Viga T.pdf

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1. Determinar el momento momento resistente nominal de la sección transversal de la viga que se muestra en la figura si f'c=210 kg/cm2 y fy=4200 kg/cm2. Analizar dos casos: (a) hf=10cm y (b) hf=15cm.

f'c ≔ 210 fy ≔ 4200 bw ≔ 20 b ≔ 50 hf ≔ 10 ϕ ≔ 0.90

Area total de acero en traccion : As ≔ 4 ⋅ 5.1 + 2 ⋅ 2 = 24.4 Peralte efectivo de la seccion: d ≔ 60 − 12 = 48 Para determinar si se trata efectivamente de una sección T se estima un valor de:

As ⋅ fy = 11.482 a ≔ ―――― 0.85 f'c ⋅ b (a) Caso I: a>hf=10cm, en este caso la seccion es T ⎛ 6000 ⎞ cb ≔ ⎜―――⎟ ⋅ d ⎝ 6000 + fy ⎠

0.65≤ β ≤0.85 1

⎛ f'c − 280 ⎞ β1 ≔ 0.85 − 0.05 ⋅ ⎜――― ⎟ = 0.9 ⎝ 70 ⎠

β1 ≔ 0.85

ab ≔ β1 ⋅ cb = 24 ⎛⎝ab ⋅ bw + ⎛⎝b − bw⎞⎠ ⋅ hf⎞⎠ = 0.0138 ρbt ≔ 0.85 ⋅ f'c ⋅ ――――――― fy ⋅ b ⋅ d Cuantia maxima:

ρmax ≔ 0.75 ⋅ ρbt = 0.0104

Cuantia colocada en traccion:

As = 0.0102 ρ ≔ ―― b⋅d

como ρ < ρmax entonces se consideran en forma separada efectos del ala y del alma Primer efecto: Acero en traccion equilibrado por la compresion en el ala 0.85 ⋅ f'c ⋅ hf ⋅ ⎛⎝b − bw⎞⎠ = 12.75 Asf ≔ ――――――― fy ⎛ hf ⎞ Mnf ≔ 0.85 ⋅ f'c ⋅ ⎛⎝b − bw⎞⎠ ⋅ hf ⋅ ⎜d − ―⎟ = 2302650 2⎠ ⎝ Segundo efecto: Acero en traccion equilibrado por la compresion en el alma Asw ≔ As − Asf = 11.65 Asw ⋅ fy = 13.706 a ≔ ―――― 0.85 f'c ⋅ bw

Asw ≔ As − Asf = 11.65 Asw ⋅ fy = 13.706 a ≔ ―――― 0.85 f'c ⋅ bw ⎛ a⎞ Mnw ≔ 0.85 ⋅ f'c ⋅ bw ⋅ a ⋅ ⎜d − ― ⎟ = 2013325.588 2⎠ ⎝ La resistencia nominal de la seccion a la flexion sera: Mn ≔ Mnf + Mnw = 4315975.588 El momento ultimo de la seccion sera: Mu ≔ ϕ ⋅ Mn = 3884378.029 T-cm (a) Caso II: a
2. La viga que se muestra en la figura forma parte de un sistema de vigas espaciadas a 3 m. que sostienen una losa de 15 cm. de espesor. La losa soporta una carga permanente de 210 kg/m2 y una sobrecarga de 300 kg/m2. Diseñar la sección central de la viga. Usar f'c=210 kg/cm2 y fy=4200 kg/cm2

Metrado de cargas: Peso propio de la viga: Peso propio de la losa: Carga permanente sobre la losa:

f'c ≔ 210 fy ≔ 4200 bw ≔ 25 b ≔ 150 hf ≔ 15 ϕ ≔ 0.90 L ≔ 6.00 s ≔ 3.00 h ≔ 40 D1 ≔ 2400 ⋅ 0.25 ⋅ 0.4 = 240 D2 ≔ 2400 ⋅ 0.15 ⋅ 2.75 = 990 D3 ≔ 210 ⋅ 3 = 630

Sobrecarga sobre la losa:

L1 ≔ 300 ⋅ 3 = 900

Wu ≔ 1.4 ⋅ ⎛⎝D1 + D2 + D3⎞⎠ + 1.7 ⋅ L1 = 4134 Wu ⋅ L 2 Mu ≔ ――― = 18603 8 El ancho de losa que contribuye con la viga para resistir las cargas aplicadas es el menor de: b ≔ bw + 16 hf = 265 b≔s=3 L b ≔ ―= 1.5 4 Peralte efectivo de la viga:

b ≔ 150 d ≔ h − 6 = 34

Suponiendo que sólo el ala contribuye a la resistencia a la compresión, el estimado inicial del área de acero requerida es: d a ≔ ―= 6.8 5

Mu ⋅ 10 2 As ≔ ――――― = 16.083 ⎛ a⎞ ϕ ⋅ fy ⋅ ⎜d − ― ⎟ 2⎠ ⎝

As ⋅ fy = 2.523 a ≔ ―――― 0.85 f'c ⋅ b

Mu ⋅ 10 2 As ≔ ――――― = 15.033 ⎛ a⎞ ϕ ⋅ fy ⋅ ⎜d − ― ⎟ 2⎠ ⎝

As ⋅ fy = 2.358 a ≔ ―――― 0.85 f'c ⋅ b

Mu ⋅ 10 2 As ≔ ――――― = 14.995 ⎛ a⎞ ϕ ⋅ fy ⋅ ⎜d − ― ⎟ 2⎠ ⎝

As ⋅ fy = 2.352 a ≔ ―――― 0.85 f'c ⋅ b

Usar: 3Φ#8

Mu ⋅ 10 2 As ≔ ――――― = 14.993 ⎛ a⎞ ϕ ⋅ fy ⋅ ⎜d − ― ⎟ 2⎠ ⎝

A's ≔ 3 ⋅ 5.07 = 15.21

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