3 DISEÑO DE LOSAS MACIZAS POR ACI

DISEÑO DE LOSAS MACIZAS POR EL METODO DE LOS COEFICIENTES DEL A.C.I. DATOS GENERALES f'c Ec fy Es

210 kg/cm2 217371 kg/cm2 4200 kg/cm2 2100000 kg/cm2

= = = =

= 2.50 cm = 2400 kg/m3 = 0.15m

recubrimiento ɣ.conc. h losa

DISEÑO DE LOSA MACIZA ARMADA EN 2 DIRECCIONES DEL SEMISOTANO CARGAS Peso propio

=

360.00 kg/m

acabados

=

100.00 kg/m

tabiqueria

=

96.66 kg/m

=

556.66 kg/m

Cm Wu.cm

=

Wu.cm

=

Wu

=

Wu

=

PAÑO 1:

s/c

1.4*Cm

Wu.cv

779.33 kg/m 1.4*Cm

CASO:

=

200.00 kg/m

Cv =

200.00 kg/m

=

Wu.cv

1.7*Cv =

340.00 kg/m

+ 1.7*Cv

1119.33 kg/m

4

B m

=

0.555828

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.400m

A

𝑀+ = 𝑊𝑢𝑚 ∗ 𝐿𝑛2 ∗ 𝐶𝑚 + 𝑊𝑢𝑣 ∗ 𝐿𝑛2 ∗ 𝐶𝑣

6.117m coeficientes para momentos negativos

coeficientes para momentos negativos: por carga muerta

m

CA

CB

m

CA

CB

0.600

0.089

0.011

0.600

0.053

0.007

0.556

0.092

0.008

0.556

0.056

0.005

0.550

0.092

0.008

0.550

0.056

0.005

coeficientes para momentos negativos: por carga viva m

CA

CB

0.600

0.067

0.009

0.556

0.071

0.007

0.550

0.072

0.007

momentos negativos MA(-) MB(-)

= =

1185.90 kg-m 349.71 kg-m

momentos positivos MA(+) MB(+)

= =

782.05 kg-m 302.94 kg-m

PAÑO 2:

CASO:

6

B m

=

0.666596

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.133m

A




m

CA

CB

m

CA

CB

0.700

0.091

0.000

0.700

0.051

0.009

0.667

0.092

0.000

0.667

0.053

0.008

0.650

0.093

0.000

0.650

0.054

0.007


CA

CB

0.700

0.060

0.013

0.667

0.063

0.011

0.650

0.064

0.010


1014.49 kg-m 0.00 kg-m

= =

PAÑO 3:

momentos positivos

CASO:

MA(+) MB(+)

= =

614.62 kg-m 271.88 kg-m

5

B m

=

0.570541

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.490m

A




m

CA

CB

m

CA

CB

0.600

0.088

0.000

0.600

0.037

0.003

0.571

0.089

0.000

0.571

0.038

0.002

0.550

0.089

0.000

0.550

0.038

0.002


CA

CB

0.600

0.059

0.007

0.571

0.061

0.006

0.550

0.063

0.005


= =

1207.78 kg-m 0.00 kg-m


= =

610.90 kg-m 243.83 kg-m

PAÑO 4:

CASO:

9

B m

=

0.78256

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

4.200m

A




m

CA

CB

m

CA

CB

0.800

0.075

0.017

0.800

0.029

0.010

0.783

0.076

0.016

0.783

0.030

0.009

0.750

0.078

0.014

0.750

0.031

0.007


CA

CB

0.800

0.042

0.017

0.783

0.043

0.016

0.750

0.046

0.013


1501.53 kg-m 514.37 kg-m

= =

PAÑO 5:

momentos positivos

CASO:

MA(+) MB(+)

