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• José Esteban Casalino Guerrero

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CURSO

DISEÑO ESTRUCTURAL EN CONCRETO

Construcción de Diagrama de Interacción, Condiciones Nominales y Ultimas.- Verificación de Diseño.

Ing. Omart Tello Malpartida

Datos: Planta

h = 0.50 m

fc’=210 k/cm2 Pu

b =0.30 m

6.25

Elevación

6.25

fy= 4200 k/cm2 As= 4 φ 1” =4(5.1)=20.40 cm2 ρ=As/b.h= 20.4/30x50=0.0136=1.36 % d’= r + [φ 1”]/2 + [φ 3/8”]

Pu Mu

d’= 4 + 1.25 + 1 = 6.25 cm d = 50 - 6.25 = 43.75 cm

¿Construir el Diagrama de Interacción de la columna?

Concreto Armado I

Ing. Omart Tello Malpartida

1.- Compresión Pura (Mn=0) h Para la condicion: b

Ast/2

Cs = (Ast/2)( fy)

Ast/2

∑ Fv = 0 Pno = Cc + 2Cs

Pno = ( 0.85.fc' )( Ac ) + ( Ast ) ( f y ) Cs

Cc

Pno = 0.85.(210)(30x50)+(20.40)(4200)

Cs

Pno = 267750 + 85680 = 353,430 = 353.43 t 0.8Pno = 0.8(353.43)= 282.74 t (Limite) 0.85f’c

Puo = φ.Pno = 0.7(353.43)= 247.40 t (columna estribada ACI) 0.8Puo = 0.8(247.40)= 197.92 t (Limite)

Concreto Armado I

Ing. Omart Tello Malpartida

2.- Falla Balanceada (εs =εy ; εc= εcu =0.003 ) h

Calculo de C b :

h/2 As

b

εy =

As’ C.P

ε’s εs Cc

Es

Cs

4200 = 0.0021; ε cu = 0.003 2 x106

Calculo de f s' :

ε' s

T = As.fy 0.85f’c

ab/2 d

=

⎛ ε cu ⎞ 0.003 ⎛ ⎞ cb = ⎜ .d = ⎜ ⎟ ⎟ .d ⎜ε +ε ⎟ + 0.003 0.0021 ⎝ ⎠ y ⎠ ⎝ cu d = 50 − 6.25 = 43.75 cm cb = 25.73 cm ; ab = β1.cb = 0.85(25.73) = 21.87 cm

Cb

εs =εy

fy

ab

Concreto Armado I

c−d'

=

ε cu

c ⎛ c−d '⎞ ⎛ 25.73 − 6.25 ⎞ = 0.003. ε 's = ε cu . ⎜ ⎟ ⎜ ⎟ 25.73 ⎠ ⎝ c ⎠ ⎝ ε 's = 0.00227 ≤ ε y ∴ ε 's = ε y fs ' = ε 's .Es = (0.0021)(2 x106 ) = 4200 k/cm 2 Ing. Omart Tello Malpartida

2.- Falla Balanceada h

Fuerzas : Cs = fs '. As ' = 4200(10.20) = 42,840 kg = 42.84 t

h/2 As

b

Cc = 0.85 fc '.b.a = 0.85(210)(30)(21.87) = 117,113.85 kg = 117.11 t T = f y. As = 4200(10.20) = 42,840 kg = 42.84 t

As’ C.P Cb

εcu =

ε’s

εs =εy

∑ Fv = 0 0.003

Pnb = Cs + Cc − T Pnb = 42.84 + 117.11 − 42.84 Pnb = 117.11 t

εs Cc

Pub = φ .Pnb = 0.70(117.11) = 81.98 t

Cs

∑ Mc.p = 0

T = As.fy 0.85f’c

ab/2 ab Concreto Armado I

h⎞ ⎛h ⎞ ⎛h a⎞ ⎛ Mnb = Cs ⎜ − d ' ⎟ + Cc ⎜ − ⎟ + T ⎜ d − ⎟ 2⎠ ⎝2 ⎠ ⎝2 2⎠ ⎝ Mnb = 42.84 ( 25 − 6.25 ) + 117.11( 25 − 21.82 2 ) + 42.84 ( 43.75 − 25 ) Mnb = 3253.65 t-cm = 32.53 t-m Mub = φ .Mnb = 0.70(32.53) = 22.77 t-m Ing. Omart Tello Malpartida

3.- Flexión Pura (Pn=0) h

Calculo de f s' :

h/2

ε cu

As

As’ C.P

ε s ' = ε cu .

