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UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.001978431373 0.003125 0.003125
280 280 280
2509980.079602 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
TIPO DE FIGURA 4
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 0.3 r(m) 0.42 175 h(m) 0.60 A(m2) 0.282743339 r(m) 0.23 210 A(m2) 0.24 I(m4) 0.006361725 A(m2) 0.3769911184 280 I 0.0072 As 0.254469 I(m4) 0.0188495565 315 As 0.2 As 0.188495566 420
VIGA 1CO,CA 0 5A (m²) 0.213 1 2 3 4 5 6Inercia (m4) 0.00386 1 92382.53 0.00 0.00 -92382.53 0.00 0.00F´C (Kg/cm2) 210 2 0.00 806.35 2015.87 0.00 -806.35 2015.87E (Tn/m²) 2173706.51 3 0.00 2015.87 6719.58 0.00 -2015.87 3359.79L (m) 5.00 4 -92382.53 0.00 0.00 92382.53 0.00 0.00Angulo 0 5 0.00 -806.35 -2015.87 0.00 806.35 -2015.87Radianes 0 6 0.00 2015.87 3359.79 0.00 -2015.87 6719.58COS 1SEN 0
TIPO DE FIGURA 4
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
VIGA 2CO,CA 0 5A (m²) 0.213 7 8 9 10 11 12Inercia (m4) 0.004 7 92382.53 0.00 0.00 -92382.53 0.00 0.00F´C (Kg/cm2) 210.000 8 0.00 806.35 2015.87 0.00 -806.35 2015.87E (Tn/m²) 2173706.512 9 0.00 2015.87 6719.58 0.00 -2015.87 3359.79L (m) 5.00 10 -92382.53 0.00 0.00 92382.53 0.00 0.00Angulo 0 11 0.00 -806.35 -2015.87 0.00 806.35 -2015.87Radianes 0.00 12 0.00 2015.87 3359.79 0.00 -2015.87 6719.58COS 1SEN 0
TIPO DE FIGURA 4
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 3
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
1 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
VIGA 3CO,CA 0 5A (m²) 0.213 13 14 15 16 17 18Inercia (m4) 0.004 13 92382.53 0.00 0.00 -92382.53 0.00 0.00F´C (Kg/cm2) 210.000 14 0.00 806.35 2015.87 0.00 -806.35 2015.87E (Tn/m²) 2173706.512 15 0.00 2015.87 6719.58 0.00 -2015.87 3359.79L (m) 5.00 16 -92382.53 0.00 0.00 92382.53 0.00 0.00Angulo 0 17 0.00 -806.35 -2015.87 0.00 806.35 -2015.87Radianes 0.00 18 0.00 2015.87 3359.79 0.00 -2015.87 6719.58COS 1SEN 0
TIPO DE FIGURA 6
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 4CO,CA 3 0A (m²) 0.813 1 2 3 7 8 9Inercia (m4) 0.037 1 35626.43 0.00 -53439.65 -35626.43 0.00 -53439.65F´C (Kg/cm2) 210.000 2 0.00 588712.18 0.00 0.00 -588712.18 0.00E (Tn/m²) 2173706.512 3 -53439.65 0.00 106879.29 53439.65 0.00 53439.65L (m) 3.00 7 -35626.43 0.00 53439.65 35626.43 0.00 53439.65Angulo 90 8 0.00 -588712.18 0.00 0.00 588712.18 0.00Radianes 1.57 9 -53439.65 0.00 53439.65 53439.65 0.00 106879.29COS 0SEN 1
TIPO DE FIGURA 6
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 5CO,CA 3 0A (m²) 0.813 7 8 9 13 14 15Inercia (m4) 0.037 7 35626.43 0.00 -53439.65 -35626.43 0.00 -53439.65F´C (Kg/cm2) 210.000 8 0.00 588712.18 0.00 0.00 -588712.18 0.00
E (Tn/m²) 2173706.512 9 -53439.65 0.00 106879.29 53439.65 0.00 53439.65L (m) 3.00 13 -35626.43 0.00 53439.65 35626.43 0.00 53439.65Angulo 90 14 0.00 -588712.18 0.00 0.00 588712.18 0.00Radianes 1.57 15 -53439.65 0.00 53439.65 53439.65 0.00 106879.29COS 0SEN 1
TIPO DE FIGURA 6
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 6CO,CA 3 0A (m²) 0.813 13 14 15 19 20 21Inercia (m4) 0.037 13 35626.43 0.00 -53439.65 -35626.43 0.00 -53439.65F´C (Kg/cm2) 210.000 14 0.00 588712.18 0.00 0.00 -588712.18 0.00E (Tn/m²) 2173706.512 15 -53439.65 0.00 106879.29 53439.65 0.00 53439.65L (m) 3.00 19 -35626.43 0.00 53439.65 35626.43 0.00 53439.65Angulo 90 20 0.00 -588712.18 0.00 0.00 588712.18 0.00Radianes 1.57 21 -53439.65 0.00 53439.65 53439.65 0.00 106879.29COS 0SEN 1
TIPO DE FIGURA 2
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 0.4 r(m) 42 175 h(m) 0.60 A(m2) 0.502654825 r(m) 23 210 A(m2) 0.24 I(m4) 0.020106193 A(m2) 37.699111844 280 I 0.0072 As 0.45238934 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 7CO,CA 3 0A (m²) 0.503 4 5 6 10 11 12Inercia (m4) 0.020 4 19424.43 0.00 -29136.64 -19424.43 0.00 -29136.64F´C (Kg/cm2) 210.000 5 0.00 364208.02 0.00 0.00 -364208.02 0.00E (Tn/m²) 2173706.512 6 -29136.64 0.00 58273.28 29136.64 0.00 29136.64L (m) 3.00 10 -19424.43 0.00 29136.64 19424.43 0.00 29136.64Angulo 90 11 0.00 -364208.02 0.00 0.00 364208.02 0.00Radianes 1.57 12 -29136.64 0.00 29136.64 29136.64 0.00 58273.28COS 0SEN 1
TIPO DE FIGURA 2
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 0.4 r(m) 42 175 h(m) 0.60 A(m2) 0.502654825 r(m) 2
3 210 A(m2) 0.24 I(m4) 0.020106193 A(m2) 37.699111844 280 I 0.0072 As 0.45238934 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 8CO,CA 3 0A (m²) 0.503 10 11 12 16 17 18Inercia (m4) 0.020 10 19424.43 0.00 -29136.64 -19424.43 0.00 -29136.64F´C (Kg/cm2) 210.000 11 0.00 364208.02 0.00 0.00 -364208.02 0.00E (Tn/m²) 2173706.512 12 -29136.64 0.00 58273.28 29136.64 0.00 29136.64L (m) 3.00 16 -19424.43 0.00 29136.64 19424.43 0.00 29136.64Angulo 90 17 0.00 -364208.02 0.00 0.00 364208.02 0.00Radianes 1.57 18 -29136.64 0.00 29136.64 29136.64 0.00 58273.28COS 0SEN 1
TIPO DE FIGURA 2
F`c 3 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 0.4 r(m) 42 175 h(m) 0.60 A(m2) 0.502654825 r(m) 23 210 A(m2) 0.24 I(m4) 0.020106193 A(m2) 37.699111844 280 I 0.0072 As 0.45238934 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 9CO,CA 3 0A (m²) 0.503 16 17 18 22 23 24Inercia (m4) 0.020 16 19424.43 0.00 -29136.64 -19424.43 0.00 -29136.64F´C (Kg/cm2) 210.000 17 0.00 364208.02 0.00 0.00 -364208.02 0.00E (Tn/m²) 2173706.512 18 -29136.64 0.00 58273.28 29136.64 0.00 29136.64L (m) 3.00 22 -19424.43 0.00 29136.64 19424.43 0.00 29136.64Angulo 90 23 0.00 -364208.02 0.00 0.00 364208.02 0.00Radianes 1.57 24 -29136.64 0.00 29136.64 29136.64 0.00 58273.28COS 0SEN 1
1 2 3 4 5 6 7 8 91 184765.05 0.00 0.00 -92382.53 0.00 0.00 -92382.53 0.00 0.002 0.00 1612.70 0.00 0.00 -806.35 2015.87 0.00 -806.35 -2015.873 0.00 0.00 13439.15 0.00 -2015.87 3359.79 0.00 2015.87 3359.794 -92382.53 0.00 0.00 #REF! #REF! #REF! 0.00 0.00 0.005 0.00 -806.35 -2015.87 #REF! #REF! #REF! 0.00 0.00 0.006 0.00 2015.87 3359.79 #REF! #REF! #REF! 0.00 0.00 0.007 -92382.53 0.00 0.00 0.00 0.00 0.00 92382.53 0.00 0.008 0.00 -806.35 2015.87 0.00 0.00 0.00 0.00 806.35 2015.879 0.00 -2015.87 3359.79 0.00 0.00 0.00 0.00 2015.87 6719.58
10 0.00 0.00 0.00 #REF! #REF! #REF! 0.00 0.00 0.0011 0.00 0.00 0.00 #REF! #REF! #REF! 0.00 0.00 0.0012 0.00 0.00 0.00 #REF! #REF! #REF! 0.00 0.00 0.00
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
Paso 04
1 0 TN2 -7.5 TN3 -6.25 TN-M4 0 TN5 -14 TN6 -0.125 TN-M
Paso 05
1 2 3 4 5 61 #REF! #REF! #REF! #REF! #REF! #REF!2 #REF! #REF! #REF! #REF! #REF! #REF!3 #REF! #REF! #REF! #REF! #REF! #REF!4 #REF! #REF! #REF! #REF! #REF! #REF!5 #REF! #REF! #REF! #REF! #REF! #REF!6 #REF! #REF! #REF! #REF! #REF! #REF!
7 8 9 10 11 121 -92382.53 0.00 0.00 0.00 0.00 0.002 0.00 -806.35 -2015.87 0.00 0.00 0.003 0.00 2015.87 3359.79 0.00 0.00 0.004 0.00 0.00 0.00 #REF! #REF! #REF!5 0.00 0.00 0.00 #REF! #REF! #REF!6 0.00 0.00 0.00 #REF! #REF! #REF!
