Hormigón Armado 1

HORMIGÓN ARMADO 1

TRABAJO N. 1

EJERCICIO N. 1

MODIFICAR VALOR DE f´c de (210 a 350) kg

cm2

DATOS:

As=(2φ28+4φ25 )+ (4 φ25+2φ22 )+(6φ22)

f ' c=350kg

cm2; f y=4200

kg

cm2

rec . libre=4cm

φde estribo=10mm

Determinar→Mn=? ?

Calculo de y

Hormigón Armado 1

As . y=As1∗y1+As2∗y2+As3∗ y3+As4∗y 4+As5∗y5

y=∑i=1

n

(As i∗y i)

∑i=1

n

(Asi)

As1=2φ28=2∗6.16 cm2=12.32cm2

y1=rec .libre+φE+ φv2

y1=4+1+2.82

=6.4 cm

As2=4φ25=4∗4.91cm2=19.64cm2

y2=4+1+2.52

=6.25cm

As3=4φ25=4∗4.91cm2=19.64cm2

y3=rec .libre+φE+φ28+φsep .+ φv2

y3=4+1+2.8+2.5+2.52

=11.55cm2

As4=2φ22=2∗3.80cm2=7.60cm2

y4=4+1+2.8+2.5+2.22

=11.40cm

As5=6 φ22=6∗3.80cm2=22.80cm2

y5=4+1+2.8+2.5+2.5+2.5+2.2 /2=16.40cm

Hormigón Armado 1

y=(12.32 cm2∗6.4 cm )+(19.64 cm2∗6.25cm )+(19.64cm2∗11.55 cm )+(7.60cm2∗11.40 cm )+(22.80cm2∗16.40cm )

(12.32+19.64+19.64+7.60+22.80 ) cm2

y=10.84 cm

Diagrama:

∑ fx=0.0

Cc−Ts=0.0

Cc=Ts

Mn=Cc∗z=Ts∗z

Cc=0.85∗f ´ c∗(AEHC )

AEHC=a∗b

Cc=0.85∗f ´ c∗a∗b

Ts=As∗fs

z=d− y

z=d−a2

Hormigón Armado 1

Calculo de a

0.85∗f ´ c∗a∗b=As∗fs

Asumimos que fs=fy

a= As∗fs0.85∗f ´ c∗b

a¿= As∗fs0.85∗f ´ c∗b

(¿ )Valores condicionado para la verificaciónde fs=fy

a¿=82.0cm2∗4200 kg

cm2

0.85∗350 kgcm2∗45cm

a¿=25.725cm

Calculo de ε s

ε s=εcuc

∗(d−c )

a=β1∗c

c= aβ1

c¿= a¿

β1

c¿=25.725 cm0.80

;( f ´ c>280∴ β1=0.85−0.05)

c¿=32.156 cm

ε s¿= 0.003∗¿32.156cm

∗(79.16−32.156 )cm ¿

ε s¿=0.00438mm

mm

Hormigón Armado 1

ε y=f yE s

ε y=4200

kg

cm2

2.1x 106kgcm2

ε y=0.002mmmm

ε s¿vs . ε y

{ εs¿>ε y

0.00438>0.002}❑⇒

ZONA PLASTICA

fs=fy=4200 kg

cm2

{a¿ , c¿ , εs¿ }SON VALORESVERDADEROS

CALCULODECc yTs

Cc=0.85∗f ´ c∗a∗b

Cc=0.85∗350 kg

cm2∗25.725cm∗45.00cm

Cc=344393.4kg

Ts=As∗fs

Ts=80.0cm2∗4200 kg

cm2

Ts=336000 kg

Hormigón Armado 1

Cc≈TsO. K

CALCULODEMn

Mn=Cc∗z

z=d−a2

z=79.16cm−25.725cm2

=66.298cm

Mn=344393.4kg∗66.298 cm

Mn=22832593.63kg−cm

Mn=228.326T−m

EJERCICIO N. 2

MODIFICAR VALOR DE f´c de (210 a 350) kg

cm2 y modificar la altura h de 90 a 110 cm

DATOS:

As=(2φ28+4φ25 )+ (4 φ25+2φ22 )+(6φ22)

f ' c=350kg

cm2; f y=4200

kg

cm2

rec . libre=4cm

φde estribo=10mm

Determinar→Mn=? ?

