ANALISIS DE UNA PLACA Y MURO DE ALBAÑILERIA Se tiene la siguiente estructura de albañileria y placa se analizara el comportamiento de los brazos rigidos dependiendo de sus rigidezes de estas y se comprobara con el porgrama sap 2000 los datos calculados
SOLUCION: como son estructuras que tienen rigidezes y trabjan como uno solo se idealizara de la siguiente manera
ELEMENTO 01: D :=
3
b :=
E :=
2173706.512
0.15
3
I := G :=
D ⋅b 12
= 0.337
0.4 ⋅ E
= 8.695 ×
5 10
A := b ⋅ D As := b ⋅ D⋅ Φ :=
5 6
= 0.375
12⋅ E⋅ I 2
G⋅ As⋅ H
= 2.556
H :=
3.25
MATRIZ DEL ELEMENTO 1:
12⋅ E⋅ I 0 3 ( 1 + Φ) ⋅ H E⋅ A 0 H −6⋅ E⋅ I 0 ( 1 + Φ) ⋅ H2 E1 := −12⋅ E⋅ I 0 ( 1 + Φ) ⋅ H3 −E⋅ A 0 H 6⋅ E⋅ I 0 2 ( 1 + Φ) ⋅ H
( 0)
( 0)
−6⋅ E⋅ I
−12⋅ E⋅ I
( 1 + Φ) ⋅ H
( 1 + Φ) ⋅ H
0
0
( 4 + Φ ) ⋅ E⋅ I
6⋅ E⋅ I
2
0
3
( 1 + Φ) ⋅ H
−E⋅ A
0
2
6⋅ E⋅ I
12⋅ E⋅ I
( 1 + Φ) ⋅ H
( 1 + Φ) ⋅ H
0
0
( 2 − Φ ) ⋅ E⋅ I
6⋅ E⋅ I
2
0
3
( 1 + Φ) ⋅ H
2
( 0)
( 1)
0
( 2 − Φ ) ⋅ E⋅ I ( 1 + Φ) ⋅ H 6⋅ E⋅ I
( 1 + Φ) ⋅ H
2
E⋅ A H
( 1 + Φ) ⋅ H
( 1 + Φ) ⋅ H
2
H
( 1 + Φ) ⋅ H
6⋅ E⋅ I
0
0
( 4 + Φ ) ⋅ E⋅ I ( 1 + Φ) ⋅ H
( 2)
( 3) ( 0)
7.211 × 0 −1.172 × −7.211 × 0 1.172 × 5 0 3.01 × 10 0 0 −3.01 × 105 0 5 5 −1.172 × 105 0 4.162 × 10 1.172 × 10 0 −3.531 × 104 E1 = 4 5 4 5 0 1.172 × 10 7.211 × 10 0 1.172 × 10 −7.211 × 10 5 0 −3.01 × 105 0 0 3.01 × 10 0 5 0 −3.531 × 104 1.172 × 105 0 4.162 × 10 1.172 × 105 4 10
5 10
4 10
ELEMENTO 02: DATOS DE LA VIGA b :=
0.3
h :=
E :=
2173706.512
A := b ⋅ h = 0.15 E = 2.174 ×
6 10
0.5
I :=
b⋅ h 12
3
= 3.125 ×
−3
10
5 10
( 0) ( 0) ( 1) ( 2) ( 3)
D :=
3
d :=
5.4
α :=
.9
β :=
0.9
a := α⋅
D 2
L :=
5.393
L :=
5.393
As := A⋅ Φ :=
5 6
= 1.35
L1 :=
D 2
− a = 0.15
Xg :=
3.893
b := β⋅ Xg = 3.504
L2 := Xg − b = 0.389
−a−b
= 0.125
12⋅ E⋅ I
G⋅ As⋅ L
2
= 0.026
MATRIZ DEL ELEMENTO 2:
E⋅ A L 0 0 E2 := −E⋅ A L 0 0
0
3
2 12⋅ E⋅ I⋅ a + 6⋅ E⋅ I ( 4 + Φ) ⋅ E⋅ I + 6⋅ E⋅ I⋅ ( 2⋅ a) + 12⋅ E⋅ I⋅ a 3 2 2 3 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L 0
0
−
−12⋅ E⋅ I ( 1 + Φ) ⋅ L
12⋅ E⋅ I⋅ a
3 ( 1 + Φ) ⋅ L
3
+
( 2)
( 3)
( 0)
−12⋅ E⋅ I
0
0
E⋅ A
( 1 + Φ) ⋅ L −
12⋅ E⋅ I⋅ a
( 1 + Φ) ⋅ L
3
2 ( 1 + Φ) ⋅ L 6⋅ E⋅ I
( 4)
( 5)
6⋅ E⋅ I
( 1 + Φ) ⋅ L
12⋅ E⋅ I
0
0
+
3
0
L
