Ejercicio Semana 10
10
Ejemplo 3.2 Considerar la barra cónica mostrada de la figura con E =210 GPa y P = 18 kN. El area de la sección transversal en el extremo izquierdo es de 0.002 m 2 y del extremo derecho 0.012 m 2 . Utilizar el método del elemento de barra lineal para determinar el desplazamiento en el extremo libre de la barra. Solución 1. Discretización del Dominio. A ( x) =0.002 + 0.01 x 3 Discretización de la barra conica Conectividad del elemento N° Elemento Nodo i Nodo j 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9
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ejercicio resuelto en matlab para calculo 4
Transcript of Ejercicio Semana 10
Ejemplo 3.2
Considerar la barra cónica mostrada de la figura con E =210 GPa y P = 18 kN. El area de la sección transversal en el extremo izquierdo es de 0.002 m2 y del extremo derecho 0.012 m2. Utilizar el método del elemento de barra lineal para determinar el desplazamiento en el extremo libre de la barra.
Solución
1. Discretización del Dominio.
A ( x )=0.002+ 0.01 x3
Discretización de la barra conica
Conectividad del elemento
N° Elemento Nodo i Nodo j
1 1 22 2 33 3 44 4 55 5 66 6 77 7 88 8 9
2. Escritura de las matrices elementales de rigidez>> E=210e6E = 210000000>> L=3/8L = 0.3750
>> A1=0.002+(0.01*0.188/3)A1 = 0.0026
>> A2=0.002+(0.01*0.563/3)A2 = 0.0039
>> A3=0.002+(0.01*0.938/3)A3 = 0.0051
>> A4=0.002+(0.01*1.313/3)A4 = 0.0064
>> A5=0.002+(0.01*1.688/3)A5 = 0.0076
>> A6=0.002+(0.01*2.063/3)A6 = 0.0089
>> A7=0.002+(0.01*2.438/3)A7 = 0.0101
>> A8=0.002+(0.01*2.813/3)A8 = 0.0114
>> k1=LinearBarElementStiffness(E,A1,L)k1 = 1.0e+006 *
1.4709 -1.4709 -1.4709 1.4709
>> k2=LinearBarElementStiffness(E,A2,L)k2 = 1.0e+006 * 2.1709 -2.1709 -2.1709 2.1709
>> k3=LinearBarElementStiffness(E,A3,L)k3 = 1.0e+006 * 2.8709 -2.8709 -2.8709 2.8709
>> k4=LinearBarElementStiffness(E,A4,L)k4 = 1.0e+006 * 3.5709 -3.5709 -3.5709 3.5709
>> k5=LinearBarElementStiffness(E,A5,L)k5 = 1.0e+006 * 4.2709 -4.2709 -4.2709 4.2709
>> k6=LinearBarElementStiffness(E,A6,L)k6 = 1.0e+006 * 4.9709 -4.9709 -4.9709 4.9709
>> k7=LinearBarElementStiffness(E,A7,L)k7 = 1.0e+006 * 5.6709 -5.6709 -5.6709 5.6709
>> k8=LinearBarElementStiffness(E,A8,L)k8 = 1.0e+006 * 6.3709 -6.3709 -6.3709 6.3709
3. Ensamble de la matriz de rigidez global>> K=zeros(9,9)K = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,1,2)K = 1.0e+006 * Columns 1 through 7 1.4709 -1.4709 0 0 0 0 0 -1.4709 1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 8 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,2,3)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,3,4)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,4,5)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,5,6)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 0 0 0 0 -1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Columns 6 through 9
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.4709 0 0 0 1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,6,7)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 0 0 0 0 -1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.4709 0 0 0 2.9419 -1.4709 0 0 -1.4709 1.4709 0 0 0 0 0 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,7,8)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 0 0 0 0 -1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.4709 0 0 0 2.9419 -1.4709 0 0 -1.4709 2.9419 -1.4709 0 0 -1.4709 1.4709 0 0 0 0 0
>> K=LinearBarAssemble(K,k1,8,9)K = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 0 0 0 0 -1.4709 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Columns 6 through 9 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1.4709 0 0 0 2.9419 -1.4709 0 0 -1.4709 2.9419 -1.4709 0 0 -1.4709 2.9419 -1.4709 0 0 -1.4709 1.4709
4. Aplicación de las condiciones de entornoF1=-18, F2=F3=F4=F5=F6=F7=F8=0, U9=0U1, U2, U3, U4, U5, U6, U6, U7, U8, todos estos desplazamientos son desconocidos.
5. Resolviendo las ecuacionesk=K(1:8,1:8)k = 1.0e+006 * Columns 1 through 5 1.4709 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 -1.4709 0 0 0 -1.4709 2.9419 0 0 0 0 -1.4709 0 0 0 0 0 0 0 0 0 0
Columns 6 through 8 0 0 0 0 0 0 0 0 0 0 0 0 -1.4709 0 0 2.9419 -1.4709 0 -1.4709 2.9419 -1.4709 0 -1.4709 2.9419
>> f=[-18;0;0;0;0;0;0;0]f = -18 0 0 0 0 0 0 0
>> u= k\fu = 1.0e-004 * -0.9790 -0.8566 -0.7342 -0.6119 -0.4895 -0.3671 -0.2447 -0.1224
6. Post procesamiento
>> U=[u;0]U = 1.0e-004 * -0.9790 -0.8566 -0.7342 -0.6119 -0.4895 -0.3671 -0.2447 -0.1224 0
>> F=K*UF = -18.0000 0.0000 0.0000 0 -0.0000 0.0000 -0.0000 0 18.000