Paso 06
1 #REF! M 7 #REF! TN2 #REF! M 8 #REF! TN3 #REF! RAD 9 #REF! TN-M4 #REF! M 10 #REF! TN5 #REF! M 11 #REF! TN6 #REF! RAD 12 #REF! TN-M
7 #REF! TN8 #REF! TN9 #REF! TN-M
10 #REF! TN11 #REF! TN12 #REF! TN-M
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑸𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑸𝒓}𝐂omplementario =
{𝑫𝒓} =
{𝑲𝒇𝒓} ={𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑹} =
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion
d
a
b
a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)
c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
10
0.000.000.00
#REF!#REF!#REF!
0.000.000.00
#REF!#REF!#REF!
789
101112
0.0000.0000.000#REF!#REF!#REF!
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.3
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.25 c(m) 0.3 c(m) 0.30.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.5
0.2125 A(m2) 0.49 A(m2) 0.690.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1770833333333 As 0.25 As 0.375
YC= 0.6 M0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.3
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.25 c(m) 0.3 c(m) 0.30.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.5
0.2125 A(m2) 0.49 A(m2) 0.690.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1770833333333 As 0.25 As 0.375
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 6
c
d
a b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
0.6 a(m) 0.3 a(m) 0.30.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.25 c(m) 0.3 c(m) 0.30.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.5
0.2125 A(m2) 0.49 A(m2) 0.690.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1770833333333 As 0.25 As 0.375
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.750.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.750.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
a a
a a
b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.750.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.750.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.75
a a
a a
b
c
a
d
b
c
a
d
a a
b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
0.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
1
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.25
0.25 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.750.25 c(m) 0.3 c(m) 0.250.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1
0.2125 A(m2) 0.49 A(m2) 0.81250.0038641237745 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.036876860.1770833333333 As 0.25 As 0.20833333
YC= 0.600 M 0.0640625
0.3014705882353 INERCIA TOTAL 0.0364 m4
a a
a a
b
c
a
d
b
c
a
d
11 12
0.00 0.000.00 0.000.00 0.00
#REF! #REF!#REF! #REF!#REF! #REF!
0.00 0.000.00 0.000.00 0.00
#REF! #REF!#REF! #REF!#REF! #REF!
0.000 M0.000 M0.000 RAD0.000 M0.000 M0.000 RAD
COMPROBACION#REF!
0-7.5
-6.25#REF!#REF!#REF!
Datos del MEP CON SU SIGNO
7 0 TN8 0 TN9 0 TN-M
10 0 TN11 0 TN12 0 TN-M
{𝑸𝒇} − {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
L 1ITEM A Y AY I d Ad2
1 0.25 0.5 0.125 0.020833 -0.161 0.006
2 0.188 0.125 0.0234375 0.000977 0.214 0.009
0.438 0.1484375 0.02181 0.015
YC= 0.339 M
INERCIA TOTAL 0.0368768601 m4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.001978431373 0.003125 0.003125
280 280 280
2509980.079602 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
TIPO DE FIGURA 1
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.20 r(m) 0.3 r(m) 0.42 175 h(m) 0.30 A(m2) 0.282743339 r(m) 0.23 210 A(m2) 0.06 I(m4) 0.006361725 A(m2) 0.3769911184 280 I 0.00045 As 0.254469 I(m4) 0.0188495565 315 As 0.05 As 0.188495566 420
VIGA 1CO,CA 0 4A (m²) 0.060 4 5 6 1 2 3Inercia (m4) 0.00045 4 45000.00 0.00 0.00 -45000.00 0.00 0.00F´C (Kg/cm2) 280 5 0.00 253.12 506.25 0.00 -253.12 506.25E (Tn/m²) 3000000.00 6 0.00 506.25 1350.00 0.00 -506.25 675.00L (m) 4.00 1 -45000.00 0.00 0.00 45000.00 0.00 0.00Angulo 0 2 0.00 -253.12 -506.25 0.00 253.12 -506.25Radianes 0 3 0.00 506.25 675.00 0.00 -506.25 1350.00COLUMNA 2CO,CA 4 0A (m²) 0.060 7 8 9 1 2 3Inercia (m4) 0.00045 7 253.12 0.00 -506.25 -253.12 0.00 -506.25F´C (Kg/cm2) 280.000 8 0.00 45000.00 0.00 0.00 -45000.00 0.00E (Tn/m²) 3000000.000 9 -506.25 0.00 1350.00 506.25 0.00 675.00L (m) 4.00 1 -253.12 0.00 506.25 253.12 0.00 506.25Angulo 90 2 0.00 -45000.00 0.00 0.00 45000.00 0.00Radianes 1.57 3 -506.25 0.00 675.00 506.25 0.00 1350.00
1 2 3 4 5 6 7 8 91 45253.13 0.00 506.25 -45000.00 0.00 0.00 -253.12 0.00 506.252 0.00 45253.13 -506.25 0.00 -253.12 -506.25 0.00 -45000.00 0.003 506.25 -506.25 2700.00 0.00 506.25 675.00 -506.25 0.00 675.004 -45000.00 0.00 0.00 45000.00 0.00 0.00 0.00 0.00 0.005 0.00 -253.12 506.25 0.00 253.12 506.25 0.00 0.00 0.006 0.00 -506.25 675.00 0.00 506.25 1350.00 0.00 0.00 0.007 -253.12 0.00 -506.25 0.00 0.00 0.00 253.12 0.00 -506.258 0.00 -45000.00 0.00 0.00 0.00 0.00 0.00 45000.00 0.009 506.25 0.00 675.00 0.00 0.00 0.00 -506.25 0.00 1350.00
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
Paso 04
1 0 TN2 -6 TN3 1.667 TN-M
Paso 05
1 2 3 0 0 01 2.214447E-05 -4.65473E-08 -4.16082E-062 -4.65473E-08 2.214447E-05 4.160816E-063 -4.16082E-06 4.160816E-06 0.000371931
4 5 6 7 8 91 -45000.00 0.00 0.00 -253.12 0.00 506.252 0.00 -253.12 -506.25 0.00 -45000.00 0.003 0.00 506.25 675.00 -506.25 0.00 675.00
Paso 06
1 -6.6568E-06 M 4 0.29955579 TN2 -0.000125931 M 5 0.33311701 TN3 0.000595044 RAD 6 0.46540683 TN-M0 M 7 -0.29955579 TN0 M 8 5.66688299 TN0 RAD 9 0.39828439 TN-M
4 0.299555793 TN5 4.333117011 TN6 3.132406826 TN-M7 -2.299555793 TN8 5.666882989 TN9 2.39828439 TN-M
-253 0 -506.2 -20 -45000 0 0
506.2 0 675 2
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑸𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑸𝒓}𝐂omplementario =
{𝑫𝒓} =
{𝑲𝒇𝒓} ={𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑹} =
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
d
a
b
456789
0.0000.0000.000
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.30.2 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.2 c(m) 0.3 c(m) 0.3
0.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.50.17 A(m2) 0.49 A(m2) 0.69
0.0019784313725 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1416666666667 As 0.25 As 0.375
YC= 0.6 M0.041
0.2411764705882 INERCIA TOTAL 0.0364 m4
c
d
a b
c
a
d
b
c
a
d
bcad b ca d
0.000 M0.000 M0.000 RAD0.000 M0.000 M0.000 RAD
COMPROBACION1
0.000-6.0001.667
Datos del MEP CON SU SIGNO
4 0 TN5 4 TN6 2.667 TN-M7 -2 TN8 0 TN9 2 TN-M
{𝑸𝒇} − {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.001978431373 0.003125 0.003125
280 280 280
2509980.079602 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
TIPO DE FIGURA 1
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 0.3 r(m) 0.42 175 h(m) 0.60 A(m2) 0.282743339 r(m) 0.23 210 A(m2) 0.24 I(m4) 0.006361725 A(m2) 0.3769911184 280 I 0.0072 As 0.254469 I(m4) 0.0188495565 315 As 0.2 As 0.188495566 420
