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- 1. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . UNIVERSIDAD POLITÉCNICA SALESIANA NOMBRE:FREDDYGUSTAVO CUCHIPARTE PASTUÑA FECHA:16/01/2022 DISEÑO DE CORTANTE CORTANTES ≔ ϕc 0.75 ≔ VA = ――― ⋅ qu Ln1 2 20.337 tonnef ≔ VBi = ――――― ⋅ 1.15 qu Ln1 2 23.387 tonnef ≔ VBd = ――― ⋅ qu Ln2 2 21.963 tonnef ≔ VB = max⎛ ⎝ , VBi VBd ⎞ ⎠ 23.387 tonnef ≔ VCi = ――― ⋅ qu Ln2 2 21.963 tonnef ≔ VCd = ――――― ⋅ 1.15 qu Ln3 2 23.387 tonnef ≔ VC = max⎛ ⎝ , VCi VCd ⎞ ⎠ 23.387 tonnef ≔ VD = ――― ⋅ qu Ln3 2 20.337 tonnef CORTANTE MODIFICADOS = Fc 1.102 ≔ VAm = ⋅ VA Fc 22.416 tonnef ≔ VBm = ⋅ VB Fc 25.778 tonnef ≔ VCm = ⋅ VC Fc 25.778 tonnef ≔ VDm = ⋅ VD Fc 22.416 tonnef ≔ VMAX = max⎛ ⎝ , , , VAm VBm VCm VDm ⎞ ⎠ 25.778 tonnef ≔ d = d2existente 47.886 cm ≔ Ln = max( ( , , Ln1 Ln2 Ln3) ) 8.1 m ≔ ϕLmemor = ϕLB2 20 mm
- 2. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . ≔ ϕLmemor = ϕLB2 20mm ≔ z = ―― VMAX ―― Ln 2 d2existente 3.048 tonnef d2existente es el menor de todos ≔ Vu = - VMAX z 22.73 tonnef ≔ λ 0.85 Hormigón liviano Hormigón de peso normal ≔ λ 1 ≔ Vc = ⋅ ⋅ ⋅ 0.53 λ ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d2existente 11.795 tonnef ≔ Vs = - ―― Vu ϕc Vc 18.511 tonnef ≔ Av_s_diseño = ――――― - ―― Vu ϕc Vc ⋅ fy d2existente 0.092 ―― cm2 cm Acero mínimo a cortante ≔ Av_s_min1 = ⋅ ⋅ 0.2 ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 ― b fy 0.022 ―― cm2 cm ―― Av s min1 ≔ Av_s_min2 = ⋅ ―― ⋅ 3.5 b fy ―― kgf cm2 0.025 ―― cm2 cm ―― Av s min2 ≔ Av_s_min = max( ( , Av_s_min1 Av_s_min2) ) 0.025 ―― cm2 cm ―― Av s min
- 3. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Av/s diseño ≔ Av_s = max( (, Av_s_diseño Av_s_min) ) 0.092 ―― cm2 cm ―― Av s diseño 1 = ≤ Vu ⋅ ― 1 2 ϕc Vc 0 2 = ≤ Vu ⋅ ϕc Vc 0 3 = > Vu ⋅ ϕc Vc 1 3.1 = ≤ Vs ⋅ ⋅ ⋅ 1.1 ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d 1 1:Estoy en el caso 3.1 3.2 = ≤ ≤ ⋅ ⋅ ⋅ 1.1 ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d Vs ⋅ ⋅ ⋅ 2.2 ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d 0 1:Estoy en el caso 3.2 3.3 = > Vs ⋅ ⋅ ⋅ 2.2 ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d 0 1:Estoy en el caso 3.3 ≔ Aest = ⋅ π ⎛ ⎜ ⎝ ―― ϕest 2 ⎞ ⎟ ⎠ 2 0.785 cm2 = ϕest 10 mm ≔ #ramales 2 ≔ Av = ⋅ #ramales Aest 1.571 cm2 ≔ s = ――――― Av Av_s_diseño 17.066 cm ≔ d d2existente Zona Confinamiento (ACI 2019) ≔ s1 = ― d 4 11.971 cm = 2 h 110 cm ≔ s2 = ⋅ 8 ϕL 17.6 cm ≔ s3 = 24 ϕest 24 cm ≔ s4 30 cm ≔ smax = min( ( , , , s1 s2 s3 s4) ) 11.971 cm
- 4. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . ≔ s = min( ( , smaxs) ) 11.971 cm ≔ s = Trunc( ( , s 1 cm) ) 11 cm 1 = ϕest 10 mm @ = s 11 cm Zona Confinamiento (NEC 15) ≔ s1 = ― d 4 11.971 cm = 2 h 110 cm ≔ s2 = ⋅ 6 ϕL 13.2 cm ≔ s3 20 cm ≔ smax = min( ( , , s1 s2 s3) ) 11.971 cm ≔ s = min( ( , smax s) ) 11 cm ≔ s = Trunc( ( , s 1 cm) ) 11 cm 1 = ϕest 10 mm @ = s 11 cm Zona No confinada ≔ s = ― d 2 23.943 cm ≔ s = Trunc( ( , s 1 cm) ) 23 cm 1 = ϕest 10 mm @ = s 23 cm CORTE POR CAPACIDAD ( ROTULAS PLASTICAS )
- 5. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . CORTE POR CAPACIDAD( ROTULAS PLASTICAS ) DATOS ≔ f'c 240 ―― kgf cm2 ≔ fy 4200 ―― kgf cm2 Dimenciones de la viga Dimenciones de columnas ≔ b 30 cm ≔ bc 50 cm ≔ bcB 50 cm ≔ h 55 cm ≔ hcA 50 cm ≔ hcB 50 cm ≔ rec 3 cm ≔ ϕest 10 mm ≔ ϕL 20 mm ≔ Lv 8.6 m dimensión de a eje de la columna ≔ d = - - - lv rec ϕest ―― ϕL 2 ? cm ≔ d 47.886 cm d correspondiente a doble capa de varillas CALCULO DE CORTANTES HIPERESTATICOS Sentido Antihorario ≔ #1 4 ≔ ϕ1 22 mm ≔ #2 3 ≔ ϕ2 20 mm ≔ AstA = + ⋅ ⋅ #1 π ⎛ ⎜ ⎝ ―― ϕ1 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ #2 π ⎛ ⎜ ⎝ ―― ϕ2 2 ⎞ ⎟ ⎠ 2 24.63 cm2 ≔ # 2 ≔ ϕ 22 mm ≔ AstB = ⋅ ⋅ π ⎛ ⎜ ⎝ ― ϕ 2 ⎞ ⎟ ⎠ 2 # 7.603 cm2
- 6. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . ≔ Ln = - - Lv ―― hcA 2 ―― hcB 2 8.1m ≔ a_traccionA = ――――― ⋅ ⋅ AstA 1.25 fy ⋅ ⋅ 0.85 f'c b 21.129 cm ≔ MprA = ⋅ ⋅ ⋅ AstA 1.25 fy ⎛ ⎜ ⎝ - d ―――― a_traccionA 2 ⎞ ⎟ ⎠ 48.26 ⋅ tonnef m ≔ a_traccionB = ――――― ⋅ ⋅ AstB 1.25 fy ⋅ ⋅ 0.85 f'c b 6.522 cm ≔ MprB = ⋅ ⋅ ⋅ AstB 1.25 fy ⎛ ⎜ ⎝ - d ――――― a_traccionB 2 ⎞ ⎟ ⎠ 17.812 ⋅ tonnef m ≔ Vpr1 = ――――― + MprA MprB Ln2 8.157 tonnef Sentido Horario SECCION B-B ≔ # 2 ≔ ϕ 22 mm ≔ AstA = ⋅ ⋅ π ⎛ ⎜ ⎝ ― ϕ 2 ⎞ ⎟ ⎠ 2 # 7.603 cm2 ≔ #1 4 ≔ ϕ1 22 mm ≔ #2 3 ≔ ϕ2 20 mm ≔ AstB = + ⋅ ⋅ #1 π ⎛ ⎜ ⎝ ―― ϕ1 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ #2 π ⎛ ⎜ ⎝ ―― ϕ2 2 ⎞ ⎟ ⎠ 2 24.63 cm2
