CÁLCULO DE MONOBLOQUES
Profesor:
Ing. Daniel E. Weber
J.T.P.:
Ing. Sebastián Romero
Cimentaci Cimentaciones ones U.T.N. U.T.N. – Facultad Facultad Regional Regional Santa Fe – 2009 E-Mail:
[email protected]
Análisis del momento reactivo lateral MS: Método de Sulzberger
M
y ⋅ tg(α )
H
dy
2 ⋅t 3
α
t G
y
Ct
F
Limos, Arenas y Arcillas no consolidadas
G PLANTA
y
Cy
MS
Cy ⋅ b y=
Ct
Mb a
b
MS
α
1 ⋅t 3
Volumen de Tensiones
t 3
Ct ⋅ b
= C t ⋅ 1 −
y t
Variación Lineal del Coeficiente de Balasto
σy
R
σ máx.
y
t 2
σ y = C t ⋅ 1 −
y ⋅ y ⋅ tg(α ) t
y = t → σy = 0 → Cy y = 0 → σy = 0 t y = → σ y = σmáx. 2
=
0
Análisis del momento reactivo lateral MS: Método de Sulzberger
dMS
=
C y ⋅ y ⋅ tg(α ) ⋅ b ⋅ dy ⋅ y = 2
= C y ⋅ b ⋅ dy ⋅ y ⋅ tg(α ) =
= C t ⋅ 1 −
MS
=
y ⋅ b ⋅ dy ⋅ y 2 ⋅ tg(α ) t
∫ dMS = Ct ⋅ b ⋅ tg(α ) ⋅
y =t
y 2 ∫Y =0 1 − t ⋅ y ⋅ dy =
y3 t 1 y 4 t b ⋅ t3 = C t ⋅ b ⋅ tg(α ) − ⋅ ⋅ C t ⋅ tg(α ) = 3 0 t 4 0 12
b ⋅ t3 ⋅ C t ⋅ tg(α ) MS = 12
Momento de inercia del diagrama de carga respecto del eje de giro del fondo
Análisis del momento reactivo lateral MS: Método de Sulzberger
Cuando H consume la resistencia de fricción de fondo, el eje de giro se traslada desde el fondo, hasta y = t/3. En este momento lo que no soporta la fricción de fondo lo toma el diagrama de reacción lateral. t t b ⋅ t3 ⋅ C t ⋅ tg(α ) F = µ ⋅ G = R ⇒ MS = R ⋅ = µ ⋅ G ⋅ = 2 2 12 tg(α1 ) =
6 ⋅ µ ⋅ G b ⋅ Ct ⋅ t 2
Este valor de a1 se analiza respecto de la deformación límite tg a1 = 0,01
b ⋅ t3 ⋅ C t ⋅ tg(α ) MS = 36
Análisis del momento reactivo lateral MS: Método de Sulzberger →
tg(α1 ) > tg(α ) = 0,01
Eje de Giro en el fondo
→
b ⋅ t3 MS = ⋅ C t ⋅ tg(α ) 12
tg a=0,01
b ⋅ t3 MS = ⋅ C t ⋅ tg(α ) 36
tg a=0,01
tg(α1 ) < tg(α ) = 0,01
Eje de Giro en y = t/3
Eje de Giro en el fondo
Mexterno
= H ⋅ (h + t )
Eje de Giro en y = t/3
Mexterno
= H ⋅ h +
2 ⋅ t 3
Análisis del momento reactivo de fondo Mb: Método de Sulzberger → λ0 > λ'
S= =
a a a (λ + λ') + 2 ⋅ (λ0 − λ') −C = − ⋅ 0 = 2 2 3 (λ0 + λ') + (λ0 − λ')
3 ⋅ λ0 − λ' a λ' a a = − ⋅ = − ⋅ − a 0 , 5 λ ⋅ 2 3 2 ⋅ λ0 2 6 0
a2 a ⋅ λ' 2 ⋅ tg(α ) a3 ⋅ b ⋅ Cb ⋅ tg(α ) = = = 6⋅G 6 ⋅ λ0 12 ⋅ G a ⋅ b ⋅ Cb a3 ⋅ b ⋅ Cb ⋅ tg(α ) → Mb = G ⋅ S = 12
