View
6
Download
0
Category
Preview:
DESCRIPTION
analisis struktur metode matrix
Citation preview
DataE = 2100 ton/cm2 P2 = 4 tonH = 3 mL = 4 mP = 4 tonM = -4 ton-m
Batang 1A = 120 cm2
I = 4000 cm4
Batang 2A = 120 cm2
PENYELESAIAN : I = 4000 cm4
1). DEGREE OF FREEDOM (D.O.F)
2). Matrik Kekakuan Dalam SUMBU LOKALa. Batang 1 (AB)
E = 21000000 t/m2 Batang 1 = 5 m sin a = 0.602A = 0.012 m2 a = 37.000 cos2 a = 0.638I = 0.00004 m4 cos a = 0.799 sin2 a = 0.362
50400 0 0 -50400 0 00 80.640 201.600 0 -80.640 201.6000 201.600 672.000 0 -201.600 672.000
[k]1= -50400 0 0 50400 0 00 -80.640 -201.600 0 80.640 -201.6000 201.600 336.000 0 -201.600 336.000
DbxDby
2
1
Rbz
L
H
P
21
C
B
A
M
P
b. Batang 2 (BC)E = 21000000 t/m2 batang 2 = 3 m sin a = 1A = 0.012 m2 a = 90 cos2 a = 4E-33I = 0.00004 m4 cos a = 6E-17 sin2 a = 1
84000 0 0 -84000 0 00 373.333 560 0 -373.333 5600 560 1120 0 -560 1120
[k]2= -84000 0 0 84000 0 00 -373.333 -560 0 373.333 -5600 560 560 0 -560 560
3). Matrik Kekakuan SUMBU GLOBAL [kg] = [T]T.[kl].[T]a. Batang 1 (AB)
a = 90 cos a = 0.00sin a = 1.00
6.12574E-17 1 0 0 0 0-1 6.12574E-17 0 0 0 0
[T]1= 0 0 1 0 0 00 0 0 6.12574E-17 1 00 0 0 -1 6.12574E-17 00 0 0 0 0 1
[kg] = [T]T.[kl].[T] :
80.640 0.000 -201.600 -80.640 0.000 -201.6000.000 50400.000 0.000 0.000 -50400.000 0.000
[Kg]1= -201.600 0.000 672.000 201.600 0.000 672.000-80.640 0.000 201.600 80.640 0.000 201.6000.000 -50400.000 0.000 0.000 50400.000 0.000
-201.600 0.000 336.000 201.600 0.000 336.000
b. Batang 2 (BC)a = 90
cos a = 0sin a = 1
0 1 0 0 0 0-1 0 0 0 0 0
[T]2= 0 0 1 0 0 00 0 0 0 1 00 0 0 -1 0 00 0 0 0 0 1
Sehingga [kg] = [T]T.[kl].[T] :
373.333 0.0E+00 -560 -373.333 0.0E+00 -560- 84000 0.0E+00 0.0E+00 -84000 0.0E+00
[Kg]2= -560 0.0E+00 1120 560 0.0E+00 1120-373.3333 0.0E+00 560 373.3333 0.0E+00 5600.0E+00 -84000 0.0E+00 0.0E+00 84000 0.0E+00
-560 0.0E+00 560 560 0.0E+00 560
4). Perakitan Matriks Kekakuan Struktur Global Tereduksi
80.640 0.000 -201.600 -80.640 0.000 -201.600 0 0 00.000 50400.000 0.000 0.000 -50400.000 0.000 0 0 0
-201.600 0.000 672.000 201.600 0.000 672.000 0 0 0[Kg]s = -80.640 0.000 201.600 453.973 0.000 -358.400 -373.333 0.000 -560.000
0.000 -50400.000 0.000 0.000 134400.000 0.000 0 -84000.000 0-201.600 0.000 336.000 -358.400 0.000 1456.000 560.000 0 1120.000