VIGA 1CO,CA 0 5A (m²) 0.240 1 2 3 4 5 6Inercia (m4) 0.00720 1 120479.04 0.00 0.00 -120479.04 0.00 0.00F´C (Kg/cm2) 280 2 0.00 1734.90 4337.25 0.00 -1734.90 4337.25E (Tn/m²) 2509980.08 3 0.00 4337.25 14457.49 0.00 -4337.25 7228.74L (m) 5.00 4 -120479.04 0.00 0.00 120479.04 0.00 0.00Angulo 0 5 0.00 -1734.90 -4337.25 0.00 1734.90 -4337.25Radianes 0 6 0.00 4337.25 7228.74 0.00 -4337.25 14457.49COS 1SEN 0
TIPO DE FIGURA 1
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.40 r(m) 1 r(m) 42 175 h(m) 0.60 A(m2) 3.141592654 r(m) 23 210 A(m2) 0.24 I(m4) 0.785398163 A(m2) 37.699111844 280 I 0.0072 As 2.82743339 I(m4) 188.49555925 315 As 0.2 As 18.84955596 420
COLUMNA 2CO,CA 3 0A (m²) 0.240 7 8 9 1 2 3Inercia (m4) 0.007 7 8031.94 0.00 -12047.90 -8031.94 0.00 -12047.90F´C (Kg/cm2) 280.000 8 0.00 200798.41 0.00 0.00 -200798.41 0.00E (Tn/m²) 2509980.080 9 -12047.90 0.00 24095.81 12047.90 0.00 12047.90L (m) 3.00 1 -8031.94 0.00 12047.90 8031.94 0.00 12047.90Angulo 90 2 0.00 -200798.41 0.00 0.00 200798.41 0.00Radianes 1.57 3 -12047.90 0.00 12047.90 12047.90 0.00 24095.81COLUMNA 3CO,CA 3 0A (m²) 0.240 10 11 12 4 5 6Inercia (m4) 0.007 10 8031.94 0.00 -12047.90 -8031.94 0.00 -12047.90F´C (Kg/cm2) 280 11 0.00 200798.41 0.00 0.00 -200798.41 0.00E (Tn/m²) 2509980.08 12 -12047.90 0.00 24095.81 12047.90 0.00 12047.90L (m) 3.00 4 -8031.94 0.00 12047.90 8031.94 0.00 12047.90Angulo 90 5 0.00 -200798.41 0.00 0.00 200798.41 0.00Radianes 1.57 6 -12047.90 0.00 12047.90 12047.90 0.00 24095.81
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
1 2 3 4 5 6 7 8 91 128510.98 0.00 12047.90 -120479.04 0.00 0.00 -8031.94 0.00 12047.902 0.00 202533.30 4337.25 0.00 -1734.90 4337.25 0.00 -200798.41 0.003 12047.90 4337.25 38553.29 0.00 -4337.25 7228.74 -12047.90 0.00 12047.904 -120479.04 0.00 0.00 128510.98 0.00 12047.90 0.00 0.00 0.005 0.00 -1734.90 -4337.25 0.00 202533.30 -4337.25 0.00 0.00 0.006 0.00 4337.25 7228.74 12047.90 -4337.25 38553.29 0.00 0.00 0.007 -8031.94 0.00 -12047.90 0.00 0.00 0.00 8031.94 0.00 -12047.908 0.00 -200798.41 0.00 0.00 0.00 0.00 0.00 200798.41 0.009 12047.90 0.00 12047.90 0.00 0.00 0.00 -12047.90 0.00 24095.81
10 0.00 0.00 0.00 -8031.94 0.00 -12047.90 0.00 0.00 0.0011 0.00 0.00 0.00 0.00 -200798.41 0.00 0.00 0.00 0.0012 0.00 0.00 0.00 12047.90 0.00 12047.90 0.00 0.00 0.00
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
Paso 04
1 0 TN2 -7.5 TN3 -6.25 TN-M4 0 TN5 -14 TN6 -0.125 TN-M
Paso 05
1 2 3 4 5 61 0.000105443 1.16487E-06 -2.82175E-05 0.000101351 -1.16487E-06 -2.66435E-052 1.16487E-06 4.9708E-06 -7.7658E-07 1.16487E-06 9.318956E-09 -7.7658E-073 -2.82175E-05 -7.7658E-07 3.455163E-05 -2.66435E-05 7.765797E-07 2.022402E-064 0.000101351 1.16487E-06 -2.66435E-05 0.000105443 -1.16487E-06 -2.82175E-055 -1.16487E-06 9.318956E-09 7.765797E-07 -1.16487E-06 4.9708E-06 7.765797E-076 -2.66435E-05 -7.7658E-07 2.022402E-06 -2.82175E-05 7.765797E-07 3.455163E-05
7 8 9 10 11 121 -8031.94 0.00 12047.90 0.00 0.00 0.002 0.00 -200798.41 0.00 0.00 0.00 0.003 -12047.90 0.00 12047.90 0.00 0.00 0.004 0.00 0.00 0.00 -8031.94 0.00 12047.905 0.00 0.00 0.00 0.00 -200798.41 0.006 0.00 0.00 0.00 -12047.90 0.00 12047.90
Paso 06
1 0.000187262 M 7 1.16150442 TN2 -3.24608E-05 M 8 6.51807125 TN3 -0.000221248 RAD 9 -0.4094679 TN-M4 0.000177621 M 10 -1.16150442 TN5 -7.46118E-05 M 11 14.9819288 TN6 -2.20067E-05 RAD 12 1.87482413 TN-M
7 1.161504425 TN8 6.518071245 TN9 -0.409467904 TN-M
10 -1.161504425 TN11 14.98192875 TN12 1.874824131 TN-M
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑸𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑸𝒓}𝐂omplementario =
{𝑫𝒓} =
{𝑲𝒇𝒓} ={𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑹} =
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
d
a
b
10
0.000.000.00
-8031.940.00
-12047.900.000.000.00
8031.940.00
-12047.90
789
101112
0.0000.0000.0000.0000.0000.000
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.30.2 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.2 c(m) 0.3 c(m) 0.3
0.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.50.17 A(m2) 0.49 A(m2) 0.69
0.0019784313725 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1416666666667 As 0.25 As 0.375
YC= 0.6 M0.041
0.2411764705882 INERCIA TOTAL 0.0364 m4
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.30.2 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.2 c(m) 0.3 c(m) 0.3
0.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.50.17 A(m2) 0.49 A(m2) 0.69
0.0019784313725 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1416666666667 As 0.25 As 0.375
YC= 0.600 M 0.041
0.2411764705882 INERCIA TOTAL 0.0364 m4
c
d
a b
c
a
d
b
c
a
d
b
c
a
d
b
c
a
d
bcad b ca d
11 12
0.00 0.000.00 0.000.00 0.000.00 12047.90
-200798.41 0.000.00 12047.900.00 0.000.00 0.000.00 0.000.00 -12047.90
200798.41 0.000.00 24095.81
0.000 M0.000 M0.000 RAD0.000 M0.000 M0.000 RAD
COMPROBACION1
0-7.5
-6.250
-14-0.125
Datos del MEP CON SU SIGNO
7 0 TN8 0 TN9 0 TN-M
10 0 TN11 0 TN12 0 TN-M
{𝑸𝒇} − {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.001978431373 0.003125 0.003125
280 280 280
2509980.079602 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
TIPO DE FIGURA 4
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(m) 0.30 r(m) 0.3 r(m) 0.42 175 h(m) 0.15 A(m2) 0.282743339 r(m) 0.23 210 A(m2) 0.045 I(m4) 0.006361725 A(m2) 0.3769911184 280 I 0.000084375 As 0.254469 I(m4) 0.0188495565 315 As 0.0375 As 0.188495566 420
BARRA 1CO,CA 0 6A (m²) 0.170 1 2 3 4 5 6Inercia (m4) 0.00198 1 71116.10 0.00 0.00 -71116.10 0.00 0.00F´C (Kg/cm2) 280 2 0.00 275.88 827.64 0.00 -275.88 827.64E (Tn/m²) 2509980.08 3 0.00 827.64 3310.55 0.00 -827.64 1655.27L (m) 6.00 4 -71116.10 0.00 0.00 71116.10 0.00 0.00Angulo 0 5 0.00 -275.88 -827.64 0.00 275.88 -827.64Radianes 0 6 0.00 827.64 1655.27 0.00 -827.64 3310.55BARRA 2CO,CA 5 0A (m²) 0.150 7 8 9 1 2 3Inercia (m4) 0.003 7 752.99 0.00 -1882.49 -752.99 0.00 -1882.49F´C (Kg/cm2) 280.000 8 0.00 75299.40 0.00 0.00 -75299.40 0.00E (Tn/m²) 2509980.080 9 -1882.49 0.00 6274.95 1882.49 0.00 3137.48L (m) 5.00 1 -752.99 0.00 1882.49 752.99 0.00 1882.49Angulo 90 2 0.00 -75299.40 0.00 0.00 75299.40 0.00Radianes 1.57 3 -1882.49 0.00 3137.48 1882.49 0.00 6274.95BARRA 3CO,CA 6 2A (m²) 0.150 10 11 12 4 5 6Inercia (m4) 0.003 10 6287.79 -17747.20 -1116.18 -6287.79 17747.20 -1116.18F´C (Kg/cm2) 280 11 -17747.20 53613.67 -372.06 17747.20 -53613.67 -372.06E (Tn/m²) 2509980.08 12 -1116.18 -372.06 4960.78 1116.18 372.06 2480.39L (m) 6.32 4 -6287.79 17747.20 1116.18 6287.79 -17747.20 1116.18Angulo 108 5 17747.20 -53613.67 372.06 -17747.20 53613.67 372.06Radianes 1.89 6 -1116.18 -372.06 2480.39 1116.18 372.06 4960.78
1 2 3 4 5 6 7 8 91 71869.10 0.00 1882.49 -71116.10 0.00 0.00 -752.99 0.00 1882.492 0.00 75575.28 827.64 0.00 -275.88 827.64 0.00 -75299.40 0.003 1882.49 827.64 9585.50 0.00 -827.64 1655.27 -1882.49 0.00 3137.484 -71116.10 0.00 0.00 77403.90 -17747.20 1116.18 0.00 0.00 0.005 0.00 -275.88 -827.64 -17747.20 53889.55 -455.58 0.00 0.00 0.006 0.00 827.64 1655.27 1116.18 -455.58 8271.33 0.00 0.00 0.007 -752.99 0.00 -1882.49 0.00 0.00 0.00 752.99 0.00 -1882.498 0.00 -75299.40 0.00 0.00 0.00 0.00 0.00 75299.40 0.009 1882.49 0.00 3137.48 0.00 0.00 0.00 -1882.49 0.00 6274.95