Análisis del momento reactivo de fondo Mb: Método de Sulzberger → λ0 = λ'
a ⋅ tg(α2 ) = 2 ⋅ λ0 → tg(α2 ) =
=
2⋅G a ⋅ b ⋅ Cb
2⋅G a 2 ⋅ b ⋅ Cb
→ λ0 < λ'
a x Mb = G ⋅ S' = G ⋅ − = 2 3 a 1 2 ⋅ G = G⋅ 2 − 3 ⋅ b ⋅ C ⋅ tg(α ) b a G → Mb = G ⋅ 2 − 0,47 ⋅ b ⋅ C ⋅ tg(α ) b
Análisis del momento reactivo lateral Mb: Método de Sulzberger
Gadh Colabora en la reducción de la excentricidad que produce el tiro horizontal de los conductores.: x = t ⋅ tg( β ) G: Área Base Mayor Pirámide Truncada g: Área Base Menor Pirámide Truncada Volumen Pirámide Truncada:
t ⋅ (G + g + G ⋅ g ) 3
Peso suelo adherente Gadh.: t Gadh. = γ S ⋅ ⋅ [a ⋅ b + (a + 2 ⋅ t ⋅ tg β ) ⋅ (b + 2 ⋅ t ⋅ tgβ )] + g ⋅ G − t ⋅ g 3
b= 3º a 5º en compresión
gs= considerar valor mínimo
Si hay sumergencia:
γ S = 1,00 t / m3
γ H = 1,20 t / m3 Densidad Hormigón Sumergido
Método de Sulzberger
Momento de Fondo tgα 2
Mb
=
2⋅G a 2 ⋅ b ⋅ Cb 2⋅G a 2 ⋅ b ⋅ Cb
>
<
=
0,01
b ⋅ a3 ⋅ Cb ⋅ tgα 12
0,01
G ⋅
a 2
− 0,47 ⋅
Momento de Empotramiento tgα1 = 6⋅µ⋅G b ⋅ t 2 ⋅ Ct 6⋅µ⋅G b ⋅ t 2 ⋅ Ct
MS >
<
=
0,01
b ⋅ t3 ⋅ C t ⋅ tgα 12
0,01
b ⋅ t3 ⋅ C t ⋅ tgα 36
b ⋅ Cb ⋅ tgα
G
Cb : Cte. Ct : Variación lineal de 0 a C t G : Peso Total de Fundación, incluye Gadh. Eje de giro en el fondo: M externo= H (h+t) Eje de giro en 2/3 t: M externo= H (h+2/3 t)
AMPLIACION E.T. LUJAN PROV. DE SAN LUIS COEFICIENTES DE BALASTO Y TENSIONES ADMISIBLES A COMPRESION PROFUNDIDAD 1,50m a 2,00m
0,25m 0,50m
σ adm = 0,25 Kg/cm
0,5 Kg/cm
2
0,66m
σ adm = 0,30 Kg/cm
3
σ adm Puntual= 0,33 Kg/cm
1,50m
2
0,7 Kg/cm
3
σ adm Puntual = 0,40 Kg/cm
2,00m
1,5 Kg/cm
2
2
3
σ adm = 0,70 Kg/cm
2
σ adm Puntual= 0,93 Kg/cm
2
1,80 Kg/cm
3
2,00 Kg/cm
3
σ adm = 0,80 Kg/cm
2
σ adm Puntual = 1,1 0 Kg/cm
2
2,4 Kg/cm
3
AMPLIACION E.T. LUJAN PROV. DE SAN LUIS COEFICIENTES DE BALASTO Y TENSIONES ADMISIBLES A COMPRESION PROFUNDIDAD 2,50m a 3,00m
0,25m 0,83m
2,50m
σ adm
0,83 Kg/cm
= 0,33Kg/cm
2
1,00m σ adm
3
= 0,40 Kg/cm
σ adm Puntual
2
1,00 Kg/cm
2
3 σ adm Puntual
3,00m
2,5 Kg/cm
= 0,50 Kg/cm
= 0,66 Kg/cm
2
3
σ adm
= 1,00 Kg/cm
2
= 1,33 Kg/cm
σ adm Puntual
3,00 Kg/cm
3
2
3,00 Kg/cm
3 σ adm
σ adm Puntual
= 1,5 Kg/cm
2
