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DISEÑO EN CONCRETO ARMADO CON MATHCAD PRIME ACI 318-14 - 6TA SESIÓN ANÁLISIS DE SECCIÓN SIMPLE Y DOBLEMENTE ARMADO ING. GABRIEL BENAVIDEZ M

ANÁLISIS DE SECCIÓN SIMPLE Y DOBLEMENTE ARMADO Ecuación de equilibrio si ε' > εy , cuando el acero empieza a fluir: As ⋅ fy ＝ 0.85 ⋅ f'c ⋅ β1 ⋅ c ⋅ b + A's ⋅ fy (1) Ecuación de equilibrio si ε' < εy , cuando el acero no fluye:

Hormigón:

lbf f'c ≔ 3000 ⋅ ―― in 2

Acero:

lbf lbf fy ≔ 60000 ――Es ≔ 29 ⋅ 10 6 ⋅ ―― in 2 in 2

Peralte:

d ≔ 27 in = 68.58 cm

Base:

b ≔ 14 in = 35.56 cm

A's ≔ 2 in 2

As ≔ 6.25 in 2

Recubrimiento: d' ≔ 2.5 in = 6.35 cm β1 = 0.85 Primera iteración se considera la ecuación (1): Distancia hacia el eje neutro: Deformación unitaria del acero superior:

⎛⎝As - A's⎞⎠ ⋅ fy c ≔ ―――――= 8.403 in 0.85 ⋅ f'c ⋅ b ⋅ β1 c - d' ε's ≔ ――⋅ 0.003 = 0.002108 c

Altura del bloque de compresión:

a ≔ β1 ⋅ c = 7.143 in

Deformación unitaria de fluencia:

fy = 0.002 εy ≔ ―― Es

if ⎛⎝ε's ≥ εy , “El acero fluye” , “El acero NO fluye”⎞⎠ = “El acero fluye”

Esfuerzo producido en el acero superior

Distancia hacia el eje neutro:

c = 8.403 in

Altura del bloque de compresión: a ≔ β1 ⋅ c = 7.143 in

Cálculo del factor de minoración:

Deformación del acero superior:

c - d' ε's ≔ ――⋅ 0.003 = 0.0021 c

Deformación del acero inferior:

d-c εt ≔ ――⋅ 0.003 = 0.0066 c

Adoptamos ϕs = 0.9

Cálculo de áreas:

εt ≥ 0.004 = 1 Comprobacion2 = “Ductil”

A's ⋅ f's = 2 in 2 As2 ≔ ――― fy

As1 ≔ As - As2 = 4.25 in 2

⎛ ⎛ ⎞ a⎞ Mu ≔ ϕs ⋅ ⎜As1 ⋅ fy ⋅ ⎜d - ―⎟ + ⎛⎝A's ⋅ f's ⋅ ((d - d'))⎞⎠⎟ = ⎛⎝668.57 ⋅ 10 3 ⎞⎠ lbf ⋅ ft 2 ⎝ ⎝ ⎠ ⎠ Mu = 906.461 kN ⋅ m

c1 ≔ 5 in ⎛⎝c1 - d'⎞⎠ As ⋅ fy ＝ 0.85 ⋅ f'c ⋅ β1 ⋅ c1 ⋅ b + A's ⋅ ―――⋅ 0.003 ⋅ Es c1 c1 ≔ find ⎛⎝c1⎞⎠ = 8.342 in

Comprobacion1 ≔ if ⎛⎝ε's ≥ εy , “El acero fluye emplear la ecuación (1)” , “El acero NO fluye emplear la ecuación (2)”⎞⎠

lbf f's = 60000 ―― in 2

0.25 = 1.04 ϕs ≔ 0.65 + ⎛⎝εt - εy⎞⎠ ⋅ ―――― 0.005 - εy

‖ || lbf β1 ≔ ‖ if f'c ≤ 4000 ⋅ ―― || ‖ in 2 || ‖ ‖ || ‖ ‖ 0.85 || ‖ lbf lbf | | ‖ else if 4000 ⋅ ―― < f'c < 8000 ⋅ ―― || ‖ in 2 in 2 | | ‖ ‖ | ⎛ lbf ⎞ | | ‖ ‖ | 0.05 ⋅ ⎜f'c - 4000 ⋅ ―― ⎟ ‖ ‖ in 2 ⎠ | | ⎝ ‖ ‖ 0.85 - ―――――――― || ‖ ‖ lbf || 1000 ⋅ ―― ‖ ‖ 2 || in ‖ ‖ || ‖ lbf || ‖ else if f'c ≥ 8000 ⋅ ―― || ‖ in 2 || ‖ ‖ || ‖ ‖ 0.65 | || ‖

ε's ≥ εy = 1

Comprobacion1 = “El acero fluye emplear la ecuación (1)” Se considera:

(2)

Solver Restricciones Valores de prueba

((c - d')) ⋅ 0.003 ⋅ Es As ⋅ fy ＝ 0.85 ⋅ f'c ⋅ β1 ⋅ c ⋅ b + A's ⋅ ――― c

lbf ‖ | | = ⎛⎝60 ⋅ 10 3 ⎞⎠ ―― f's ≔ ‖ if ε's ≥ εy | 2 | in ‖ ‖f | | y ‖ ‖ || ‖ else if ε's < εy || ‖ ‖ || c1 - d' ‖ ‖ || ← ⋅ 0.003 ε' ――― ‖ ‖ s || c1 ‖ ‖ || ‖ ‖ ε's ⋅ Es | || ‖ ‖

‖ | | = 8.403 in c ≔ ‖ if ε's ≥ εy || ‖ ‖c || ‖ ‖ || ‖ else if ε's < εy| | ‖ ‖ || ‖ ‖ c ← c1 || ‖ ‖ c1 || | ‖ ‖ |

Comprobacion2 ≔ if ⎛⎝εt ≥ 0.004 , “Ductil” , “Fragil”⎞⎠ = “Ductil” ‖ || ϕs ≔ ‖ if ϕs < 0.65 || ‖ ‖ 0.65 || ‖ ‖ || ‖ else if 0.65 ≤ ϕs ≤ 0.9| | ‖ ‖ || ‖ ‖ ϕs || ‖ else if ϕ > 0.9 || s ‖ || ‖ ‖ 0.9 | || ‖ ‖