10 0.00 0.00 0.00 -6287.79 17747.20 -1116.18 0.00 0.00 0.0011 0.00 0.00 0.00 17747.20 -53613.67 -372.06 0.00 0.00 0.0012 0.00 0.00 0.00 1116.18 372.06 2480.39 0.00 0.00 0.00
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
Paso 04
1 0.000 TN2 0.000 TN3 0.000 TN-M4 0.000 TN5 0.000 TN6 0.000 TN-M
Paso 05
1 2 3 4 5 61 0.001159254 4.416852E-06 -0.000178378 0.001152745 0.00037607 -9.95885E-052 4.416852E-06 1.327166E-05 -1.62291E-06 4.420659E-06 1.486021E-06 -1.51789E-063 -0.000178378 -1.62291E-06 0.000135726 -0.000176674 -5.61598E-05 -6.2514E-064 0.001152745 4.420659E-06 -0.000176674 0.001160274 0.000378565 -0.0001008085 0.00037607 1.486021E-06 -5.61598E-05 0.000378565 0.000142101 -3.21686E-056 -9.95885E-05 -1.51789E-06 -6.2514E-06 -0.000100808 -3.21686E-05 0.000134134
7 8 9 10 11 121 -752.99 0.00 1882.49 0.00 0.00 0.002 0.00 -75299.40 0.00 0.00 0.00 0.003 -1882.49 0.00 3137.48 0.00 0.00 0.004 0.00 0.00 0.00 -6287.79 17747.20 1116.185 0.00 0.00 0.00 17747.20 -53613.67 372.066 0.00 0.00 0.00 -1116.18 -372.06 2480.39
Paso 06
1 0 M 7 0 TN2 0 M 8 0 TN3 0 RAD 9 0 TN-M4 0 M 10 0 TN5 0 M 11 0 TN6 0 RAD 12 0 TN-M
7 0 TN8 0 TN9 0 TN-M
10 0 TN11 0 TN12 0 TN-M
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑸𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑸𝒓}𝐂omplementario =
{𝑫𝒓} =
{𝑲𝒇𝒓} ={𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑹} =
Seccion a(m)b(m)c(m)d(m)
A(m2)I(m4)
As
MxYg
10
0.000.000.00
-6287.7917747.20-1116.18
0.000.000.00
6287.79-17747.20
-1116.18
d
a
b
dab
dab
789
101112
0.0000.0000.0000.0000.0000.000
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
4 Seccion 5 ITEM A Y AY I d Ad2 Seccion 60.6 a(m) 0.3 a(m) 0.30.2 b(m) 0.4 1 0.3 0.5 0.15 0.025 0.1 0.003 b(m) 0.40.2 c(m) 0.3 c(m) 0.3
0.25 d(m) 1 2 0.12 0.85 0.102 0.0009 -0.25 0.0075 d(m) 1.50.17 A(m2) 0.49 A(m2) 0.69
0.0019784313725 I(m4) 0.0364 0.42 0.252 0.0259 0.0105 I(m4) 0.119380260.1416666666667 As 0.25 As 0.375
YC= 0.6 M0.041
0.2411764705882 INERCIA TOTAL 0.0364 m4
11 12
0.00 0.000.00 0.000.00 0.00
17747.20 1116.18-53613.67 372.06
-372.06 2480.390.00 0.000.00 0.000.00 0.00
-17747.20 -1116.1853613.67 -372.06
-372.06 4960.78
c
d
a b
c
a
d
b
c
a
d
cda bcad b ca d
cda bcad b ca d
0.000 M0.000 M0.000 RAD0.000 M0.000 M0.000 RAD
COMPROBACION1
0.0000.0000.0000.0000.0000.000
Datos del MEP CON SIGNO CAMBIADO
7 0 TN8 0 TN9 0 TN-M
10 0 TN11 0 TN12 0 TN-M
{𝑸𝒇} − {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒓}𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒐 =
L 1ITEM A Y AY I d Ad2
1 0.45 0.75 0.3375 0.084375 -0.126 0.007
2 0.12 0.15 0.018 0.0009 0.474 0.027
0.57 0.3555 0.085275 0.034
YC= 0.624 M
INERCIA TOTAL 0.1193802632 m4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.0019784314 0.003125 0.003125
280 280 280
2509980.0796 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
TIPO DE FIGURA 4
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(cm) 0.30 r(cm) 0.3 R(cm) 0.42 175 h(cm) 0.15 A(cm2) 0.282743339 r(cm) 0.23 210 A(cm2) 0.045 I(cm4) 0.006361725 A(cm2) 0.3769911184 280 I 0.000084375 As 0.254469 I(cm4) 0.0188495565 315 As 0.0375 As 0.188495566 420
BARRA 1CO,CA 0 6A (m²) 0.170 1 2 3 4 5 6Inercia (m4) 0.00198 1 71116.10 0.00 -827.64 -71116.10 0.00 0.00F´C (Kg/cm2) 280 2 0.00 275.88 827.64 0.00 -275.88 827.64E (Tn/m²) 2509980.08 3 -827.64 827.64 3310.55 0.00 -827.64 1655.27L (m) 6.00 4 -71116.10 0.00 0.00 71116.10 -275.88 0.00Angulo 0 5 0.00 -275.88 -827.64 -275.88 275.88 -827.64Radianes 0 6 0.00 827.64 1655.27 0.00 -827.64 3310.55COS 1SEN 0
TIPO DE FIGURA 1
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(cm) 0.30 r(cm) 1 R(cm) 42 175 h(cm) 0.50 A(cm2) 3.141592654 r(cm) 23 210 A(cm2) 0.15 I(cm4) 0.785398163 A(cm2) 37.699111844 280 I 0.003125 As 2.82743339 I(cm4) 188.49555925 315 As 0.125 As 18.84955596 420
BARRA 2CO,CA 5 0A (m²) 0.150 1 2 3 7 8 9Inercia (m4) 0.003 1 0.000 0.000 -1882.485 -752.994 0.000 -1882.485F´C (Kg/cm2) 280.000 2 0.000 75299.402 0.000 0.000 -75299.402 0.000E (Tn/m²) 2509980.080 3 -1882.485 0.000 6274.950 1882.485 0.000 3137.475L (m) 5.00 7 -752.994 0.000 1882.485 752.994 0.000 1882.485Angulo 90 8 0.000 -75299.402 0.000 0.000 75299.402 0.000Radianes 1.57 9 -1882.485 0.000 3137.475 1882.485 0.000 6274.950COS 0SEN 1
TIPO DE FIGURA 1
F`c 4 Rectangulo 1 Circulo 2 S. Tubular 31 140 b(cm) 0.30 r(cm) 1 R(cm) 4
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
2 175 h(cm) 0.50 A(cm2) 3.141592654 r(cm) 23 210 A(cm2) 0.15 I(cm4) 0.785398163 A(cm2) 37.699111844 280 I 0.003125 As 2.82743339 I(cm4) 188.49555925 315 As 0.125 As 18.84955596 420
BARRA 3CO,CA 6 2A (m²) 0.150 4 5 6 10 11 12Inercia (m4) 0.003 4 5952.940 -17858.821 -1176.553 -6287.793 17747.204 -1116.176F´C (Kg/cm2) 280 5 -17858.821 53613.670 -372.059 17747.204 -53613.670 -372.059E (Tn/m²) 2509980.08 6 -1176.553 -372.059 4960.784 1116.176 372.059 2480.392L (m) 6.32 10 -6287.793 17747.204 1116.176 6287.793 -17741.166 1116.176Angulo 108 11 17747.204 -53613.670 372.059 -17741.166 53613.670 372.059Radianes 1.89 12 -1116.176 -372.059 2480.392 1116.176 372.059 4960.784COS 0SEN 1
1 2 3 4 5 6 7 8 9 10 11 121 71116.102 0.000 -2710.122 -71116.102 0.000 0.000 -752.994 0.000 -1882.485 0.000 0.000 0.0002 0.000 75575.281 827.637 0.000 -275.879 827.637 0.000 -75299.402 0.000 0.000 0.000 0.0003 -2710.122 827.637 9585.499 0.000 -827.637 1655.274 1882.485 0.000 3137.475 0.000 0.000 0.0004 -71116.102 0.000 0.000 77069.043 -18134.700 -1176.553 0.000 0.000 0.000 -6287.793 17747.204 -1116.1765 0.000 -275.879 -827.637 -18134.700 53889.549 -1199.696 0.000 0.000 0.000 17747.204 -53613.670 -372.0596 0.000 827.637 1655.274 -1176.553 -1199.696 8271.333 0.000 0.000 0.000 1116.176 372.059 2480.3927 -752.994 0.000 0.000 0.000 0.000 0.000 752.994 0.000 1882.485 0.000 0.000 0.0008 0.000 -75299.402 0.000 0.000 0.000 0.000 0.000 75299.402 0.000 0.000 0.000 0.0009 -1882.485 0.000 0.000 0.000 0.000 0.000 1882.485 0.000 6274.950 0.000 0.000 0.00010 0.000 0.000 -6287.793 -6287.793 17747.204 1116.176 0.000 0.000 0.000 6287.793 -17741.166 1116.17611 0.000 0.000 17747.204 17747.204 -53613.670 372.059 0.000 0.000 0.000 -17741.166 53613.670 372.05912 0.000 0.000 0.000 -1116.176 -372.059 2480.392 0.000 0.000 0.000 1116.176 372.059 4960.784
Paso 04
1 2 3 4 5 6 9 7 8 10 11123456978101112
1 2.000 Tn 7 2.000 Tn2 -4.020 Tn 8 0.000 Tn3 -1.560 Tn 10 0.000 Tn4 0.000 Tn 11 0.000 Tn5 -10.980 Tn 12 0.000 Tn6 -0.020 Tn9 -2.000 Tn
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇} {𝒌𝒇𝒓}{𝒌𝒓𝒇} {𝒌𝒓𝒓}
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
{𝑸𝒇} =
{𝑫𝒓} =
1 2 3 4 5 6 91 71116.102 0.000 -2710.122 -71116.102 0.000 0.000 -1882.4852 0.000 75575.281 827.637 0.000 -275.879 827.637 0.0003 -2710.122 827.637 9585.499 0.000 -827.637 1655.274 3137.4754 -71116.102 0.000 0.000 77069.043 -18134.700 -1176.553 0.0005 0.000 -275.879 -827.637 -18134.700 53889.549 -1199.696 0.0006 0.000 827.637 1655.274 -1176.553 -1199.696 8271.333 0.0009 -1882.485 0.000 0.000 0.000 0.000 0.000 6274.950
