Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Hormigón Armado
Resumen y Ejercicios Propuestos Curso de Diseño en Hormigón Armado
Leonardo Brescia Norambuena Estudiante de Ingeniería Civil en Obras Civiles Versión 2.0
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Universidad Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Índice
Pagina Prologo
I
Bibliografía
II
1-. Compresión Simple
2
Ejercicio 1
3
Ejercicio 2
4
2-. Flexión Simple
6
Ejercicio 1
9
Ejercicio 2
12
Ejercicio 3
16
3-. Diseño Armadura de Corte
18
Ejercicio 1
20
4-. Flexo - Compresión
22
Ejercicio 1
31
Ejercicio 2
36
Ejercicio 3
41
5-. Diseño de Losas
44
Ejercicio 1
51
Ejercicio 2
53
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Universidad Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Prologo
Los siguientes apuntes son un resumen de la materia que se ve en el curso de Hormigón armado de la carrera de ingeniería civil en obras civiles de la USACH, están basados en los apuntes de clase de la profesora Silvana Cominetti y ACI 2008. Estos solo representan un resumen de los pasos para poder desarrollar el diseño de elementos de H.A., lo que no indica que sean la única manera de realizar dichos diseños, y se recomienda revisar factores y consideraciones de acuerdo a los códigos de diseños que estén vigentes en el país. Se recuerda que se presentan métodos basados diseño LRFD y bajo indicaciones de ACI. También se sugiere revisar manuales de diseño que entregan empresas dedicadas al tema. Se recomienda al estudiante desarrollar sus propias planillas digitales para el diseño de elementos, ya un buen uso de herramientas computacionales puede favorecer el trabajo a realizar. Finalmente se recuerda a los alumnos que esta versión puede estar sujeta a mejoras, por lo que es importante que quienes quienes puedan realizar su aporte a esta, estarán ayudando a los futuros alumnos de la carrera y que bajo ningún punto de vista, es suficiente s uficiente para la aprobación de la asignatura.
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Universidad Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Bibliografía
Apunte de Hormigón Armado, Profesora Silvana Cominetti Cotti-Cometti ACI 318 – 318 – versión 2005 ACI 318 – 318 – versión 2008
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
1-. Compresión Simple
Consideraciones Previas - Toda ecuación aquí presentada puede ser deducida por equilibrio de fuerzas y deformaciones - Se está diseñando bajo método LRFD, puede que valores de factores cambien de acuerdo a la versión del código ACI Vigente. (1.1) Pu P Pu: Carga obtenida ya mayorada a través de combinaciones de norma NCh3171 y efectos del pandeo Pn: Carga Nominal que puede resistir el elemento (Carga que resisten materiales) ϕ: Factor de seguridad. ϕ=0.65 Pn
0.85A 0.85Ac c f´c f´c As f
(1.2)
Pn
f´c As fy 0.85(Ag As )f´c
(1.3)
Ag: Área global del elemento (área transversal) Ac: Área de Concreto (Área Global menos Área Acero, (2) equivalente a (3)) As: Área de Acero (transformar al final a áreas comerciales) f´c: Resistencia a compresión del hormigón en probeta cilíndrica fy: Limite de fluencia del acero Nota: Pu debe mayorarse por un factor ω, para prevenir problemas de pandeo. Este factor ω se obtiene de
tablas aunque en algunos textos den ecuaciones para determinarlo, y además el ACI para evitar riesgos por posibles excentricidades se disminuye la resistencia del hormigón a un 80%, por lo que la ecuación (1) puede reescribirse como: (1.4) Pu 0.8 P
Tabla 1.1: Factor ω por pandeo
También es bueno verificar que se esté bajo la carga crítica de pandeo, dado que esta dependerá de condiciones geométricas del elemento. (1.5) Pu Pcrit 2
Pcrit
Ec I ( K l)
(1.6)
2
Ec: Modulo de elasticidad del Hormigón (Se puede aproximar entre 15000*f´c^.5 a 15400*f´c^.5) I: Inercia de la Sección K: Factor de Largo de Empotramiento l: Largo del elemento
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Ejercicio 1
Para la estructura mostrada en la figura determine la máxima sobrecarga que puede resistir el elemento central indicado en esta si se está utilizando H30 y A63-42. No considere falla por funcionamiento de la losa y verifique la carga crítica de pandeo para el caso más favorable y desfavorable.