0 0 0 -373.333 0 560.000 373.333 0 560.0000 0 0 0 -84000.000 0 0 84000.000 00 0 0 -560.000 0 560.000 560.000 0 560.000
SETELAH DIRDUKSI :
453.973 0.000 -358.400[Kg]s = 0.000 134400.000 0.000
-358.400 0.000 1456.000
5). Vektor Perpindahan {D} =[Kg]SR-1 . [{F} + {Fo}]
4 453.973 0.000 -358.400 Dbx4 = 0.000 134400.000 0.000 x Dby-4 -358.400 0.000 1456.000 Rbz
Dbx 0.0027341 0.0000000 0.0006730 4Dby = 0.0000000 0.0000074 0.0000000 x 4Rbz 0.0006730 0.0000000 0.0008525 -4
Dbx 0.0082 mDby = 0.0000 mRbz -0.0007 m
6). Vektor perpindahan bagi masing-masing elemen {d}i = [T]i . {D}a. Batang 1
00
{d}1 = 00.000030-0.008244
-0.000718
b. Batang 2
0.000030-0.008244
{d}2 = -0.000718000
7). Element Forces : {f} = [k]i {d}i - {fo}ia. Batang 1 (AB)
f1 50400 0 0 -50400 0 0 0.000000 0f2 0 81 202 0 -81 202 0.000000 0f3 = 0 202 672 0 -202 672 x 0.000000 - 0f4 -50400 0 0 50400 0 0 0.000030 4f5 0 -81 -202 0 81 -202 -0.008244 4f6 0 202 336 0 -202 336 -0.000718 -4
f1 -1.500 0 -1.500 tonf2 0.520 0 0.520 tonf3 = 1.180 - 0 = 1.180 ton-mf4 1.500 4 -2.500 tonf5 -0.520 4 -4.520 tonf6 1.421 -4 5.421 ton-m
b. Batang 2 (BC)
f1 84000 0 0 -84000 0 0 0.00003 4 Dbxf2 0 373 560 0 -373 560 -0.00824 4 Dbyf3 = 0 560 1120 0 -560 1120 x -0.00072 - -4 Rbzf4 -84000 0 0 84000 0 0 0.00000 0 Dcxf5 0 -373 -560 0 373 -560 0.00000 0 Dcyf6 0 560 560 0 -560 560 0.00000 0 Rcz
f1 2.500 4 -1.500 tonf2 -3.480 4 -7.480 tonf3 = -5.421 - -4 = -1.421 ton-mf4 -2.500 0 -2.500 tonf5 3.480 0 3.480 tonf6 -5.019 0 -5.019 ton-m
8). Reaksi Tumpuan {F} = [Kg]s.{D} - ({F}+{Fo})
80.640 0.000 -201.600 -80.640 0.000 -201.600 0 0 0 0 00.000 50400.000 0.000 0.000 -50400.000 0.000 0 0 0 0 0
-201.600 0.000 672.000 201.600 0.000 672.000 0 0 0 0 0[F] = -80.640 0.000 201.600 453.973 0.000 -358.400 -373.333 0.000 -560.000 0.00824435 4
0.000 -50400.000 0.000 0.000 134400.000 0.000 0.000 -84000.000 0.000 x 2.9762E-05 - 4-201.600 0.000 336.000 -358.400 0.000 1456.000 560.000 0.000 1120.000 -0.0007179 -4
0 0 0 -373.333 0.000 560.000 373.333 0.000 560.000 0 00 0 0 0.000 -84000.000 0.000 0.000 84000.000 0.000 0 00 0 0 -560.000 0.000 560.000 560.000 0.000 560.000 0 0
-0.520 0 -0.520 ton-1.500 0 -1.500 ton1.180 0 1.180 ton-m
[F] = 4.000 - 4 = 0 ton4.000 4 0 ton-4.000 -4 0 ton-m-3.480 0 -3.480 ton-2.500 0 -2.500 ton-5.019 0 -5.019 ton-m
8). GAMBAR
-2.500 5.421
-1.500 -7.480 -1.421-4.520
0.520-1.500 1.180
-2.500 3.480-5.019
NFD SFD BMD
Recommended