Contoh Soal Analisa Matriks (Portal Hal.153)

7/25/2019 Contoh Soal Analisa Matriks (Portal Hal.153)

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CONTOH 4.1

Properties penampang

Batang 1 :

b1 = 0.35 m

h1 = 0.35 m

A1 = 0.1225 m2

I1 = 0.0012505208 m4

L1 = 3.5 m

Batang 2 :

b2 = 0.20 m

h2 = 0.45 m

A2 = 0.09 m2


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%& 1.3 = 0 = 0

%& 2.3 = .EI / L2 = 1282.5

%& 3.3 = 4 EI / L = 30012.5

Akibat (i#)la*ement arah 4 ++, -4 $tran#la#i arah #b lkal 4'

%& 1.4 = + E.A / L = +735000

%& 2.4 = 0 = 0

%& 3.4 = 0 = 0


%& 1.5 = 0 = 0

%& 2.5 = +12.EI/L3 = +7350

%& 3.5 = +.EI / L2 = +1282.5

Akibat (i#)la*ement arah ++, - $rta#i arah #b lkal '

%& 1. = 0 = 0


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0 0 0 0

ran#)#e matrik ta#i

0.00 -1 0 0

= 1 0.00 0 0

0 0 1 0

0 0 0 0.00

0 0 0 1

0 0 0 0

&atrik# kekak"an #tr"kt"r men6a(i :

0.00 +1.00 0.00 0.00

1.00 0.00 0.00 0.00

0.00 0.00 1.00 0.00%&% 1 = 0.00 0.00 0.00 0.00

0.00 0.00 0.00 1.00

0.00 0.00 0.00 0.00

4.50242E+011 +7350 +1282.5 +4.502E+011

735000 4.502E+013 7.8792E+013 +735000

0 1282.5 30012.5 0

%&% = . %& .


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%& 1.3 = 0 = 0

%& 2.3 = .EI / L2 = 754.5

%& 3.3 = 4 EI / L = 25515


%& 1.4 = + E.A / L = +378000

%& 2.4 = 0 = 0

%& 3.4 = 0 = 0


%& 1.5 = 0 = 0

%& 2.5 = +12.EI/L3 = +301.8

%& 3.5 = +.EI / L2 = +754.5

Akibat (i#)la*ement arah ++, - $rta#i arah #b lkal '

%& 1. = 0 = 0

%& 2 EI / L2 754 5


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ran#)#e matrik ta#i

1.00 0 0 0= 0 1.00 0 0

0 0 1 0

0 0 0 1.00

0 0 0 0

0 0 0 0

&atrik# kekak"an #tr"kt"r men6a(i :

1.00 0.00 0.00 0.00

0.00 1.00 0.00 0.00

0.00 0.00 1.00 0.00

%&% 2 = 0.00 0.00 0.00 1.000.00 0.00 0.00 0.00

0.00 0.00 0.00 0.00

378000 0 0 +378000

0 301.8 754.5 0

0 754.5 25515 0

+378000 0 0 378000

%&% = . %& .


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MATRIKS KEKAKUAN STRUKTUR

-1 -2 -3

7 8 9385350 0 1283 7

%&% = 0 73802 755 8

1283 755 55528 9

Batang 1

Gbr AM 1 =

Batang 2

Gbr AM 2 =

Hitung rekasi di ujung batang / freebody (AML) --- Beban dimasukkan ke!uali beban di joint

AML1

AML2

AML3

AML4

AML5

AML6

AML1

AML2

AML3


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0.00 0.00 0.00 0.00

0.00 0.00 0.00 1.00

0.00 0.00 0.00 0.00

0 2 1

0 1 2

= 0 3 3

0 5 7 D1

0 4 8 D2

0 9 D3

SB GLOBAL

Batang 2

0 1 7

0 2 8AM 2 = 0 3 9

0 4 4

0 5 5

0 6

SB LOKAL

AMS1


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4 0 0

5 0 0

+AE = 0 0

-1 7 0 0-2 8 0 0

-3 9 0 0

SB GLOBAL

B#b$% &$'$ ()*%+ ,$%/% !A("

1 0

2 0

3 0

4 0

A6 = 5 0

0

-1 7 40-2 8 0

-3 9 0

SB GLOBAL

BEBAN OINT GABNGAN !A"


