DIESÑO ELASTICO

ϵckd

= ϵsd−kd

ϵckd

= ϵsd (1−k )

fcϵckd

=

fsϵs

d (1−k )fcϵc

=fs

ϵs (1−k )

Despejar

fc= fs∗ϵc∗kdϵs∗d (1−k )

fc= fs∗kn (1−k )

fcfs

= kn (1−k )

fsfc

=r

1r= kn (1−k )

nr= k1−k

rn=1−k

k

rn=1k− kk

rn=1k−1

rn+1=1

k

r+nn

=1k

nr−n

=kbal

d= k3d+ jd

1= k3+ j

J=1− k3

C=T

fc∗kd2

∗b= fs∗As

ρ= Asbd

fc∗kd∗b2

= fs∗ρ∗b∗d

fc∗kd2

=fs∗ρ

ρbal=fc∗k2∗fs

ρbal=k2 r

Momento resistente del concreto

MC=fc∗kd∗b

2×Jd= fc∗kd∗J d2

2=Mc

Momento resistente del acero

M s=fs∗ρ∗b∗d∗ j∗d=fs∗ρ∗b∗ j∗d2=M S

M=K∗b∗d2

KC=fc∗k2

∗ j

KS=fs∗ρ∗ j

√d2=√ MKb

√d=√ MKb

d=K2√Mb√Mb =K2

METODO DE FLEXION

Y=3×3.87cm2×(4cm+1cm+ 12×78×2.54 cm)

+2×2.84 cm2×(4 cm+1cm+ 12×34×2.54cm)

3×3.87cm2+2×2.89cm3

y=6.04cm

Y=7×1.27×(5cm+ 12×12×2.54 cm)

+7×1.27×(5cm+ 12×2.54+5+ 1

2×12×2.54)

14×1.27

d=h− y

d=53.60cm

Ms=110 KNM

Mc=105 KNM

M res=M act

105KN .m=WR ¿¿

WR=105¿¿

Mc=12fc∗k∗ j∗b∗d2

Ms=T∗ j∗d=As∗Fs∗ j∗d

J=1−K3

J=1−5.7073

J=17.234

Mc=12∗21∗51.707∗17234

n=200000Mpa33900

n=51.282

k=−np+√¿¿

51282∗0.83+√¿¿

k=51707

Ejemplo

F´c=21Mpa

h≤2b

hm×l6=36≈0.4m

l18.5

W = 3.7 T/m

W = 5.75m

Mact=w l2

8

Mact=3.7 Tm× ¿¿

Mact=15.35m

k= M

bd2

k=153Kn∗m0.25∗¿¿

ρ K

0.006 923.7

0.007 1070.20

0.008 1215.6

0.0090 1359.1

1359.1−1215.60.001

=1255−1215.6X

x=0.00027

ρ=0.00827

As=pbd

As=0.00827×35cm×59cm=17. .08cm2

Cuantas varillas?

AsAt

=17.081.27

=14 varillas

Cuantía

ρreal= Asrealbd

= 17.0835×59

=0.0083

Ejercicio 2

F´c = 21MPa

Fy = 35MPa

L = 6m

ω=¿ ?

Altura útil

d=h− y

y=Σ(Ay)ΣA

Y=4×2.84×(5cm+ 12×34×2.54cm)

+4×1.19cm2×(5cm+ 12×82×2.54)

4×2.84 cm2+4×1.99cm2

Cuantía

ρ= Asba

= 19.32cm2

25cm×41.27cm=0.018

0.009555=8×2.84+4×1.99cm35cm×d

h=105cmb=55cm

p=8×2.84+4×1.99cm2

35cm×(194 cm)

p=0.0059<Pbal

k=916.5KN /M 2

K=M

w=105.3KN /m

w=105T /m

Mc=12fcbKdjd=1

2fcbkj d2

Ms=Asfcjd=ρddfsdd=ρbd2 jfs

Deducción de ecuación

Relación de triángulos

Dibujo

ϵckd

= ϵcd−kd

= ϵs ´kd−d ´

Fcϵckd

=

Fsϵs

d (1−k )=

Fs ´ϵs

kd−k ´

fcϵc∗kd

= ϵcϵs∗d (1−k )

= fs´ϵs(kd−d´ )

fs´ϵs(kd−d ´ )

= fcϵc∗kd

fs´=ϵs (kd−d ´ ) fc

ϵc∗kd

fs´=n (kd−d ´ ) fc

kd≤ fs

fs´ϵs(kd−d ´ )

= fcϵs∗d (1−k )

fs´=(kd−d´ )ϵsd (1−k)

≤ fs

Momento actual

Mact=Mbal+∆M Acero en compresión

Mact−Mbal=∆M

M=F∗d

M=fsAsjd

M=fsρbdjd

Ms=fspbjd2=Mρ

Área de acero

As ´= ∆Mfs ´ (d−d ´ )

Viga

Apoyo continuo

l12

=6m12

=0.5

Voladizo

l5=6m12

=0.5

h=0.5b=30

30

50cm

f ´ c=21Mpa

fy=420Mpa

Momento máximo = 300KN.M

Hallar K

K=300kn .m0.3× ¿¿

Momento para cuantía balanceada

M 1bal=(170000Mpa)(0.009555)(0.3m)(0.8860)¿

M 1bal=83.52kn .m

∆M=300Kn .m−83.58Kn.m

∆M=216.42Kn .m

As ´= ∆Mfs ´ (d−d ´ )

→As´= 216.42Kn .m

(71025Kn /m2)(0.44m−0.09m)

2 As ´=(0.3420×0.44m−0.09m )×170000 kn/m2

0.44m(1−0.3920)

2 As ´=71025kn /m2

As ´=0.0087m2→87.05cm2

Asρbal=0.009555×30cm×44 cm

Asρbal=12.61cm2

Viga en T

2.1

0.85 0.40 0.85

F ´ c=24.5Mpa

Fy=350Mpa

Hallas x

(2.1−1.7 )=x

A∈∪→ Areaestatica util

b x2

2=nAs(d− x)

Figura por figura

2.1×0.12× ( x−0.06m )+0.4×(x−0.12)( x−0.122

)→acero

¿10.36×3.87 cm2(71cm−x )

210×12× ( x−6m )+20׿

250×−1520+20 X2−480 x−2880=22772.5−320.74 x

0.12

20 X2+2360.74 x−40772=0

x=15.3cm

Ix−x= (20 X2+2040 x−18000 ) 23x+320.74¿

Ix−x=1.342849cm4/¿

Momentos

Mc= fc∗Ic−xx

=11.02×100

T

m2×0.011m4

0.153=79.22T .m

Ms= fs∗Ic−xx

=170×100

T

m2×0.011m4

10.36 (0.71−0.15 )m=32.40T .m

→se escoge elmomentomenor

M=W l2

8

W=5T /m

W=5T /m

W pp= (0.12×2.1+0.68×0.4 )×2.4T /m3

Wpp=1.25T /m

W ext=5 Tm

−1.25 Tm

=3.73T /m

Sección transformada u homogénea

Fc

As Fs,ϵs

ϵs=ϵ t

fsϵs

= ftϵ c

F s= f ϵ ϵsϵc

LFS=∩ ft

FS=f s /∩

T s=T t

As*Fs=At* ft

Asfs=Atfsn

LAS=n∗As

r= fsn

fc=Mc× xIx−×

x

d-x