tugas ke 2 anstruk iii (03101001129)
DESCRIPTION
tugasTRANSCRIPT
Nama Harry HutagalungNIM 03101001129
SOAL 4.1. KONSTRUKSI BALOK MENERUSdata :
q 600 kg/m R BC 2400 M CB 3200L1 10 m R CB 2400L2 8 m M AB -5000
R AB 3000 M BA 5000R BA 3000 M BC -3200
Dari gambar, bisa didapatkan kekakuan dari masing-masing elemen AB dan BC :
12EI/L^3 6EI/L^2 - 12 EI/L^3 6EI/L^2 0.012 0.06 -0.012 0.066EI/L^2 4EI/L -6EI/L^2 2EI/L 0.06 0.4 -0.06 0.2 EI[KAB]i = [KAB]i =
12EI/L^3 -6EI/L^2 12EI/L^3 -6EI/L^2 -0.012 -0.06 0.012 -0.06- 12 EI/L^3 2EI/L -6EI/L^2 4EI/L 0.06 0.2 -0.06 0.4
0.023 0.09375 -0.023 0.093750.09375 0.5 -0.09375 0.25 EI-0.023 -0.09375 0.023 -0.09375
0.09375 0.25 -0.09375 0.5
sekarang dapat disuperposisikan KAB dan KBC menjadi suatu matriks kesatuan struktur:
0.012 0.06 -0.012 0.060.06 0.4 -0.06 0.2
Ks = -0.012 -0.06 0.035 0.03375 -0.023 0.093750.06 0.2 0.03375 0.9 -0.09375 0.25
-0.023 -0.09375 0.023 -0.093750.09375 0.25 -0.09375 0.5
selanjutnya : Kff KfbKbf Kbb
Q4 0.9 0.06 0.2 0.03375 -0.09375 0.25 D4Q1 0.06 0.012 0.06 -0.012 D1Q2 = 0.2 0.06 0.4 -0.06 x D2Q3 0.03375 -0.012 -0.06 0.035438 -0.023438 0.09375 D3Q5 -0.09375 -0.023 0.023438 -0.09375 D5Q6 0.25 0.09375 -0.09375 0.5 D6
1800 = EI x 0.9 Df
[KBC]i =
[EI]
[Qf/Qb] = {Df/Db}
[EI]
{Qf} = [Kff]{Df}
Df = 1.111111 x 1800 /EID4 = 2000 /EI
Q1 D1 0.012 0.06 -0.012 0.06 0Q2 = [KAB]i D2 = 0.06 0.4 -0.06 0.2 x 0 EIQ3 D3 -0.012 -0.06 0.012 -0.06 0Q4 D4 0.06 0.2 -0.06 0.4 2000 /EI
untuk batang BC :
Q3 D3 0.023438 0.09375 -0.023438 0.09375 0Q4 = [KBC]i D4 = 0.09375 0.5 -0.09375 0.25 x 2000 /EI =Q5 D5 -0.023438 -0.09375 0.023438 -0.09375 0Q6 D6 0.09375 0.25 -0.09375 0.5 0
maka :MA 400 - -5000 = 5400 kgm
MBA 800 - 5000 = -4200 kgmMBC 1000 - -3200 = 4200 kgmMC 500 - 3200 = -2700 kgmRA 120 + 3000 = 3120 kg
RBA -120 + 3000 = 2880 kgRBC 187.5 + 2400 = 2587.5 kgRC -187.5 + 2400 = 2212.5 kg
{Qi} = [Ki][Di]
120= 400
-120800
187.51000
-187.5500
SOAL 4.1. KONSTRUKSI PORTAL BIDANG TANPA PERGOYANGAN DIMANA DEFORMASI AXIAL DIABAIKAN
soal :q 300 kg/mEI 2P AB 600 KgP CD 600 KgL AD 5 mL1 3 mL2 2 m
struktur dasar :
0 0 0 0 0 0 2[KAB]i =
0 0.096 0.24 0 -0.096 0.240 0.24 0.8 0 -0.24 0.40 0 0 0 0 00 -0.096 -0.24 0 0.096 -0.240 0.24 0.4 0 -0.24 0.8