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+.08E+007 +1.885E+007 1.817E+005

A.A-A.A "ALAM 0 #L#M#0

Batang 1

D#)r$* +r/+/r '$,$ $r$ SMB GLOBAL '* b$+$% 1: bb ;

0 1

0 2

, 0 30.000105 7 D1

0.000000 8 D2

-0.000024 9 D3

D#)r$* +r/+/r '$,$ $r$ SMB LOKAL '* b$+$% 1: bb ;

0 1 0 0

-1 0 0 0

, 0 0 1 0

0 0 0 0

0 0 0 -1

"M'1

M 1 = R . DMS1

M 1


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0 6

D#)r$* +r/+/r '$,$ $r$ SMB LOKAL '* b$+$% 1: bb ;

1 0 0 0

0 1 0 0

, 0 0 1 0

0 0 0 1

0 0 0 0

0 0 0 0

G$$ /(/% b$+$%

0 378000 0

0 0 3061.8

AM 2 , 0 0 7654.5

0 -378000 0

0 0 -3061.8

0 0 7654 5

M 2 = R . DMS2

M 2

AM 2 = AML2 SM 2 .

M 2


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Data ean

; = 40 k

< = 0 k/m

E!emen "a!am matri#s #e#a#uan atang

Batang EA / L 4.EI / L 2.EI / L .EI / L2 12.EI / L3

1 735000 30012.5 1500.25 1282.5 7350

2 378000 25515 12757.5 754.5 301.8

3


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%& 4.3 = 0 = 0

%& 5.3 = +.EI / L2 = +1282.5

%& .3 = 2 EI / L = 1500.25

%& 4.4 = E.A / L = 735000

%& 5.4 = 0 = 0

%& .4 = 0 = 0

%& 4.5 = 0 = 0

%& 5.5 = 12.EI/L3 = 7350

%& .5 = +.EI / L2 = +1282.5

%& 4. = 0 = 0


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0 1 0 0

0 0

0 0

0 0

-1 0

0.00 0

0 1

0.00 0.00 735000 0 0 +735000

0.00 0.00 0 7350 1282.5 0

0.00 0.00 0 1282.5 30012.5 0

+1.00 0.00 +735000 0 0 735000

0.00 0.00 0 +7350 +1282.5 0

0.00 1.00 0 1282.5 1500.25 0

7350 +1282.5 0 1 0 0

+4.502E+013 7.879E+013 +1 0 0 0

+1282.5 1500.25 0 0 1 0


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%& 4.3 = 0 = 0

%& 5.3 = +.EI / L2 = +754.5

%& .3 = 2 EI / L = 12757.5

%& 4.4 = E.A / L = 378000

%& 5.4 = 0 = 0

%& .4 = 0 = 0

%& 4.5 = 0 = 0

%& 5.5 = 12.EI/L3 = 301.8

%& .5 = +.EI / L2 = +754.5

%& 4. = 0 = 0

%& 5 EI / L2 754 5


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0 00 0

0 0

0 0

1.00 0

0 1

0.00 0.00 378000 0 0 +378000

0.00 0.00 0 301.8 754.5 0

0.00 0.00 0 754.5 25515 0

0.00 0.00 +378000 0 0 378000

1.00 0.00 0 +301.8 +754.5 0

0.00 1.00 0 754.5 12757.5 0

0 0 1 0 0 0

+301.8 754.5 0 1 0 0

+754.5 12757.5 0 0 1 0

0 0 0 0 0 1


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-1 2.1529E+00 .29192E+009 +.07E+007

-2 = .29192E+009 1.358E+00 +1.885E+007

-3 +.079E+007 +1.8850E+007 1.817E+005

0 1

0 2

= 0 3

0 4

0 5

0

SB LOKAL

0 1

0 2

= 0 3

%&%+1


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-1.00 0.00 = 0.00

0.00 0.00 0.00

0.00 1.00 0.00

AMS1


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0 0

= 0 AE = 0

0 0

0 00 0

0 0


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0 +2.427E+005 -3 -0.0000243

0 0 0 0 1

0 0 0 0 2

0 0 0 , 0.000 3

1 0 0.0001046117 0.000 4

0 0 2 516768E-007 0 000 5


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0 0 0.0001046117 0.0001046117 7