= =

668.53 kg-m 503.13 kg-m

6

B m

=

0.898635

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.883m

A




m

CA

CB

m

CA

CB

0.900

0.079

0.000

0.900

0.039

0.021

0.899

0.079

0.000

0.899

0.039

0.021

0.850

0.083

0.000

0.850

0.042

0.017


CA

CB

0.900

0.042

0.025

0.899

0.042

0.025

0.850

0.046

0.022


= =

1335.12 kg-m 0.00 kg-m


= =

675.10 kg-m 520.76 kg-m

PAÑO 6:

CASO:

9

B m

=

0.487609

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

2.617m

A


losa en una direccion

5.367m

se tomara los coeficientes para m=0.5

coeficientes para momentos negativos


m

CA

CB

m

CA

CB

0.488

0.088

0.003

0.488

0.038

0.002


CA

CB

0.488

0.067

0.004


674.60 kg-m 96.73 kg-m

= =

PAÑO 7:

momentos positivos

CASO:

MA(+) MB(+)

= =

358.83 kg-m 128.97 kg-m

4

B m

=

0.650829

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.493m

A




m

CA

CB

m

CA

CB

0.700

0.081

0.019

0.700

0.046

0.011

0.651

0.085

0.015

0.651

0.050

0.009

0.650

0.085

0.015

0.650

0.050

0.009


CA

CB

0.700

0.057

0.014

0.651

0.062

0.011

0.650

0.062

0.011


= =

1159.94 kg-m 485.77 kg-m


= =

731.65 kg-m 356.26 kg-m

PAÑO 8:

CASO:

8

B m A

=

0.683764

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

1.533m 𝑀+ = 𝑊𝑢𝑚 ∗ 𝐿𝑛2 ∗ 𝐶𝑚 + 𝑊𝑢𝑣 ∗ 𝐿𝑛2 ∗ 𝐶𝑣



m

CA

CB

m

CA

CB

0.700

0.068

0.036

0.700

0.040

0.011

0.684

0.070

0.032

0.684

0.041

0.010

0.650

0.074

0.024

0.650

0.044

0.009


CA

CB

0.700

0.054

0.014

0.684

0.056

0.013

0.650

0.059

0.011


= =

momentos positivos

184.00 kg-m 180.63 kg-m

MA(+) MB(+)

= =

120.08 kg-m 73.29 kg-m

DIAGRAMA DE MOMENTOS DIRECCIÓN X-X

1014.49 kg-m RD.Mtos

FRANJA 5-6

349.71 kg-m

D

B'

A

302.94 kg-m

614.62 kg-m

cuando el momento negativo a un lado del apoyo sea menor que le 80% del momento en el otro lado la diferencia será distribuida en proporción a sus rigideces relativas, caso contrario se puede tomar el promedio REDISTRIBUCION DE MOMENTOS APOYO B' Mi =

349.707Kg.m

Li =

6.117m

Mj =

1014.494Kg.m

Lj =

3.133m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.163479 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

0.319183

𝑀′ 1 =

Δ𝑀 ∗ 𝐾𝑖 = 𝐾𝑖 + 𝐾𝑗

225.1654

𝑀′ 2 =

Δ𝑀 ∗ 𝐾𝑗 = 𝐾𝑖 + 𝐾𝑗

439.6223

664.788Kg.m

M1 = Mi + M'1

=

574.872Kg.m

M2 = Mj - M'2

=

574.872Kg.m

MOMENTOS FINALES FRANJA 5-6

574.87 kg-m

D

B'

A

302.94 kg-m

614.62 kg-m

DISEÑO POR FLEXION d=h-rec-ɸ/2

𝐴𝑠 =

Asmin =ƿmin*b*d 𝑆=

𝐴𝑏 ∗ 100 𝐴𝑠. 𝑟𝑒𝑞

asumiendo Ø d ƿmin

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

3/8''

= = =

𝐴𝑏 ∗ 100 𝑆

0.71 cm2

=

12.31 cm 0.0018

3/8'' tomando Ø entonces

As.min S @ 30.00 cm As.min amin Mu.min

se colocara solo acero minimo

Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm

FRANJA 5-4 D

B' 243.83 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

3/8'' asumiendo Ø = 12.31 cm d = 0.0018 ƿmin =

0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


1335.12 kg-m RD.Mtos FRANJA 4-3

514.37 kg-m

D

B'