- 7. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . ≔ ln = - - Lv ―― hcA 2 ―― hcB 2 8.1m ≔ a_traccionA = ――――― ⋅ ⋅ AstA 1.25 fy ⋅ ⋅ 0.85 f'c b 6.522 cm ≔ MprA = ⋅ ⋅ ⋅ AstA 1.25 fy ⎛ ⎜ ⎝ - d ―――― a_traccionA 2 ⎞ ⎟ ⎠ 17.812 ⋅ tonnef m ≔ a_traccionB = ――――― ⋅ ⋅ AstB 1.25 fy ⋅ ⋅ 0.85 f'c b 21.129 cm ≔ MprB = ⋅ ⋅ ⋅ AstB 1.25 fy ⎛ ⎜ ⎝ - d ――――― a_traccionB 2 ⎞ ⎟ ⎠ 48.26 ⋅ tonnef m ≔ Vpr2 = ――――― + MprA MprB Ln 8.157 tonnef ≔ Vpr = max( ( , Vpr1 Vpr2) ) 8.157 tonnef Corte Gravitacional - Isostatico ≔ quT 5.999 ――― tonnef m ≔ VA = ――― ⋅ quT Ln 2 24.296 tonnef = ―― Ln 2 4.05 m ≔ y = ⋅ ―― VA ―― Ln 2 d 2.873 tonnef = d 47.886 cm Cortante de diseño
- 8. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Cortante de diseño ≔ Vdiseño= + Vpr Vu 30.887 tonnef 1: Vc=0 0: ≔ Vc ⋅ ⋅ ⋅ 0.53 λ ‾‾‾‾‾‾‾ ⋅ f'c ―― kgf cm2 b d = ≥ Vpr ――― Vdiseño 2 0 = Vc 11.795 tonnef ≔ Av_s_diseño = ――――― - ――― Vdiseño ϕc Vc ⋅ fy d 0.146 ―― cm2 cm ≔ Aest = ⋅ π ⎛ ⎜ ⎝ ―― ϕest 2 ⎞ ⎟ ⎠ 2 0.008 ⋅ m ―― cm2 cm = ϕest 1 cm ≔ #ramales 2 ≔ Av = ⋅ #ramales Aest 1.571 cm2 ≔ s = ――――― Av Av_s_diseño 10.75 cm Zona Confinamiento (ACI 2019) ≔ s1 = ― d 4 11.972 cm = 2 h 110 cm ≔ s2 = ⋅ 8 ϕL 16 cm ≔ s3 = 24 ϕest 24 cm ≔ s4 30 cm ≔ smax = min( ( , , , s1 s2 s3 s4) ) 11.972 cm ≔ s = min( ( , smax s) ) 10.75 cm ≔ s = Trunc( ( , s 1 cm) ) 10 cm 1 = ϕest 10 mm @ = s 10 cm Zona Confinamiento (NEC 15) ≔ s1 = ― d 4 11.972 cm = 2 h 110 cm ≔ s2 = ⋅ 6 ϕL 12 cm ≔ s3 20 cm ≔ smax = min( ( , , s1 s2 s3) ) 11.972 cm ≔ s = min( ( , smax s) ) 10 cm ≔ s = Trunc( ( , s 1 cm) ) 10 cm 1 = ϕest 10 mm @ = s 10 cm
