ADS PR 2
58
Problem 1 fc' = 20 Mpa determine - stiffness and mass of the structure - frequency of structure Answer Moment inertia of structure 0.0010667 m4 Modulus elasticity 21019.039 Mpa 21019039 KN/m2 EI 22420.308 kNm2 Area 0.08 m2 Local stiffness for element 1 0.444444444 0.6666667 -0.444444 0.6666667 0.666666667 1.3333333 -0.666667 0.6666667 EI -0.44444444 -0.666667 0.444444 -0.666667 0.666666667 0.6666667 -0.666667 1.3333333 mass matrix for element 1 468 198 162 -117 198 108 117 -81 162 117 468 -198 -117 -81 -198 108 Local stiffness for element 2 1.5 1.5 -1.5 -1.5 1.5 2 -1.5 1 -1.5 -1.5 1.5 -1.5 EI -1.5 1 -1.5 2 mass matrix for element 2 312 88 108 -52 88 32 52 -24 108 52 312 -88 -52 -24 -88 32 Combine the first element and second element Stiffness matrix of the structure 0.444444444 0.6666667 -0.444444 0.6666667 0 0 0 0 0.666666667 1.3333333 -0.666667 0.6666667 0 0 0 0 -0.44444444 -0.666667 0.444444 -0.666667 0 0 0 0 0.666666667 0.6666667 -0.666667 1.3333333 0 0 0 0 EI 0 0 0 0 1.5 1.5 -1.5 -1.5 0 0 0 0 1.5 2 -1.5 1 0 0 0 0 -1.5 -1.5 1.5 -1.5 0 0 0 0 -1.5 1 -1.5 2 = 2.4 t/m3 s/420 s/420 3 m 2 m 40 cm 20 cm
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Transcript of ADS PR 2
Problem 1
fc' = 20 Mpa
determine - stiffness and mass of the structure- frequency of structure
AnswerMoment inertia of structure 0.0010667 m4Modulus elasticity 21019.039 Mpa
21019039 KN/m2EI 22420.308 kNm2Area 0.08 m2
Local stiffness for element 10.444444444 0.6666667 -0.444444 0.66666670.666666667 1.3333333 -0.666667 0.6666667 EI-0.44444444 -0.666667 0.444444 -0.6666670.666666667 0.6666667 -0.666667 1.3333333mass matrix for element 1
468 198 162 -117198 108 117 -81162 117 468 -198-117 -81 -198 108
Local stiffness for element 21.5 1.5 -1.5 -1.51.5 2 -1.5 1-1.5 -1.5 1.5 -1.5 EI-1.5 1 -1.5 2
mass matrix for element 2312 88 108 -5288 32 52 -24
108 52 312 -88-52 -24 -88 32
Combine the first element and second elementStiffness matrix of the structure0.444444444 0.6666667 -0.444444 0.6666667 0 0 0 00.666666667 1.3333333 -0.666667 0.6666667 0 0 0 0-0.44444444 -0.666667 0.444444 -0.666667 0 0 0 00.666666667 0.6666667 -0.666667 1.3333333 0 0 0 0 EI
0 0 0 0 1.5 1.5 -1.5 -1.50 0 0 0 1.5 2 -1.5 10 0 0 0 -1.5 -1.5 1.5 -1.50 0 0 0 -1.5 1 -1.5 2
= 2.4 t/m3
s/420
s/420
3 m 2 m40 cm
20 cm
Mass matrix of the structure468 198 162 -117 0 0 0 0198 108 117 -81 0 0 0 0162 117 468 -198 0 0 0 0-117 -81 -198 108 0 0 0 0
0 0 0 0 312 88 108 -520 0 0 0 88 32 52 -240 0 0 0 108 52 312 -880 0 0 0 -52 -24 -88 32
Restraintu1 0 0 0 0u2 0 0 0 0u3 1 0 0 0 u3u4 = 0 1 0 0 u4u5 0 0 1 0 u5u6 0 0 0 1 u6u7 0 0 0 0u8 0 0 0 0
Global stiffness of stucture