Desarrollo
1-. Se debe determinar el facto ω, para esto se debe determinar el caso más desfavorable, dado que no hay mayor información, para esto se considerara K=0.5 (si la losa aporta rigidez suficiente junto a los otros elementos para empotrar la columna en la parte superior) y K=2 (si la columna se comporta como voladizo) b 30cm
l 2.2
Si
K ( l K)
b
Si
0.
( l K)
b
0.6
Caso más favorable ==>
3.667
K 2
h 30cm
1
(de tabla 1.1)
Caso más desfavorable
14.667
==>
1
(de tabla 1.1)
2-. Calculo de Pn 2
Ag 900 900cm
2
As 42.54cm
fy 4200
kg 2
cm
f´c 250
kg 2
cm
Pn 0.85(Ag As)f´c f´c As fy Pn
231763 kg
3-. Calculo de SC max: Viendo las combinaciones de la NCh3171of2010, se tiene que el caso más desfavorable será:
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
horm 2500
kg 3
m 2
Plosa 9m 0.15m horm
Plosa 3375kg Ppilar 495 495kg
Ppilar 0.3m0.3m2.2mhorm
PP Plos losa Ppila pila Puma umax 0.8P 0.8Pn SCmax
se considera peso pilar porque es el caso más desfavorable para este
PP 3870kg
Pumax 120516.76kg
( Puma Pumax 1.2PP 1.2PP) 1.6
SCmax 72420.475kg
Pero la SC debe darse distribuida en el Área Tributaria SCmax
SC
kg SC 8046.719
2
2
m
9m
4-. Calculo de Cargas Críticas de Pandeo 3
I b
h
4
I 67500 cm
12
kg
Ec 15000
2
cm
2
f´c
cm
kg
Ec 237170.825
kg 2
cm
K 0. 2
Pcrit1
Ec I (K l)
2
Pcrit1 13058082.596 kg
K 2 2
Pcrit1
Ec I (K l)
2
Pcrit1 816130.162 kg
Nota: Se ve que la carga crítica de pandeo es muy superior a la carga que se resiste por materiales
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Pu Ast
0.8
c 0.85 f ´ At 2
Ast 51.506cm
fy 0.85f ´ c
2
Asmin 0.01At
Asmin 9 cm
Asmax 0.08A
Asmax 72cm
2
Por lo que se está dentro de los rangos permitidos. La armadura se escogerá de la siguiente tabla obtenida de Gerdau Aza
Con 12 ϕ28 se tiene un área de: 2
2
124.83cm 57.96cm
Se obtiene con un recubrimiento de 2 cm una separación libre entre refuerzos de 5cm aproximadamente, con el siguiente esquema de diseño:
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
2-. Flexión Simple
Consideraciones Previas - Toda ecuación aquí presentada puede ser deducida por equilibrio de fuerzas y deformaciones - Se está diseñando bajo método LRFD, puede que valores de factores cambien de acuerdo a la versión del código ACI Vigente. Armadura Simple
I) Caso 1: Cálculo de Resistencia a) Calculo de Equilibrio de fuerzas (2.1) (2.2)
(1) Rectángulo de compresión según ACI (2) Área de Hormigón Comprimida (cambia para viga no rectangular)
T: Tracción C: Compresión As: Área de Acero (de manuales comerciales) fy: Limite de fluencia del Acero f´c: Resistencia en probeta cilíndrica del Hormigón a: Altura que alcanza la compresión del Hormigón b: Ancho Viga b) Determinar a y con esto c (C=a/β1, donde β1 depende de f´c)
c) Determinar el momento nominal de la viga Mn
d a As fy 2
(2.3)
Mn: Momento nominal que resiste la viga d: Altura libre de Grieta de la viga (sección transversal). d = H - recubrimiento d) Calcular el Momento Ultimo de la viga (2.4) M ϕ: Factor de reducción que minora resistencia de la viga. ϕ=0.9 Mu
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
II) Caso 2: Diseñar la Armadura a) Determinar Mu al cual está sometida la viga. Se recomienda mayoría los momentos para cada punto y no las cargas, con los factores dados por NCh3171 b) Con la ecuación dada a continuación obtener armadura necesaria (y la cuantía): As