- 9. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Zona No confinada ≔ s= ― d 2 23.943 cm ≔ s = Trunc( ( , s 1 cm) ) 23 cm 1 = ϕest 10 mm @ = s 23 cm CASO PRECTICO CON VIGAS BANDA ≔ rec 3 cm ≔ ϕest 8 mm ≔ ϕL 12 mm ≔ d = - - - 25 cm rec ϕest ―― ϕL 2 20.6 cm Zona Confinamiento (ACI 2019) ≔ s1 = ― d 4 5.15 cm = 2 h 110 cm ≔ s2 = ⋅ 8 ϕL 9.6 cm ≔ s3 = 24 ϕest 19.2 cm ≔ s4 30 cm ≔ smax = min( ( , , , s1 s2 s3 s4) ) 5.15 cm ≔ s = min( ( , smax s) ) 5.15 cm ≔ s = Trunc( ( , s 1 cm) ) 5 cm 1 = ϕest 8 mm @ = s 5 cm Zona Confinamiento (NEC 15) ≔ s1 = ― d 4 5.15 cm = 2 h 110 cm ≔ s2 = ⋅ 6 ϕL 7.2 cm ≔ s3 20 cm ≔ smax = min( ( , , s1 s2 s3) ) 5.15 cm ≔ s = min( ( , smax s) ) 5 cm ≔ s = Trunc( ( , s 1 cm) ) 5 cm 1 = ϕest 8 mm @ = s 5 cm
- 10. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Zona No confinada ≔ s= ― d 2 10.3 cm ≔ s = Trunc( ( , s 1 cm) ) 10 cm 1 = ϕest 8 mm @ = s 10 cm
- 11. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . DISEÑO POR TORSION DATOS ≔ Tu⋅ 30 kN m ≔ #3 2 ≔ ϕL 22 mm ≔ #B1 2 = ΦLB1 22 mm ≔ #B2 3 = ΦLB2 20 mm ≔ Φt 0.75 ≔ Φc 0.75 requerido por Mu(-) ≔ As = + + ⋅ ⋅ 2 π ⎛ ⎜ ⎝ ―― ϕL 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ 2 π ⎛ ⎜ ⎝ ―― ΦLB1 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ 3 π ⎛ ⎜ ⎝ ―― ΦLB2 2 ⎞ ⎟ ⎠ 2 24.63 cm2 Caluculo Se necesita refuerzo de torsión ≔ Acp = ⋅ b h 227500 mm2 ≔ Pcp = 2 ( ( + b h) ) 1700 mm ≔ λ 0.85 Hormigón liviano ≔ λ 1 Hormigón de peso normal Compruebo si se puede obiviar los efectos de la torsión Para Estructura Isostática (T. Equilibrio) = ⋅ ―――――― ⋅ ⋅ Φt λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 12 ―― Acp2 Pcp 9.231 ⋅ kN m 0: No se cumple y se debe tomar en cuenta la torsión 1: Se puede obiar los efectos de la torsión = < Tu ⋅ ―――――― ⋅ ⋅ Φt λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 12 ―― Acp2 Pcp 0 Para Estructura Hiperestatica (T. Compatibilidad) = ⋅ ―――――― ⋅ ⋅ Φt λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 ⎛ ⎜ ⎝ ―― Acp2 Pcp ⎞ ⎟ ⎠ 36.925 ⋅ kN m 0: No se cumple y se debe tomar en cuenta la torsión 1: Se puede obiar los efectos de la torsión = < Tu ⋅ ―――――― ⋅ ⋅ Φt λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 ⎛ ⎜ ⎝ ―― Acp2 Pcp ⎞ ⎟ ⎠ 1