0 0 1 0 0 0 0 00 0 0 1 0 0 0 00 0 0 0 1 0 0 00 0 0 0 0 1 0 0
-0.44444444 -0.666667 0.444444 -0.666667 0 0 0 00.666666667 0.6666667 -0.666667 1.3333333 0 0 0 0 EI
0 0 0 0 1.5 1.5 -1.5 -1.50 0 0 0 1.5 2 -1.5 1
Hence stiffness of the structure :0.444444444 -0.666667 0 0-0.66666667 1.3333333 0 0 EI
0 0 1.5 1.50 0 1.5 2
develop the mass matrix of structure
0 0 1 0 0 0 0 00 0 0 1 0 0 0 0
s/420
0 0 0 0 1 0 0 00 0 0 0 0 1 0 0
162 117 468 -198 0 0 0 0-117 -81 -198 108 0 0 0 0
0 0 0 0 312 88 108 -520 0 0 0 88 32 52 -24
Hence mass matriks of the structure468 -198 0 0
-198 108 0 00 0 312 880 0 88 32
Develop the global displacement which induce the stiffness of structureu3 1 0u4 0 1 U1u5 1 0 U2u6 0 1
Develop the stifness matrix of structure1 0 1 0 0.44444444 -0.666667 0 ###0 1 0 1 -0.6666667 1.333333 0 ###
0 0 1.5 ##0 0 1.5 ###
0.444444444 -0.666667 1.5 1.5 1 0-0.66666667 1.3333333 1.5 2 0 1
1 00 1
Global stiffness1.944444444 0.8333333 EI = 43595.0438 18683.590.833333333 3.3333333 18683.5902 74734.36
Global mass of structure1 0 1 0 468 -198 0 ###0 1 0 1 -198 108 0 ###
0 0 312 ##0 0 88 ##
468 -198 312 88 1 0-198 108 88 32 0 1
1 0
s/420
s/420
0 1Global mass
780 -110 3.56571429 -0.502857-110 140 -0.5028571 0.64
Egeinvalue equation[K] - l2 [m]
43595.04383 18683.59 - 3.56571429 -0.50285718683.59021 74734.361 -0.5028571 0.64determinant of the following matrix equal to zero
43595.04383 - 3.565714 18683.5902 + 0.502857
18683.59021 + 0.502857 74734.3608 - 0.64
3258047737 -294382.2 2.2820571 -349076543 -18790.35 -0.253
2908971193 -313172.6 2.0291918 = 0
144406.37 380.00838 Hz9927.2733 99.635703 Hz
Problem 2
fc' = 20 Mpa
determine - stiffness and mass of the structure- frequency of structure
spring in the midle of structureglobal stiffness matrix of structure global mass matrix of structure
54805.19796 18683.59 - 3.56754286 -0.50285718683.59021 74734.361 -0.5028571 0.64
determinant of the following matrix equal to zero
54805.19796 - 3.567543 18683.5902 - -0.502857
18683.59021 - -0.502857 74734.3608 - 0.64
4095831440 -301693.4 2.2832274 -349076543 -18790.35 0
s/420
2
2 2
2 + 4 2
2 + 4
^2 =^2 =
= 2.4 t/m3
2
2 2
2 2
2 4 2
40 cm
3 m 2 m
20 cm
m=sL/210
k=4EI/L3
3746754897 -320483.7 2.0303621
145130.39 380.95982 Hz12715.207 112.76173 Hz
2 4
2= =2= =
0.444444 0.666667 -0.444444 0.666667 0 0 0 00.666667 1.333333 -0.666667 0.666667 0 0 0 0-0.444444 -0.666667 0.444444 -0.666667 0 0 0 00.666667 0.666667 -0.666667 1.333333 0 0 0 0 EI
0 0 0 0 1.5 1.5 -1.5 -1.50 0 0 0 1.5 2 -1.5 10 0 0 0 -1.5 -1.5 1.5 -1.50 0 0 0 -1.5 1 -1.5 2
0 0 0 00 0 0 01 0 0 00 1 0 00 0 1 00 0 0 10 0 0 00 0 0 0
468 198 162 -117 0 0 0 0198 108 117 -81 0 0 0 0