T AB = 0 0 0 0-sin α 0 0 0 0
0 0 1 0 0 0 =0 0 0 00 0 0 -sin α 00 0 0 0 0 1
[K AB]s = [T AB] transpose . [K AB]i. [T AB]
0 -1 0 0 0 01 0 0 0 0 0
= 0 0 1 0 0 0 x0 0 0 0 -1 00 0 0 1 0 00 0 0 0 0 1
0 -0.096 -0.24 0 0.096 -0.240 0 0 0 0 0
[K AB]s = 0 0.24 0.8 0 -0.24 0.4 x0 0.096 0.24 0 -0.096 0.240 0 0 0 0 00 0.24 0.4 0 -0.24 0.8
selanjutnya diturunkan elemen kekakuan BC
2
1 0 0 0 0 00 1 0 0 0 0
T BC = 0 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1
[K BC]s = [T BC] transpose . [K BC]i. [T BC]
1 0 0 0 0 00 1 0 0 0 0
= 0 0 1 0 0 0 x0 0 0 1 0 0
cos α sin αcos α
cos α sin αcos α
[K BC]i = 2[K AB]i
0 0 0 0 1 00 0 0 0 0 1
0 0 0 0 0 00 0.192 0.48 0 -0.192 0.48
= 0 0.48 1.6 0 -0.48 0.8 x0 0 0 0 0 00 -0.192 -0.48 0 0.192 -0.480 0.48 0.8 0 -0.48 1.6
0 0 0 0 0 00 0.192 0.48 0 -0.192 0.48
= 0 0.48 1.6 0 -0.48 0.8 EI0 0 0 0 0 00 -0.192 -0.48 0 0.192 -0.480 0.48 0.8 0 -0.48 1.6
Akhirnya diturunkan kekakuan elemen CD :
T CD = 0 -1 0 0 0 01 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
[KCD]s = [T CD] transpose . [K CD]i. [T CD]
0 1 0 0 0 0-1 0 0 0 0 0
= 0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1
0 0 0 0 0 00 0.096 0.24 0 -0.096 0.24
= 0 0.24 0.8 0 -0.24 0.40 0 0 0 0 00 -0.096 -0.24 0 0.096 -0.240 0.24 0.4 0 -0.24 0.8
0 0 0 0 0 00.096 0 0.24 -0.096 0 0.24
= 0.24 0 0.8 -0.24 0 0.4
[K CD]i = [K AB]i
0 0 0 0 0 0-0.096 0 -0.24 0.096 0 -0.240.24 0 0.4 -0.24 0 0.8
Superposisi : 0.096 0 -0.24 -0.096 0 -0.240 0 0 0 0 0
-0.24 0 0.8 0.24 0 0.4-0.096 0 0.24 0.096 0 0.24 0
0 0 0 0 0.192 0.48 0-0.24 0 0.4 0.24 0.48 2.4 0
0 0 0 00 -0.133632 -0.33408 00 0.33408 0.7552 0.24
-0.0960
0.24
untuk batang AB :
D1 0 1 0 0 0 0D2 -1 0 0 0 0 0D3 = 0 0 1 0 0 0D4 0 0 0 0 1 0D5 0 0 0 -1 0 0D6 0 0 0 0 0 1
Q1 0 0 0 0 0 0Q2 0 0.096 0.24 0 -0.096 0.24Q3 = 0 0.24 0.8 0 -0.24 0.4Q4 0 0 0 0 0 0Q5 0 -0.096 -0.24 0 0.096 -0.24Q6 0 0.24 0.4 0 -0.24 0.8
Untuk batang BC
D1 1 0 0 0 0 0D2 0 1 0 0 0 0D3 = 0 0 1 0 0 0D4 0 0 0 1 0 0D5 0 0 0 0 1 0D6 0 0 0 0 0 1
Q1 0 0 0 0 0 0Q2 0 0.192 0.48 0 -0.192 0.48Q3 = 0 0.48 1.6 0 -0.48 0.8Q4 0 0 0 0 0 0
Q5 0 -0.192 -0.48 0 0.192 -0.48Q6 0 0.48 0.8 0 -0.48 1.6
Untuk batang CD:
D1 0 -1 0 0 0 0D2 1 0 0 0 0 0D3 = 0 0 1 0 0 0D4 0 0 0 0 -1 0D5 0 0 0 1 0 0D6 0 0 0 0 0 1