Hormigón Armado 1

Calculo de y

As . y=As1∗y1+As2∗y2+As3∗ y3+As4∗y 4+As5∗y5

y=∑i=1

n

(As i∗y i)

∑i=1

n

(Asi)

As1=2φ28=2∗6.16 cm2=12.32cm2

y1=rec .libre+φE+ φv2

y1=4+1+2.82

=6.4 cm

As2=4φ25=4∗4.91cm2=19.64cm2

y2=4+1+2.52

=6.25cm

As3=4φ25=4∗4.91cm2=19.64cm2

y3=rec .libre+φE+φ28+φsep .+ φv2

Hormigón Armado 1

y3=4+1+2.8+2.5+2.52

=11.55cm2

As4=2φ22=2∗3.80cm2=7.60cm2

y4=4+1+2.8+2.5+2.22

=11.40cm

As5=6 φ22=6∗3.80cm2=22.80cm2

y5=4+1+2.8+2.5+2.5+2.5+2.2 /2=16.40cm

y=(12.32cm2∗6.4 cm )+(19.64cm2∗6.25cm )+(19.64cm2∗11.55 cm )+(7.60cm2∗11.40 cm )+(22.80cm2∗16.40cm )

(12.32+19.64+19.64+7.60+22.80 ) cm2

y=10.84 cm

Diagrama:

∑ fx=0.0

Hormigón Armado 1

Cc−Ts=0.0

Cc=Ts

Mn=Cc∗z=Ts∗z

Cc=0.85∗f ´ c∗(AEHC )

AEHC=a∗b

Cc=0.85∗f ´ c∗a∗b

Ts=As∗fs

z=d− y

z=d−a2

Calculo de a

0.85∗f ´ c∗a∗b=As∗fs

Asumimos que fs=fy

a= As∗fs0.85∗f ´ c∗b

a¿= As∗fs0.85∗f ´ c∗b

(¿ )Valores condicionado parala verificaciónde fs=fy

a¿=82.0cm2∗4200 kg

cm2

0.85∗350 kgcm2∗45cm

a¿=25.725cm

Calculo de ε s

ε s=εcuc

∗(d−c )

a=β1∗c

Hormigón Armado 1

c= aβ1

c¿= a¿

β1

c¿=25.725 cm0.80

;( f ´ c>280∴ β1=0.85−0.05)

c¿=32.156 cm

ε s¿= 0.003∗¿32.156cm

∗(99.16−32.156 ) cm¿

ε s¿=0.00625mm

mm

ε y=f yE s

ε y=4200

kg

cm2

2.1x 106kgcm2

ε y=0.002mmmm

ε s¿vs . ε y

{ εs¿>ε y

0.00625>0.002}❑⇒

ZONA PLASTICA

fs=fy=4200 kg

cm2

{a¿ , c¿ , εs¿ }SON VALORESVERDADEROS

Hormigón Armado 1

CALCULODECc yTs

Cc=0.85∗f ´ c∗a∗b

Cc=0.85∗350 kg

cm2∗25.725cm∗45.00cm

Cc=344393.4kg

Ts=As∗fs

Ts=80.0cm2∗4200 kg

cm2

Ts=336000 kg

Cc≈TsO. K

CALCULODEMn

Mn=Cc∗z

z=d−a2

z=99.16 cm−25.725cm2

=86.298cm

Mn=344393.4kg∗86.298 cm

Mn=29720289.44kg−cm

Mn=297.203T−m

EJERCICIO N. 3

MODIFICAR VALOR DE f´c de (210 a 350) kg

cm2 y modificar la altura h de 90 a 70 cm

DATOS:

Hormigón Armado 1

As=(2φ28+4φ25 )+ (4 φ25+2φ22 )+(6φ22)

f ' c=350kg

cm2; f y=4200

kg

cm2

rec . libre=4cm

φde estribo=10mm

Determinar→Mn=? ?