0 0 2.516768E-007 2.5167678E-007 8

0 0 -2.42672E-005 , 0.000 9

0 0 0 0.000 41 0 0 0.000 5

0 1 0 0.000 6

0 -378000 0 0 0.0001046117

7654.5 0 -3061.8 7654.5 2.5167678E-007

25515 0 -7654.5 12757.5 0.000 ,

0 378000 0 0 0.000

-7654.5 0 3061.8 -7654.5 0.000

12757 5 0 7654 5 25515 0 000


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0 0 0 1

0 0 0 1 0 0

+7350 1282.5 +1 0 0 0

+1282.5 1500.25 0 0 1 0

0 0 0 0 0 0

7350 +1282.5 0 0 0 +1

+1282.5 30012.5 0 0 0 0

0 0 7350 0 +1283 +7350

0 0 0 735000 0 0

0 0 = 12 83 0 30 013 12 83


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0 0 1 0 0 0

+301.8 754.5 0 1 0 0

+754.5 12757.5 0 0 1 0

0 0 0 0 0 1

301.8 +754.5 0 0 0 0

+754.5 25515 0 0 0 0

0 0 378000 0 0 +378000

0 0 0 302 755 0

0 0 = 0 755 25515 0

0 0 378 000 0 0 378 000


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0.00 0.00 0.00 -1.00 0.00 0

0.00 0.00 1.00 0.00 0.00 0

0.00 0.00 0.00 0.00 1.00 0


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D2

D1


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D1

D2

D3

00

0

0 39.543 39.54 7 D1

0 -0.185 -0.18 8 D2

0 -0.617 , -0.62 9 D3

0 -39.543 -39.54 4 0

0 0.185 0.18 5 0

0 0 308 0 31 6 0


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0 0

0 0

0 0

1 00 0

0 1

0 +1283

+735000 0

0 15 00


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0 0

0 0

0 0

0 0

1 00 1

0 0

+302 755

+755 12758

0 0


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39.54k

+0.18 k

+0.2 k.m

+39.54k

0.18k

0 31 k


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Properties penampang @atatan ;a = /m2

Bahan / material :

E = 2.10E07 k/m2

Batang 1 : Batang 2 :

0.35m

0.35m

1.23E+01m2

E = 2.10E07 k/m2

1.25E+03m4

3.5 m

90 (egree#

E!emen "a!am matri#s #e#a#uan atang E!emen "a!am

EA / L 4.EI / L 2.EI / L .EI / L2 12.EI / L3

7.35E05 3.00E04 1.50E04 1.29E04 7.35E03

Re#ap E!emen "a!am Matri#s Ke#a#uan $atang

Batang EA / L 4.EI / L 2.EI / L .EI / L2 12.EI / L3 theta

1 7.35E05 3.00E04 1.50E04 1.29E04 7.35E03 90 (eg

2 3.78E05 2.55E04 1.28E04 7.5E03 3.0E03 0 (eg

%%%& su"a' DA(AM SUM$U STRUKTUR)*(O$A(

Batang = 1

90 (egree#

b1=

h1=

A1=

I1=

L1=

theta1=

Hitung matri#s #e#a#uan atang 1 +SM1,

theta1=

EA / L 4 EI / L 2 EI / L EI / L2 12 EI / L3


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7.35E05 3.00E04 1.50E04 1.29E04 7.35E03

