Ejercicio de diseño de una viga
3
Datos: Mu = 41.72 t-m f´c = 280 kg/cm 2 fy = 4200 kg/cm 2 FR = 0.90 d´ = 3.74 cm Pmin= 0.70 √ f´c fy = 0.70 √ 280 kg / cm 2 4200 kg / cm 2 =0.0028 f ¿ c=0.80 f´c=0.80 ( 280 kg / cm 2 ) =224.0 kg / cm 2 f c=0.85 {f} ^ {* } c=0.85 (224 kg/c {m} ^ {2} )= 190.40 kg/c {m} ^ {2 Pb= f c} over {fy} {{β} rsub {{1} ^ <?>} 6000} over {(fy +6000)} = left Pmax =0.75 Pb=0.75 ( 0.0227 )=0.0170 Qu = Mu FRbd 2 f c} = {4172000.0} over {(0.90)(25 cm)(36.26 cm {)} ^ {2} (190.4k q=1− √ 1−2 Q=1− √ 1−2 ( 0.74) =error ( p−p´ ) ≥f c} over {fy} {{β} rsub {{1} ^ <?>} 6000} over {left (6000-fy rig ( p−p´ ) = f¨c fy β 1 ❑ 6000 ( 6000−fy ) d´ d = ( 190.40 kg / cm 2 4200 kg/ cm 2 )( ( 0.85 )( 6000) 6000−4200 kg/ cm 2 ) ( 3.74 cm 36.26 cm ) =0.0133 Se propone un ( p−p´ ) pro=0.015 ( As− As´) =( p−p´ ) pro ( b )( d )= ( 0.015 )( 25 cm)( 36.26 cm ) =13.60 cm 2 a= ( As− As´ ) fy bf c} = {(13.60 c {m} ^ {2} )(4200 kg/c {m} ^ {2} )} over {(25)(190.4kg Viga doblemente armada
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Diseño de una viga de concreto reforzado.
Transcript of Ejercicio de diseño de una viga
Datos:
Mu = 41.72 t-m
f´c = 280 kg/cm2
fy = 4200 kg/cm2
FR = 0.90
d´ = 3.74 cm
Pmin=0.70√ f ´ cfy
=0.70 √280kg /c m2
4200 kg/c m2 =0.0028
f ¿c=0.80 f ´ c=0.80 (280kg /cm2 )=224.0kg/c m2
f c=0.85 {f} ^ {* } c=0.85 (224 kg/c {m} ^ {2} )= 190.40 kg/c {m} ^ {2
Pb=f c} over {fy} {{β} rsub {{1} ^ <?>} 6000} over {(fy +6000)} = left ({190.40 kg/c {m} ^ {2}} over {4200kg/c {m} ^ {2}} right ) left ({(0.85)(6000)} over {(4200kg/c {m} ^ {2} +6000)} right ) = 0.022 ¿
Pmax=0.75Pb=0.75 (0.0227 )=0.0170
Qu= Mu
FRb d2 f c} = {4172000.0} over {(0.90)(25 cm)(36.26 cm {)} ^ {2} (190.4kg/c {m} ^ {2} )} = 0.7 ¿
q=1−√1−2Q=1−√1−2 (0.74 )=error
( p−p ´ )≥ f c} over {fy} {{β} rsub {{1} ^ <?>} 6000} over {left (6000-fy right )} {d´} over {d} ¿
( p−p ´ )= f ¨ cfy
β1❑ 6000
(6000−fy )d´d
=( 190.40kg/c m2
4200kg /cm2 )( (0.85)(6000)6000−4200kg /cm2 )( 3.74cm
36.26cm )=0.0133
Se propone un ( p−p ´ ) pr o=0.015
( As−As ´ )=(p−p´ ) pro (b ) (d )=(0.015 ) (25cm ) (36.26 cm )=13.60cm2
a=(As−As ´ ) fy
bf c} = {(13.60 c {m} ^ {2} )(4200 kg/c {m} ^ {2} )} over {(25)(190.4kg/c {m} ^ {2} )} = 12.0 c¿
M r1=( As−As ´ ) fy (d−12a)=(13.60c m2 ) ( 4200kg /cm2 ) ( 36.26cm−0.5 (12cm ) )=1728451.20kg−cm
Mr=MuFR
=4172000kg−cm0.90
=4635555.56 kg−cm
Viga doblemente armada
M r2=Mr−Mr1= (4635555.56kg−cm ) (−1728451.20kg−cm )=2907104.36 kg−cm
As ´=Mr 2
fy (d−d´ )= 2907104.36kg−cm
4200kg /c m2(36.26 cm−3.74cm)=21.28cm2
( As−As ´ )=13.60cm2
As=13.60+As ´=13.60cm2+21.28c m2=34.88cm2
No .Vs= AsAv
y No.Vs= As ´Av
Numero de varillas:
Ø varilla (pulg) Área de la varilla cm2
As=34.88 cm2 As´= 21.28 cm2
1” 5.07 6.87 vs ≈ 7 vs 4.19 vs≈ 4Vs
Revisión
( As ´ )dado=No .deVs∗Avarilla=(4 ) (5.07c m2 )=20.28cm2
( As )dado=No .deVs∗Avarilla=(7 ) (5.07c m2)=35.49c m2
( As−As ´ )dado=As−As ´=35.49cm2−20.28c m2=15.21cm2
a=(As−As ´ ) fy
bf c} = {(15.21 c {m} ^ {2} )(4200 kg/c {m} ^ {2} )} over {(25)(190.4kg/c {m} ^ {2} )} = 13.42 c¿
7 Vs de Ø 1”
4 Vs de Ø 1”
( p−p ´ )=(As−As ´ )dado
bd= 15.21cm2
(25)(36.26)=0.0168 , si cumple
Mr=(As−As ´ ) fy (d−12a)+As ´ fy (d−d ´ )
Mr=(15.21 ) (4200 ) (36.26−0.5 (13.42 ) )+(20.28 ) (4200 ) (36.26−3.74 )
Mr=4657636.6 kg−cm
MR=FRMr=0.90 ( 4657636.6kg−m )=4191872.94kg−cm
MR=41.91 tn−m
Mu<MRsi cumple .
