Pregunta 2.rtf
Transcript of Pregunta 2.rtf
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 1/8
Problema 2
tonf 1000kgf :=
Pu 1.2 0.65 10⋅ tonf ( )⋅ 1.6 0.35 10⋅ tonf ( )⋅+ 13.4 tonf ⋅=:=
qu 1.2 0.65 4⋅ tonf
m
⋅ 1.6 0.35 4⋅ tonf
m
⋅+ 5.36tonf
m⋅=:=
Solicitaciones
Mu1 48.24tonf ⋅:=
Vu1 29.48tonf :=
Mu2 32.16tonf ⋅:=
Vu2 21.44tonf :=
Perfil
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 2/8
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 3/8
Desarrollo
C"R#E
$a Viga esta Sometida adems de corte % fle!i&n a una fuera a!ial de compresi&n por tanto seutilia la e!presi&n Vc como (
#g h b⋅ 64.34cm2
− 1435.66 cm2
⋅=:= )r*a +ruta del Perfil , -o inclu%e el area de losespacios de las barras de acero.
Vc 1 "u
14 #g⋅+
fc
6⋅ b⋅ d⋅:= Para MPa
"u1
"u
kgf 0=:= fc1
fc
kgf
cm2
250=:=
#g1
#g
cm2
1435.66=:=
f 1
f
kgf
cm2
4200=:=
b1 b
cm25=:=
d1d
cm=:=
d
Vc 0.53 1 "u1
140 #g1⋅+ fc1
6⋅ b1⋅ d1⋅
kgf := d1 Vc kgf ⋅=Vc
$um%&' ()*( 0.85 Vc⋅ Vu>if
(+, qu' dis',r !stribos( oth'r/is'
:= Vc
$um%&' =$um%&'
Dise/o Estribos
φ 0.85:=
φ Vc Vs+( ) Vu≥
Vs
Vu φ Vc⋅−
φ:=
Vc
Vs kgf ⋅=Vs
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 4/8
)rea Re0uerida
Separaci&n M!ima para obtener un ) minima
d
2c⋅=
d
s1 min 60cmd
2
, c⋅=:=
ds
d
2
≤
s3 23c:=Separaci&n m!ima
#.1
Vu
φ Vc−
f d⋅cm
2
m⋅=:=
Vc )rea Re0uerida
φ ' 10m:=
s
2 π⋅φ '
2
2
f ⋅ d⋅
Vs
c⋅=:=Vs
E 1314ϕ
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 5/8
5le!i&n
β1 0.85 fc
kgf
cm2
300≤if
0.85 0.008 fc
kgf
cm2
300−
⋅−
fc
kgf
cm2
300>if
:=
β1 0.85=
Refuero Re0uerido por esfuero de armadura para fle!i&n
φ c 0.9:=
#sr'q
0.85 fc⋅ b⋅ di⋅
f
1 12 Mu⋅
φ c 0.85⋅ fc⋅ b⋅ di2
⋅−−
⋅:=
#sr'q 28.538 cm2⋅=
Cuantias
ρ b,&,nc'
0.85 fc⋅ β1
⋅
f
0.003!
sf 0.003!s+⋅:=
ρ b,&,nc' 0.026=
ρm, 0.5 ρ b,&,nc'⋅ 0.019=:=
#sm, ρm, b⋅ di⋅ 2.094 cm2⋅=:=
ρ#s
r'q b di⋅
0.02=:=
cond (( ρm, ρ≥if
(#,dir !nfi'rr,dur, 'n &, P,rt' com%rimid,( oth'r/is'
:=
cond #,dir !nfi'rr,dur, 'n &, P,rt' com%rimid,=
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 6/8
Refuero m6nimo en fle!i&n
#min2
14 b⋅ di⋅
f
kgf
cm2
:=#min1
0.85
fc
kgf
cm2
⋅
f
kgf
cm2
b⋅ di⋅:=
#min1 4.48 cm2⋅= #min2 4.66 cm
2⋅=
#smin
m, #min1
#min2
,( )
4.66 cm2
⋅=:=
Compatibilidad de de formaciones
Con )cero M!imoc1
#sm, f ⋅
0.85 fc⋅ b⋅ β1⋅ 25.2 c⋅=:=
Mn φ 0.85⋅ β1⋅ fc⋅ b⋅ c1⋅ di
β1 c1⋅
2−
⋅ 43.80 tonf ⋅⋅=:=
Mu 48.24 tonf ⋅⋅=
φ f 0.9:=
#ss
Mu
φ f di ds−( )⋅ f ⋅
Mn
di ds−( ) f ⋅− 4.52 cm
2⋅=:=
εs
di c1−( )c1
0.003⋅ 0.004=:=
εs1 0.003c1 ds−
c1
⋅ 0.002=:=
εs 0.005> Controla en tensi&n
εs 0.004< -o es Ductil
εs1 0.002≥ )cero en compresi&n 5lu%e
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 7/8
#ss 4.52 cm2
⋅= −− > #smin 4.66 cm2
⋅=
#si #sm, 2.094 cm2⋅=:=
,do
φ db 1m:=
"1 4:=
#sm, "1 π
φ db2
4⋅
ind φ db( ) 29.36 m⋅= Diametro
sdb m, φ db 2.5cm,( ) 2.5 c⋅=:=
con4 (( "1 1−( ) 2+ sdb⋅ "1 φ db⋅+ b≤if
(Pon'r ob&' corrid, d' 'nfi'rr,dur,( oth'r/is'
:=
con4 =
Para la primera secci&n de la iga
8/16/2019 Pregunta 2.rtf
http://slidepdf.com/reader/full/pregunta-2rtf 8/8