Sexta Práctica
Sexta Práctica
Sexta Práctica
CALCULO POR
PROFESOR: ING. EDWIN ABREGU
ELEMENTOS FINITOS
ALUMNO: MEZA RAMOS WALTER
(MC516C)
ANALISIS POR EL METODO DE ELEMENTO FINITOS
𝛾: 77.0085𝑁/𝑚𝑚3
𝐸 = 195𝐺𝑃𝑎
A temperatura ambiente
160 + 136
𝑏1 = = 148𝑚𝑚
2
136 + 112
𝑏2 = = 124𝑚𝑚
2
112 + 88
𝑏3 = = 100𝑚𝑚
2
88 + 64
𝑏4 = = 76𝑚𝑚
2
64 + 40
𝑏5 = = 52𝑚𝑚
2
CUADRO DE CONECTIVIDAD:
4 4 5 Q4 Q5 100 76 4536.45979178
5 5 6 Q5 Q6 100 52 2123.71663383
2) GRADOS DE LIBERTAD NODALES (Vector desplazamiento)
0
Q 2
Q3
El cálculo el vector de desplazamiento será: Q
Q 4
Q 5
Q 6
3) VECTOR CARGA
1 0
1 1
1 1
1 1
𝐴1 𝐿1 𝛾 0 𝐴2 𝐿2 𝛾 1
𝑓1 = 0 = 66.241 𝑓2 = 0 = 46.499
2 0 2 0
0 0
[0] 0 [ 0] 0
[0] [0]
0 0
1 1
0 0
1 1
𝐴3 𝐿3 𝛾 1 𝐴4 𝐿4 𝛾 0
𝑓3 = 2 0 = 30.241 𝑓4 = 2 0 = 17.468
1 1
0 0
[0] 0 [ 0] 1
[0] [0]
0
1
0
1
𝐴 𝐿 𝛾 0
𝑓5 = 525 0 = 8.177
0
0
[0] 1
[1]
𝑅
0
0
Fuerzas producidas por cargas puntuales y fuerza de reacción: 𝑓′ =
0
0
[50000]
𝑅 + 66.241
112.74
76.74
El vector carga quedara como sigue: 𝐹= 𝑁
47.71
25.645
[ 50008.177 ]
4) MATRIZ DE RIGIDEZ
1 −1 0 0 0 0 1 −1 0 0 0 0
−1 1 0 0 0 0 −1 1 0 0 0 0
𝐴𝐸 0 0 0 0 0 0 0 0 0 0 0 0
𝑘1 = ( ) = 33546554.6734
𝐿 1 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 1 −1 0 0 0 0 1 −1 0 0 0
𝐴𝐸 0 −1 1 0 0 0 0 −1 1 0 0 0
𝑘2 = ( ) = 23548750.2128
𝐿 2 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
𝐴𝐸 0 0 1 −1 0 0 0 0 1 −1 0 0
𝑘3 = ( ) = 15315264.1863
𝐿 3 0 0 −1 1 0 0 0 0 −1 1 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
𝐴𝐸 0 0 0 0 0 0 0 0 0 0 0 0
𝑘4 = ( ) = 8846096.59397
𝐿 4 0 0 0 1 −1 0 0 0 0 1 −1 0
0 0 0 −1 1 0 0 0 0 −1 1 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
𝐴𝐸 0 0 0 0 0 0 0 0 0 0 0 0
𝑘5 = ( ) = 4141247.43597
𝐿 5 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 −1 0 0 0 0 1 −1
[0 0 0 0 −1 1 ] [0 0 0 0 −1 1 ]
Sumando obtenemos K:
𝑘1 + 𝑘2 + 𝑘3 + 𝑘4 + 𝑘5
33546554.6734 −33546554.6734 0 0 0 0
−33546554.6734 5709534.89 −23548750.2 0 0 0
0 −23548750.2 38864014.4 −15315264.2 0 0
=
0 0 −15315264.2 24161360.78 −8846096.59 0
0 0 0 −8846096.59 12987344.03 −4141247.43597
[ 0 0 0 0 −4141247.43597 4141247.43597 ]
𝐹=𝐾𝑋𝑄
𝑄2 = 0.0015
𝑄3 = 0.00363
𝑄4 = 0.0069
𝑄5 = 0.0126
𝑄6 = 0.02464
Para obtener la reacción en el empotramiento tomamos la siguiente submatriz:
