Arch Dams
Translated from the slides of
Prof. Dr. Recep YURTAL (Ç.Ü.)
by his kind courtesy
ercan kahya
Recep YURTAL
Ç.Ü. İnş.Müh.Böl.
Arch Dams
•
•
Curved in plan and carry most of the water thrust
horizontally to the side abutments by arch action.
A certain percentage of water load is vertically
transmitted to the foundation by cantilever action.
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Arch Dams
!
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Ç.Ü. İnş.Müh.Böl.
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Arch Dams
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Where & why we design it:
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Appropriate when “Width / Height of valley” (B/H) < 6
Rocks at the base and hillsides should be strong enough
with high bearing capacity.
To save in the volume of concrete.
Stresses are allowed to be as high as allowable stress of
concrete.
Connection to the slope of hillsides should be 45o at least.
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Arch Dams
Base width
Arch Width
Upstream
face
Crest
Downstream
face
Base
width
Height
radiousp
Toe
Axis
Central angle
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Downstream
Face
Base
Arch Dam Types
Constant Center Arch Dam (variable angle)
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Good for U-shaped valleys
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Easy construction
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Vertical upstream face
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Appropriate for middle-high
dams
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Arch Dam Types
Variable Center Arch Dams (constant angle)
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2
q
Good for V-shaped valleys
4
1
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Limited to the ratio of B/H=5
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Best center angle: 133o 34´
3
3-4
q
To obtain arch action at
the bottom parts
1-2
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Arch Dam Types
Variable Center –
Variable Angle Arch
Dams
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q
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crest
Combination of the two
above.
a
Its calculation based on
shell theory applied to
arch dams
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a
a
a
Arch Dam Types
⇐
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Upstream
Arch Dams
n
3.6.2 DESIGN OF ARCH DAMS
Structural Design: à Load distribution in the dam body
& beyond scope of this course
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Independent ring method
Trial load method
Elasticity theory
Shell theory
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Arch Dams
n
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Hydraulic Design
Determination of thickness at any elevation
Effect of uplift force – ignored
Stresses due to ice & temp changes-important
Arch action - near the crest of dam
Cantilever action - near the bottom of dam
Horizontal hydrostatic pressure is assumed to be
taken by arch action
Arch Dams (design principles)
n
Independent Ring method:
h
t
t
t/2
t/2
rm
rm = rd – t/2
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rd
O
y
Hh
dFV´
Differential force
for infinitely small
elemental piece with
center angle of dφ :
dFV
φ
p
p
t
R
θa
dφ
φ
x
R
dFV = p . r . dφ
r
Vertical component:
Ba
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ʹ′
dFV = p . r . dφ . Sinφ
Arch Dams (design principles)
▪ Total
horizontal force:
H h = 2γ .h.r . Sin
θa
2
h : height of arch lib from the reservoir surface
r : radius of arch
θa: central angle
▪ Equilibrium of forces in the flow direction (y):
Hh = 2 Ry
Ry: reaction force at the sides in y direction [= R sin (θa/2)]
y
Hh
dFV´
dFV
φ
Reaction of the sides (R):
p
p
t
R
θa/2
Ry
θa
dφ
φ
R = γ .h.r
r
Rx
R
x Ry
Rx
Ba
σall : allowable working stress
for concrete in compression