162 117 468 -198 0 0 0 0-117 -81 -198 108 0 0 0 0
0 0 0 0 312 88 108 -520 0 0 0 88 32 52 -240 0 0 0 108 52 312 -880 0 0 0 -52 -24 -88 32
0 0 0 00 0 0 01 0 0 00 1 0 00 0 1 00 0 0 10 0 0 00 0 0 0
s/420
4
fc' = 15 Mpa
E = 18203.02 Mpa
L kolom = 4 mL balok = 5 mI balok = 0.00045 m4I kolom = 0.000133 m4
EI Balok = 8191.36
EI Kolom= 2427.07
Matriks kekakuan lokal untuk elemen:elemen 1 dan 3EI/L3 37.92296
U1 V1 U2 V224.271 0.000 0.000 -24.271 0.000 0.000 U10.000 455.076 910.151 0.000 -455.076 910.151 V10.000 910.151 2427.070 0.000 -910.151 1213.535
-24.271 0.000 0.000 24.271 0.000 0.000 U20.000 -455.076 -910.151 0.000 455.076 -910.151 V20.000 910.151 1213.535 0.000 -910.151 2427.070
elemen 2EI/L3 65.53088
U1 V1 U2 V298.296 0.000 0.000 -98.296 0.000 0.000 U10.000 786.371 1965.926 0.000 -786.371 1965.926 V10.000 1965.926 6553.088 0.000 -1965.926 3276.544
-98.296 0.000 0.000 98.296 0.000 0.000 U20.000 -786.371 -1965.926 0.000 786.371 -1965.926 V20.000 1965.926 3276.544 0.000 -1965.926 6553.088
Matriks massa untuk elemen:Elemen 1 dan 3
0.000914U1 V1 U2 V2
0.128 0.000 0.000 0.064 0.000 0.000 U10.000 0.143 0.080 0.000 0.049 -0.048 V1
= 2.4 t/m3
KNm4
KNm4
1 2
1
2
1 2
1
2
sl/4201 2
5 m
1
2
3
30 cm
20 cms=20/204 m
Pot. Melintang Balok
E=4700√ fc '
0.000 0.080 0.059 0.000 0.048 -0.0440.064 0.000 0.000 0.128 0.000 0.000 U20.000 0.049 0.048 0.000 0.143 -0.080 V20.000 -0.048 -0.044 0.000 -0.080 0.059
Elemen 20.001714
U1 V1 U2 V20.240 0.000 0.000 0.120 0.000 0.000 U10.000 0.267 0.189 0.000 0.093 -0.111 V10.000 0.189 0.171 0.000 0.111 -0.1290.120 0.000 0.000 0.240 0.000 0.000 U20.000 0.093 0.111 0.000 0.267 -0.189 V20.000 -0.111 -0.129 0.000 -0.189 0.171
Transformasi koordinat lokal menjadi koordinat global
matriks tranformasi [T] adalah0 1 0 0 0 0-1 0 0 0 0 00 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1
Tranformasi matriks kekakuan:
0 -1 0 0 0 0 24.271 0.0001 0 0 0 0 0 0.000 455.0760 0 1 0 0 0 0.000 910.1510 0 0 0 -1 0 -24.271 0.0000 0 0 1 0 0 0.000 -455.0760 0 0 0 0 1 0.000 910.151
455.07554 0 -910.1511 -455.0755 0 -910.15110 24.2707 0 0 -24.270696 0
-910.15109 0 2427.0696 910.1511 0 1213.5348-455.07554 0 910.15109 455.0755 0 910.15109
0 -24.2707 0 0 24.270696 0-910.15109 0 1213.5348 910.1511 0 2427.0696
Tranformasi matriks massa elemen
0 -1 0 0 0 0 0.128 0.0001 0 0 0 0 0 0.000 0.1430 0 1 0 0 0 0.000 0.0800 0 0 0 -1 0 0.064 0.0000 0 0 1 0 0 0.000 0.049
1
2
sl/4201 2
1
2
- untuk elemen 1 dan 3 dengan = 90
Ke1
G = TT Ke T
0 0 0 0 0 1 0.000 -0.048
0.1426286 0 -0.080457 0.049371 0 0.04754290 0.128 0 0 0.064 0
-0.0804571 0 0.0585143 -0.047543 0 -0.0438860.0493714 0 -0.047543 0.142629 0 0.0804571