2
0.85 f´c b d
0.85 f´c b d fy
fy
As
( 1.89 Mu b f´c)
(2.5)
2
fy
(2.6)
bd
As: Área de armadura necesaria para satisfacer el momento que solicita la viga fy: Limite de fluencia del Acero f´c: Resistencia en probeta cilíndrica del Hormigón b: Ancho Viga d: Altura libre de Grieta de la viga (sección transversal). d = H - recubrimiento ρ: Cuantía
Nota: - Si As es mayor que 0, se debe verificar los límites máximos y mínimos de armadura - Si As es menor o igual a 0, no requiere armadura, no obstante se utiliza armadura mínima - Si As es un n° complejo, se requerirá refuerzo de acero a compresión. c) Determinar la cuantía de balance bal
0.85 f´c 1
fy
0.003Es
fy 0.003Es
(2.7)
ρbal: Cuantía de Balance
Es: Modulo de elasticidad del Acero d) Comparar Cuantía obtenida 0. 8 f´c 14 0.8 fy fy
min max
(2.8)
max
0.75 bal if CasoEst asoEsta atico ico 0.025 if CasoSismico
(2.9)
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Armadura Doble
Nota: Se mantiene Nomenclatura utilizada para Armadura Simple y principios utilizados en esta. I) Caso 1: Cálculo de Resistencia La resistencia de la viga se lograra mediante un proceso iterativo, del cual se describen los pasos a continuación. a) Realizar equilibrio de fuerzas y con esto determinar a Cc
Cs
0.85 f´c a b As´ fs´
a
(2.10)
T As fs
( As fs
As´ fs´) 0.85f´c b
(2.11) (2.12)
Nota: aquellos términos con ´ en su final son para denotar la armadura que trabaja a compresión, en este caso además, se desconoce la tensión de trabajo del acero. Para la primera iteración se ingresa con fs=fs´=fy b) Determinar c=a/β1
c) Realizar diagrama de deformación unitaria, y obtener deformación unitaria del acero. Se recomienda usar congruencia de triángulos
Fig. 2.2 ( a)
( b)
Fig. 2.2 (a): Esquema básico viga con armadura doble.
( c)
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
II) Caso 2: Determinar Armadura de la viga vi ga (método recomendado) a) Determinar una armadura base sobre la cual se comienza a trabajar, esta debe estar contenida entre As máxima y As mínima para armadura simple. b) Determinar el momento nominal que resiste la viga con esta armadura, con ecuaciones de armadura simple c) Determinar el ΔMn requerido
Mn
Mu Mn
(2.14)
d) Determinar armadura que falta para cumplir con momento requerido Asr
Mn ( d d´) fy
(2.15)
Asr: Armadura necesaria Agregar, esta se agrega tanto en zona de compresión como en zona de tracción. d´: Distancia entre borde viga y armadura de refuerzo a compresión (generalmente igual a recubrimiento) e) Verificar: As
2Asr
14
(2.16)
fy
c t 0.75ba
ρc: Cuantía Armadura a compresión
(2.17)
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Ejercicios
1-. Determine el momento máximo que puede resistir una viga V30/40, cuyo recubrimiento es de 2 cm, una armadura inferior de 14cm2, y superior de 8cm2, considere H30 con acero A63-42 Es=2.1x10^6 As
2
14cm
As´
d 38cm
2
8cm
b 30cm
d1 2cm f´c 250
kg
fy 4200
2
cm Es
2
cm
a
( As fs
As´ fs´) 0.85f´c b ( d c)
es
c
fs
es Es
c
c
a
1 (c
es´
c
fs´ fs´
es´ fs
d1) c
2
cm
6 kg
2.110
kg
c 0.00
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