- 12. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . 0: No secumple y se debe tomar en cuenta la torsión 1: Se puede obiar los efectos de la torsión = < Tu ⋅ ―――――― ⋅ ⋅ Φt λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 ⎛ ⎜ ⎝ ―― Acp2 Pcp ⎞ ⎟ ⎠ 1 Calculo de las propiedades de la siccióin ≔ x1 = - - b ( ( ⋅ 2 rec) ) ϕest 230 mm ≔ y1 = - - h 2 rec ϕest 480 mm ≔ Aoh = ⋅ x1 y1 110400 mm2 ≔ Ph = 2 ( ( + x1 y1) ) 1420 mm ≔ Ao = ⋅ 0.85 Aoh 93840 mm2 ≔ d = d2existente 47.886 cm ¿Es la sección de hormigón suficientemente grande como para soportaar torsión? ≔ Vc = ⋅ ⋅ ――――― ⋅ λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 6 b d 116.156 kN ≔ Vc = ⋅ ⋅ ―――――― ⋅ 0.53 λ ‾‾‾‾‾‾‾ ⋅ f'c MPa 6 b d 95.638 kN Para secciónes solidas = ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + ⎛ ⎜ ⎝ ―― Vu ⋅ b d ⎞ ⎟ ⎠ 2 ⎛ ⎜ ⎝ ―――― ⋅ Tu Ph ⋅ 1.7 Aoh2 ⎞ ⎟ ⎠ 2 2.576 ―― N mm2 = + ⋅ Φc ―― Vc ⋅ b d ⋅ Φt ――――― ⋅ 2 ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 3.032 ―― N mm2 1: La sección SI es suficiente grande 0: La sección NO es suficiente grande = ≤ ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ + ⎛ ⎜ ⎝ ―― Vu ⋅ b d ⎞ ⎟ ⎠ 2 ⎛ ⎜ ⎝ ―――― ⋅ Tu Ph ⋅ 1.7 Aoh2 ⎞ ⎟ ⎠ 2 + ⋅ Φc ―― Vc ⋅ b d ⋅ Φt ――――― ⋅ 2 ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 1 Para secciones huecas = ≤ + ⎛ ⎜ ⎝ ―― Vu ⋅ b d ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ―――― ⋅ Tu Ph ⋅ 1.7 Aoh2 ⎞ ⎟ ⎠ + ⋅ Φc ―― Vc ⋅ b d ⋅ Φt ――――― ⋅ 2 ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 0 ≔ espesor 150 mm = ―― Aoh Ph 77.746 mm = < espesor ―― Aoh Ph 0 = ≤ + ⎛ ⎜ ⎝ ―― Vu ⋅ b d ⎞ ⎟ ⎠ ⎛ ⎜ ⎝ ―――――― Tu ⋅ ⋅ 1.7 Aoh espesor ⎞ ⎟ ⎠ + ⋅ Φc ―― Vc ⋅ b d ⋅ Φt ――――― ⋅ 2 ‾‾‾‾‾‾‾ ⋅ f'c MPa 3 1
- 13. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Determinación del refuerzotransversal requerido por torsión ≔ Tn = ―― Tu Φt 40 ⋅ kN m ≔ θ ° 45 No debe ser menor que 30° o mayor que 60° ≔ At_s = ―――――― Tn ⋅ ⋅ ⋅ 2 Ao fy cot( (θ) ) 0.517 ―― mm2 mm para 1 rama de los estribos Determinación del refuerzo transversal requerido por cortante = ≥ Vdiseño ⋅ ⋅ ― 1 2 Φc Vc 1 1: Estribo mecesario 0:no se necesita estribos ≔ Vs = ―――――― - Vdiseño ⋅ Φc Vc Φc 287.708 kN ≔ Av_s = ―― Vs ⋅ fy d 0.146 ―― cm2 cm para 2 ramas de estribos ≔ #ramales 2 = + ⋅ #ramales At_s Av_s 0.249 ―― cm2 cm ≔ Aest = ⋅ π ⎛ ⎜ ⎝ ―― ϕest 2 ⎞ ⎟ ⎠ 2 0.785 cm2 ≔ Av = ⋅ #ramales Aest 1.571 cm2 ≔ s = ―――――――― Av + #ramales At_s Av_s 6.299 cm
