compuertas - universidad veracruzana
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CompuertasCompuertas
F R=∫AdF=∫A
PdA=∫Aρ gydA
P=ρ gh
P prom=ρ gh2
F R=P prom A=ρghA
2
Como la presión solo varía en y
F R=∫0
h
ρ gy a dy
A=ah
F R=ρgah2
2
F R=aρ g∫0
h
y dy
F R=∫AdF=∫A
PdA=∫Aρ gydA
F R=∫0
3
ρ gy a dy
F R=aρ g∫0
3
y dy
F R=1.5m 1000kg
m3 9.81m
s2 3m 2
2 =66.22 kN
F R=aρ g [ y2
2 ]0
3
Fuerza sobre áreas planas: Fuerza sobre áreas planas: superficies inclinadassuperficies inclinadas
αO
dF h
Superficie libre
y
hsen =α
αysenh =
∫=AdAgysenF αρ
y
F R=∫AdF=∫
APdA=∫
Aρ gy sen α dA
FR=aρg∫0
1.5
1 ysen α dy
F R=ρ gysen α∫AdA
F R=aρg [ yy2
2sen α ]
0
1 . 5
F R=3m 1000kg
m3 9.81m
s2 [1.5m1.5m 2
2sen 30 ]=60.7 kN
a=ancho de la compurta=3m
F R=∫AdF=∫A
PdA=∫Aρ gydA
m5.1=y
1.3m)30cos(5.1 ==x
F R=1000kg
m3 9.81m
s2 [1m1.5m
2sen 30 ]3m 1.5m =60.7 kN
F R=1000kg
m3 9.81m
s2 [1m0.75m
2 ]3m 1.5m=60.7 kN
h=1.5msen 30=0.75
30.35kN)7.60)(60cos( ==xF
52.56kN)7.60)(60sin( ==yF
wvvy gVgVwFF ρρ +=+=
52.6kN14.34kNkN25.38 =+=yFF R=60.7 kN
F x=30.35 kN
F y=52.56 kN
F x=F H= 1000kg
m3 9.81m
s2 [1m0.75m
2 ]3m 0.75m =30.35 kN
Fv= 1000kg
m3 9.81m
s2 1.3m1m 3m =38 .26 kN
w= 1000kg
m3 9 .81m
s2 [ 1.3m 0.75m
2 ]3m =14 .34 kN
F R=∫AdF=∫A
PdA=∫Aρ gydA
F x=∫Aρ gydA y
F y=∫Aρ gydA x
F x=∫0
1
ρ gy a dy
F x=aρ g∫0
1
y dy
F x=aρ g [ y2
2 ]0
1
F x=0.6m 1000kg
m3 9.81m
s2 [ 12
2 ]=2943 NF x=aρg∫0
1
1− y dy
F x=aρg [ y−y2
2 ]0
1
F x=0.6m 1000kg
m3 9.81m
s2 [1−12
2 ]=2943 N
F y=aρg∫0
1
1− y dx
Para la dirección y
F y=aρg∫0
1
1−x2 dx
F y=aρg [ x− x3
3 ]0
1
F y=0 .6m 1000kg
m3 9 .81m
s2 [1−13
3 ]=3924 N
F y=aρg∫0
1
y dy
F y=aρg [ 2y3 /2
3 ]0
1
F y=0 .6 1000kg
m3 9.81m
s2 [ 213/2
3 ]=3924 N
VgwFy ρ==
F y=w= 1000kg
m3 9. 81m
s2 21m 1m
3 0 .6m =3924N
V=A a
A=2h x
3
F y=w=ρ g2h x
3a
F R=F x2F y
2
F R=294323924 2
=4905N
o1.53N 2943
N 3924 tan 1 ==−θ
Centroides y Momentos de Inercia de Figuras comunes
Por estática
ydFFycp =
Por estática
ydFFycp =
Pero
∴=PdAdF
PdAyFyAcp ∫=
αρgysenP =
∫=Acp dAgsenyFy αρ2
∫=Acp dAygsenFy 2αρ
Pero
∫=Ax dAyI 2 Segundo momento de área ó
Momento de inercia de área
I x= I y2 A
ycp
F=ρ gsen α I y2 A
ycp ρg y sen αA =ρ gsen α I y2 A
yCP= yI
y A
Por estática
∫=Acp xPdAFx
∫=Acp dAgsenxyFx αρ
∫=Axy xydAI
AyxII xyxy +=
Producto de Inercia de área
xcp ρg y sen αA =ρ gsen α I xyx y A
xcp= xI xy
y A
∑ =0M
F1 .5−∫AyPdA=0
F1 .5=∫0
1 . 5
yρ g y a dy
F1 .5=a ρg [ y 3