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θa/2
The required thickness of
the rib (t) when t << r :
σ≈ R/t
t=
γ .h.r
σ all
Arch Dams (design principles)
The volume of concrete for unit height for a single arch:
V=Lt
L: arch length (L = r θa)
(note that θa is in radians)
γ .h 2
V=
r θa
σ all
Arch Dams (design principles)
The required thickness of the rib (t):
n
Pressure
p = γ × h:
p × rd
t=
σ all
t
t/2
t/2
rm
rm = rd – t/2
rd
O
Reduction factor for base pressure at arch dams is zero (m=0)
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Arch Dams (design principles)
▪ The optimum θa for minimum volume of arch rib:
dV / dθ = 0
à θa = 133o 34’
► This is the reason why the constant-angle dams
require less concrete that the constant-center dams
► Formwork is more difficult
► In practice; 100o < θa < 140o for the constant-angle
dams
Arch Dams (design principles)
n
Displacement in an arch ring:
rm
δ = σ all
E
t
E = Concrete elasticity modulus
rm = Radius from ring axis
For the relation between
center angle (2φ), beam
length (L) & arch radius (r):
L
r=
2 × Sin φ
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t/2
t/2
rm
rd
O
rm = rd – t/2
Örnek
n
Yüksekliği 120 m olan bir barajın ağırlık ve kemer-ağırlık türünde
yapılması halleri için taban genişliklerini ve birim kalınlıktaki hacimlerini
karşılaştırınız. Barajda taban su basıncı küçültme faktörü 0.8 dir. γb =
2.4 t/m³
Çözüm:
b
=
h
1
γʹ′ − m
Hacim=
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Ağırlık baraj için:
b
120
=
γb
γʹ′ =
γw
1
2.4 − 0.8
120 × 95 ×
b = 120 ×
1
= 5700 m³
2
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γʹ′ =
1
2.4 − 0.8
2.4
= 2.4
1.0
b= 95 m.
n
Kemer-Ağrlık baraj için:
m=0
bʹ′
=
h
1
γʹ′
bʹ′ = 120 ×
1
= 78 m
2. 4
1
Hacim = 120 × 78 × = 4680 m ³
2
Hacim azalması :
5700 − 4680
= % 18
5700
120
95
78
Ağırlık yerine kemer-ağırlık seçmeyle beton hacminde %18 azalma olacaktır.
Bu seçim ancak yamaçlar yeterince sağlam olduğunda yapılabilir.
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Örnek
n
Yüksekliği 80 m olacak bir sabit yarıçaplı (silindirik)
kemer barajda tabandaki maksimum kemer
kalınlığının, en fazla baraj yüksekliğinin ¼ üne eşit
olması istendiğine gore kemer yarıçapını
hesaplayınız. Kemer halkalarının emniyet gerilmesi
220 t/m², betonun elastisite modülü 2×106 t/m²
olduğuna göre tepede anahtar noktasında ortaya
çıkacak sehimi bulunuz.
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Çözüm
σh
rd = t ×
p
t = h/4 alalım
:
σh
⎛ h ⎞
rd = ⎜ ⎟ ×
⎝ 4 ⎠ (γ × h )
p=γ×h
rm = rd −
t = 80 / 4 = 20 m
t
2
= 55 −
20
= 45 m
2
δ = σh
1 σ
1 220
rd = × h = ×
= 55 m
4 γ 4 1
rm
E
0.5 cm sehim oluşur.
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δ = 220 ×
45
= 0.005 m
6
2 × 10
Örnek
n
Yüksekliği 100 m ve tepe genişliği 5 m olacak
bir kemer barajın tepeden itibaren 0, 25, 50,
75 ve 100 m derinliklerdeki eksen yarıçapları
sıra ile 95, 80, 63, 58, 50 m dir. Betonun
emniyet gerilmesi 300 t/m² ve elastisite
modülü 2×106 t/m² dir. Her seviyede kemer
halkasının kalınlık ve sehimini bulunuz.
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Kemer Barajlar (Hesap Esasları)
n
Bağımsız Halkalar Yöntemi
h
t
t
0
25
25
25
t/2
t/2
rm
rd
50
25
75
100
25
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rm = rd – t/2
O
Çözüm
2
2
σ h = σ bem = × 300 = 200 t / m ²
3
3
rd





p

p
t ⎞ 
⎛

p × rd
(
γ × h )× ⎜ rm + ⎟
t
t=
rd = rm +
2 ⎠ (γ × h )× (rm + 0.5 × t )
⎝
t
=
=
2
σh
σh
σh
γ × h × rm + γ × h × 0.5 × t
t=
σh
γ = 1 t/m³ , σh = 200 t/m²
Sehim:
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için:
γ × h × rm
t=
(σ h − γ × h × 0.5)
h × rm
t=
(200 − h × 0.5)
rm
δ = σh
E
Ç.Ü. İnş.Müh.Böl.
bulunur.
t=
h × rm
(200 − h × 0.5)
δ = σh
Sıra
No
Baraj tepesinden itibaren
derinlik
h (m)
1
0
95
0,00
0,95
2
25
80
10,67
0,80
3
50
63
18,00
0,63
4
75
58
26,77
0,58
5
100
50
33,33
0,50
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Eksen yarıçapı Kemer kalınlığı
rm (m)
t (m)
rm
E
Sehim
δ (cm)