A

503.13 kg-m

675.10 kg-m


514.374Kg.m

Li =

5.367m

Mj =

1335.116Kg.m

Lj =

3.883m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


344.534

𝑀′ 2 =


476.2076

820.742Kg.m

0.257533

M1 = Mi + M'1

=

858.908Kg.m

M2 = Mj - M'2

=

858.908Kg.m


858.91 kg-m

D

B'

A

503.13 kg-m

675.10 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

0.71 cm2

=

12.31 cm 0.0018




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


96.73 kg-m

D

B'

A

128.97 kg-m

120.08 kg-m


96.726Kg.m

Li =

5.367m

Mj =

184.000Kg.m

Lj =

1.533m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


19.39013

𝑀′ 2 =


67.88442

87.275Kg.m

0.652316

M1 = Mi + M'1

=

116.116Kg.m

M2 = Mj - M'2

=

116.116Kg.m


116.12 kg-m

D

B'

A

128.97 kg-m

120.08 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

0.71 cm2

=

12.31 cm 0.0018




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm

FRANJA 1-2 485.77 kg-m D

B' 356.26 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


DIRECCION Y-Y FRANJA D-B

1185.90 kg-m

NRD Mtos

1207.78 kg-m

1207.78 kg-m

6

5

782.05 kg-m

NRD Mtos

1501.53 kg-m

1501.53 kg-m

RD.Mtos

674.60 kg-m

674.60 kg-m

4'

1159.94 kg-m

3'

610.90 kg-m

RD.Mtos

2'

668.53 kg-m

1

358.83 kg-m

731.65 kg-m

cuando el momento negativo a un lado del apoyo sea menor que le 80% del momento en el otro lado la diferencia será distribuida en proporción a sus rigideces relativas, caso contrario se puede tomar el promedio REDISTRIBUCION DE MOMENTOS APOYO 3' Mi =

1501.530Kg.m

Li =

4.200m

Mj =

674.601Kg.m

Lj =

2.617m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.238095 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


-317.453

𝑀′ 2 =


-509.477

-826.929Kg.m

0.382117

M1 = Mi + M'1

=

1184.078Kg.m

M2 = Mj - M'2

=

1184.078Kg.m

APOYO 2' Mi =

674.601Kg.m

Li =

2.617m

Mj =

1159.936Kg.m

Lj =

3.493m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.382117 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


277.459

𝑀′ 2 =


207.8758

485.335Kg.m

0.286287

M1 = Mi + M'1

=

952.060Kg.m

M2 = Mj - M'2

=

952.060Kg.m

MOMENTOS FINALES FRANJA D-B 1196.84 kg-m

6

5

782.05 kg-m

1354.66 kg-m

1184.08 kg-m

4'

3'

610.90 kg-m

668.53 kg-m

358.83 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8'' 12.31 cm 0.0018

=

0.71 cm2

As.min

=

952.06 kg-m

2'

2.22 cm2

1

731.65 kg-m


@

S 30.00 cm As.min amin Mu.min

se colocara solo acero minimo PARA Mu

S PARA Mu

27.01 cm

=

30.00 cm

entonces

23.78 cm

=

@

0.71 cm2

=

= 1354.656Kg.m 0.71 cm a = 3.00 cm2 As = 3/8'' tomando Ø = S

32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m

3/8''

Ø

= 1196.842Kg.m 0.62 cm a = 2.64 cm2 As = 3/8'' tomando Ø =

=

Ø

3/8''

@

25.00 cm

Ø

3/8''

@

22.50 cm

0.71 cm2

= entonces

FRANJA B'-A B'

