$ 2E+012 +$04# 120-$

1 = 1

1 = 1

1 = 1


119/157

2 = 0

2 = +3

2 = 300

3 = 0

3 = +3

3 = +300

4 = 4

4 = 4

4 = 0

U1 = 0

1 = 0

@1 = 0

U2 = U2

2 = 2

@2 = @2

U3 = U3

3 = 3

&entukla/ 6ektor displacement berdasarkan kondisi kekangan joint dan tumpuan


120/157

=

2

1

2

1

U

U

d c

ba

P

P

0 0 0 0 $04# +1E+012 201$ 4032

U2 0.000$ 0.00045 +0.000# 0.0005' +0.0002 +0.000$ +0.000$


121/157

=

2 0.00045 0.00144 +0.000# 0.000-4 0.00042 +0.001 +0.000-

@2 +0.000# +0.000# 0.0015# +0.000- 0.00034 0.000-5 0.000-2

U3 0.0005' 0.000-4 +0.000- 0.00204 +0.0004 +0.001- +0.0021

3 +0.0002 0.00042 0.00034 +0.0004 0.000#3 0.0004 0.0003$

@3 +0.000$ +0.001 0.000-5 +0.001- 0.0004 0.00202 0.001#

@4 +0.000$ +0.000- 0.000-2 +0.0021 0.0003$ 0.001# 0.00253

U2

=

+0.0'1#

2 0.03#1

@2 0.1-253

U3 0.2#102

3 +0.0212

@3 +0.3202

@4 +0.2$14

eng/itung Beaksi Tumpuan

1

=

+4#3#.4 3E+013 +241-.2 0 0 0 0

1 3E+013 +403.2 1E+013 0 0 0 0

1 241-.2 +1E+013 #0$.4 0 0 0 0

U2 = +0.0'1#

2 = 0.03#1

@2 = 0.1-253


122/157

U3 = 0.2#102

3 = +0.0212

@3 = +0.3202

U4 = 0

4 = 0

@4 = +0.2$14

Tentukan ga7a+ga7a pada ujung batang mulai batang 1 sampai m

&atang 1

f1 $E+01' 1 0 0 0 0

g1 +1 $E+01' 0 0 0 0

m1 0 0 1 0 0 0

[ ] [ ] [ ]U g k T

m

g

f

m

g

f

⋅⋅=

11

2

2

2

1

1

1

g1

=

11#.53$

m1 +1#.3$1

f2 15.3$0-


123/157

g2 +11#.54

m2 13$.#-'