0
0.0015
[𝑅 + 132.481] = 106 [33.55 −33.55 0 0 0 0.0036
0]
0.0069
0.0126
[0.0246]
𝑅 = −50337.241𝑁 = −50.337241 𝐾𝑁
𝐸 𝑄
𝜎𝑒 = ( ) [−1 1] [ 𝑒 ]
𝐿 𝑒 𝑄𝑒+1
195 103 0
𝜎1 = ( ) [−1 1] [ ] = 2.922𝑀𝑃𝑎
100 1 0.0015
RESULTADOS:
R −50.337241 𝐾𝑁
𝜎1 2.922𝑀𝑃𝑎
𝜎2 4.153𝑀𝑃𝑎
𝜎3 6.377𝑀𝑃𝑎
𝜎4 11.029𝑀𝑃𝑎
𝜎5 23.547𝑀𝑃𝑎
DIAGRAMA DE FLUJO
Para i=1:n
x=zeros(n+1)
x(i,i)=1 x(i+1,i)=-1
x(i,i+1)=-1 x(i+1,i+1)=1
Para i=1:n
Calculo de esfuerzos:
Imprime Reaccion,
desplazamientos y
esfuerzos
FIN
PROGRAMA:
%datos de entrada
h=input('Ingrese altura(mm):')
y=input('Ingrese densidad(N/mm3):')
for i=1:n
b(i)=((2*n*bo)+((2*i-1)*(bn-bo)))/(2*n);
a(i)=((b(i))^2)*(pi/4);
v(i)=(a(i)*h)/n;
end
disp('areas:')
disp(a')
disp('volumenes:')
disp(v')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for i=1:n
Fn(i)=y*v(i)/2;
end
F(1)=Fn(1);
for i=2:n
F(i)=Fn(i)+Fn(i-1);
end
F(n+1)=Fn(n);F(1)=F(1)+pa
disp(F')
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
k=zeros(n+1);
for i=1:n
x=zeros(n+1);
x(i,i)=1;x(i+1,i)=-1;x(i,i+1)=-1;x(i+1,i+1)=1;
k=k+(a(i)*E/(h/n))*x;
end
disp(k)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%calculo de desplazamientos
Q=inv(k(1:n,1:n))*F(1:n);
Q=[Q; 0];
disp(Q)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%calculo de la reaccion
k(n+1,:)*Q
R1=k(n+1,:)*Q-F(n+1)’
disp(R1)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%calculo de esfuerzos
for i=1:n
end
disp(e');
bo =40
bn =160
Ingrese altura(mm):500
h =500
n =5
E =195000
pa =50000
Ingrese densidad(N/mm3):77.0085*1.0e-006
y = 7.7009e-005
areas:
1.0e+004 *
0.2124
0.4536
0.7854
1.2076
1.7203
volumenes:
1.0e+006 *
0.2124
0.4536
0.7854
1.2076
1.7203
el vector de fuerzas:
1.0e+004 *
5.0008
0.0026
0.0048
0.0077
0.0113
0.0066
1.0e+007 *
0.4141 -0.4141 0 0 0 0
0 0 0 0 -3.3547 3.3547
0.0246
0.0126
0.0069
0.0036
0.0015
R1 =
-5.0337e+004
-5.0337e+004
23.5475
11.0293
6.3766
4.1535
2.9222
Conclusiones:
BIBLIOGRAFIA:
1) CHANDRUPATLA T. Introducción al estudio de los elementos finitos en ingeniería
Prentice Hall, 1999
2) Calculo por elementos finitos ( separatas del Ing. Cueva)
http://www.thyssenkrupp.cl/Documentos/fichasT/Aceros%20maquinaria%20barras/SAE%201045.
pdf (CATALOGO COMERCIAL DE ACEROS SAE1045 THYSSENKRUPP ACEROS Y SERVICIOS)