0 0.064 0 0 0.128 00.0475429 0 -0.043886 0.080457 0 0.0585143
Untuk elemen dua tidak ditranformasikan karena koordinat lokal elemen,juga merupakan koordinat global struktur
Matriks kekakuan struktur
455.07554 0 -910.1511 -455.0755 0 -910.1511 0 0 00 24.2707 0 0 -24.270696 0 0 0 0
-910.15109 0 2427.0696 910.1511 0 1213.5348 0 0 0-455.07554 0 910.15109 455.0755 0 910.15109 0 0 0
0 -24.2707 0 0 24.270696 0 0 0 0-910.15109 0 1213.5348 910.1511 0 2427.0696 0 0 0
0 0 0 0 0 0 98.29632 0 00 0 0 0 0 0 0 786.3705 1965.92630 0 0 0 0 0 0 1965.926 6553.08780 0 0 0 0 0 -98.29632 0 00 0 0 0 0 0 0 -786.3705 -1965.92630 0 0 0 0 0 0 1965.926 3276.54390 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
Matriks massa struktur
0.1426286 0 -0.080457 0.049371 0 0.0475429 0 0 00 0.128 0 0 0.064 0 0 0 0
-0.0804571 0 0.0585143 -0.047543 0 -0.043886 0 0 00.0493714 0 -0.047543 0.142629 0 0.0804571 0 0 0
0 0.064 0 0 0.128 0 0 0 00.0475429 0 -0.043886 0.080457 0 0.0585143 0 0 0
0 0 0 0 0 0 0.24 0 00 0 0 0 0 0 0 0.267429 0.18857140 0 0 0 0 0 0 0.188571 0.17142860 0 0 0 0 0 0.12 0 00 0 0 0 0 0 0 0.092571 0.11142860 0 0 0 0 0 0 -0.111429 -0.12857140 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
Kondisi tumpuanu1 0 0 0 0 0 0 0 0u2 0 0 0 0 0 0 0 0u3 0 0 0 0 0 0 0 0u4 1 0 0 0 0 0 0 0u5 0 1 0 0 0 0 0 0u6 0 0 1 0 0 0 0 0u7 0 0 0 1 0 0 0 0u8 0 0 0 0 1 0 0 0u9 0 0 0 0 0 1 0 0u10 0 0 0 0 0 0 1 0u11 0 0 0 0 0 0 0 1u12 0 0 0 0 0 0 0 0u13 0 0 0 0 0 0 0 0u14 0 0 0 0 0 0 0 0u15 0 0 0 0 0 0 0 0u16 0 0 0 0 0 0 0 0u17 0 0 0 0 0 0 0 0u18 0 0 0 0 0 0 0 0
Solusi dari Kekakuan struktur adalah:
0 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
455.07554 0 910.15109 0 0 0 0 0 00 24.2707 0 0 0 0 0 0 0
910.15109 0 2427.0696 0 0 0 0 0 0
[K] = [B]T [KG] [B]
0 0 0 98.29632 0 0 -98.29632 0 00 0 0 0 786.37054 1965.9263 0 -786.3705 1965.92630 0 0 0 1965.9263 6553.0878 0 -1965.926 3276.54390 0 0 -98.29632 0 0 98.29632 0 00 0 0 0 -786.37054 -1965.926 0 786.3705 -1965.92630 0 0 0 1965.9263 3276.5439 0 -1965.926 6553.08780 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
Matriks massa struktur
0 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 00 0 0 0 0 0 1 0 00 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0.1426286 0 0.0804571 0 0 0 0 0 00 0.128 0 0 0 0 0 0 0
0.0804571 0 0.0585143 0 0 0 0 0 00 0 0 0.24 0 0 0.12 0 00 0 0 0 0.2674286 0.1885714 0 0.092571 -0.11142860 0 0 0 0.1885714 0.1714286 0 0.111429 -0.12857140 0 0 0.12 0 0 0.24 0 00 0 0 0 0.0925714 0.1114286 0 0.267429 -0.18857140 0 0 0 -0.1114286 -0.128571 0 -0.188571 0.17142860 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
u4 1 0 0u5 0 1 0u6 0 0 1
