anstruk matriks
13
SOAL : b (ton) BEBAN a = 2.0 a (ton) b = 3.0 BATANG 1 A = 40 I = 2000 L = 300 300 BATANG 2 A = 30 I = 2000 L = 424.26 300 E = 2100 90 315.00 45.00 PENYELESAIAN : MATRIK KEKAKUAN ELEMEN BATANG 1 (ELEMEN 1) [ T ] 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 1 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 0 0 0 1 280 0 0 -280 0 0 0 1.867 280 0 -1.867 280 0 280 56000 0 -280 28000 -280 0 0 280 0 0 0 -1.867 -280 0 1.867 -280 0 280 28000 0 -280 56000 α btg 1 = α btg 2 = α btg 3 = [ T ] T [ Kl ] 2 1 3 1 2 β
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Transcript of anstruk matriks
SOAL :
b (ton) BEBANa = 2.0 ton
a (ton) b = 3.0 ton
BATANG 1A = 40 cm²I = 2000L = 300 cm
300 BATANG 2A = 30 cm²I = 2000L = 424.26 cm
300
E = 2100 t/cm²90 derajad
315.00 derajad45.00 derajad
PENYELESAIAN :
MATRIK KEKAKUAN ELEMEN
BATANG 1 (ELEMEN 1)
[ T ]
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 -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
280 0 0 -280 0 00 1.867 280 0 -1.867 2800 280 56000 0 -280 28000
-280 0 0 280 0 00 -1.867 -280 0 1.867 -2800 280 28000 0 -280 56000
cm^4
cm^4
α btg 1 =α btg 2 =α btg 3 =
[ T ] T
[ Kl ]
2
1 3
1 2
β
=
0.00 -1.87 -280.00 0.00 1.87 -280.00280.00 0.00 0.00 -280.00 0.00 0.00
0 280 56000 0 -280 280000.00 1.87 280.00 0.00 -1.87 280.00
-280.00 0.00 0.00 280.00 0.00 0.000 280 28000 0 -280 56000
[ T ] =
1.87 0 -280 -1.87 0 -2800 280 0 0 -280 0
= -280 0 56000 280 0 28000
-1.87 0 280 1.87 0 2800 -280 0 0 280 0
-280 0 28000 280 0 56000
BATANG 2 (ELEMEN 2)
[ T ]
0.7071 -0.7071 0 0 0 00.7071 0.7071 0 0 0 0
0 0 1 0 0 00 0 0 0.7071 -0.7071 00 0 0 0.7071 0.7071 00 0 0 0 0 1
0.7071 0.7071 0 0 0 0-0.7071 0.7071 0 0 0 0
0 0 1 0 0 00 0 0 0.7071 0.7071 00 0 0 -0.7071 0.7071 00 0 0 0 0 1
148.49 0 0 -148.49 0 00 0.66 140.00 0 -0.66 140.000 140.00 39597.99 0 -140.00 19798.99
-148.49 0 0 148.49 0 00 -0.66 -140.00 0 0.66 -140.000 140.00 19798.99 0 -140.00 39597.99
=
105.00 0.47 98.99 -105.00 -0.47 98.99-105.00 0.47 98.99 105.00 -0.47 98.99
0 140 39598 0 -140 19799-105.00 -0.47 -98.99 105.00 0.47 -98.99105.00 -0.47 -98.99 -105.00 0.47 -98.99
0 140 19799 0 -140 39598
[ T ] T [ Kl(1) ]
[ Kg(1) ] [ T ] T [ Kl(1) ]
[Kg(1)]
[ T ] T
[ Kl ]
[ T ] T [ Kl(2) ]
[ T ] =
74.58 -73.92 98.99 -74.58 73.92 98.99-73.92 74.58 98.99 73.92 -74.58 98.99
= 98.99 98.99 39597.99 -98.99 -98.99 19798.99
-74.58 73.92 -98.99 74.58 -73.92 -98.9973.92 -74.58 -98.99 -73.92 74.58 -98.9998.99 98.99 19798.99 -98.99 -98.99 39597.99
BATANG 3 (ELEMEN 3)
[ T ]
0.7071 0.7071 0 0 0 0-0.7071 0.7071 0 0 0 0
0 0 1 0 0 00 0 0 0.7071 0.7071 00 0 0 -0.7071 0.7071 00 0 0 0 0 1
0.7071 -0.7071 0 0 0 00.7071 0.7071 0 0 0 0
0 0 1 0 0 00 0 0 0.7071 -0.7071 00 0 0 0.7071 0.7071 00 0 0 0 0 1
210.00 0 0 -210.00 0 00 1.87 280.00 0 -1.87 280.000 280.00 56000.00 0 -280.00 0.00
-210.00 0 0 210.00 0 00 -1.87 -280.00 0 1.87 -280.000 280.00 0.00 0 -280.00 56000.00
=
148.49 -1.32 -197.99 -148.49 1.32 -197.99148.49 1.32 197.99 -148.49 -1.32 197.99
0 280 56000 0 -280 0-148.49 1.32 197.99 148.49 -1.32 197.99-148.49 -1.32 -197.99 148.49 1.32 -197.99
0 280 0 0 -280 56000
[ T ] =
105.93 104.07 -197.99 -105.93 -104.07 -197.99104.07 105.93 197.99 -104.07 -105.93 197.99