Q1 0 0 0 0 0 0Q2 0 0.096 0.24 0 -0.096 0.24Q3 = 0 0.24 0.8 0 -0.24 0.4Q4 0 0 0 0 0 0Q5 0 -0.096 -0.24 0 0.096 -0.24Q6 0 0.24 0.4 0 -0.24 0.8
Maka :
MA -48.25 - -288 = 239.75 kgmMBA -96.5 - 432 = -528.5 kgmMBC -96.5 - -625 = 528.5 kgmMCB 96.5 - 625 = -528.5 kgmMCD 96.5 - -432 = 528.5 kgmMD 48.25 - 288 = -239.75 kgmRAH -28.95 + 211.2 = 182.25 kgRAV 0 + 750 = 750 kgRDH -28.95 + 211.2 = 182.25 kgRDV 0 + -750 = -750 kg
SOAL 4.1. KONSTRUKSI PORTAL BIDANG TANPA PERGOYANGAN DIMANA DEFORMASI AXIAL DIABAIKAN
0 0 0 0 0 0
0 0.192 0.48 0 -0.192 0.480 0.48 1.6 0 -0.48 0.80 0 0 0 0 00 -0.192 -0.48 0 0.192 -0.480 0.48 0.8 0 -0.48 1.6
0 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
0 0 0 0 0 0 0 1 00 0.096 0.24 0 -0.096 0.24 -1 0 00 0.24 0.8 0 -0.24 0.4 x 0 0 10 0 0 0 0 0 0 0 00 -0.096 -0.24 0 0.096 -0.24 0 0 00 0.24 0.4 0 -0.24 0.8 0 0 0
0 1 0 0 0 0 0.096 0 -0.24-1 0 0 0 0 0 0 0 00 0 1 0 0 0 = -0.24 0 0.80 0 0 0 1 0 -0.096 0 0.240 0 0 -1 0 0 0 0 00 0 0 0 0 1 -0.24 0 0.4
[K BC]i = 2[K AB]i = 0 0 0 0 0 00 0.192 0.48 0 -0.192 0.480 0.48 1.6 0 -0.48 0.80 0 0 0 0 00 -0.192 -0.48 0 0.192 -0.480 0.48 0.8 0 -0.48 1.6
0 0 0 0 0 0 1 0 00 0.192 0.48 0 -0.192 0.48 0 1 00 0.48 1.6 0 -0.48 0.8 x 0 0 10 0 0 0 0 0 0 0 0
0 -0.192 -0.48 0 0.192 -0.48 0 0 00 0.48 0.8 0 -0.48 1.6 0 0 0
1 0 0 0 0 00 1 0 0 0 00 0 1 0 0 00 0 0 1 0 00 0 0 0 1 00 0 0 0 0 1
0 0 0 0 0 0 0 -10 0.096 0.24 0 -0.096 0.24 1 0
x 0 0.24 0.8 0 -0.24 0.4 x 0 00 0 0 0 0 0 0 00 -0.096 -0.24 0 0.096 -0.24 0 00 0.24 0.4 0 -0.24 0.8 0 0
0 -1 0 0 0 01 0 0 0 0 0
x 0 0 1 0 0 00 0 0 0 -1 00 0 0 1 0 00 0 0 0 0 1
EI
0 0-0.133632 0.33408-0.33408 0.7552 EI
0 0 -0.096 0 0.240.133632 -0.33408 0 0 0-0.33408 1.7152 -0.24 0 0.4
0 -0.24 0.096 0 -0.240 0 0 0 00 0.4 -0.24 0 0.8
0 00 0
x 0 = 00 00 0
-120.625 /EI -120.625 /EI
0 00 -28.95
EI x 0 = -48.250 00 28.95
-120.625 /EI -96.5
0 00 0
x -120.625 /EI = -120.625 /EI0 00 0
120.625 /EI 120.625 /EI
0 00 0
EI x -120.625 /EI = -96.50 0
0 0120.625 /EI 96.5
0 00 0
x 120.625 /EI = 120.625 /EI0 00 00 0
0 00 28.95
EI x 120.625 /EI = 96.50 00 -28.950 48.25
0 0 00 0 00 0 00 1 0-1 0 00 0 1
-0.096 0 -0.240 0 0
0.24 0 0.4 EI0.096 0 0.24
0 0 00.24 0 0.8
0 0 00 0 00 0 01 0 0
0 1 00 0 1
0 0 0 00 0 0 01 0 0 00 0 -1 00 1 0 00 0 0 1