fss y fs´s de la fila i se copian en fsi y fs´i de la fila i+1 respectivamente, y se itera nuevamente hasta converger
Se ve con cada iteración se acerca más a un valor, lo cual se puede observar en los siguientes gráficos
f´s (kg/cm2) vs N° Iteraciones Iteraciones 4050.000 4000.000 3950.000 3900.000 3850.000 3800.000 3750.000 3700.000 3650.000 3600.000
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
2-. Para una viga V25/45 determine la armadura necesaria si se usa hormigón H30 y Acero A63-42, si se presenta la siguiente tabla de momentos punto
1
2
3
4
5
Comb1
320711
416732
873529
750316
537703
Comb2
300916
169501
50133
132
139669
Comb3
903512
50713
163021
290332
330735
*Momentos en Kg*cm Desarrollo
f´c 250
kg
fy
2
kg
4200
b 25cm
2
d 43cm
cm
cm
d1 2cm
1 0.8
Es
6 kg
2.110
2
cm
a) Se calculara armadura base:
Asmin smin max 14
0.8
kg 2
b
d fy
f´c
kg 2
cm fy
b d
Asmin
2
3.583 cm
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Mn As f y d
a
5
5
6.958 10 kg cm Mn 6.958
Mu 0.9M
1
2
3
4
5
903512
416732
873529
750316
537703
2
6.263 10 kg cm Mu 6.263
c) La envolvente queda: punto Mu
Por lo que se requeriría refuerzo en los puntos 1, 3 y 4 Mn1 Mn3 Mn4
903512 0.9
kg cm
873529 0.9
kg cm
750316 0.9
kg cm
5
Mn
Mn1 3.081 3.081 10 kg cm
Mn
Mn3 2.748 2.748 10 kg cm
Mn
Mn4 1.378 1.378 10 kg cm
5
5
Asr1
Mn1 ( d d1) fy
Asr1
1.789 cm
Asr3
Mn3 ( d d1) fy
Asr3
1.596 cm
2
2
Asr1
116
Asr3
116
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
3-. Para la viga mostrada en la figura, y en función de los datos entregados determine el valor de t para que el área comprimida quede solo en la parte superior de la viga. Determine además el momento último que resiste. Datos: kg
f´c 250
2
cm fy 2800
kg 2
cm
1 0.8
2
As1
10cm
As2
8cm
As3
6cm
r
2
2cm
2
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Se obtiene la siguiente tabla de iteraciones i ter
fs 1 i
fs 2 i
fs 3 i
a
c
t
d
es 1
es 2
es 3
fs 1 s
fs 2 s
fs 3 s
Mu
1
4200 4200 4200 4200 4200. 4200.00 00 3.95 3.9529 29 4.65 4.6505 05
5
53
0.03 0.0311 119 9 0.02 0.0279 796 6 0.00 0.0017 171 1 6549 65498. 8.44 44 58725. 58725.00 00 3590. 3590.63 331990 3319907 7
2
4200 4200 4200 4200 3590. 3590.63 63 4.23 4.2397 97 4.98 4.9879 79
5
53
0.02 0.0288 888 8 0.02 0.0258 587 7 0.00 0.0018 180 0 6064 60642. 2.14 14 54326. 54326.85 85 3773. 3773.88 331300 3313009 9
3
4200 4200 4200 4200 3773. 3773.88 88 4.15 4.1535 35 4.88 4.8864 64
5
53
0.02 0.0295 954 4 0.02 0.0264 647 7 0.00 0.0017 177 7 6203 62032. 2.06 06 55585. 55585.64 64 3721. 3721.43 331518 3315183 3
4
4200 4200 4200 4200 3721. 3721.43 43 4.17 4.1781 81 4.91 4.9155 55
5
53
0.02 0.0293 935 5 0.02 0.0263 630 0 0.00 0.0017 178 8 6162 61628. 8.39 39 55220. 55220.05 05 3736. 3736.66 331456 3314569 9
5 6
4200 4200 4200 4200 3736. 3736.66 66 4.17 4.1710 10 4.90 4.9070 70 4200 4200 4200 4200 3732. 3732.26 26 4.17 4.1731 31 4.90 4.9095 95
5 5
53 53
0.02 0.0294 940 0 0.02 0.0263 635 5 0.00 0.0017 178 8 6174 61745. 5.14 14 55325. 55325.78 78 3732. 3732.26 331474 3314748 8 0.02 0.0293 939 9 0.02 0.0263 633 3 0.00 0.0017 178 8 6171 61711. 1.33 33 55295. 55295.17 17 3733. 3733.53 331469 3314697 7
7 8
4200 4200 4200 4200 3733. 3733.53 53 4.17 4.1725 25 4.90 4.9088 88 4200 4200 4200 4200 3733. 3733.17 17 4.17 4.1726 26 4.90 4.9090 90
5 5
53 53