12⋅ E⋅ I⋅ b + 6⋅ E⋅ I ( 2 − Φ) ⋅ E⋅ I + 6⋅ E⋅ I⋅ ( a + b ) + 12⋅ E⋅ I⋅ a⋅ b 3 2 2 3 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L
( 0)
0
L
12⋅ E⋅ I⋅ a + 6⋅ E⋅ I 3 2 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L
12⋅ E⋅ I
( 1 + Φ) ⋅ L
−E⋅ A
0
( 1 + Φ) ⋅ L −
12⋅ E⋅ I⋅ b
( 1 + Φ) ⋅ L
3
+
3 6⋅ E⋅ I
( 1 + Φ) ⋅ L
6.046 × 104 0 0 −6.046 × 104 0 0 3 0 506.622 2.05 × 10 0 −506.622 3.141 × 103 3 3 0 2.05 × 10 9.555 × 10 0 −2.05 × 103 1.145 × 104 E2 = 4 0 0 6.046 × 104 0 0 −6.046 × 10 0 −506.622 −2.05 × 103 0 506.622 −3.141 × 103 3 4 0 3.141 × 10 1.145 × 10 0 −3.141 × 103 2.074 × 104 ELEMENTO 03: DATOS DEL MURO L := t :=
m
5.4
m
0.3
b1 := h :=
m
0.40
3.25
Ec :=
ton
2173706.512
m2 ton
Em :=
325000.00
m2
Ln := L − b1 = 5 n :=
Ec Em
A :=
I :=
= 6.688
.15 ⋅
3 5
+
12
0.4 ⋅ Em
As := A⋅ Φ :=
n ⋅ t = 2.006
1.55
I = 4.395103 G :=
m
1
K
.15 ⋅ 5⋅ 1.393
m4 = 1.3 ×
5 10
= 0.75
12⋅ Em⋅ I
G⋅ As⋅ h
2
= 16.644
2
+
.4
3 2 ⋅ 12
+ .4 ⋅ 2⋅ 1.307 2
K :=
2.066666666667
( 0) ( 2) ( 3) ( 0) ( 4) ( 5)
MATRIZ DEL ELEMENTO 3: −6⋅ Em⋅ I 12⋅ Em⋅ I 0 3 2 ( 1 + Φ) ⋅ H ( 1 + Φ) ⋅ H Em⋅ A 0 0 H ( 4 + Φ ) ⋅ Em⋅ I −6⋅ Em⋅ I 0 2 ( 1 + Φ) ⋅ H ( 1 + Φ) ⋅ H E3 := 6⋅ Em⋅ I −12⋅ Em⋅ I 0 2 ( 1 + Φ) ⋅ H3 ( 1 + Φ) ⋅ H −Em⋅ A 0 0 H 6⋅ Em⋅ I ( 2 − Φ ) ⋅ Em⋅ I 0 2 ( 1 + Φ) ⋅ H ( 1 + Φ) ⋅ H
( 0)
( 0)
−12⋅ Em⋅ I ( 1 + Φ) ⋅ H
3
0 6⋅ Em⋅ I
( 1 + Φ) ⋅ H
2
12⋅ Em⋅ I
( 1 + Φ) ⋅ H
3
0 6⋅ Em⋅ I
( 1 + Φ) ⋅ H
2
( 0)
( 1 + Φ) ⋅ H −Em⋅ A 0 H ( 2 − Φ ) ⋅ Em⋅ I 0 ( 1 + Φ) ⋅ H 6⋅ Em⋅ I 0 2 ( 1 + Φ) ⋅ H Em⋅ A 0 H ( 4 + Φ ) ⋅ Em⋅ I 0 ( 1 + Φ) ⋅ H 6⋅ Em⋅ I
0
( 1)
2
( 4)
( 5)
2.83 × 104 0 −4.599 × 104 −2.83 × 104 0 4.599 × 104 5 0 1.55 × 10 0 0 −1.55 × 105 0 −4.599 × 104 0 5.142 × 105 4.599 × 104 0 −3.648 × 105 E3 = 4 4 4 4 0 4.599 × 10 2.83 × 10 0 4.599 × 10 −2.83 × 10 5 0 −1.55 × 105 0 0 1.55 × 10 0 5 0 −3.648 × 105 4.599 × 104 0 5.142 × 10 4.599 × 104
( 0) ( 0) ( 0) ( 1) ( 4) ( 5)
MATRIZ DE RIGIDEZ DEL SISTEMA K :
E13 , 4 E13 , 5 E33 , 4 E33 , 5 E13 , 3 + E33 , 3 E14 , 3 E14 , 4 + E21 , 1 E14 , 5 + E21 , 2 E21 , 4 E21 , 5 E15 , 3 E15 , 4 + E22 , 1 E15 , 5 + E22 , 2 E22 , 4 E22 , 5 K := E34 , 3 E24 , 1 E24 , 2 E24 , 4 + E34 , 4 E24 , 5 + E34 , 5 E35 , 3 E25 , 1 E25 , 2 E25 , 4 + E35 , 4 E25 , 5 + E35 , 5
( 1) 1.004 × 0 K = 1.172 × 0 4.599 ×