A 271.88 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=


As.min S 30.00 cm @ As.min amin Mu.min


Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


FRANJA B-A B

A 520.76 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


DISEÑO DE LOSA MACIZA ARMADA EN 2 DIRECCIONES DEL 1ER NIVEL CARGAS Peso propio

=

360.00 kg/m

acabados

=

100.00 kg/m

tabiqueria

=

214.65 kg/m

=

674.65 kg/m

Cm Wu.cm

=

Wu.cm

=

Wu

=

Wu

=

PAÑO 1:

s/c

1.4*Cm

Wu.cv

944.50 kg/m 1.4*Cm

CASO:

=

200.00 kg/m

Cv =

200.00 kg/m

=

Wu.cv

1.7*Cv =

340.00 kg/m

+ 1.7*Cv

1284.50 kg/m

4

B m

=

0.555828

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.400m

A




m

CA

CB

m

CA

CB

0.600

0.089

0.011

0.600

0.053

0.007

0.556

0.092

0.008

0.556

0.056

0.005

0.550

0.092

0.008

0.550

0.056

0.005


CA

CB

0.600

0.067

0.009

0.556

0.071

0.007

0.550

0.072

0.007


1360.90 kg-m 401.31 kg-m

= =

PAÑO 2:

momentos positivos

CASO:

MA(+) MB(+)

= =

888.31 kg-m 347.65 kg-m

6

B m

=

0.666596

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.133m

A




m

CA

CB

m

CA

CB

0.700

0.091

0.000

0.700

0.051

0.009

0.667

0.092

0.000

0.667

0.053

0.008

0.650

0.093

0.000

0.650

0.054

0.007


CA

CB

0.700

0.060

0.013

0.667

0.063

0.011

0.650

0.064

0.010


1164.20 kg-m 0.00 kg-m

= =

PAÑO 3:

momentos positivos

CASO:

MA(+) MB(+)

= =

700.56 kg-m 312.00 kg-m

5

B m

=

0.570541

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.490m

A




m

CA

CB

m

CA

CB

0.600

0.088

0.000

0.600

0.037

0.003

0.571

0.089

0.000

0.571

0.038

0.002

0.550

0.089

0.000

0.550

0.038

0.002


CA

CB

0.600

0.059

0.007

0.571

0.061

0.006

0.550

0.063

0.005


1386.01 kg-m 0.00 kg-m

= =

PAÑO 4:

momentos positivos

CASO:

MA(+) MB(+)

= =

686.52 kg-m 279.81 kg-m

9

B m

=

0.78256

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

4.200m

A




m

CA

CB

m

CA

CB

0.800

0.075

0.017

0.800

0.029

0.010

0.783

0.076

0.016

0.783

0.030

0.009

0.750

0.078

0.014

0.750

0.031

0.007


CA

CB

0.800

0.042

0.017

0.783

0.043

0.016

0.750

0.046

0.013


1723.11 kg-m 590.28 kg-m

= =

PAÑO 5:

momentos positivos

CASO:

MA(+) MB(+)

= =

755.06 kg-m 577.37 kg-m

6

B m

=

0.898635

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.883m

A




m

CA

CB

m

CA

CB

0.900

0.079

0.000

0.900

0.039

0.021

0.899

0.079

0.000

0.899

0.039

0.021

0.850

0.083

0.000

0.850

0.042

0.017


CA

CB

0.900

0.042

0.025

0.899

0.042

0.025

0.850

0.046

0.022


1532.14 kg-m 0.00 kg-m

= =

PAÑO 6:

momentos positivos

CASO:

MA(+) MB(+)

= =

772.43 kg-m 597.61 kg-m

9

B m

=

0.487609

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

2.617m

A


5.367m





m

CA

CB

m

CA

CB

0.488

0.088

0.003

0.488

0.038

0.002


CA

CB

0.488

0.067

0.004


774.15 kg-m 111.00 kg-m

= =

PAÑO 7:

momentos positivos

CASO:

MA(+) MB(+)