Paso 05
1 2 3 4 5 6 91 -0.000656942 2.229142E-06 -0.000190162 -0.000661057 -0.000227351 -8.91749E-05 -0.0001022 1.643969E-06 1.324914E-05 -4.70486E-07 1.642617E-06 5.930674E-07 -9.11893E-07 7.28434E-073 -0.000123256 -5.28124E-07 7.252313E-05 -0.000124081 -4.1493E-05 -3.81288E-05 -7.3238E-054 -0.00066109 2.231566E-06 -0.000191415 -0.000651079 -0.000223964 -8.70138E-05 -0.000102625 -0.000226632 7.931409E-07 -6.44392E-05 -0.000223229 -5.81587E-05 -2.73723E-05 -3.577E-056 -0.000102406 -7.87565E-07 -5.10406E-05 -0.000100323 -3.20489E-05 0.000112274 -5.2015E-069 -0.000197083 6.687426E-07 -5.70485E-05 -0.000198317 -6.82052E-05 -2.67525E-05 0.00012876
7 8 10 11 121 -752.994 0.000 0.000 0.000 0.0002 0.000 -75299.402 0.000 0.000 0.0003 1882.485 0.000 0.000 0.000 0.0004 0.000 0.000 -6287.793 17747.204 -1116.1765 0.000 0.000 17747.204 -53613.670 -372.0596 0.000 0.000 1116.176 372.059 2480.3929 1882.485 0.000 0.000 0.000 0.000
Paso 06
1 0.112315436 m2 0.001447443 m3 -0.182683952 m4 0.113071512 m5 0.036334364 m6 0.057765656 m9 -0.566624097 m
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑲𝒇𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑲𝒇𝒓} = {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒇} - {𝑲𝒇𝒓}∗{𝑫𝒓}=
7 #VALUE! Tn8 #VALUE! Tn10 #VALUE! Tn11 #VALUE! Tn12 #VALUE! Tn
7 #VALUE! Tn8 #VALUE! Tn10 #VALUE! Tn11 #VALUE! Tn12 #VALUE! Tn
{𝑸𝒓}complemtario =
{𝑹} =
Seccion a(cm)b(cm)c(cm)d(cm)
A(cm2)I(cm4)
As
MxYg
Seccion a(cm)b(cm)c(cm)d(cm)
A(cm2)I(cm4)
As
MxYg
Seccion a(cm)
d
a
b
d
a
b
d
a
b
b(cm)c(cm)d(cm)
A(cm2)I(cm4)
As
MxYg
12
-1505.9880.000
3764.9700.0000.0000.000
3764.970
1507.988-4.020
-3766.5300.000
-10.980-0.020
-3766.970
{𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒇} - {𝑲𝒇𝒓}∗{𝑫𝒓}=
4 Seccion 5 ITEM A Y AY I d Ad20.6 a(cm) 300.2 b(cm) 40 1 3000 50 150000 2500000 10 3000000.2 c(cm) 30
0.25 d(cm) 100 2 1200 85 102000 90000 -25 750000
0.17 A(cm2) 49000.0019784313725 I(cm4) 3640000 4200 252000 2590000 10500000.1416666666667 As 2500
YC= 60 cm0.041
0.2411764705882 INERCIA TOTAL 3640000 cm4
4 Seccion 5 ITEM A Y AY I d Ad20.6 a(cm) 300.2 b(cm) 40 1 3000 50 150000 2500000 10 3000000.2 c(cm) 30
0.25 d(cm) 100 2 1200 85 102000 90000 -25 750000
0.17 A(cm2) 49000.0019784313725 I(cm4) 3640000 4200 252000 2590000 10500000.1416666666667 As 2500
YC= 60 cm0.041
0.2411764705882 INERCIA TOTAL 3640000 cm4
4 Seccion 5 ITEM A Y AY I d Ad20.6 a(cm) 30
c
d
a b
c
a
d
c
d
a b
c
a
d
c
d
a b
c
a
d
0.2 b(cm) 40 1 3000 50 150000 2500000 10 3000000.2 c(cm) 30
0.25 d(cm) 100 2 1200 85 102000 90000 -25 750000
0.17 A(cm2) 49000.0019784313725 I(cm4) 3640000 4200 252000 2590000 10500000.1416666666667 As 2500
YC= 60 cm0.041
0.2411764705882 INERCIA TOTAL 3640000 cm4
L 100Seccion 6 ITEM A Y AY I d Ad2
a(cm) 30b(cm) 40 1 4500 75 337500 8437500 -12.632 718005.540c(cm) 30d(cm) 150 2 1200 15 18000 90000 47.368 ###
A(cm2) 6900I(cm4) 11938026.3 5700 355500 8527500 ###
As 3750YC= 62.3684211 cm
INERCIA TOTAL 11938026.3 cm4
L 100Seccion 6 ITEM A Y AY I d Ad2
a(cm) 30b(cm) 40 1 4500 75 337500 8437500 -12.632 718005.540c(cm) 30d(cm) 150 2 1200 15 18000 90000 47.368 ###
A(cm2) 6900I(cm4) 11938026.3 5700 355500 8527500 ###
As 3750YC= 62.3684211 cm
INERCIA TOTAL 11938026.3 cm4
L 100Seccion 6 ITEM A Y AY I d Ad2
a(cm) 30
b
c
a
d
b
c
a
d
b
c
a
d
b(cm) 40 1 4500 75 337500 8437500 -12.632 718005.540c(cm) 30d(cm) 150 2 1200 15 18000 90000 47.368 ###
A(cm2) 6900I(cm4) 11938026.3 5700 355500 8527500 ###
As 3750YC= 62.3684211 cm
INERCIA TOTAL 11938026.3 cm4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.240 0.071 0.126
0.368 0.000398 0.001257
210 210 210
2173706.5119 2173706.5119 2173706.5119
L (m) 6.00 4.00 7.00
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
4 Tn
3 Tn/m
2m
2m
3m
6m 2m2m
4m
Ø= 0.30m
Ø= 0.40m
1
3
2
2m
2m
3m
6m 2m2m
4m
4
5
1
2
7
8
10
11
3 6
9
12
0.20m
0.40m
0.30m
0.60m
Ø= 0.30m
Ø= 0.40m
0.20m
0.40m
0.30m
0.60m
Angulo (grados) 0 270 270
Grados (Rad) 0.00000000 4.71238898 4.71238898
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
BARRA 1A (m²) 0.240 1 2 3 4 5 6Inercia (m4) 0.368 1 86948.260 0.000 -133320.666 -86948.260 0.000 0.000F´C (Kg/cm2) 210.000 2 0.000 44440.222 133320.666 0.000 -44440.222 133320.666E (Tn/m²) 2173706.512 Tn/m² 3 -133320.666 133320.666 533282.664 0.000 -133320.666 266641.332L (m) 6.00 m 4 -86948.260 0.000 0.000 86948.260 -44440.222 0.000Angulo 0 5 0.000 -44440.222 -133320.666 -44440.222 44440.222 -133320.666Radianes 0 6 0.000 133320.666 266641.332 0.000 -133320.666 533282.664COS 1SEN 0
BARRA 2A (m²) 0.071 1 2 3 7 8 9Inercia (m4) 0.000 1 0.000 0.000 -324.106 -162.053 0.000 324.106F´C (Kg/cm2) 210.000 2 0.000 38412.565 0.000 0.000 -38412.565 0.000E (Tn/m²) 2173706.512 Tn/m² 3 -324.106 0.000 864.283 -324.106 0.000 432.141L (m) 4.00 m 7 -162.053 0.000 -324.106 162.053 0.000 -324.106Angulo 270 8 0.000 -38412.565 0.000 0.000 38412.565 0.000Radianes 4.71 9 324.106 0.000 432.141 -324.106 0.000 864.283COS 0SEN -1
BARRA 3A (m²) 0.126 4 5 6 10 11 12Inercia (m4) 0.001 4 0.000 0.000 -334.477 -95.565 0.000 334.477
F´C (Kg/cm2) 210.000 5 0.000 39022.288 0.000 0.000 -39022.288 0.000E (Tn/m²) 2173706.512 Tn/m² 6 -334.477 0.000 1560.892 -334.477 0.000 780.446L (m) 7.00 m 10 -95.565 0.000 -334.477 95.565 0.000 -334.477Angulo 270 11 0.000 -39022.288 0.000 0.000 39022.288 0.000Radianes 4.71 12 334.477 0.000 780.446 -334.477 0.000 1560.892COS 0SEN -1
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
cm²cm4
Kg/cm²
cm²cm4
Kg/cm²
cm²cm4
Kg/cm²
1 2 3 4 5 6 9 7 8 10 11 121 86948.260 0.000 -133644.772 -86948.260 0.000 0.000 324.106 -162.053 0.000 0.000 0.000 0.0002 0.000 82852.787 133320.666 0.000 -44440.222 133320.666 0.000 0.000 -38412.565 0.000 0.000 0.0003 -133644.772 133320.666 534146.947 0.000 -133320.666 266641.332 432.141 -324.106 0.000 0.000 0.000 0.0004 -86948.260 0.000 0.000 86948.260 -44440.222 -334.477 0.000 0.000 0.000 -95.565 0.000 334.4775 0.000 -44440.222 -133320.666 -44440.222 83462.510 -133320.666 0.000 0.000 0.000 0.000 -39022.288 0.0006 0.000 133320.666 266641.332 -334.477 -133320.666 534843.556 0.000 0.000 0.000 -334.477 0.000 780.4469 324.106 0.000 0.000 0.000 0.000 0.000 864.283 -324.106 0.000 0.000 0.000 0.0007 -162.053 0.000 0.000 0.000 0.000 0.000 -324.106 162.053 0.000 0.000 0.000 0.0008 0.000 -38412.565 0.000 0.000 0.000 0.000 0.000 0.000 38412.565 0.000 0.000 0.00010 0.000 0.000 -95.565 -95.565 0.000 -334.477 0.000 0.000 0.000 95.565 0.000 -334.47711 0.000 0.000 0.000 0.000 -39022.288 0.000 0.000 0.000 0.000 0.000 39022.288 0.00012 0.000 0.000 334.477 334.477 0.000 780.446 -334.477 0.000 0.000 -334.477 0.000 1560.892
Paso 04
1 2 3 4 5 6 9 7 8 10 11 12123456978101112
1 2.000 Tn 7 2.000 Tn2 -4.020 Tn 8 0.000 Tn3 -1.560 Tn 10 0.000 Tn4 0.000 Tn 11 0.000 Tn5 -10.980 Tn 12 0.000 Tn6 -0.020 Tn9 -2.000 Tn