( 2) 5 10
5 10
( 3)
0 3.015
×
5 10
2.05
×
3 10
−506.622 104 3.141
×
( 4)
( 5)
1.172
×
5 10
0
4.599
×
4 10
2.05
×
3 10
−506.622
3.141
×
3 10
4.257
×
5 10
−2.05 ×
3 10
1.145
×
4 10
−2.05 ×
3 10
1.555
×
5 10
−3.141 ×
×
104
−3.141 ×
103 1.145
103
5.35
×
3 10
105
( 1) ( 2) ( 3) ( 4) ( 5)
VECTOR DE FUERZAS EXTERNAS DEL SISTEMA f:
7 0 f := 0 0 0
ton
VECTOR DE DESPLAZAMIENTO DEL SISTEMA u :
−1
U := K
⋅f
1.082 × 10− 4 2.903 × 10− 7 U = −2.954 × 10− 5 −5.637 × 10− 7 −8.67 × 10− 6
m
( 2)
m
( 1)
( 3)
rad m
( 4)
rad
( 5)
VECTOR FUERZAS INTERNAS DE LOS ELEMENTOS f(e) : ELEMENTO 01:
0 0 0 E11 := E1⋅ U 0, 0 U1 , 0 U2 , 0
−11.262 −0.087 = 13.718 4.338 0.087 0.38
0 U1 , 0 U2 , 0 E22 := E2 0 U3 , 0 U4 , 0
0 −0.087 −0.38 = 0 0.087 −0.515
ton
ton ton − m ton ton ton − m
ELEMENTO 02:
ton ton ton − m ton ton ton − m
ELEMENTO 03:
0 0 0 E33 := E3⋅ U 0, 0 U3 , 0 U4 , 0
−3.46 0.087 = 8.137 2.662 −0.087 0.515
ton ton ton − m ton ton ton − m
2
FUERZAS EN LOS EXTREMOS DE LA BARRA FLEXIBLE: CARAS DE APOYO
1 0 0 1 0 −a T2 := 0 0 0 0 0 0
1 0 0 1 = 0 −1.35 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0
b
1
0 −0.087 −0.262 τ := T2 ⋅ E22 = 0 0.087 −0.209
0 0
0
0
0 0
0
0
1 0
0
0
0 1
0
0
0 0
1
0
0 0 3.504 1
ton ton ton − m ton ton ton − m
DESPLAZAMIENTOS EN LOS EXTREMOS DE LA BARRA FLEXIBLE: CARAS DE APOYO
1 0 0 1 0 0 U2 := 0 0 0 0 0 0
0 0 0 a 0 0 1 0 0 0 1 0 0 0 1 0 0 0
1 0 0 0 = 0 0 −b 0 1 0 0
0
0 0
0
0
1 1.35 0 0
0
0
1
0 0
0
0
0
1 0
0
0
0
0 1
−3.504
0
0
0 0
1
0
m
0 U1 , 0 U2 , 0 μ := U2 0 U3 , 0 U4 , 0
0 −5 −3.959 × 10 −2.954 × 10− 5 = 0 2.981 × 10− 5 −8.67 × 10− 6
m rad m m rad
−12⋅ E⋅ I 12⋅ E⋅ I⋅ b 6⋅ E⋅ I 0 + 3 3 2 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L 12⋅ E⋅ I⋅ a 6⋅ E⋅ I ( 2 − Φ) ⋅ E⋅ I + 6⋅ E⋅ I⋅ ( a + b ) + 12⋅ E⋅ I⋅ a⋅ b 0 − + 3 2 2 3 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L E⋅ A 0 0 L 12⋅ E⋅ I 12⋅ E⋅ I⋅ b 6⋅ E⋅ I 0 − + 3 3 2 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L 2 12 ⋅ E ⋅ I ⋅ b 6 ⋅ E ⋅ I ( 4 + Φ ) ⋅ E ⋅ I 6 ⋅ E ⋅ I ⋅ ( 2 ⋅ b ) 12 ⋅ E ⋅ I ⋅ b 0 − + + + 3 2 2 3 ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L ( 1 + Φ) ⋅ L
−E⋅ A L
3 Φ) ⋅ L
⋅ I⋅ a
2
Φ) ⋅ L
⋅ I⋅ a⋅ b
3
0
0
RAS DE APOYO