= =

401.82 kg-m 148.00 kg-m

4

B m

=

0.650829

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.493m

A




m

CA

CB

m

CA

CB

0.700

0.081

0.019

0.700

0.046

0.011

0.651

0.085

0.015

0.651

0.050

0.009

0.650

0.085

0.015

0.650

0.050

0.009


CA

CB

0.700

0.057

0.014

0.651

0.062

0.011

0.650

0.062

0.011


1331.10 kg-m 557.45 kg-m

= =

PAÑO 8:

momentos positivos

CASO:

MA(+) MB(+)

= =

832.29 kg-m 408.84 kg-m

8

B m A

=

0.683764

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖




m

CA

CB

m

CA

CB

0.700

0.068

0.036

0.700

0.040

0.011

0.684

0.070

0.032

0.684

0.041

0.010

0.650

0.074

0.024

0.650

0.044

0.009


CA

CB

0.700

0.054

0.014

0.684

0.056

0.013

0.650

0.059

0.011


= =

momentos positivos

211.15 kg-m 207.28 kg-m

MA(+) MB(+)

= =

136.11 kg-m 84.10 kg-m


1164.20 kg-m RD.Mtos

FRANJA 5-6

401.31 kg-m

D

B'

A

347.65 kg-m

700.56 kg-m

cuando el momento negativo a un lado del apoyo sea menor que le 80% del momento en el otro lado la diferencia será distribuida en proporción a sus rigideces relativas, caso contrario se puede tomar el promedio

REDISTRIBUCION DE MOMENTOS APOYO B' Mi =

401.312Kg.m

Li =

6.117m

Mj =

1164.200Kg.m

Lj =

3.133m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.163479 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


258.3924

𝑀′ 2 =


504.4962

762.889Kg.m

0.319183

M1 = Mi + M'1

=

659.704Kg.m

M2 = Mj - M'2

=

659.704Kg.m


659.70 kg-m

D

B'

347.65 kg-m

A

700.56 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

3/8''

= = =

𝐴𝑏 ∗ 100 𝑆

0.71 cm2

=

12.31 cm 0.0018




3/8''

Ø

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm

FRANJA 5-4 D

B' 279.81 kg-m

DISEÑO POR FLEXION d=h-rec-ɸ/2 Asmin =ƿmin*b*d

𝐴𝑠 =

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=


𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


=

0.71 cm2

As.min S 3/8'' @ 30.00 cm tomando Ø entonces As.min amin Mu.min 3/8'' se colocara solo acero minimo Ø

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @ 30.00 cm


590.28 kg-m

D

B'

577.37 kg-m

A

772.43 kg-m


590.279Kg.m

Li =

5.367m

Mj =

1532.135Kg.m

Lj =

3.883m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


395.376

𝑀′ 2 =


546.4804

941.856Kg.m

0.257533

M1 = Mi + M'1

=

985.655Kg.m

M2 = Mj - M'2

=

985.655Kg.m


985.65 kg-m

D

B'

A

577.37 kg-m

772.43 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

=

0.71 cm2

12.31 cm 0.0018


= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @ 30.00 cm


111.00 kg-m

D

B'

148.00 kg-m

A

136.11 kg-m


110.999Kg.m

Li =

5.367m

Mj =

211.153Kg.m

Lj =

1.533m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


22.25148

𝑀′ 2 =


77.90195

100.153Kg.m

0.652316

M1 = Mi + M'1

=

133.251Kg.m

M2 = Mj - M'2

=

133.251Kg.m


133.25 kg-m

D

B'

A

148.00 kg-m

136.11 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

0.71 cm2

=

12.31 cm 0.0018




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


B' 408.84 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm



1360.90 kg-m

NRD Mtos

1386.01 kg-m

1386.01 kg-m

6

5

888.31 kg-m

NRD Mtos

1723.11 kg-m

1723.11 kg-m

RD.Mtos

774.15 kg-m

774.15 kg-m

4'

1331.10 kg-m

3'

686.52 kg-m

RD.Mtos

2'

755.06 kg-m

1

401.82 kg-m

832.29 kg-m


1723.107Kg.m

Li =

4.200m

Mj =

774.150Kg.m

Lj =

2.617m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.238095 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