u7 1 0 0 U1u8 0 1 0 U2u9 0 0 1 U3u10 1 0 0u11 0 1 0u12 0 0 1u13 1 0 0u14 0 1 0u15 0 0 1
1 0 0 1 0 0 1 0 00 1 0 0 1 0 0 1 00 0 1 0 0 1 0 0 1
Matriks kekakuan struktur
910.15109 0 00 48.54139 00 0 24513.403
1 0 0 1 0 0 1 0 00 1 0 0 1 0 0 1 00 0 1 0 0 1 0 0 1
0.1426286 0 0.0804571 0.36 0 0 0.36 0 0
[K] = [B]T [K] [B]
[M] = [B]T [M] [B]
0 0.128 0 0 0.36 0.3 0 0.36 -0.30.0804571 0 0.0585143 0 0.0771429 0.0428571 0 -0.077143 0.0428571
Matriks massa struktur adalah sebagai berikut:1.0052571 0 0
0 0.976 00 0 0.2027429
910.1511 0 0 1.005257 0
0 48.541391 0- 0 0.976
0 0 24513.4 0 0
Solusi dari matriks tersebut adalah:
905.3913 30.089721 Hz
49.73503 7.0523069 Hz
120908.8 347.71948 HzAdditionalProblem 2
Nilai kekakuan pegas adalah 65.53088 kNmDengan adanya pegas akan menambah nilai kekakuan struktur pada arah horizontalMaka kekakuan struktur menjadi
910.1510+65.5309 0 00 48.541391 00 0 24513.4
[ K -2 M ] =0
=(910.1511-1.005257)((48.54139-0.976).(24513.4-0.2027))=0
12 =
1 =
22 =
2 =
32 =
3 =
5 m
1
2
3s=20/20
k x=EI
L3
975.6819 0 00 48.5414 00 0 24513.4
Matriks massa struktur adalah sebagai berikut:1.0052571 0 0
0 0.976 00 0 0.2027429
0 00 00 0
Nilai determinan dari matrix di atas harus sama dengan nol
970.5794 31.154124 Hz
49.73504 7.0523075 Hz
120908.8 347.71946 Hz
Penambahan pegas hanya mempengaruhi kekakuan arah horizontal Pada persamaan ini mode 1 dari kedua struktur akan berbeda. Pada struktur kedua lebih kaku untuk mendukung goyangan horizontal
[ K -2 M ] =0
975.6819-1.00525748.5414-0.976
24513.4-0.20274
sehingga diperoleh nilai adalah
12 =
1 =
22 =
2 =
32 =
3 =
0.000 -24.271 0.000 0.000 0 1 0 0910.151 0.000 -455.076 910.151 -1 0 0 0
2427.070 0.000 -910.151 1213.535 0 0 1 00.000 24.271 0.000 0.000 0 0 0 0
-910.151 0.000 455.076 -910.151 0 0 0 -11213.535 0.000 -910.151 2427.070 0 0 0 0
0.000 0.064 0.000 0.000 0 1 0 00.080 0.000 0.049 -0.048 -1 0 0 00.059 0.000 0.048 -0.044 0 0 1 00.000 0.128 0.000 0.000 0 0 0 00.048 0.000 0.143 -0.080 0 0 0 -1
-0.044 0.000 -0.080 0.059 0 0 0 0
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
-98.296317 0 0 0 0 0 0 0 00 -786.3705 1965.9263 0 0 0 0 0 00 -1965.926 3276.5439 0 0 0 0 0 0
98.2963173 0 0 0 0 0 0 0 00 786.3705 -1965.9263 0 0 0 0 0 00 -1965.926 6553.0878 0 0 0 0 0 00 0 0 455.075543 0 -910.1511 -455.0755 0 -910.15110 0 0 0 24.2707 0 0 -24.2707 00 0 0 -910.15109 0 2427.07 910.1511 0 1213.5350 0 0 -455.07554 0 910.1511 455.0755 0 910.15110 0 0 0 -24.2707 0 0 24.2707 00 0 0 -910.15109 0 1213.535 910.1511 0 2427.07
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0.12 0 0 0 0 0 0 0 00 0.092571 -0.1114286 0 0 0 0 0 00 0.111429 -0.1285714 0 0 0 0 0 0
