Transición longitudinal

TRANSICIÓN LONGITUDINAL
SECCIÓN 2

tgθ=11.5
θ=tg-111.5
θ=33.69
tg θ=yx
tg 33.69=0.28x
x=0.8863m
tg 12.50=0.8863L
L= 3.9978
L= 4.00m
Manning
Q=1n AR2/3S1/2
Q=10.030(b+zy)y*(b+zy)yb+2y1+z22/3 *0.0035
Y=0.5909m
Y=0.60m
SECCIÓN 12

tgθ=12
θ=tg-112
θ=26.57
Manning
Q=1n AR2/3S1/2
Q=10.015 (b*Y)*(b*y)b+2y2/3 *0.0185
Y= 0.32
tgθ=0.32x
X= 0.64
tg1.25=xL
L=2.886m

CAMBIO DEALINEAMIENTO

T= R*tg∆2
R= T/ tg∆2
R= 395tg85.402
R=426.06m

C= 2R sen ∆2
C= 2(426.06) sen 85.402
C= 579.33m
E= R* (sen ∆2 - 1)
E=426.06* (sen85.402 - 1)
E= 154.93m
F=R*(1 – cos ∆2 )
F=426.06*(1 – cos 85.402 )
F=113.62m
Longitud curva
G10 =572.96R
G10 =1.345
L=∆G10 *(10)
L= 637.03m

T= R*tg∆2R= T/ tg∆2
R= 115tg24.302
R=529.6579

C= 2R sen ∆2
C= 2(529.6579) sen 24.302
C= 224.7632m
E= R* (sen ∆2 - 1)
E= 529.6579* (sen824.302 - 1)
E= 12.3407m
F=R*(1 –cos ∆2 )
F=529.6579*(1 – cos 24.302 )
F=12.0597m
Longitud curva
G10 =572.96R
G10 =1.0818
L=∆G10 *(10)
L= 226.4838m
Pendiente sección 8
Q=1n AR2/3S1/2
Q=1n (b*y3)*(b*y3)b+2y32/3 *S1/2Q=10.015 (3.5*1.26)*(3.5*1.26)3.5+2*1.262/3 *S1/2
S = 0.00053
CALCULOS DE VERTEDERO
h0=1.85m Q=4.25 m3/s C= 2.2
∆cotas=2.90mL= 3.5m
Chequeo C
S/ fig. 14.4 de Ven Te Chow
CCd=1.0
Cd =C1.0=2.2
Q= C*L*He1.5
4.25= 2.2*3.5* He1.5
He =0.67m
Q=C*L*(Hd+4.2522*9.81*3.52*(1.85+Hd)21.5
4.25= 2.2*3.5*(Hd+4.2522*9.81*3.52*(1.85+Hd)21.5

Por solver
He=0.66m = 2.16ft
He = Hd + Ha
Ha = He- Hd
Ha = (0.67 -0.60) = 0.01m
Va =QHp + Hd*L = 4.251.85 + 0.66*3.5 =
Va = 0.48 m/sHeHd=0.670.66=1.02
HpHd=1.850.66=2.80
Z= ∆z+HP +Hd = 2.90+1.85+0.66
Z =5.41m
Calculo de V …. La cresta
V = QA V = QL*Hd
V = 4.253.5*0.66 = 1.84 m/s
VC =3Q*gα*L...