= 1,99 Kg/cm
2
3,6 Kg/cm
3
MOMENTO REACTIVO DE BASE M b [tm]
MOMENTO REACTIVO DE EMPOTRAMIENTO M S [tm]
tg α 2 = 2∗∗ GT / (a2∗ b∗∗ cb)
tg α 1 = 6∗∗ µ∗ GT / (b∗∗ t2∗ ct)
tg α 2 = 0.00572
tg α 1 = 0.00107
tg α 2 > 0,01 M b =(b∗∗ a3∗ cb∗ tagα α ) / 12 Mb =
tg α 1 > 0,01 M s =(b∗∗ t3∗ ct∗ tagα α ) / 12 Ms =
No corresponde
0,5 tg α 2 < 0,01 M b=GT∗ [ (0,5∗∗ a)-0,47∗∗ (GT/( b∗∗ cb∗ tagα α )) ]
M b = 1.399 FUNDACION INTERRUPTORES Mf
Q L
β
2t/3
γ γ horm t
γ γ s GT
Ct
µ
Cb
tg α 1 < 0,01 M s =(b∗∗ t3∗ ct∗ tagα α ) / 36 M s = 9.77
N s
No corresponde
a [m]= b [m]= t [m]= s [m]=
0.90 0.90 2.50 0.25
Mf pto giro[tm]= 5.983
D [m]= 0.35
M s [tm] = 9.766
3
cb [t/m ]= 3000 ct [t/m3]= 2500
M b [tm] = 1.399
µ=
Ms/Mb = 6.98
0.40
β [G sex.]= 0 3
γ γ s [t/m ]= 1.60 3
b
D
1.350 0.95 4.40 6.25
L [m]= 1.00
do
a
N [ton]= Q [ton]= Mf [tm]= GT [ton]=
Ms+Mb = 11.165
GIRO=( Mserv/(Ms+Mb) ) ∗ 0,01
= 0.005359
Fs m in = 1.50
γ γ horm [t/m ]= 2.20 tag β = 0.00000 a+(2t∗ ∗ tag β ) = 0.900
σh t/3 [Kg/cm2]= 3.72
b+(2t∗ ∗ tag β ) = 0.900
σv a t [Kg/cm 2]= 11.52
Fs real = 1.87
σh a t [Kg/cm 2]= 11.16 3
Ghorm [ton]= 4.90
Volumen de excavacion [m ]= 2.025
Gsuelo [ton]= 0.00
Volumen de Hormigon [m ] = 2.131
3
N= 0.15
Solicitaciones finales
ton
(
η
Q= 0.30 ton
x max
( + W ) [ton ] =
K H = = 0.80 K V
(
4
∗
p1 6
E1
∗
H L
∗ η ∗ H L
=
R Suelo arcilloso CL
pm ax
= 37.5
3
B ∗
4 ∗η ∗
suelo Kv =
1875 t/m
L 3 12
B ∗ H 3 12
50
2
3 L + 4 ∗ η ∗ H 3
N + W
1.82
3
6 ∗ η ∗ H 2
2H/3 e pm in
η
H = 2.00 m
W
3.836
Valores Auxiliares
sf = 0.40 m
α
α
)
M + Q ∗ H + s f =
Suelo Kh = 1500 t/m
)
W = B ∗ L ∗ H + s f ∗ 2.4 = 3.686
M= 1.10 tm
= 0.735
= 0.034 = 1.707
3
L
e ≤
L= 0.80 m
e=
[ (
B= 0.80 m
[
(
1 ∗
3 H 1+ 4∗η ∗ L
+W
E 1 =
= 0.13
Cáculos
)]
M + Q∗ H + s f
6
6 ∗η ∗ H L
3
L
0.009
2
(
)
* + W ∗ e = 1.338
6 ∗η ∗ H 2
)]
R = M + Q ∗ H + s f ∗
=
3 + 4 ∗η ∗
3
H
− Q = 1.038
tg α =
x
max
[ (
M + Q ∗ H + s f
)]
B ∗ L 3 3 + 4 ∗η ∗ B ∗ H K v ∗ 12 12
p1 =
= H ∗ tg α = 0.001
p max =
( + W ) B ∗ L
∗ 1 + 6 ∗
= 0.00056
e = 6.413 L
Verif = ( p max + p min )∗
L 2
[
η ∗ M + Q ∗
3 B ∗ L 3 + 4 ∗ η ∗ B ∗ H 12 12
p min =
∗ B =3.8364
( H + s f )]∗ H
= 1.673
( + W ) e ∗ 1 − 6 ∗ = 5.576 B ∗ L L