- 14. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Separación máxim permisibleentre estribos ≔ s1 = ―― Ph 8 177.5 mm ≔ s2 300 mm ≔ smax = min( ( , s1 s2) ) 177.5 mm Zona Confinamiento (NEC 15) ≔ s1 = ― d 4 11.971 cm = 2 h 110 cm ≔ s2 = ⋅ 6 ϕLB2 12 cm ≔ s3 20 cm ≔ smax = min( ( , , s1 s2 s3) ) 11.971 cm ≔ s = min( ( , smax s) ) 6.299 cm ≔ s = Trunc( ( , s 1 cm) ) 6 cm 1 = ϕest 10 mm @ = s 6 cm Zona No confinada = ⋅ #ramales At_s 0.103 ―― cm2 cm ≔ Aest = ⋅ π ⎛ ⎜ ⎝ ―― ϕest 2 ⎞ ⎟ ⎠ 2 0.785 cm2 ≔ Av = ⋅ #ramales Aest 1.571 cm2 ≔ s = ――――― Av #ramales At_s 15.178 cm ≔ smax = ― d 2 23.943 cm ≔ s = min( ( , smax s) ) 15.178 cm ≔ s = Trunc( ( , s 1 cm) ) 15 cm 1 = ϕest 10 mm @ = s 15 cm
- 15. C r e a d o c o n P T C M a t h c a d E x p r e s s . C o n s u l t e w w w . m a t h c a d . c o m p a r a o b t e n e r m á s i n f o r m a c i ó n . Área mínima deestribos = ⋅ ⋅ ― 1 16 ‾‾‾‾‾‾‾ ⋅ f'c MPa ―― ⋅ b s fy 33.127 mm2 = ⋅ ― 1 3 ―― ⋅ b s fy MPa 36.418 mm2 = < ⋅ ⋅ ― 1 16 ‾‾‾‾‾‾‾ ⋅ f'c MPa ―― ⋅ b s fy ⋅ ― 1 3 ―― ⋅ b s fy MPa 1 Selección del refuerzo longitudinal por torsión ≔ A_L = ⋅ ⋅ ⋅ At_s Ph ⎛ ⎜ ⎝ ― fy fy ⎞ ⎟ ⎠ ( (cot( (θ) )) ) 2 734.785 mm2 = h 55 cm ≔ #lechos 3 ≔ espaciamiento = ―――――――――― - - - h ⋅ 2 rec ϕest ⋅ #lechos ϕL - #lechos 1 20.7 cm = < espaciamiento 30 cm 1 ≔ ALecho1 = + ―― A_L 3 As 27.079 cm2 ≔ ALecho2 = ―― A_L 3 2.449 cm2 ≔ #3 2 ≔ ϕL 22 mm ≔ #B1 2 = ΦLB1 22 mm ≔ ALecho3 = ―― A_L 3 2.449 cm2 ≔ #B2 3 = ΦLB2 20 mm Lecho 1 ≔ #1 1 ≔ ϕ1 20 mm ≔ A1 = + + + ⋅ ⋅ π ⎛ ⎜ ⎝ ―― ϕ1 2 ⎞ ⎟ ⎠ 2 #1 ⋅ ⋅ 2 π ⎛ ⎜ ⎝ ―― ϕL 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ 2 π ⎛ ⎜ ⎝ ―― ΦLB1 2 ⎞ ⎟ ⎠ 2 ⋅ ⋅ 3 π ⎛ ⎜ ⎝ ―― ΦLB2 2 ⎞ ⎟ ⎠ 2 27.772 cm2 = ≤ ALecho1 A1 1 Lecho 2 ≔ #2 2 ≔ ϕ2 14 mm ≔ A2 = ⋅ π ⎛ ⎜ ⎝ ―― ϕ2 2 ⎞ ⎟ ⎠ 2 #2 3.079 cm2 = ≤ ALecho2 A2 1 Lecho 3 ≔ #3 2 ≔ ϕ3 ϕL ≔ A3 = ⋅ ⋅ π ⎛ ⎜ ⎝ ―― ϕ3 2 ⎞ ⎟ ⎠ 2 #3 7.603 cm2 = ≤ ALecho3 A3 1