= -197.99 197.99 56000.00 197.99 -197.99 0.00-105.93 -104.07 197.99 105.93 104.07 197.99-104.07 -105.93 -197.99 104.07 105.93 -197.99
[ Kg(2) ] [ T ] T [ Kl(2) ]
[Kg(2)]
[ T ] T
[ Kl ]
[ T ] T [ Kl(3) ]
[ Kg(3) ] [ T ] T [ Kl(2) ]
[Kg(3)]
-197.99 197.99 0.00 197.99 -197.99 56000.00MENYUSUN OVERALL STIFFNESS MATRIX
=
1.87 0 -280 -1.87 0 -280 0 0 00 280 0 0 -280 0 0 0 0
-280 0 56000 280 0 28000 0 0 0-1.87 0 280 1.87 0 280 0 0 0
0 -280 0 0 280 0 0 0 0-280 0 28000 280 0 56000 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
=
0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 00 0 0 0 0 0 0 0 01 1 0 74.58 -73.92 98.99 -74.58 73.92 98.990 0 1 -73.92 74.58 98.99 73.92 -74.58 98.990 0 0 98.99 98.99 39597.99 -98.99 -98.99 19798.990 0 0 -74.58 73.92 -98.99 74.58 -73.92 -98.990 0 0 73.92 -74.58 -98.99 -73.92 74.58 -98.990 0 0 98.99 98.99 19798.99 -98.99 -98.99 39597.99
1.87 0.00 -280.00 -1.87 0 -280 0 0 00 280 0 0 -280 0 0 0 0
-280 0 56000 280 0 28000 0 0 0-1.16 0.71 280 76.44 -73.92 378.99 -74.58 73.92 98.99
0 -280 1 -73.92 354.58 98.99 73.92 -74.58 98.99-280 0 28000 378.99 98.99 95597.99 -98.99 -98.99 19798.99
0 0 0 -74.58 73.92 -98.99 74.58 -73.92 -98.990 0 0 73.92 -74.58 -98.99 -73.92 74.58 -98.990 0 0 98.99 98.99 19798.99 -98.99 -98.99 39597.99
UNKNOWN DISPLACEMENT AND REACTIONS
H 2.0 76.44 -73.92 378.99V -3.0 = -73.92 354.58 98.99M 0 378.99 98.99 95597.99
-176.44 -73.92 378.99 2.0
= -73.92 354.58 98.99 -3.0378.99 98.99 95597.99 0
= 0.016846 0.003531 -0.000070 2.00.003531 0.003561 -0.000018 -3.0
-0.000070 -0.000018 0.000011 0
U2 0.02310
[Kg(3)]
[Kg(1)]
[Kg(2)]
U2V2θ2
U2V2θ2
V2 = -0.00362θ2 -0.00009
F1 -1.87 0.00 -280.00 -0.019G1 0.00 -280.00 0.00 0.02310 1.014M1
=280.00 0.00 28000.00 -0.00362
=4.008
F3 -74.58 73.92 -98.99 -0.00009 -1.981G3 73.92 -74.58 -98.99 1.986M3 98.99 98.99 19798.99 0.189
MEMBER FORCES
ELEMEN 1
00
[ U ] =0
0.02310-0.00362-0.00009
[ T ]
0 280 0 0 -280 0-1.8667 0 280 1.8667 0 280
=-280 0 56000 280 0 28000
0 -280 0 0 280 01.8667 0 -280 -1.8667 0 -280
-280 0 28000 280 0 56000
f1 1.01399g1 0.01853m1
= [ T ] [ U ] =4.00828
f2 -1.01399g2 -0.01853m2 1.54940
ELEMEN 2
0.02310-0.00362
[ U ] =-0.00009
000
[ Kg(1) ]
[ Kg(1) ]
[ T ]105.00 -105.00 0.00 -105.00 105.00 0.00
0.47 0.47 140.00 -0.47 -0.47 140.00
=98.99 98.99 39597.99 -98.99 -98.99 19798.99
-105.00 105.00 0.00 105.00 -105.00 0.00-0.47 -0.47 -140.00 0.47 0.47 -140.0098.99 98.99 19798.99 -98.99 -98.99 39597.99
f2 #VALUE!g2 #VALUE!m2
= [ T ] [ U ] =#VALUE!
f3 #VALUE!g3 #VALUE!m3 #VALUE!
FREE BODY
#VALUE!1.01399 #VALUE!
0.01853
1.54940 #VALUE!
#VALUE!
0.01853 4.00828#VALUE!
1.01399 #VALUE!
β = 45.00 derajad
H1 = -0.019 ton
V1 = 1.014 ton
M1 = 4.008 ton m
[ Kg(2) ]
[ Kg(2) ]
H2 = = #VALUE! ton
V2 = = #VALUE! ton
M2 = 1.54940 #VALUE! = #VALUE! ton m
H3 = = #VALUE! ton
V3 = = #VALUE! ton
M3 = #VALUE! ton m
#VALUE!
#VALUE!
0.019 #VALUE! #VALUE!4.008
1.014 #VALUE!
( 0,01287 + 6,59747 cos β - 0,00436 sin β ) =
( -2,33644 - 6,59747 sin β - 0,00436 cos β ) =
m2(1) + m2(2) =
(- 6,59747 cos β + 0,00436 sin β ) =
( 6,59747 sin β + 0,00436 cos β ) =
BATANG 3A = 30 cm²I = 2000L = 300.00 cm
cm^4
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