@ = *# theta = .125742E+017

@ = #in theta = 1

"1 1 )hi1 "2 2 )hi2"1 7350 4.45739E+011 +1282.5 +7350 +4.4574E+011 +1282.5

1 4.4573937E+011 735000 7.88E+13 +4.4574E+011 +735000 7.88E+13

)hi1 +1282.5 7.88E+13 3.00E04 1.29E04 +7.8792E+013 1.50E04

"2 +7350 +4.45739E+011 1.29E04 7.35E03 4.45740E+011 1.29E04

2 +4.4573937E+011 +735000 +7.87923001E+013 4.45740E+011 7.35E05 +7.8792E+013

)hi2 +1282.5 7.88E+13 1.50E04 1.29E04 +7.8792E+013 3.00E04

%%%& su"a' DA(AM SUM$U STRUKTUR)*(O$A(

Batang = 2

0 (egree#


3.78E05 2.55E04 1.28E04 7.5E03 3.0E03

@ = *# theta = 1

@ = #in theta = 0

"2 2 )hi2 "3 3 )hi3

"2 378000 0 0 +378000 0 0

2 0 301.8 7.5E03 0 +301.8 7.5E03

)hi2 0 7.5E03 2.55E04 0.00E00 +754.5 1.28E04

"3 +378000 0 0.00E00 3.78E05 0 0.00E00

3 0 +301.8 +754.5 0 3.0E03 +754.5

)hi3 0 7.5E03 1.28E04 0.00E00 +754.5 2.55E04

%&1 =

Hitung matri#s #e#a#uan atang - +SM-,

theta1=

%&1 =

MATRIKS KEKAKUAN STRUKTUR +SM ,


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(1 (2 (3

"2 2 )hi2 "1 1 )hi1 "3

"2 385350 4.45739E+011 1282.5 +7350 +4.4574E+011 1282.5 +378000

2 4.4573937E+011 73801.8 754.5 +4.4574E+011 +735000 +7.8792E+013 0

)hi2 1282.5 754.5 55527.5 +1282.5 7.87924E+013 1500.25 0

"1 +7350 +4.45739E+011 +1282.5 7350 4.45740E+011 +1282.5 0

1 +4.4573937E+011 +735000 7.87923000E+013 4.45740E+011 735000 013 0

)hi1 1282.5 +7.87923E+013 1500.25 +1282.5 7.87924E+013 30012.5 0

"3 +378000 0 0 0 0 0 378000

3 0 +301.8 +754.5 0 0 0 0

)hi3 0 754.5 12757.5 0 0 0 0

eneral e


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1= kF(1k#F(2

2= k#F(1k##F(2

reGrite 1: kF(1=1+k#F(2

(1=iner#e$k'F$1+k#F(2'

-i#)la*ement#: "2 = 0.000104117 m

2 = 2.51778402E+007 m

)hi2 = +2.4271575E+005 ra(

*ek : #b lkal #b glbal

ea*tin#: !1 = +0.457599155 +0.18 +0.4

1 = +0.184982433 0.4 +0.18

&1 = 0.9814094302 0.98 0.98

!3 = +39.5432400845 +39.54324008 +39.5432400845

3 = 0.184982433 0.18498243 0.184982433&3 = +0.30719071 +0.3071907 +0.30719071


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0.20m

0.45m

9.00E+02m2

E = 2.10E07 k/m2

1.52E+03m4

5 m

0 (egree#

atri#s #e#a#uan atang


3.78E05 2.55E04 1.28E04 7.5E03 3.0E03

b2=

h2=

A2=

I2=

L2=

theta2=


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*ek :7350 0 +1283 +7350 0 +1283 +

0 735000 0 0 +735000 0 +

+1283 0 30013 1283 0 1500 +

+7350 0 1283 7350 0 1283 +

0 +735000 0 0 735000 0 +

+1283 0 1500 1283 0 30013 +

*ek :

378000 0 0 +378000 0 0 +

0 302 755 0 +302 755 +

0 755 25515 0 +755 12758 +

+378000 0 0 378000 0 0 +

0 +302 +755 0 302 +755 +

0 755 12758 0 +755 25515 +


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3 )hi3 *ek :

0 0 (1 385350 4.4.E+11 1283 +

+301.8 754.5 (2 4.4.E+11 73802 755 +

+754.5 12757.5 (3 1283 755 55528 +

0 0

0 0

0 0

0 0

301.8 +754.5

+754.5 25515

"1 1 )hi1 "3 3 )hi3+7350 +4.45740E+011 1282.5 +378000 0 0 "2

+4.45740E+011 +735000 +7.87924E+013 0 +301.8 754.5 2

+1282.5 7.87924E+013 1500.25 0 +754.5 12757.5 )hi2

7350 4.45740E+011 +1282.5 0 0 0 0

4.45739E+011 735000 7.87924E+013 0 0 0 0

+1282.5 7.87924E+013 30012.5 0 0 0 0

0 0 0 378000 0 0 0

0 0 0 0 301.8 +754.5 0

0 0 0 0 +754.5 25515 0

(1

(2


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*ek :

0.000104117

2.51778402E+007

+2.4271575E+005

+

+

+

+

++


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+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +

+ + + + +


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+ +

+ +

+ +


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Contoh Soal Analisa Matriks (Portal Hal.153)