0.02 0.0293 939 9 0.02 0.0263 634 4 0.00 0.0017 178 8 6172 61721. 1.12 12 55304. 55304.03 03 3733. 3733.17 331471 3314712 2 0.02 0.0293 939 9 0.02 0.0263 633 3 0.00 0.0017 178 8 6171 61718. 8.28 28 55301. 55301.46 46 3733. 3733.27 331470 3314707 7
9
4200 4200 4200 4200 3733. 3733.27 27 4.17 4.1726 26 4.90 4.9089 89
5
53
0.02 0.0293 939 9 0.02 0.0263 633 3 0.00 0.0017 178 8 6171 61719. 9.10 10 55302. 55302.21 21 3733. 3733.24 331470 3314708 8
10 4200 4200 4200 3733. 3733.24 4.172 4.1726 6 4.9089 4.9089 11 4200 4200 4200 3733. 3733.25 4.172 4.1726 6 4.9089 4.9089
5 5
53 53
0.029 0.02939 39 0.026 0.02633 33 0.001 0.00178 78 61718. 61718.87 87 55301. 55301.99 99 3733. 3733.25 3314708 3314708 0.029 0.02939 39 0.026 0.02633 33 0.001 0.00178 78 61718. 61718.94 94 55302. 55302.05 05 3733. 3733.25 3314708 3314708
12 4200 4200 4200 3733. 3733.25 4.172 4.1726 6 4.9089 4.9089
5
53
0.029 0.02939 39 0.026 0.02633 33 0.001 0.00178 78 61718. 61718.92 92 55302. 55302.04 04 3733. 3733.25 3314708 3314708
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
3-. Diseño Armadura de Corte
Como es habitual, en el diseño a corte también debemos verificar que el corte que resiste el elemento Vn (Corte nominal), al ser disminuido por un factor de seguridad sea mayor a la máxima solicitación, lo cual queda expresado por la ecuación 2.1 (3.1) Vu Vn Vu: Corte Ultimo que existe en el elemento Vn: Corte Nominal que resiste el elemento ϕ: Factor de Seguridad que reduce resistencia del elemento Nota: El diseño de armadura de corte debe ser determinada para cada uno de los puntos de interés (según se explicara a continuación), respetando las consideraciones de la combinación que predomine para el diseño. Consideraciones especiales a tomar
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
ii)
si
1.25 si
id
1.25 si
xy
==>
fy
(3.14.a)
f´c fy
sd
1.25 si
fy
(3.14.b)
ii
f´c
1.25 si
fy
(3.14.c)
f´c
(3.14.d)
f´c
x: Superior o Inferior y: Izquierda o Derecha
iii) Calculo de Momento Plástico MPdsi
2
id b d f´c ( 1 0. 59id) 2
(3.15.a)
MPisi
2
si b d f´c (1 0. 59si ) 2
(3.15.b)
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Ejercicio
Para la viga mostrada en la figura se pide diseñar la armadura de corte. Use H30 y A63-42
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
iv) Se debe tener el Corte Para cada punto de relevancia, se tomaran los apoyos, fin de las ZRP y en el centro. (Para los cortes estáticos en este caso se pueden obtener por congruencia de triángulos)
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
Punto
Apoyo 1
ZRP1
Apoyo2
Comentario
Vu
7373.05
6890.41
7373.05 Maxi mo Corte
Vn
9830.74
9187.21
9830.74 V e r 2.1 d
d
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
4-. Flexo - Compresión 4.1-. Flexo- Compresión (Métodos Analíticos) Consideraciones Consideraciones Especiales:
Siempre se prefiere una falla dúctil sobre una falla frágil Toda Ecuación puede ser determinada a partir del equilibrio de fuerzas y momentos
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Universidad de Santiago de Chile Departamento de Ingeniería en Obras Civiles Diseño en Hormigón Armado
ii.1) Si se está bajo el punto de balance: Pu
a
(4.4)
** fs=fy
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