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Ejercicos de secciones sometidas a flexión... con modificaciones y graficos...
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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
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