0.24 0 0 0 0 0 0 0 00 0.267429 -0.1885714 0 0 0 0 0 00 -0.188571 0.1714286 0 0 0 0 0 00 0 0 0.14262857 0 -0.080457 0.049371 0 0.0475430 0 0 0 0.128 0 0 0.064 00 0 0 -0.0804571 0 0.058514 -0.047543 0 -0.043886
0 0 0 0.04937143 0 -0.047543 0.142629 0 0.0804570 0 0 0 0.064 0 0 0.128 00 0 0 0.04754286 0 -0.043886 0.080457 0 0.058514
0 0 0 00 0 0 00 0 0 00 0 0 0 u40 0 0 0 u50 0 0 0 u60 0 0 0 u70 0 0 0 u80 0 0 0 u90 0 0 0 u100 0 0 0 u111 0 0 0 u120 1 0 0 u130 0 1 0 u140 0 0 1 u150 0 0 00 0 0 00 0 0 0
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 0
0 0 00 0 00 0 0
0 0 00 0 00 0 00 0 00 0 00 0 0
455.075543 0 -910.151090 24.2707 0
-910.15109 0 2427.0696
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 01 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 00 0 1 0 0 0 0 0 00 0 0 1 0 0 0 0 00 0 0 0 1 0 0 0 00 0 0 0 0 1 0 0 0
0 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 00 0 0
0.14262857 0 -0.08045710 0.128 0
-0.0804571 0 0.0585143
1 0 0 455.0755 0 910.1511 0 00 1 0 0 24.2707 0 0 00 0 1 910.1511 0 2427.07 0 0
0 0 0 98.29632 00 0 0 0 786.37050 0 0 0 1965.9260 0 0 -98.29632 00 0 0 0 -786.37050 0 0 0 1965.9260 0 0 0 00 0 0 0 00 0 0 0 0
1 0 0 0.142629 0 0.080457 0 00 1 0 0 0.128 0 0 00 0 1 0.080457 0 0.058514 0 0
0 0 0 0.24 00 0 0 0 0.2674290 0 0 0 0.1885710 0 0 0.12 00 0 0 0 0.0925710 0 0 0 -0.1114290 0 0 0 00 0 0 0 0
0.14262857 0 -0.0804571 0 0 0 0 0
0 0.128 0-0.0804571 0 0.0585143
0
00.20274286
Pada persamaan ini mode 1 dari kedua struktur akan berbeda. Pada struktur kedua lebih kaku untuk mendukung goyangan horizontal
0 00 00 01 00 00 1
0 00 00 01 00 0
0 1
455.0755 0 -910.1511 -455.0755 0 -910.1511 0 0 00 24.2707 0 0 -24.2707 0 0 0 0
-910.1511 0 2427.07 910.1511 0 1213.535 0 0 0-455.0755 0 910.1511 455.0755 0 910.1511 0 0 0
0 -24.2707 0 0 24.2707 0 0 0 0-910.1511 0 1213.535 910.1511 0 2427.07 0 0 0
0 0 0 0 0 0 98.29632 0 00 0 0 0 0 0 0 786.3705 1965.9260 0 0 0 0 0 0 1965.926 6553.0880 0 0 0 0 0 -98.29632 0 00 0 0 0 0 0 0 -786.3705 -1965.9260 0 0 0 0 0 0 1965.926 3276.5440 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0.142629 0 -0.080457 0.049371 0 0.047543 0 0 00 0.128 0 0 0.064 0 0 0 0
-0.080457 0 0.058514 -0.047543 0 -0.043886 0 0 00.049371 0 -0.047543 0.142629 0 0.080457 0 0 0
0 0.064 0 0 0.128 0 0 0 00.047543 0 -0.043886 0.080457 0 0.058514 0 0 0
0 0 0 0 0 0 0.24 0 00 0 0 0 0 0 0 0.267429 0.1885710 0 0 0 0 0 0 0.188571 0.1714290 0 0 0 0 0 0.12 0 00 0 0 0 0 0 0 0.092571 0.1114290 0 0 0 0 0 0 -0.111429 -0.1285710 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 10 0 0 0 0 0 0 0 00 -98.29632 0 0 0 0 0 1 0
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