# Arch Dams

```Arch Dams
Translated from the slides of
Prof. Dr. Recep YURTAL (&Ccedil;.&Uuml;.)
by his kind courtesy
ercan kahya
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
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.
Recep YURTAL
Arch Dams
!
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
Arch Dams
n
Where &amp; why we design it:
q
q
q
q
q
Appropriate when “Width / Height of valley” (B/H) &lt; 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.
Recep YURTAL
Arch Dams
Base width
Arch Width
Upstream
face
Crest
Downstream
face
Base
width
Height
Toe
Axis
Central angle
Recep YURTAL
Downstream
Face
Base
Arch Dam Types
Constant Center Arch Dam (variable angle)
n
q
Good for U-shaped valleys
q
Easy construction
q
Vertical upstream face
q
Appropriate for middle-high
dams
Recep YURTAL
Arch Dam Types
Variable Center Arch Dams (constant angle)
n
2
q
Good for V-shaped valleys
4
1
q
Limited to the ratio of B/H=5
q
Best center angle: 133o 34&acute;
3
3-4
q
To obtain arch action at
the bottom parts
1-2
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
Arch Dam Types
Variable Center –
Variable Angle Arch
Dams
n
q
q
Recep YURTAL
crest
Combination of the two
above.
a
Its calculation based on
shell theory applied to
arch dams
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
a
a
a
Arch Dam Types
⇐
Recep YURTAL
Upstream
Arch Dams
n
3.6.2 DESIGN OF ARCH DAMS
Structural Design: &agrave; Load distribution in the dam body
&amp; beyond scope of this course
q
q
q
q
Independent ring method
Elasticity theory
Shell theory
Recep YURTAL
Arch Dams
n
&sup2;
&sup2;
&sup2;
&sup2;
&sup2;
&sup2;
Hydraulic Design
Determination of thickness at any elevation
Effect of uplift force – ignored
Stresses due to ice &amp; 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
Recep YURTAL
rd
O
y
Hh
dFV&acute;
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
Recep YURTAL
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
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&acute;
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
Recep YURTAL
θa/2
The required thickness of
the rib (t) when t &lt;&lt; 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 &times; h:
p &times; 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)
Recep YURTAL
Arch Dams (design principles)
▪ The optimum θa for minimum volume of arch rib:
dV / dθ = 0
&agrave; 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 &lt; θa &lt; 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) &amp; arch radius (r):
L
r=
2 &times; Sin φ
Recep YURTAL
t/2
t/2
rm
rd
O
rm = rd – t/2
&Ouml;rnek
n
Y&uuml;ksekliği 120 m olan bir barajın ağırlık ve kemer-ağırlık t&uuml;r&uuml;nde
yapılması halleri i&ccedil;in taban genişliklerini ve birim kalınlıktaki hacimlerini
karşılaştırınız. Barajda taban su basıncı k&uuml;&ccedil;&uuml;ltme fakt&ouml;r&uuml; 0.8 dir. γb =
2.4 t/m&sup3;
&Ccedil;&ouml;z&uuml;m:
b
=
h
1
γʹ′ − m
Hacim=
Recep YURTAL
Ağırlık baraj i&ccedil;in:
b
120
=
γb
γʹ′ =
γw
1
2.4 − 0.8
120 &times; 95 &times;
b = 120 &times;
1
= 5700 m&sup3;
2
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
γʹ′ =
1
2.4 − 0.8
2.4
= 2.4
1.0
b= 95 m.
n
Kemer-Ağrlık baraj i&ccedil;in:
m=0
bʹ′
=
h
1
γʹ′
bʹ′ = 120 &times;
1
= 78 m
2. 4
1
Hacim = 120 &times; 78 &times; = 4680 m &sup3;
2
Hacim azalması :
5700 − 4680
= % 18
5700
120
95
78
Ağırlık yerine kemer-ağırlık se&ccedil;meyle beton hacminde %18 azalma olacaktır.
Bu se&ccedil;im ancak yama&ccedil;lar yeterince sağlam olduğunda yapılabilir.
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
&Ouml;rnek
n
Y&uuml;ksekliği 80 m olacak bir sabit yarı&ccedil;aplı (silindirik)
kemer barajda tabandaki maksimum kemer
kalınlığının, en fazla baraj y&uuml;ksekliğinin &frac14; &uuml;ne eşit
olması istendiğine gore kemer yarı&ccedil;apını
hesaplayınız. Kemer halkalarının emniyet gerilmesi
220 t/m&sup2;, betonun elastisite mod&uuml;l&uuml; 2&times;106 t/m&sup2;
olduğuna g&ouml;re tepede anahtar noktasında ortaya
&ccedil;ıkacak sehimi bulunuz.
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
&Ccedil;&ouml;z&uuml;m
σh
rd = t &times;
p
t = h/4 alalım
:
σh
⎛ h ⎞
rd = ⎜ ⎟ &times;
⎝ 4 ⎠ (γ &times; h )
p=γ&times;h
rm = rd −
t = 80 / 4 = 20 m
t
2
= 55 −
20
= 45 m
2
δ = σh
1 σ
1 220
rd = &times; h = &times;
= 55 m
4 γ 4 1
rm
E
0.5 cm sehim oluşur.
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
δ = 220 &times;
45
= 0.005 m
6
2 &times; 10
&Ouml;rnek
n
Y&uuml;ksekliği 100 m ve tepe genişliği 5 m olacak
bir kemer barajın tepeden itibaren 0, 25, 50,
75 ve 100 m derinliklerdeki eksen yarı&ccedil;apları
sıra ile 95, 80, 63, 58, 50 m dir. Betonun
emniyet gerilmesi 300 t/m&sup2; ve elastisite
mod&uuml;l&uuml; 2&times;106 t/m&sup2; dir. Her seviyede kemer
halkasının kalınlık ve sehimini bulunuz.
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
Kemer Barajlar (Hesap Esasları)
n
Bağımsız Halkalar Y&ouml;ntemi
h
t
t
0
25
25
25
t/2
t/2
rm
rd
50
25
75
100
25
Recep YURTAL
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
rm = rd – t/2
O
&Ccedil;&ouml;z&uuml;m
2
2
σ h = σ bem = &times; 300 = 200 t / m &sup2;
3
3
rd





p

p
t ⎞ 
⎛

p &times; rd
(
γ &times; h )&times; ⎜ rm + ⎟
t
t=
rd = rm +
2 ⎠ (γ &times; h )&times; (rm + 0.5 &times; t )
⎝
t
=
=
2
σh
σh
σh
γ &times; h &times; rm + γ &times; h &times; 0.5 &times; t
t=
σh
γ = 1 t/m&sup3; , σh = 200 t/m&sup2;
Sehim:
Recep YURTAL
i&ccedil;in:
γ &times; h &times; rm
t=
(σ h − γ &times; h &times; 0.5)
h &times; rm
t=
(200 − h &times; 0.5)
rm
δ = σh
E
&Ccedil;.&Uuml;. İnş.M&uuml;h.B&ouml;l.
bulunur.
t=
h &times; rm
(200 − h &times; 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
Recep YURTAL
Eksen yarı&ccedil;apı Kemer kalınlığı
rm (m)
t (m)
rm
E
Sehim
δ (cm)
```