analisis 2
DESCRIPTION
determinacion de reaciones y desplazaminetospor metodo de rigidezTRANSCRIPT
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DATOS
Ingrese Distancias
c a = 8 b = 3 c = 7
a bIngrese fuerzas
F1 = 0 F2 = 0
F3 = 10 x y
F4 = 0 1 ( 0 0 )
F5 = 5 2 ( 11 0 )
F6 = 0 3 ( 8 7 )
ELEMENTO N° K1 E = 1 A = 2 λx = 1
λy = 0
1 2 3 4
1 1 0 -1 0
K1 =2 0 0 0 0 2
3 -1 0 1 0 11
4 0 0 0 0
1 2 3 4
1 0.182 0.000 -0.182 0.000
K1 =2 0.000 0.000 0.000 0.0003 -0.182 0.000 0.182 0.0004 0.000 0.000 0.000 0.000
1
2
3
4
1 2
3
3
4
1
2
5
6
1
2 3
F1
F2F4
F3
F5
F6
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x y1 ( 0 0 ) E = 4
2 ( 11 0 ) A = 3
3 ( 8 7 )
ELEMENTO N° K2
λx = 0.753 λy = 0.659
1 2 5 6
1 0.57 0.50 -0.57 -0.50
K2 =2 0.50 0.43 -0.50 -0.43 12
5 -0.57 -0.50 0.57 0.50 10.630
6 -0.50 -0.43 0.50 0.43
1 2 5 61 0.639 0.559 -0.639 -0.559
K2 = 2 0.559 0.490 -0.559 -0.4905 -0.639 -0.559 0.639 0.5596 -0.559 -0.490 0.559 0.490
1
2 3
4
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x y1 ( 0 0 ) E = 6
2 ( 11 0 ) A = 5
3 ( 8 7 )
ELEMENTO N° K3
λx = -0.394 λy = 0.919
3 4 5 6
3 0.16 -0.36 -0.16 0.36
K3 =4 -0.36 0.84 0.36 -0.84 30
5 -0.16 0.36 0.16 -0.36 7.616
6 0.36 -0.84 -0.36 0.84
3 4 5 63 0.611 -1.426 -0.611 1.426
K3 = 4 -1.426 3.328 1.426 -3.3285 -0.611 1.426 0.611 -1.4266 1.426 -3.328 -1.426 3.328
3
4
56
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0.821 0.496 -0.182 0.000 -0.639 -0.5590.559 0.490 0.000 0.000 -0.559 -0.490
K =-0.182 0.000 0.793 -1.426 -0.611 1.4260.000 0.000 -1.426 1.426 1.426 -3.328-0.639 -0.559 -0.611 1.426 0.611 -0.867-0.559 -0.490 1.426 -3.328 -1.426 3.328
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0.793 -0.611 1.426 K33 = -0.611 0.611 -0.867
1.426 -1.426 3.328
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q = k´ * d
56.463 -58.950 73.628 3.829 17.899 -22.243 0-85.916 89.024 -114.119 -6.292 -29.407 33.608 0
q =81.140 -83.933 113.712 6.039 28.228 -34.048
*10
5.505 -5.573 7.715 -0.033 2.304 -2.633 07.018 -8.892 9.835 0.628 1.373 -3.357 5
-29.407 29.772 -41.212 -2.633 -10.521 12.025 0
q1 = 825.77 q2 = -1288.23 q3 = 1278.25 q4 = 88.67 q5 = 105.21 q6 = -464.73
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q = k´ * d
5.500 0.000 -2.357 10 q = 5.500 4.171 -1.271 * 5
0.000 1.788 0.766 0
q3 = 55.00 q5 = 75.85 q6 = 8.94