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇} {𝒌𝒇𝒓}{𝒌𝒓𝒇} {𝒌𝒓𝒓}
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
{𝑸𝒇} =
{𝑫𝒓} =
1 2 3 4 5 6 91 86948.260 0.000 -133644.772 -86948.260 0.000 0.000 324.1062 0.000 82852.787 133320.666 0.000 -44440.222 133320.666 0.0003 -133644.772 133320.666 534146.947 0.000 -133320.666 266641.332 432.1414 -86948.260 0.000 0.000 86948.260 -44440.222 -334.477 0.0005 0.000 -44440.222 -133320.666 -44440.222 83462.510 -133320.666 0.0006 0.000 133320.666 266641.332 -334.477 -133320.666 534843.556 0.0009 324.106 0.000 0.000 0.000 0.000 0.000 864.283
Paso 05
1 2 3 4 5 6 91 -3.02661E-06 1.743479E-06 -3.35659E-06 -9.35793E-06 -1.23734E-05 -1.85139E-06 2.81327E-062 1.745598E-06 2.498219E-05 -2.38416E-06 5.412197E-06 7.198174E-06 -3.24104E-06 5.37479E-073 -3.35899E-06 -2.38512E-06 1.182493E-06 -5.17407E-06 -3.54459E-06 -8.81779E-07 6.68377E-074 -9.36049E-06 5.407118E-06 -5.16946E-06 -1.39199E-06 -6.90789E-06 -4.93452E-07 6.09491E-065 -1.23784E-05 7.19239E-06 -3.54029E-06 -6.91289E-06 1.067368E-05 2.62844E-06 6.41206E-066 -1.85196E-06 -3.24201E-06 -8.80944E-07 -4.93667E-07 2.629144E-06 3.772088E-06 1.13496E-069 1.134977E-06 -6.53805E-07 1.25872E-06 3.509224E-06 4.640041E-06 6.942702E-07 0.00115597
7 8 10 11 121 -162.053 0.000 0.000 0.000 0.000 -324.1062 0.000 -38412.565 0.000 0.000 0.000 0.0003 -324.106 0.000 0.000 0.000 0.000 -648.2124 0.000 0.000 -95.565 0.000 334.477 0.0005 0.000 0.000 0.000 -39022.288 0.000 0.0006 0.000 0.000 -334.477 0.000 780.446 0.0009 -324.106 0.000 0.000 0.000 0.000 -648.212
326.106-4.020
646.6520.000
-10.980-0.020
646.212
Paso 06
1 -0.00121068 m2 -0.00080454 m3 0.000149712 m4 -0.00240263 m5 -0.00232863 m6 -0.00045609 m9 0.748139947 m
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑲𝒇𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑲𝒇𝒓} = {𝑲𝒇𝒓}∗{𝑫𝒓}=
{𝑸𝒇} - {𝑲𝒇𝒓}∗{𝑫𝒓}=
7 81.83 Tn8 30.90 Tn10 0.37 Tn11 90.87 Tn12 -251.34 Tn
7 83.826 Tn8 30.905 Tn10 0.368 Tn11 90.869 Tn12 -251.345 Tn
{𝑸𝒓}complemtario =
{𝑹} =
Paso 01
Barra Und. 1 2 3 4 5E 2.00E+06 2.00E+06 2.00E+06 2.00E+06 2.00E+06
A M² 0.05 0.05 0.05 0.05 0.05
L M 10 10 10.00 10.00 14.14
Angulo GRADOS 0 0 90.00 90.00 45.00
Grados RADIANES 0.00000000 0.00000000 1.57079633 1.57079633 0.78539816
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
KN/M²
10m
10m
3
4
1
2
6
5
7
8
1
3 4
65
2
Paso 02
Paso 03
Coordenadas Globales ( Barra de Armadura )
BARRA 1E 2.00E+06 KN/M² 1A 0.05 M² 1 1L 10.00 M k1 = 10000.00 2 0Angulo 0 6 -1Radianes 0 5 0COS 1SEN 0
BARRA 2E 2.00E+06 KN/M² 3A 0.05 M² 3 1L 10.00 M k1 = 10000.00 4 0Angulo 0 7 -1Radianes 0 8 0COS 1SEN 0
BARRA 3E 2.00E+06 KN/M² 1A 0.05 M² 1 3.752472E-33L 10.00 M k1 = 10000.00 2 6.125742E-17Angulo 90.00 3 -3.75247E-33Radianes 1.570796327 4 -6.12574E-17COS 0.000000000SEN 1.000000000
BARRA 4E 2.00E+06 KN/M² 6A 0.05 M² 6 3.752472E-33L 10.00 M k1 = 10000.00 5 6.125742E-17Angulo 90.00 7 -3.75247E-33
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de una armadura
Podemos encontrar esta matriz para todas las barras de la armadura de la sgte manera:
Radianes 1.570796327 8 -6.12574E-17COS 0.00000000SEN 1.000000000
BARRA 5E 2.00E+06 KN/M² 1A 0.05 M² 1 0.5L 14.14 M k1 = 7071.07 2 0.5Angulo 45.00 7 -0.5Radianes 0.785398163 8 -0.5COS 0.70710678SEN 0.707106781
BARRA 6 6E 2.00E+06 KN/M² 6A 0.05 M² 6 0.5L 14.14 M k1 = 7071.07 5 -0.5Angulo 135.00 3 -0.5Radianes 2.356194490 4 0.5COS -0.70710678SEN 0.707106781
1 2 31 13535.53 3535.53 0.002 3535.53 13535.53 0.003 0.00 0.00 13535.534 0.00 -10000.00 -3535.535 0.00 -10000.00 3535.536 0.00 0.00 -3535.537 -3535.53 -3535.53 -10000.008 -3535.53 -3535.53 0.00
Paso 04
1 2 312345678
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇}{𝒌𝒓𝒇}
[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
1 0.00 KN2 0.00 KN3 2.00 KN4 -4.00 KN5 0.00 KN
Paso 05
1 2 31 0.000100 -0.000100 -0.0000172 -0.000100 0.000383 0.0000663 0.000000 0.000000 0.0000834 -0.000100 0.000383 0.0000835 -0.000100 0.000383 0.000049
6 0.00 M
Paso 06
1 0.000297056 M2 -0.00113726 M3 9.70563E-05 M4 -0.00143431 M5 -0.0012402 M
Para invertir la matriz (kff) seleccionamos las celdas (E170:H173) la funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
{𝑸𝒇} =
{𝑲𝒇𝒇}¯¹ =
{𝑫𝒇} =
{𝑫𝒓} =
6 -1.03 KN7 2.00 KN8 2.97 KN
{𝑸𝒓} =
RELACION S - R - CS R C
180 PI 200
62.00E+06
0.05
14.14
135.00
2.35619449
libres y restringidos de la estructura
1 23 45 67 8
2 6 5 1 20 -1 0 1 10000 00 0 0 = 2 0 00 1 0 6 -10000 00 0 0 5 0 0
4 7 8 3 40 -1 0 3 10000 00 0 0 = 4 0 00 1 0 7 -10000 00 0 0 8 0 0
2 3 4 1 26.125742E-17 -3.75247E-33 -6.12574E-17 1 0.00 0.00
1 -6.12574E-17 -1 = 2 0.00 10000.00-6.12574E-17 3.752472E-33 6.125742E-17 3 0.00 0.00
-1 6.125742E-17 1 4 0.00 -10000.00
5 7 8 6 56.125742E-17 -3.75247E-33 -6.12574E-17 6 0.00 0.00
1 -6.12574E-17 -1 = 5 0.00 10000.00-6.12574E-17 3.752472E-33 6.125742E-17 7 0.00 0.00
para la barra de una armadura
matriz para todas las barras de la armadura de la sgte manera:
-1 6.125742E-17 1 8 0.00 -10000.00
2 7 8 1 20.5 -0.5 -0.5 1 3535.53 3535.530.5 -0.5 -0.5 = 2 3535.53 3535.53-0.5 0.5 0.5 7 -3535.53 -3535.53-0.5 0.5 0.5 8 -3535.53 -3535.53
5 3 4 6 5-0.5 -0.5 0.5 6 3535.53 -3535.530.5 0.5 -0.5 = 5 -3535.53 3535.530.5 0.5 -0.5 3 -3535.53 3535.53-0.5 -0.5 0.5 4 3535.53 -3535.53
4 5 6 7 8
0.00 0.00 -10000.00 -3535.53 -3535.53-10000.00 0.00 0.00 -3535.53 -3535.53
-3535.53 3535.53 -3535.53 -10000.00 0.0013535.53 -3535.53 3535.53 0.00 0.00-3535.53 13535.53 -3535.53 0.00 -10000.003535.53 -3535.53 13535.53 0.00 0.00
0.00 0.00 -10000.00 13535.53 3535.530.00 0.00 0.00 3535.53 13535.53
4 5 6 7 8
matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇} {𝒌𝒇𝒓}{𝒌𝒓𝒇} {𝒌𝒓𝒓}
10m
3
4
1
2
1
3
5
2
4 5
-0.000083 -0.0000170.000317 0.0000660.000017 -0.0000170.000400 0.0000830.000334 0.000149
funcion es =MINVERSA(E161:H164) y sin soltar presionamos "CTROL" + "SHIFT", sin soltar damos enter
Para Multiplicar una matriz con el vector de cargas nodales seleccionamos las celdas (E187:E190) la funcion es =MMULT(E170:H173,E149:E152) y sin CTROL" + "SHIFT", sin soltar damos enter
3
4
2
10m
1
2
1
3
65
6 5-10000 0
0 010000 0
0 0
7 8-10000 0
0 010000 0
0 0
3 40.00 0.000.00 -10000.000.00 0.000.00 10000.00
7 80.00 0.000.00 -10000.000.00 0.00
0.00 10000.00
7 8-3535.53 -3535.53-3535.53 -3535.533535.53 3535.533535.53 3535.53
3 4-3535.53 3535.533535.53 -3535.533535.53 -3535.53-3535.53 3535.53
10m
10m
6
5
7
8
1
4
6
2
7
8
10m
6
5
4
CURSO: ANALISIS ESTRUCTURAL II
EJERCICIO PRACTICO DE MUROS DE CORTE
Paso 01
10000
425
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
10
3
21
5
3
5
1
2 4
7
500
150 150
75 75 75 75
Paso 03
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 150.003 210 A(cm2) 45004 280 I 84375005 315 As 37506 420
PLACA 13750
G 86948.2605 6 7
ø 0.373702422 6 208706.745
210 7 2301571.601
217370.6512 8 -44350183.2
4500 5 -208706.745
8437500 1 -2301571.601H de placa (cm) 425 2 -44350183.2
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