-364.298

𝑀′ 2 =


-584.659

-948.957Kg.m

0.382117

M1 = Mi + M'1

=

1358.809Kg.m

M2 = Mj - M'2

=

1358.809Kg.m

APOYO 2' Mi =

774.150Kg.m

Li =

2.617m

Mj =

1331.105Kg.m

Lj =

3.493m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.382117 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


318.4029

𝑀′ 2 =


238.5515

556.954Kg.m

0.286287

M1 = Mi + M'1

=

1092.553Kg.m

M2 = Mj - M'2

=

1092.553Kg.m


6

5

888.31 kg-m

1554.56 kg-m

1358.81 kg-m

4'

3'

686.52 kg-m

755.06 kg-m

401.82 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8'' 12.31 cm 0.0018

=

0.71 cm2

As.min

=

1092.55 kg-m

2'

2.22 cm2

1

832.29 kg-m


@

S 30.00 cm As.min amin Mu.min

se colocara solo acero minimo PARA Mu

S PARA Mu

23.44 cm

=

30.00 cm

entonces

20.63 cm

=

@

0.71 cm2

=

= 1554.559Kg.m 0.81 cm a = 3.45 cm2 As = 3/8'' tomando Ø = S

32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m

3/8''

Ø

= 1373.457Kg.m 0.72 cm a = 3.04 cm2 As = 3/8'' tomando Ø =

=

Ø

3/8''

@

22.50 cm

Ø

3/8''

@

20.00 cm

0.71 cm2

= entonces

FRANJA B'-A B'

A 312.00 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=


As.min S 30.00 cm @ As.min amin Mu.min


Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


FRANJA B-A B

A 597.61 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


DISEÑO DE LOSA MACIZA ARMADA EN 2 DIRECCIONES DEL 5TO NIVEL CARGAS Peso propio

=

360.00 kg/m

acabados

=

100.00 kg/m

tabiqueria

=

0.00 kg/m

=

460.00 kg/m

Cm Wu.cm

=

Wu.cm

=

Wu

=

Wu

=

PAÑO 1:

s/c

1.4*Cm

Wu.cv

644.00 kg/m 1.4*Cm

CASO:

=

100.00 kg/m

Cv =

100.00 kg/m

=

Wu.cv

1.7*Cv =

170.00 kg/m

+ 1.7*Cv

814.00 kg/m

4

B m

=

0.555828

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.400m

A




m

CA

CB

m

CA

CB

0.600

0.089

0.011

0.600

0.053

0.007

0.556

0.092

0.008

0.556

0.056

0.005

0.550

0.092

0.008

0.550

0.056

0.005


CA

CB

0.600

0.067

0.009

0.556

0.071

0.007

0.550

0.072

0.007


862.41 kg-m 254.31 kg-m

= =

PAÑO 2:

momentos positivos

CASO:

MA(+) MB(+)

= =

554.65 kg-m 220.31 kg-m

6

B m

=

0.666596

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.133m

A




m

CA

CB

m

CA

CB

0.700

0.091

0.000

0.700

0.051

0.009

0.667

0.092

0.000

0.667

0.053

0.008

0.650

0.093

0.000

0.650

0.054

0.007


CA

CB

0.700

0.060

0.013

0.667

0.063

0.011

0.650

0.064

0.010


737.76 kg-m 0.00 kg-m

= =

PAÑO 3:

momentos positivos

CASO:

MA(+) MB(+)

= =

439.64 kg-m 197.72 kg-m

5

B m

=

0.570541

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.490m

A




m

CA

CB

m

CA

CB

0.600

0.088

0.000

0.600

0.037

0.003

0.571

0.089

0.000

0.571

0.038

0.002

0.550

0.089

0.000

0.550

0.038

0.002


CA

CB

0.600

0.059

0.007

0.571

0.061

0.006

0.550

0.063

0.005


878.33 kg-m 0.00 kg-m

= =

PAÑO 4:

momentos positivos

CASO:

MA(+) MB(+)

= =

421.90 kg-m 177.32 kg-m

9

B m

=

0.78256

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

4.200m

A




m

CA

CB

m

CA

CB

0.800

0.075

0.017

0.800

0.029

0.010

0.783

0.076

0.016

0.783

0.030

0.009

0.750

0.078

0.014

0.750

0.031

0.007


CA

CB

0.800

0.042

0.017

0.783

0.043

0.016

0.750

0.046

0.013


1091.95 kg-m 374.06 kg-m

= =

PAÑO 5:

momentos positivos

CASO:

MA(+) MB(+)

= =

467.50 kg-m 365.89 kg-m

6

B m

=

0.898635

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.883m

A




m

CA

CB

m

CA

CB

0.900

0.079

0.000

0.900

0.039

0.021

0.899

0.079

0.000

0.899

0.039

0.021

0.850

0.083

0.000

0.850

0.042

0.017


CA

CB

0.900

0.042

0.025

0.899

0.042

0.025

0.850

0.046

0.022


970.93 kg-m 0.00 kg-m

= =

PAÑO 6:

momentos positivos

CASO:

MA(+) MB(+)

= =

487.42 kg-m 378.71 kg-m

9

B m

=

0.487609

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

2.617m

A


5.367m





m

CA

CB

m

CA

CB

0.488

0.088

0.003

0.488

0.038

0.002


CA

CB

0.488

0.067

0.004


490.59 kg-m 70.34 kg-m

= =

PAÑO 7:

momentos positivos

CASO:

MA(+) MB(+)

= =

245.61 kg-m 93.79 kg-m

4

B m

=

0.650829

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖

3.493m

A




m

CA

CB

m

CA

CB

0.700

0.081

0.019

0.700

0.046

0.011

0.651

0.085

0.015

0.651

0.050

0.009

0.650

0.085

0.015

0.650

0.050

0.009


CA

CB

0.700

0.057

0.014

0.651

0.062

0.011

0.650

0.062

0.011


843.53 kg-m 353.26 kg-m

= =

PAÑO 8:

momentos positivos

CASO:

MA(+) MB(+)

= =

520.78 kg-m 259.08 kg-m

8

B m A

=

0.683764

𝑀− = 𝑊𝑢 ∗ 𝐿𝑛2 ∗ 𝐶𝑖




m

CA

CB

m

CA

CB

0.700

0.068

0.036

0.700

0.040

0.011

0.684

0.070

0.032

0.684

0.041

0.010

0.650

0.074

0.024

0.650

0.044

0.009


CA

CB

0.700

0.054

0.014

0.684

0.056

0.013

0.650

0.059

0.011


= =

momentos positivos

133.81 kg-m 131.36 kg-m

MA(+) MB(+)

= =

84.73 kg-m 53.30 kg-m


737.76 kg-m RD.Mtos

FRANJA 5-6

254.31 kg-m

D

B'

A

220.31 kg-m

439.64 kg-m

cuando el momento negativo a un lado del apoyo sea menor que le 80% del momento en el otro lado la diferencia será distribuida en proporción a sus rigideces relativas, caso contrario se puede tomar el promedio

REDISTRIBUCION DE MOMENTOS APOYO B' Mi =

254.315Kg.m

Li =

6.117m

Mj =

737.763Kg.m

Lj =

3.133m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.163479 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


163.7454

𝑀′ 2 =


319.7033

483.449Kg.m

0.319183

M1 = Mi + M'1

=

418.060Kg.m

M2 = Mj - M'2

=

418.060Kg.m


418.06 kg-m

D

B'

220.31 kg-m

A

439.64 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

3/8''

= = =

𝐴𝑏 ∗ 100 𝑆

0.71 cm2

=

12.31 cm 0.0018




3/8''

Ø

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm

FRANJA 5-4 D

B' 177.32 kg-m

DISEÑO POR FLEXION d=h-rec-ɸ/2 Asmin =ƿmin*b*d

𝐴𝑠 =

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=


𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


=

0.71 cm2


= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @ 30.00 cm


374.06 kg-m

D

B'