Calculo de y

As . y=As1∗y1+As2∗y2+As3∗ y3+As4∗y 4+As5∗y5

y=∑i=1

n

(As i∗y i)

∑i=1

n

(Asi)

As1=2φ28=2∗6.16 cm2=12.32cm2

y1=rec .libre+φE+ φv2

y1=4+1+2.82

=6.4 cm

Hormigón Armado 1

As2=4φ25=4∗4.91cm2=19.64cm2

y2=4+1+2.52

=6.25cm

As3=4φ25=4∗4.91cm2=19.64cm2

y3=rec .libre+φE+φ28+φsep .+ φv2

y3=4+1+2.8+2.5+2.52

=11.55cm2

As4=2φ22=2∗3.80cm2=7.60cm2

y4=4+1+2.8+2.5+2.22

=11.40cm

As5=6 φ22=6∗3.80cm2=22.80cm2

y5=4+1+2.8+2.5+2.5+2.5+2.2 /2=16.40cm

y=(12.32cm2∗6.4 cm )+(19.64cm2∗6.25cm )+(19.64cm2∗11.55 cm )+(7.60cm2∗11.40 cm )+(22.80cm2∗16.40cm )

(12.32+19.64+19.64+7.60+22.80 ) cm2

y=10.84 cm

Diagrama:

Hormigón Armado 1

∑ fx=0.0

Cc−Ts=0.0

Cc=Ts

Mn=Cc∗z=Ts∗z

Cc=0.85∗f ´ c∗(AEHC )

AEHC=a∗b

Cc=0.85∗f ´ c∗a∗b

Ts=As∗fs

z=d− y

z=d−a2

Calculo de a

0.85∗f ´ c∗a∗b=As∗fs

Asumimos que fs=fy

a= As∗fs0.85∗f ´ c∗b

a¿= As∗fs0.85∗f ´ c∗b

(¿ )Valores condicionado parala verificaciónde fs=fy

Hormigón Armado 1

a¿=82.0cm2∗4200 kg

cm2

0.85∗350 kgcm2∗45cm

a¿=25.725cm

Calculo de ε s

ε s=εcuc

∗(d−c )

a=β1∗c

c= aβ1

c¿= a¿

β1

c¿=25.725 cm0.80

;( f ´ c>280∴ β1=0.85−0.05)

c¿=32.156 cm

ε s¿= 0.00332.156cm

∗(59.16−32.156 )cm

ε s¿=0.00252mm

mm

ε y=f yE s

ε y=4200

kg

cm2

2.1x 106kgcm2

ε y=0.002mmmm

Hormigón Armado 1

ε s¿vs . ε y

{ ε s¿>ε y

0.00252>0.002}❑⇒

ZONA PLASTICA

fs=fy=4200 kg

cm2

{a¿ , c¿ , εs¿ }SON VALORESVERDADEROS

CALCULODECc yTs

Cc=0.85∗f ´ c∗a∗b

Cc=0.85∗350 kg

cm2∗25.725cm∗45.00cm

Cc=344393.4kg

Ts=As∗fs

Ts=80.0cm2∗4200 kg

cm2

Ts=336000 kg

Cc≈TsO. K

CALCULODEMn

Mn=Cc∗z

z=d−a2

z=59.16cm−25.725cm2

=46.298cm

Hormigón Armado 1

Mn=344393.4kg∗46.298cm

Mn=15944553.44kg−cm

Mn=159.446T−m