C

εcu =

ε’s

εs =εy

0.003

εs Cc

εs '

c−d ( c − d ') ; c = a ε s ' = ε cu . β1 c c

b

=

Cs

T = As.fy 0.85f’c

( a − β1.d ') a

kg ⎞ ⎛ ⎛ ⎛ a − β1.d ' ⎞ ⎞ f s ' = Es .ε s ' = ⎜ 2 x106 2 ⎟ ⎜ 0.003 ⎜ ⎟⎟ cm ⎠ ⎝ a ⎝ ⎝ ⎠⎠ ⎛ a − β1.d ' ⎞ f s ' = 6000 ⎜ ⎟ ; β1.d ' = 0.85(6.25) = 5.31 a ⎝ ⎠ Calculo de Fuerzas : ⎡ ⎛ a − β1.d ' ⎞ ⎤ ⎛ a - 5.31 ⎞ Cs = As '. fs ' = 10.20 ⎢6000 ⎜ ⎟ ⎥ = 61200 ⎜ ⎟ a ⎝ a ⎠ ⎝ ⎠⎦ ⎣ Cc = 0.85 fc '.a.b = 0.85(210)(a)(30) = 5355a T = As. fs = As. fy = 10.20(4200) = 42480 Para Flexion Pura ∑ Fv = 0

a/2 a Concreto Armado I

Pn = Cs + Cc − T = 0 ⎛ a − 5.31 ⎞ 61200 ⎜ ⎟ + 5355a − 42480 = 0 ⎝ a ⎠ 61200 ( a − 5.31) + 5355a 2 − 42480a = 0 5355a 2 + 18360a − 325125 = 0 a = 6.264 cm

3.- Flexión Pura h h/2 As

b

Reemplazando : ⎛ a − 5.31 ⎞ Cs = 61200 ⎜ ⎟ = 9296.27 = 9.30 t ⎝ a ⎠ Cc = 5355a = 33543.73 = 33.54 t T = 42480 = 42.84 t

As’ C.P C

εcu =

ε’s

εs =εy

0.003

εs Cc

Cs

0.85f’c

a Concreto Armado I

Pn = Cs + Cc − T = 0 Pn = 9.30 + 33.54 − 42.84 = 0

∑ Mc.p = 0

T = As.fy

a/2

∑ Fv = 0

h⎞ ⎛h ⎞ ⎛h a⎞ ⎛ Mn = Cs ⎜ − d ' ⎟ + Cc ⎜ − ⎟ + T ⎜ d − ⎟ 2⎠ ⎝2 ⎠ ⎝2 2⎠ ⎝ 6.25 ⎞ ⎛ Mn = 9.30 ( 25 − 6.25 ) + 33.54 ⎜ 25 − ⎟ + 42.84 ( 43.75 − 25 ) 2 ⎠ ⎝ Mn = 1711.14 t-cm = 17.11 t-m Mu = φ .Mn = 0.90(17.11) = 15.40 t-m Ing. Omart Tello Malpartida

4.- Falla por Tensión (c < cb=25.73 cm) Asumimos: c= 12.5 cm < cb ; a=β1.c =0.85(12.5)= 10.63 cm h

Calculo fs' :

h/2 As

b

⎛ c − d '⎞ ⎟ ⎝ c ⎠

ε s ' = ε cu ⎜

As’ C.P c

εc =

ε’s

εs >εy

0.003

⎛ c − d '⎞ ⎛ c − d '⎞ 6) fs ' = E s .ε s ' = E s .ε cu ⎜ ⎟ = (2 x10 (0.003) ⎜ ⎟ ⎝ c ⎠ ⎝ c ⎠ ⎛ c − d '⎞ fs ' = 6000 ⎜ ⎟ ≤ fy ⎝ c ⎠ ⎛ 12.5 − 6.25 ⎞ 2 fs ' = 6000 ⎜ ⎟ = 3000 kg/cm ≤ f y 12.5 ⎝ ⎠ Fuerzas : Cs = As '. fs ' = 10.20(3000) = 30, 600 = 30.60 t Cc = 0.85 fc '.a .b = 0.85(210)(10.63)(30) = 56, 923.65 = 56.924 t T = As . f y = 10.20(4200) = 42, 840 = 42.84 t

εs Cc

∑ Fv = 0

Cs

T = As.fy 0.85f’c

Pn Pn Pn Pu

= Cs + Cc − T = 30.60 + 56.92 − 42.84 = 44.68t = φ .Pn = 0.70(44.68) = 31.28 t

∑ M c.p = 0 a/2 a Concreto Armado I

h⎞ ⎛h ⎞ ⎛h a⎞ ⎛ Mn = Cs ⎜ − d ' ⎟ + Cc ⎜ − ⎟ + T ⎜ d − ⎟ 2⎠ ⎝2 ⎠ ⎝2 2⎠ ⎝ 10.63 Mn = 30.60 ( 25 − 6.25 ) + 56.92 ( 25 − 2 ) + 42.84 ( 43.75 − 25 ) Mn = 2497.47 t-cm = 24.97 t -m Mu = φ .M n = 0.70(24.97) = 17.48 t -m

5.- Falla por Compresión (c > cb=25.73 cm) Asumimos: c= 45 cm > cb ; a=β1.c =0.85(45) =38.25 cm h h/2 As

b

As’

εs cb ; a=β1.c =0.85(45) =38.25 cm h h/2

Calculo fs' : ⎛ c − d '⎞ ⎟ ⎝ c ⎠

ε s ' = ε cu ⎜

As

b

As’

⎛ c − d '⎞ ⎛ c − d '⎞ 6) fs ' = E s .ε s ' = E s .ε cu ⎜ ⎟ = (2 x10 (0.003) ⎜ ⎟ ⎝ c ⎠ ⎝ c ⎠ ⎛ c − d '⎞ ⎛ 45 − 6.25 ⎞ 2 fs ' = 6000 ⎜ ⎟ = 6000 ⎜ ⎟ = 5167 kg/cm ≤ f y 45 ⎝ c ⎠ ⎝ ⎠

C.P c

εcu =

ε’s

εs