METODO DE RIGIDEZ-CAP7

Ejercicio Nº 9
Calcular los esfuerzos en las barras del reticulado sometido a la carga exterior P. Las barras
poseen sección circular con un diámetro   0.020 m .
Datos:
y
1
x
P
P  11 kN
E  2.10 108 kN m 2
A   2 4
b
2
3
a
 3.1416  104 m 2
EA  65973 kN
a  1.0 m
b  1.5 m
4
a
Conectividades
γ1
γ2
(x )  ( y )
ij
x l
 y ij l 
k = EA / L
[kN/m]
1.5
1.803
0.555
0.832
36595
0
1.5
1.500
0
1.000
43982
1–4
1
1.5
1.803
0.555
0.832
36595
2–3
1
0
1.000
1.000
0
65973
3–4
1
0
1.000
1.000
0
65973
Barra
xij [m]
yij [m]
1–2
1
1–3
L [m]
ij 2
ij 2
Matriz de rigidez elemental (coordenadas globales)
i
2
1
Ki j

i  
EA   1 2

L   12
j
  1 2
j
 1 2
 22
 1 2
 22

2
1
 1 2
 12
 1 2
 1 2 

 22 
 1 2 

 22 
1
Barra 1 – 2
1
K1 2

2
 11260 16890 11260 16890 
1
16890
25335 16890 25335 
 
 11260 16890 11260 16890 
2

25335 
 16890 25335 16890
Barra 1 – 3
1
K13

3
0
0
1
0 43982
 
0
0
3
 0 43982
0
0 
0 43982 
0
0 

0 43982 
Barra 1 – 4
K1 4

1
4
11260

16890

11260
16890 

1
16890 25335 16890 25335 
 
 11260 16890 11260 16890 
4

 16890 25335 16890 25335 
Barra 2 – 3
K 2 3

2
 65973
2
0
 
 65973
3
 0
0
0
0
0
3
65973
0
65973
0
0
0 
0

0
0
0
0
0
4
65973
0
65973
0
0
0 
0

0
Barra 3 – 4
K 3 4

3
 65973
3
0
 
 65973
4
 0
2
Sistema global de ecuaciones (ensamblaje)
1
2
3
4
0
11260 16890
0
0
11260 16890  U1x   P1x 
 22520
1
   
94652 16890 25335
0
43982 16890 25335  U1y   0 
 0
 11260 16890 77233 16890 65973
0
0
0  U 2x   0 
   
2
25335
0
0
0
0   0   R2y 
 16890 25335 16890

 0
0
65973
0
131946
0
65973
0  U 3x   0 
   
3
43982
0
0
0
43982
0
0  U 3y   0 
 0
 11260 16890
0
0
65973
0
77233 16890   0   R4x 

   
4
16890 25335
0
0
0
0
16890 25335   0   R4y 

  
U
P
K



Sistema reducido (eliminación de filas y columnas asociadas a desplazamientos conocidos)
1
2
3
0
11260
0
0  U1x  11
 22520
1
 
94652 16890
0
43982 U1y   0 
 0
0  U 2x    0 
2  11260 16890 77233 65973

   
0
0
65973 131946
0  U 3x   0 

3
 0
43982
0
0
43982  U 3y   0 

U1x  5.718 104 
 y 
5 
U1  5.558 10 
U 2x   1.667 104 
 x 
5 
U 3  8.337 10 
U y  5.558 105 

 3 
Determinación de reacciones (utilización de filas eliminadas del sistema global)
1
2
3
4
 R2y 
0
0
0
0 
2  16890 25335 16890 25335
 
 11260 16890
0
0
65973 0 77233 16890  U   R4x 
4

 R4y 
 16890 25335
0
0
0
0 16890 25335 
 
 R2y   8.25
 x 

 R4    11 
 R4y   8.25 
 
  F x  P1x  R4x  0

Verificación:  F y  R2y  R4y  0

x
y
  M 4  1.5  P1  2  R2  0
Determinación de esfuerzos en las barras
N i  j  ki  j ei  j
 
ti  j   1 

 2 
ei j  ti  j  U i  j

x
x
 (U j  U i ) 
U i  j   y
y 

(U j  U i ) 
N i  j  ki  j  1 (U jx  U ix )   2 (U jy  U iy ) 
3
N12  36595   0.555  (1.667 10 4  5.718 10 4 )  0.832  (0  5.558 105 )   9.915 kN
N13  43982  0  (8.337 105  5.718 10 4 )  1.000  (5.558 10 5  5.558 105 )   0 kN
N14  36595   0.555  (0  5.718 10 4 )  0.832  (0  5.558 10 5 )   9.915 kN
N 23  65973  1.000  (8.337 10 5  1.667 10 4 )  0  (5.558 10 5  0)   5.500 kN
N 3 4  65973  1.000  (0  8.337 10 5 )  0  (0  5.558 10 5 )   5.500 kN
4