Inercia (cm4)
9
10
116
7
8
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 150.003 210 A(cm2) 45004 280 I 84375005 315 As 37506 420
PLACA 23750
G 86948.2605 9 10
ø 0.373702422 9 208706.745
210 10 2301571.601
217370.6512 11 -44350183.2
4500 5 -208706.745
8437500 3 -2301571.601H de placa (cm) 425 4 -44350183.2
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 50.003 210 A(cm2) 15004 280 I 3125005 315 As 12506 420
VIGA 31250
G 86948.2605 1 2
ø 0.030000000 1 6521.12 2119363.85210 2 2119363.85 824649907.96
217370.6512 3 -6521.12 -2119363.85
1500 4 2119363.85 552936593.97
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
Inercia (cm4)
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
312500Long de viga (cm) 500
a 75b 75
1 2 3 41 2308093 2119364 -6521 21193642 2119364 14564510598 -2119364 5529365943 -6521 -2119364 2308093 -21193644 2119364 552936594 -2119364 145645105985 0 44350183 0 443501836 0 -44350183 0 07 -2301572 0 0 08 0 5108967187 0 09 0 0 0 -4435018310 0 0 -2301572 011 0 0 0 5108967187
Paso 04
1 2 3 41234569781011
Inercia (cm4)
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇}{𝒌𝒓𝒇}[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
[𝑴𝒂𝒕𝒓𝒊𝒛 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒆 𝑲𝒇𝒇−𝟏]
1 2 3 41 4.33557E-07 -1.6105559E-10 9.2916688E-10 -1.6105559E-102 -1.61056E-10 1.23631341E-10 1.6105559E-10 5.22617719E-113 9.29167E-10 1.61055593E-10 4.3355656E-07 1.61055593E-104 -1.61056E-10 5.22617719E-11 1.6105559E-10 1.23631341E-105 3.42243E-08 -1.8688643E-08 -3.4224314E-08 -1.8688643E-08
1 02 03 04 05 10000
6 7 8 96 0.00 -44350183.24 0.00 0.007 -2301571.60 0.00 0.00 0.008 0.00 5108967187.06 0.00 0.009 0.00 0.00 0.00 -44350183.2410 0.00 0.00 -2301571.60 0.0011 0.00 0.00 0.00 5108967187.06
6 -5000 kg7 -787.697082 kg8 1868998.449 kg9 -5000 kg10 787.6970815 kg11 1868998.449 kg
{𝑸𝒇} =
[𝑴𝒂𝒕𝒓𝒊𝒛 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒆 𝑲𝒇𝒇−𝟏]
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝑸𝒇] [𝑫𝒆𝒔𝒑𝒍𝒂𝒛𝒂𝒎𝒊𝒆𝒏𝒕𝒐𝒔 𝑳𝒊𝒃𝒓𝒆𝒔]
[𝑴𝒂𝒕𝒓𝒊𝒛 𝑲𝒓𝒇]
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝒍𝒂𝒔 𝒓𝒆𝒂𝒄𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆𝒍 𝒆𝒔𝒕𝒂𝒅𝒐 𝑪𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝒚 𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒂]
{𝑲𝒓𝒇} =
{𝑹 𝒄𝒐𝒎} =
{𝑹 𝒑𝒓𝒊𝒎} =
6 -5000 kg7 -787.697082 kg8 1868998.449 kg.cm9 -5000 kg10 787.6970815 kg11 1868998.449 kg.cm
{𝑹 𝒇𝒊𝒏𝒂𝒍} =
EJERCICIO PRACTICO DE MUROS DE CORTERELACION S - R - C
S R180 PI
50
150
barras libres y restringidos de la estructura
TIPO DE FIGURA 1
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
8 5 1 2
-44350183.2 -208706.745 -44350183.2
-2301571.601
13739860690 44350183.24 5108967187
44350183.24 208706.7447 44350183.24
2301571.6015108967187 44350183.24 13739860690
TIPO DE FIGURA 1
matriz para todas las barras del portico de la sgte manera:
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
11 5 3 4
-44350183.2 -208706.745 -44350183.2
-2301571.601
13739860690 44350183.24 5108967187
44350183.24 208706.7447 44350183.24
2301571.6015108967187 44350183.24 13739860690
TIPO DE FIGURA 1
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
3 4
-6521.12 2119363.85
-2119363.85 552936593.97
6521.12 -2119363.85
-2119363.85 824649907.96
5 6 7 8 9 10 11
0 0 -2301572 0 0 0 044350183 -44350183 0 5108967187 0 0 0
0 0 0 0 0 -2301572 044350183 0 0 0 -44350183 0 5108967187
417413 -208707 0 44350183 -208707 0 44350183-208707 208707 0 -44350183 0 0 0
0 0 2301572 0 0 0 044350183 -44350183 0 13739860690 0 0 0
-208707 0 0 0 208707 0 -443501830 0 0 0 0 2301572 0
44350183 0 0 0 -44350183 0 13739860690
5 6 9 7 8 10 11
matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇} {𝒌𝒇𝒓}{𝒌𝒓𝒇} {𝒌𝒓𝒓}
5
3.4224314E-08-1.868864E-08-3.422431E-08-1.868864E-086.3670427E-06
FUERZADESPLZAMIENTO
1 0.0003422431 cm2 -0.000186886 cm K LATERAL DEL PORTICO3 -0.000342243 cm4 -0.000186886 cm5 0.0636704269 cm
10
-208706.740.00
44350183.24-208706.74
0.0044350183.24
6 0 kg7 0 kg8 0 kg9 0 kg10 0 kg11 0 kg
[𝑫𝒆𝒔𝒑𝒍𝒂𝒛𝒂𝒎𝒊𝒆𝒏𝒕𝒐𝒔 𝑳𝒊𝒃𝒓𝒆𝒔]
{𝑫𝒇} =
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝒍𝒂𝒔 𝒓𝒆𝒂𝒄𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆𝒍 𝒆𝒔𝒕𝒂𝒅𝒐 𝑪𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝒚 𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒂]
{𝑹 𝒑𝒓𝒊𝒎} =
6 -5 Tn7 -0.78769708152 Tn8 18.6899844851 Tn.m9 -5 Tn10 0.78769708152 Tn11 18.6899844851 Tn.m
RELACION S - R - CC
200
Seccion 4a(cm) 0.6b(cm) 0.2c(cm) 0.4d(cm) 0.3
A(cm2) 0.24I(cm4) 0.0074
As 0.2
Mx 0.084Yg 0.35
c
d
a
b
cd
a
b
Seccion 4a(cm) 0.6b(cm) 0.2c(cm) 0.4d(cm) 0.3
A(cm2) 0.24I(cm4) 0.0074
As 0.2
Mx 0.084Yg 0.35
Seccion 4a(cm) 0.6b(cm) 0.2c(cm) 0.4d(cm) 0.3
A(cm2) 0.24I(cm4) 0.0074
As 0.2
Mx 0.084Yg 0.35
cd
a
b
10000 kgDESPLZAMIENTO 0.063670426908 cm
K LATERAL DEL PORTICO 157058.78671825 kg/cm1570587.8671825 Tn/m
CURSO: ANALISIS ESTRUCTURAL II
EJERCICIO PRACTICO DE MUROS DE CORTE
Paso 01
10000
425
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
10
3
21
5
3
5
1
2 4
7
500
150 150
75 75 75 75Paso 03
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 150.003 210 A(cm2) 45004 280 I 84375005 315 As 37506 420
PLACA 12500
G 86948.2605 6 7
ø 0.161217993 6 106513.091
210 7 2506155.743
217370.6512 8 -22634031.9
4900 5 -106513.091
3640000 1 -2506155.743H de placa (cm) 425 2 -22634031.9
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
Inercia (cm4)
9
10
116
7
8
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 150.003 210 A(cm2) 45004 280 I 84375005 315 As 37506 420
PLACA 218.84955592154
G 86948.2605 9 10
ø 0.000008349 9 6.405
210 10 19281.601
217370.6512 11 -1361.0
38 5 -6.405
188 3 -19281.601H de placa (cm) 425 4 -1361.0
F`c 3 Rectangulo 11 140 b(cm) 30.002 175 h(cm) 50.003 210 A(cm2) 15004 280 I 3125005 315 As 12506 420
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
Inercia (cm4)
VIGA 3
1250G 86948.2605 1 2
ø 0.030000000 1 6521.12 2119363.85210 2 2119363.85 824649907.96
217370.6512 3 -6521.12 -2119363.85
1500 4 2119363.85 552936593.97
312500Long de viga (cm) 500
a 75b 75
1 2 3 41 2512677 2119364 -6521 21193642 2119364 7496097383 -2119364 5529365943 -6521 -2119364 25803 -21193644 2119364 552936594 -2119364 8250355385 0 22634032 0 13616 0 -22634032 0 07 -2506156 0 0 08 0 2948016085 0 09 0 0 0 -136110 0 0 -19282 011 0 0 0 192814
Paso 04
1 2 3 412345
As (cm2)
f´c (Kg/cm2)
E (Kg/cm2)
Area (cm2)
Inercia (cm4)
A partir de las matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇}
69781011
1 2 3 41 3.98886E-07 -1.1531185E-10 1.7135904E-08 -9.0340334E-102 -1.15312E-10 4.36254196E-10 1.4987835E-08 -2.5342628E-103 1.71359E-08 1.49878351E-08 4.9635647E-05 1.17421237E-074 -9.03403E-10 -2.5342628E-10 1.1742124E-07 1.68577995E-095 2.45138E-08 -9.2695204E-08 -3.1862238E-06 5.38283055E-08
1 02 03 04 05 10000
6 7 8 96 0.00 -22634031.91 0.00 0.007 -2506155.74 0.00 0.00 0.008 0.00 2948016085.03 0.00 0.009 0.00 0.00 0.00 -1361.0410 0.00 0.00 -19281.60 0.0011 0.00 0.00 0.00 192813.60
