05lecture-ArchDams

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Recep YURTAL Ç.Ü. İnş.Müh.Böl.

Arch Dams

Translated from the slides of

Prof. Dr. Recep YURTAL (Ç.Ü.)by his kind courtesy

ercan kahya



Recep YURTAL

Arch Dams

• Curved in plan and carry most of the water thrusthorizontally to the side abutments by arch action.

• A certain percentage of water load is verticallytransmitted to the foundation by cantilever action.



!

Arch Dams









Recep YURTAL

n Where & why we design it:

q Appropriate when “Width / Height of valley” (B/H) < 6

q Rocks at the base and hillsides should be strong enoughwith high bearing capacity.

q To save in the volume of concrete.

q Stresses are allowed to be as high as allowable stress of

concrete.q Connection to the slope of hillsides should be 45o at least.

Arch Dams



Recep YURTAL

Base

width radiousp

Central angle

Arch Width

Downstream

Face

Toe

Downstreamface

Crest Upstream

face

Base width

Height

Base

Axis

Arch Dams



Recep YURTAL

Arch Dam Types

n Constant Center Arch Dam (variable angle )

q Good for U-shaped valleys

q Easy construction

q Vertical upstream face

q Appropriate for middle-highdams




n Variable Center Arch Dams (constant angle)

q Good for V-shaped valleys

q Limited to the ratio of B/H=5

q Best center angle: 133o 34´

q To obtain arch action atthe bottom parts

3-4

1-2

1

2

3

4

Arch Dam Types




n Variable Center –Variable Angle ArchDams

q Combination of the twoabove.

q Its calculation based onshell theory applied to

arch dams

crest

a a

a a

Arch Dam Types



Recep YURTAL

⇐ Upstream

Arch Dam Types



Recep YURTAL

n 3.6.2 DESIGN OF ARCH DAMS

Structural Design: à Load distribution in the dam body

& beyond scope of this course

q Independent ring method

q Trial load method

q Elasticity theory

q Shell theory

Arch Dams



n Hydraulic Design

² Determination of thickness at any elevation

² Effect of uplift force – ignored² Stresses due to ice & temp changes-important

² Arch action - near the crest of dam

² Cantilever action - near the bottom of dam

² Horizontal hydrostatic pressure is assumed to betaken by arch action

Arch Dams



Recep YURTAL

Arch Dams (design principles)

n Independent Ring method:

h

t

t/2t/2

t

r m r d

r m = r d – t/2O



Recep YURTAL

R R

Hh

Ba

r

dFV

x

y

p p

dFV´ φ

θa φ dφ

t

Differential forcefor infinitely small

elemental piece withcenter angle of dφ :

φd r pdF V ..

Vertical component:

φ Sind r pdF V ...

ʹ



▪ Total horizontal force:

h : height of arch lib from the reservoir surfacer : radius of archθa: central angle

▪ Equilibrium of forces in the flow direction (y):

H h = 2 R y

Ry: reaction force at the sides in y direction [= R sin ( θ a /2)]


2...2

ah Sinr h H

θγ



Recep YURTAL

R R

Hh

θa/2

Ba

r

dFV

x

y

p p

dFV´ φ

θa φ dφ

t

Reaction of the sides (R):

θa/2 Rx Ry

Ry

Rx

r h R ..

The required thickness of the rib (t) when t << r :

t/R ≈σ

t =γ .h.r

σ allσall : allowable working stress

for concrete in compression




V =

γ .h

σ all

r

2

θ a

The volume of concrete for unit height for a single arch:

V = L t

L: arch length (L = r θa) (note that θa is in radians)



Recep YURTAL

n Pressure p = γ × h:

t = p× rd

σ all

t/2t/2

t

r m r d

r m = r d – t/2

O

Reduction factor for base pressure at arch dams is zero (m=0)


The required thickness of the rib (t):




▪ The optimum θa for minimum volume of arch rib:

d V / d θ = 0 à θa = 133o 34’

► This is the reason why the constant-angle damsrequire less concrete that the constant-center dams