3 ]0
1 . 5
F1 .5=a ρg∫0
1 .5
y2 dy
F1 .5=1.5ρg [ y3
3 ]0
1 . 5
F=1000kg
m3 9.81m
s2 [ 1.53
3 ]=11.03 kN
∑ =0M
F1 .5−∫AyPdA=0
F1 .5=∫0
1 . 5
yρ g 3 .6−1.5 y a dy
F1 .5=a ρg [2.1y 2
2
y 3
3 ]0
1 . 5
F1 .5=a ρg∫0
1 . 5
2.1y y 2 dy
F1 .5=1.5ρg [2.1y2
2
y3
3 ]0
1 . 5
F=1000kg
m3 9.81m
s2 [2.11 .52
2
1.53
3 ]=34 .21 kN
b = 2 m
a = 2.5 m
water8 m
4 m
FR
abA π=
F R=ρ gysen α∫AdA
F R=ρg 8 ysen α πb∫0
5 dy2
F R=ρgπb [4yy 2
4sen α ]
0
5
F R=1000kg
m3 9.81m
s2 π 2 [4 5 52
4sen 53.1 ]=1540. 82kN
F R=9800 N
m3 8 m2m π 2 .5 m 2 m =1540 .82kN
Ancho =1m2m
ghAFF Hx ρ==
4m
1m
2m
gVghAwFF vy ρρ +=+=
arAaV 2
41 π==
109.3kN30.8kNkN5.78 =+=yF
kN9.146109.398.1 22 =+=rF
F x=F H= 1000kg
m3 9.81m
s2 4112=98 .1kN
Fv= 1000kg
m3 9.81m
s2 4 12=78 .5 kN
w=1000kg
m3 9 .81m
s2 1422 1=30.8 kN
98.1kN=xF
109.3kN=yF kN9.146=rF
AyI
yycp +=
Ay
Ixx xy
cp +=
o4898.1kN
kN3.109 tan 1 ==−θ
cpy
cpx
m067.5)2)(1(4
22
12)2(1
422
3
=
+
+
+=cpy
1.067m4m-m067.5 ==cpy
42
+=Hy
12
3bHI =
Por estática
xwxFFx vvycp +=
y
vvcp F
xwxFx
+=
m957.03.109
)849.0(8.30)1(5.78 =+=cpx
m849.0)2(3
4
3
4 ===ππ
rx
1m2
2
2=== B
xv
o480.957m
m067.1 tan 1 ==−θ
98.1kN=xF
109.3kN=yF kN9.146=rF
m957.0=cpx
m067.1=cpy
Ancho =1m
2m
1m
2m
19.62kN)2)(1)(1(sm
81.9mkg
1000 23 =
== Hx FF
30.8kN)1()2(41
sm
81.9mkg
1000 223 =
== πwFy
kN51.3630.819.62 22 =+=rF
19.62kN=xF
30.8kN=yF
kN51.36=rF
o5.5719.62kN
kN8.30 tan 1 ==−θ
cpy
cpx
m333.1)2)(1(
22
12)2(1
22
3
=
+
=cpy
m849.08.30
)849.0(8.30 ==cpx
o5.570.849m
1.333m tan 1 ==−θ
2m 4m
1m
98.1kN)2)(1)(14(sm
81.9mkg
1000 23 =+
== Hx FF
gVghAwFF vy ρρ +=+=
78.5kN)2)(1)(4(sm
81.9mkg
1000 23 =
=vF
19.62kN)1(2
)2)(2(sm
81.9mkg
1000 23 =
=w
aBH
AaV2
==
98.1kN19.62kNkN5.78 =+=yF
kN7.13898.198.1 22 =+=rF
Ancho =1m
98.1kN=xF
98.1kN=yF
kN7.138=rF
AyI
yycp +=
Ay
Ixx xy
cp +=
o4598.1kN
kN1.98 tan 1 ==−θ
cpy
cpx
m067.5)2)(1(4
22
12)2(1
422
3
=
+
+
+=cpy
1.067m4m-m067.5 ==cpy
42
+=Hy
12
3bHI =
Por estática
xwxFFx vvycp +=
y
vvcp F
xwxFx
+=
m067.11.98
)33.1(6.19)1(5.78 =+=cpx
m33.13
)2(2
3
2 === Bx
1m2
2
2=== B
xv
AgysenF αρ=
Ancho=3 m
m5.1=y
1.3m)30cos(5.1 ==x
F R=1000kg
m3 9.81m
s2 [1m1.5m
2sen 30 ]3m 1.5m =60.7 kN
F R=1000kg
m3 9.81m
s2 [1m0.75m
2 ]3m 1.5m=60.7 kN
h=1.5msen 30=0.75
m4.1)75.0)(3)(
2
75.01(
12)75.0(3
)275.0
1(
3
=+
++=CPy
Por estática
xwxFFx vvycp +=
y
vvcp F
xwxFx
+=
m7098.06.52
)866.0(34.14)65.0(26.38 =+=cpx
m866.03
)3.1(2
3
2 === Bx
0.65m2
3.1
2=== B
xv