6 -9997.40458 kg
{𝒌𝒓𝒇}[𝑲𝒈𝒍𝒐𝒃𝒂𝒍] =
{𝑸𝒇} =
[𝑴𝒂𝒕𝒓𝒊𝒛 𝒊𝒏𝒗𝒆𝒓𝒔𝒂 𝒅𝒆 𝑲𝒇𝒇−𝟏]
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝑸𝒇] [𝑫𝒆𝒔𝒑𝒍𝒂𝒛𝒂𝒎𝒊𝒆𝒏𝒕𝒐𝒔 𝑳𝒊𝒃𝒓𝒆𝒔]
[𝑴𝒂𝒕𝒓𝒊𝒛 𝑲𝒓𝒇]
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝒍𝒂𝒔 𝒓𝒆𝒂𝒄𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆𝒍 𝒆𝒔𝒕𝒂𝒅𝒐 𝑪𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝒚 𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒂]
{𝑲𝒓𝒇} =
7 -614.354964 kg8 3850169.642 kg9 -2.59541751 kg10 614.354964 kg11 499.6314252 kg
6 -9997.40458 kg7 -614.354964 kg8 3850169.642 kg.cm9 -2.59541751 kg10 614.354964 kg11 499.6314252 kg.cm
{𝑹 𝒄𝒐𝒎} =
{𝑹 𝒇𝒊𝒏𝒂𝒍} =
{𝑹 𝒑𝒓𝒊𝒎} =
EJERCICIO PRACTICO DE MUROS DE CORTERELACION S - R - C
S R180 PI
50
150
barras libres y restringidos de la estructura
TIPO DE FIGURA 5
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
8 5 1 2
-22634031.9 -106513.091 -22634031.9
-2506155.743
6671447475 22634031.91 2948016085
22634031.91 106513.0913 22634031.91
2506155.7432948016085 22634031.91 6671447475
matriz para todas las barras del portico de la sgte manera:
TIPO DE FIGURA 3
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
11 5 3 4
-1361.0 -6.405 -1361.0
-19281.601
385630 1361.04 192814
1361.04 6.4049 1361.04
19281.601192814 1361.04 385630
TIPO DE FIGURA 1
Circulo 2 S. Tubular 3r(cm) 1 R(cm) 4
A(cm2) 3.1415926536 r(cm) 2I(cm4) 0.7853981634 A(cm2) 37.69911184
As 2.827433388 I(cm4) 188.4955592As 18.8495559
3 4
-6521.12 2119363.85
-2119363.85 552936593.97
6521.12 -2119363.85
-2119363.85 824649907.96
5 6 7 8 9 10 11
0 0 -2506156 0 0 0 022634032 -22634032 0 2948016085 0 0 0
0 0 0 0 0 -19282 01361 0 0 0 -1361 0 192814
106519 -106513 0 22634032 -6 0 1361-106513 106513 0 -22634032 0 0 0
0 0 2506156 0 0 0 022634032 -22634032 0 6671447475 0 0 0
-6 0 0 0 6 0 -13610 0 0 0 0 19282 0
1361 0 0 0 -1361 0 385630
5 6 9 7 8 10 11
matrices de todas las barras se obtiene esta matriz de toda la estructura sumando las componentes de cada matriz en función de sus grados de libertad con lo cual tenemos una matriz de 8x8 que luego particionamos como se indica
{𝒌𝒇𝒇} {𝒌𝒇𝒓}
5
2.4513838E-08-9.26952E-08
-3.186224E-065.3828306E-082.9083812E-05
FUERZADESPLZAMIENTO
1 0.0002451384 cm2 -0.000926952 cm K LATERAL DEL PORTICO3 -0.031862238 cm4 0.0005382831 cm5 0.2908381151 cm
10
-106513.090.00
22634031.91-6.400.00
1361.04
6 0 kg
{𝒌𝒓𝒇} {𝒌𝒓𝒓}
[𝑫𝒆𝒔𝒑𝒍𝒂𝒛𝒂𝒎𝒊𝒆𝒏𝒕𝒐𝒔 𝑳𝒊𝒃𝒓𝒆𝒔]
{𝑫𝒇} =
[𝑪𝒂𝒍𝒄𝒖𝒍𝒐 𝒅𝒆 𝒍𝒂𝒔 𝒓𝒆𝒂𝒄𝒄𝒊𝒐𝒏𝒆𝒔 𝒅𝒆𝒍 𝒆𝒔𝒕𝒂𝒅𝒐 𝑪𝒐𝒎𝒑𝒍𝒆𝒎𝒆𝒏𝒕𝒂𝒓𝒊𝒐 𝒚 𝑷𝒓𝒊𝒎𝒂𝒓𝒊𝒂]
7 0 kg8 0 kg9 0 kg10 0 kg11 0 kg
6 -9.99740458249 Tn7 -0.61435496399 Tn8 38.5016964198 Tn.m9 -0.00259541751 Tn10 0.61435496399 Tn11 0.00499631425 Tn.m
{𝑹 𝒑𝒓𝒊𝒎} =
RELACION S - R - CC
200
Seccion 4 Seccion 5 ITEM A Ya(cm) 0.6 a(cm) 30b(cm) 0.2 b(cm) 40 1 3000 50c(cm) 0.2 c(cm) 30d(cm) 0.25 d(cm) 100 2 1200 85
A(cm2) 0.17 A(cm2) 4900I(cm4) 0.0019784313725 I(cm4) 3640000 4200
As 0.1416666666667 As 2500YC= 60
Mx 0.041Yg 0.2411764705882 INERCIA TOTAL
c
d
a
b
a
b
c
a
d
Seccion 4a(cm) 0.6b(cm) 0.2c(cm) 0.4d(cm) 0.3
A(cm2) 0.24I(cm4) 0.0074
As 0.2
Mx 0.084Yg 0.35
Seccion 4a(cm) 0.6b(cm) 0.2c(cm) 0.4d(cm) 0.3
A(cm2) 0.24I(cm4) 0.0074
As 0.2
Mx 0.084Yg 0.35
cd
b
cd
a
b
10000 kgDESPLZAMIENTO 0.2908381151137 cm
K LATERAL DEL PORTICO 34383.388835029 kg/cm343833.88835029 Tn/m
L 100AY I d Ad2 Seccion 6 ITEM A Y
a(cm) 30150000 2500000 10 300000 b(cm) 40 1 4500 75
c(cm) 30102000 90000 -25 750000 d(cm) 150 2 1200 15
A(cm2) 6900252000 2590000 1050000 I(cm4) 11938026.3 5700
As 3750cm YC= 62.368421053
3640000 cm4 INERCIA TOTAL
b
c
a
d
AY I d Ad2
337500 8437500 -12.632 718005.540
18000 90000 47.368 2692520.776
355500 8527500 3410526.316
cm
11938026.316 cm4
UNIVERSIDAD ALAS PERUANASCARRERA: INGENIERIA CIVILCURSO: ANALISIS ESTRUCTURAL IIALUMNO: GARCIA CONDORCALLO, ALAN
EJERCICIO PRACTICO DE PORTICO POR EL METODO DE RIGIDEZ
RELACION S - R - CS R C
180 PI 200
Paso 01
Barra 1 2 30.170 0.150 0.150
0.0019784314 0.003125 0.003125
280 280 280
2509980.0796 2509980.0796 2509980.0796
L (m) 6.00 5.00 6.32
Angulo (grados) 0 90 108.43
Grados (Rad) 0.00000000 1.57079633 1.89254688
Numeramos las barras Numeramos los Grados de Libertad ( DOF ) libres y restringidos de la estructura
A (m²)Inercia (m4)F´C (Kg/cm2)
E (Tn/m²)
1m
6m 2m
0.50m
5m
1
3
2
4
5
1
2
3 6
7
8
9
10
11
12
0.20m
0.20m
0.25m
0.60m
0.30m
Paso 02
Paso 03
Coordenadas Globales ( Barra del Portico )
BARRA 1A (m²) 0.170 1 2 3 4 5 6Inercia (m4) 0.002 1 71116.102 0.000 -827.637 -71116.102 0.000 0.000F´C (Kg/cm2) 280.000 2 0.000 275.879 827.637 0.000 -275.879 827.637E (Tn/m²) 2509980.080 Tn/m² 3 -827.637 827.637 3310.549 0.000 -827.637 1655.274L (m) 6.00 m 4 -71116.102 0.000 0.000 71116.102 -275.879 0.000Angulo 0 5 0.000 -275.879 -827.637 -275.879 275.879 -827.637Radianes 0 6 0.000 827.637 1655.274 0.000 -827.637 3310.549COS 1SEN 0
BARRA 2A (m²) 0.150 1 2 3 7 8 9Inercia (m4) 0.003 1 0.000 0.000 -1882.485 -752.994 0.000 -1882.485F´C (Kg/cm2) 280.000 2 0.000 75299.402 0.000 0.000 -75299.402 0.000E (Tn/m²) 2509980.080 Tn/m² 3 -1882.485 0.000 6274.950 1882.485 0.000 3137.475L (m) 5.00 m 7 -752.994 0.000 1882.485 752.994 0.000 1882.485Angulo 90 8 0.000 -75299.402 0.000 0.000 75299.402 0.000Radianes 1.57 9 -1882.485 0.000 3137.475 1882.485 0.000 6274.950COS 0SEN 1
BARRA 3A (m²) 0.150 4 5 6 10 11 12Inercia (m4) 0.003 4 5952.940 -17858.821 -1176.553 -6287.793 17747.204 -1116.176
F´C (Kg/cm2) 280.000 5 -17858.821 53613.670 -372.059 17747.204 -53613.670 -372.059E (Tn/m²) 2509980.080 Tn/m² 6 -1176.553 -372.059 4960.784 1116.176 372.059 2480.392L (m) 6.32 m 10 -6287.793 17747.204 1116.176 6287.793 -17741.166 1116.176Angulo 108 11 17747.204 -53613.670 372.059 -17741.166 53613.670 372.059Radianes 1.89 12 -1116.176 -372.059 2480.392 1116.176 372.059 4960.784COS 0SEN 1
1 2 3 4 5 6 7 8 9 10 11 121 71116.102 0.000 -2710.122 -71116.102 0.000 0.000 -752.994 0.000 -1882.485 0.000 0.000 0.0002 0.000 75575.281 827.637 0.000 -275.879 827.637 0.000 -75299.402 0.000 0.000 0.000 0.0003 -2710.122 827.637 9585.499 0.000 -827.637 1655.274 1882.485 0.000 3137.475 0.000 0.000 0.0004 -71116.102 0.000 0.000 77069.043 -18134.700 -1176.553 0.000 0.000 0.000 -6287.793 17747.204 -1116.1765 0.000 -275.879 -827.637 -18134.700 53889.549 -1199.696 0.000 0.000 0.000 17747.204 -53613.670 -372.0596 0.000 827.637 1655.274 -1176.553 -1199.696 8271.333 0.000 0.000 0.000 1116.176 372.059 2480.3927 -752.994 0.000 0.000 0.000 0.000 0.000 752.994 0.000 1882.485 0.000 0.000 0.0008 0.000 -75299.402 0.000 0.000 0.000 0.000 0.000 75299.402 0.000 0.000 0.000 0.0009 -1882.485 0.000 0.000 0.000 0.000 0.000 1882.485 0.000 6274.950 0.000 0.000 0.00010 0.000 0.000 -6287.793 -6287.793 17747.204 1116.176 0.000 0.000 0.000 6287.793 -17741.166 1116.17611 0.000 0.000 17747.204 17747.204 -53613.670 372.059 0.000 0.000 0.000 -17741.166 53613.670 372.059
Esta es la matriz de RIGIDEZ en coordenadas globales para la barra de un portico
Podemos encontrar esta matriz para todas las barras del portico de la sgte manera:
cm²cm4
Kg/cm²
cm²cm4
Kg/cm²
cm²cm4
Kg/cm²