+33$ #E+014 0 33$ +#E+014 0 0.2#102

+1E+012 +4032 +201$ 1E+012 4032 +201$ +0.0212

5E+013 201$ $'2 +5E+013 +201$ 1344 +0.3202


124/157

=

+11#.54

+1#.4

1$3.103

11#.53$

1#.3$0-

+1#1.4$

120-$ +2E+012 $04# +120-$ 2E+012 $04# 0.2#102

+2E+012 100# +1E+012 2E+012 +100# +1E+012 +0.0212


125/157

$04# +1E+012 4032 +$04# 1E+012 201$ +0.3202+120-$ 2E+012 +$04# 120-$ +2E+012 +$04# 0

2E+012 +100# 1E+012 +2E+012 100# 1E+012 0

$04# +1E+012 201$ +$04# 1E+012 4032 +0.2$14

=

+2E+013 +100# 0 2E+013 100# 0 0.2#102

120-$ +2E+012 $04# +120-$ 2E+012 $04# +0.0212

$04# +1E+012 4032 +$04# 1E+012 201$ +0.32022E+013 100# 0 +2E+013 +100# 0 0

+120-$ 2E+012 +$04# 120-$ +2E+012 +$04# 0

$04# +1E+012 201$ +$04# 1E+012 4032 +0.2$14

=

21.3$0-

+11#.54

+11#.54

+21.3$1

11#.53$

+5E+013


126/157


127/157

I


128/157


129/157


130/157


131/157


132/157

+100# +1E+012

@31E+012 201$

2E 012 $04# U4


133/157

+2E+012 +$04# U4

100# 1E+0124

1E+012 4032@4

4 @4

0 0 U1

0 0 1

0 0 @1

0 0 U2

0 0 20 0 @2

2E+012 $04# U3

+100# +1E+012 3

1E+012 201$ @3

+2E+012 +$04# U4


134/157


135/157

@4

0

+3


136/157

+3300

0

+3

+300

0

+0.0'1#

0.03#1

0.1-253


137/157


138/157


139/157


140/157


141/157


142/157


143/157


144/157


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146/157


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148/157


149/157


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&entuk matri keseimbangan struktur

U1 1 @1 U2 2 @2 U3 3 @3 U4 4 @4

1

=

4#3#.4 +3E+013 +241-.2 +4#3#.4 3E+013 +241-.2 0 0 0 0 0 0 U1

"

0

1 +3E+013 403.2 1E+013 3E+013 +403.2 1E+013 0 0 0 0 0 0 1 0

1 +241-.2 1E+013 1$12.# 241-.2 +1E+013 #0$.4 0 0 0 0 0 0 @1 0

0 +4#3#.4 3E+013 241-.2 51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0 0 0 U2 U2

+3 3E+013 +403.2 +1E+013 $E+013 4435.2 201$ +-E+013 +4032 201$ 0 0 0 2 2

300 +241-.2 1E+013 #0$.4 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0 0 0 @2 @2

0 0 0 0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# +120-$ 2E+012 $04# U3 U3

+3 0 0 0 +-E+013 +4032 +201$ +1E+012 5040 +201$ 2E+012 +100# +1E+012 3 3

[ ] [ ] [ ]U K P ⋅=


156/157

+300 0 0 0 5E+013 201$ $'2 $04# +201$ 53'$ +$04# 1E+012 201$ @3 @3

4 0 0 0 0 0 0 +120-$ 2E+012 +$04# 120-$ +2E+012 +$04# U4 0

4 0 0 0 0 0 0 2E+012 +100# 1E+012 +2E+012 100# 1E+012 4 0

0 0 0 0 0 0 0 $04# +1E+012 201$ +$04# 1E+012 4032 @4 @4

0

=

51'4.4 $E+013 241-.2 +33$ +-E+013 5E+013 0 +4#3#.4 3E+013 241-.2 0 0 U2

+3 $E+013 4435.2 201$ +-E+013 +4032 201$ 0 3E+013 +403.2 +1E+013 0 0 2

300 241-.2 201$ 2-5$.# +5E+013 +201$ $'2 0 +241-.2 1E+013 #0$.4 0 0 @2

0 +33$ +-E+013 +5E+013 12432 +1E+012 $04# $04# 0 0 0 +120-$ 2E+012 U3

+3 +-E+013 +4032 +201$ +1E+012 5040 +201$ +1E+012 0 0 0 2E+012 +100# 3

+300 5E+013 201$ $'2 $04# +201$ 53'$ 201$ 0 0 0 +$04# 1E+012

@3

0 0 0 0 $04# +1E+012 201$ 4032 0 0 0 +$04# 1E+012 @4

1 +4#3#.4 3E+013 +241-.2 0 0 0 0 4#3#.4 +3E+013 +241-.2 0 0 01 3E+013 +403.2 1E+013 0 0 0 0 +3E+013 403.2 1E+013 0 0 0

1 241-.2 +1E+013 #0$.4 0 0 0 0 +241-.2 1E+013 1$12.# 0 0 0


=

2

1

2

1

U

U

d c

ba

P

P

4 0 0 0 +120-$ 2E+012 +$04# +$04# 0 0 0 +2E+012 100# 0

4 0 0 0 2E+012 +100# 1E+012 1E+012 0 0 0 +$04# 1E+012 0


157/157

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