Doc Estructuras
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Transcript of Doc Estructuras
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8/18/2019 Doc Estructuras
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FECHA:10/03/2016
Introducción
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FECHA:10/03/2016
Armadura 1
1. REACCIONES: ( ∑ MA-
∑ MF)
∑ MA= RE (4)+1(4)+5(4)-8(1)-2
RE = (4)+4+20-8-214
4=3.5
RE = -3.5
∑ME = -RA (4)-8(3)-2
RA = (4)-24-2
−264
=6.5 RA=6.5
ANGULOS: ∝1=tg−1
(
2
1
)=63.43
∝2=tg−1(23)=33.69
Nodo “A”
∑FV =+ RA− AB ∙ sen63.43
¿6.5− AB ∙ sen63.43
¿ 6.5
sen63.43=7.27
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FECHA:10/03/2016
•
AB=7.27COMPRESION
∑FH =7.27cos63.43+ AC
AC =3.25TRACCION
Nodo “B”
∑FV =−BC +7.27 sen63.43
• −BC +6.50
−BC =−6.5
BC =6.5TENSION
∑FH =BD−3.25+2.70cos33.69
BD−3.25+2.25=0
BD−1
BD=1TENCION
Nodo “C”∑FV =CDsen33.69+6.5−8
CDsen53.31−1.5
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FECHA:10/03/2016
• CD=
1.5
sen33.69=2.70
CD=2.70
∑FH =−CE−2+7.27cos63.43
−CE−2+3.25
−CE+1.25
−CE=−1.25
CE=1.25COMPRESION
Nodo “E” ∑FH =−1+ E” X ”
EX=1
∑FV = ED−3.5
ED=3.5 TRACCION
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FECHA:10/03/2016
Armadura 2
1. REACCIONES:
∑MA−∑MF
∑MF = RA (3 )−2+6 (5 )
−4+30
26
3 =8.67
RA ¿−8.67
∑MA= RF (3 )+2 (2 )−6 (5 )+4−30
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FECHA:10/03/2016
RF =26
3
RF =8.67
ANGULOS:∝1=tg
−1(22)=45
∝2=tg−1(21)=6.43
∝3=tg−1
(3
2
)=56.31
Nodo “C”∑FH =6−CEcos56.31
−CE cos50.31=−6
CE= −6cos56.31
CE=10.82 COMPRESION
•
∑FV =−CB+10.82 sen56.31
−CB+9
−CB=−9
CB=9
TRACCION
Nodo “B” ∑FH =¿ 2!"E
BE=2TRACCION
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FECHA:10/03/2016
• ∑EV =9−BA
−BA=−9
BA=9TRACCION
Nodo “A”
∑FV =9−8.67− AE sen45 0.33− AEsen 45
• − AE=
−0.33sen45
− AE=−0.47
AE=0.47 COMPRESION
∑FH =−0.47cos45+ AD
−0.33+ AD
AD=0.33TRACCION
Nodo “D”
∑FV =¿
DE=0
∑FH =−0.33+ FD
DF =0.33TRACCION
•
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FECHA:10/03/2016
Nodo “F”
∑FV =8.67− FE sen63.43
• − FE sen 63.43=−8.67
− FE= −8.67sen63.43
=−9.69
FE=9.69COMPRESION
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FECHA:10/03/2016
Armadura 3
REACCIONES: ( ∑ MA-
∑ MF#
∑MA= RH ( 4)+5 (4 )−2 (3)−3 (2 )+2(2)
RH (4 )+20−6−6+4
RH =12
4=3
RH =−3
∑MH = RA−3+5−2−3
RA−3
RA=3
AN$%&OS: ∝1=2
1=63.43
Nodo “A”
∑FV =3+ AC sen63.43
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sen63.43=3.35
• AC =3.35
COMPRESION
∑FH = AB−3.35cos63.43
AB=1.50TENSION
Nodo “C”
∑FV =3.35 sen63.43+CB
•CB=3TRACCION
∑FH =3.35cos63.43−2+CE
CE=0.5TRACCION
Nodo “B”
∑FV =3+BEsen63.43
¿
3
sen63.43
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FECHA:10/03/2016
BE=3.35COMPRESION
∑FH =−3.35cos63.43−1.5+BD
BD=3TRACCION
Nodo “E”
∑FV =3
•
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Armadura 4
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FECHA:10/03/2016
Armadura 5
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