► Formwork is more difficult

► In practice; 100o < θa < 140o for the constant-angledams



Recep YURTAL

t/2t/2

t

r m r d

r m = r d – t/2

O

n Displacement in an arch ring:

δ =σ all

rm

E

E = Concrete elasticity modulus

r m = Radius from ring axis

For the relation betweencenter angle (2φ), beam

length (L) & arch radius (r ):

φ×=

Sin2

Lr





Örnek

n Yüksekliği 120 m olan bir barajın ağırlık ve kemer-ağırlık türündeyapılması halleri için taban genişliklerini ve birim kalınlıktaki hacimlerinikar şılaştırınız. Barajda taban su basıncı küçültme faktörü 0.8 dir. γb =2.4 t/m³

Ağırlık baraj için:Çözüm:

w

b

γ

γ=γʹ

4.20.1

4.2==γʹ

m

1

h

b

−γʹ=

8.04.2

1

120−

=

b

8.04.2

1120

−

×=b

Hacim= ³57002

195120 m =××

b= 95 m.




n Kemer-Ağrlık baraj için:

m = 0γʹ

=

ʹ 1

h

b

m784.2

1120 b =×=ʹ

Hacim = ³m46802

178120 =××

Hacim azalması : 18%5700

46805700=

−

Ağırlık yerine kemer-ağırlık seçmeyle beton hacminde %18 azalma olacaktır.

Bu seçim ancak yamaçlar yeterince sağlam olduğunda yapılabilir.

95

120

78




Örnek

n Yüksekliği 80 m olacak bir sabit yarıçaplı (silindirik)kemer barajda tabandaki maksimum kemer kalınlığının, en fazla baraj yüksekliğinin ¼ üne eşitolması istendiğine gore kemer yarıçapınıhesaplayınız. Kemer halkalarının emniyet gerilmesi220 t/m², betonun elastisite modülü 2×106 t/m² olduğuna göre tepede anahtar noktasında ortayaçıkacak sehimi bulunuz.




Çözüm

t = h/4 alalım t = 80 / 4 = 20 m :

ptr h

d

σ

×=

p = γ × h( )h4

hr

h

d×γ

σ×⎟

⎠

⎞⎜⎝

⎛= m55

1

220

4

1

4

1r

h

d=×=

γ

σ×=

m t

r r d m 452

2055

2=−=−=

E

r m h σ=δ m005.0

102

45220

6=

××=δ

0.5 cm sehim oluşur.




Örnek

n Yüksekliği 100 m ve tepe genişliği 5 m olacakbir kemer barajın tepeden itibaren 0, 25, 50,75 ve 100 m derinliklerdeki eksen yarıçapları

sıra ile 95, 80, 63, 58, 50 m dir. Betonunemniyet gerilmesi 300 t/m² ve elastisitemodülü 2×106 t/m² dir. Her seviyede kemer halkasının kalınlık ve sehimini bulunuz.




Kemer Barajlar (Hesap Esasları)

n Bağımsız Halkalar Yöntemi

h

t

t/2t/2

t

r m r d

r m = r d – t/2O

25

25

25

25

0

25

50

75100




Çözüm

²m/t200300

3

2

3

2

em bh =×=σ=σ

2

t r r m d

+=

h

dr pt

σ

×

=

( )( ) ( )

h

m

p

h

dr

m

p

t5.0r h2

tr h

t

σ

×+××γ=

σ

⎟⎠

⎞⎜⎝

⎛+××γ

=

h

mt5.0hr h

tσ

×××γ+××γ=

( )5.0h

r ht

h

m

××γ−σ

××γ=

γ = 1 t/m³ ,σh = 200 t/m² için: ( )5.0h200

r h

tm

×−

×

= bulunur.

Sehim:

E

r m

hσ=δ



Recep YURTAL

SıraNo Baraj tepesinden itibaren

derinlik

h (m)

Eksen yarıçapırm (m)

Kemer kalınlığıt (m)

Sehimδ (cm)

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 0,585 100 50 33,33 0,50

E

r m

hσ=δ( )5.0h200

r ht

m

×−

×

=

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05lecture-ArchDams