365.89 kg-m

A

487.42 kg-m


374.064Kg.m

Li =

5.367m

Mj =

970.927Kg.m

Lj =

3.883m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


250.553

𝑀′ 2 =


346.3091

596.862Kg.m

0.257533

M1 = Mi + M'1

=

624.617Kg.m

M2 = Mj - M'2

=

624.617Kg.m


624.62 kg-m

D

B'

A

365.89 kg-m

487.42 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

=

0.71 cm2

12.31 cm 0.0018


2.22 cm2 32.15 cm

= =

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @ 30.00 cm


70.34 kg-m

D

B'

93.79 kg-m

A

84.73 kg-m


70.341Kg.m

Li =

5.367m

Mj =

133.809Kg.m

Lj =

1.533m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.186324 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


14.10094

𝑀′ 2 =


49.3671

63.468Kg.m

0.652316

M1 = Mi + M'1

=

84.442Kg.m

M2 = Mj - M'2

=

84.442Kg.m


84.44 kg-m

D

B'

A

93.79 kg-m

84.73 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8''

0.71 cm2

=

12.31 cm 0.0018




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


B' 259.08 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm



862.41 kg-m

NRD Mtos

878.33 kg-m

878.33 kg-m

6

5

554.65 kg-m

NRD Mtos

1091.95 kg-m

1091.95 kg-m

RD.Mtos

490.59 kg-m

490.59 kg-m

4'

843.53 kg-m

3'

421.90 kg-m

RD.Mtos

2'

467.50 kg-m

1

245.61 kg-m

520.78 kg-m


1091.947Kg.m

Li =

4.200m

Mj =

490.585Kg.m

Lj =

2.617m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.238095 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


-230.859

𝑀′ 2 =


-370.503

-601.362Kg.m

0.382117

M1 = Mi + M'1

=

861.088Kg.m

M2 = Mj - M'2

=

861.088Kg.m

APOYO 2' Mi =

490.585Kg.m

Li =

2.617m

Mj =

843.532Kg.m

Lj =

3.493m

ΔM

= Mj - Mi = 1 𝐾𝑖 = = 0.382117 𝐿𝑖 𝐾𝑗 =

1 = 𝐿𝑗

𝑀′ 1 =


201.7745

𝑀′ 2 =


151.172

352.947Kg.m

0.286287

M1 = Mi + M'1

=

692.360Kg.m

M2 = Mj - M'2

=

692.360Kg.m


6

5

554.65 kg-m

985.14 kg-m

861.09 kg-m

4'

3'

421.90 kg-m

467.50 kg-m

245.61 kg-m


𝐴𝑠 =




𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2

= = =

𝐴𝑏 ∗ 100 𝑆

3/8'' 12.31 cm 0.0018

=

0.71 cm2

As.min

=

692.36 kg-m

2'

2.22 cm2

1

520.78 kg-m


@


S 30.00 cm As.min amin Mu.min 3/8'' Ø

=

32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @ 30.00 cm

FRANJA B'-A B'

A 197.72 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=


𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm

FRANJA B-A B

A 378.71 kg-m


𝐴𝑠 =

Asmin =ƿmin*b*d

𝐴𝑏 ∗ 100 𝑆

𝑎𝑚𝑖𝑛 =

𝐴𝑠 ∗ 𝑓𝑦 0.85 ∗ 𝑓 ′ 𝑐 ∗ 𝑏

𝑆=

𝑎 𝑀𝑢. 𝑚𝑖𝑛 = 0.9 ∗ 𝐴𝑠 ∗ 𝑓𝑦 ∗ (𝑑 − ) 2


0.71 cm2

=




Ø

3/8''

= =

2.22 cm2 32.15 cm

2.38 cm2 = 0.56 cm = = 1080.36 kg-m @

30.00 cm


3 DISEÑO DE LOSAS MACIZAS POR ACI

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