Arch Dams. Prof. Dr. Recep YURTAL (Ç.Ü.) Translated from the slides of. by his kind courtesy. ercan kahya

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1 Arch Dams Translated from the slides of Prof. Dr. (Ç.Ü.) by his kind courtesy ercan kahya Ç.Ü. İnş.Müh.Böl.

2 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.

3 Arch Dams!

4 Ç.Ü. İnş.Müh.Böl.

5 Ç.Ü. İnş.Müh.Böl.

6 Arch Dams 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 enough with 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 45 o at least.

7 Arch Dams Base width Arch Width Crest Upstream face Downstream face Height Base width radiousp Toe Axis Central angle Downstream Face Base

8 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-high dams

9 Arch Dam Types n Variable Center Arch Dams (constant angle) 2 q Good for V-shaped valleys 1 4 q Limited to the ratio of B/H=5 3 q Best center angle: 133 o 34 q To obtain arch action at the bottom parts Ç.Ü. İnş.Müh.Böl.

10 Arch Dam Types n Variable Center Variable Angle Arch Dams crest q Combination of the two above. a a q Its calculation based on shell theory applied to arch dams a a Ç.Ü. İnş.Müh.Böl.

11 Arch Dam Types Upstream

12 Arch Dams n 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

13 Arch Dams n Hydraulic Design ² ² ² ² ² ² 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

14 Arch Dams (design principles) n Independent Ring method: t h t t/2 t/2 r m r d r m = r d t/2 O

15 p t H h y df V φ df V p Differential force for infinitely small elemental piece with center angle of dφ : R θ a dφ φ x R df V = p. r. dφ r Vertical component: B a df ʹ V = p. r. dφ. Sinφ

16 Arch Dams (design principles) Total horizontal force: H h = θ γ. h. r. Sin 2 2 a h : height of arch lib from the reservoir surface r : radius of arch θ a : central angle Equilibrium of forces in the flow direction (y): H h = 2 R y R y : reaction force at the sides in y direction [= R sin (θ a /2)]

17 y H h df V φ df V Reaction of the sides (R): p p θ a /2 R Rx Ry t θ a dφ φ x r Ry R θ a /2 Rx R = γ. h. r The required thickness of the rib (t) when t << r : B a σ R / t σall : allowable working stress for concrete in compression t = γ.h.r σ all

18 Arch Dams (design principles) The volume of concrete for unit height for a single arch: V = L t L: arch length (L = r θ a ) (note that θ a is in radians) V = γ.h σ all r 2 θ a

19 Arch Dams (design principles) The required thickness of the rib (t): n Pressure p = γ h: t t = p r d σ all t/2 t/2 r m r m = r d t/2 O r d Reduction factor for base pressure at arch dams is zero (m=0)

20 Arch Dams (design principles) The optimum θ a for minimum volume of arch rib: dv / dθ = 0 à θ a = 133 o 34 This is the reason why the constant-angle dams require less concrete that the constant-center dams Formwork is more difficult In practice; 100 o < θ a < 140 o for the constant-angle dams

21 Arch Dams (design principles) n Displacement in an arch ring: δ = σ all r m E E = Concrete elasticity modulus r m = Radius from ring axis t For the relation between center angle (2φ), beam length (L) & arch radius (r): r = L 2 Sin φ t/2 t/2 r m r m = r d t/2 O r d

22 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. γ b = 2.4 t/m³ Çözüm: Ağırlık baraj için: γ γ b γʹ = γʹ = = 2. 4 w b h = 1 γʹ m b 1 = b = b= 95 m. Hacim= = 5700 m³ 1 2 Ç.Ü. İnş.Müh.Böl.

23 n Kemer-Ağrlık baraj için: m = 0 b ʹ 1 = h γʹ 1 b ʹ = 120 = Hacim = = 4680 m³ 78 m Hacim azalması : = % Ağırlık yerine kemer-ağırlık seçmeyle beton hacminde %18 azalma olacaktır. Bu seçim ancak yamaçlar yeterince sağlam olduğunda yapılabilir. Ç.Ü. İnş.Müh.Böl.

24 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 t/m² olduğuna göre tepede anahtar noktasında ortaya çıkacak sehimi bulunuz. Ç.Ü. İnş.Müh.Böl.

25 Çözüm r d = t σ p h t = h/4 alalım t = 80 / 4 = 20 m : p = γ h r d h σ 1 σ h h = rd = = = 55 m 4 ( γ h) 4 γ 4 1 t 20 rm = rd = 55 = m δ = r σ m h E δ = = m 0.5 cm sehim oluşur. Ç.Ü. İnş.Müh.Böl.

26 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 t/m² dir. Her seviyede kemer halkasının kalınlık ve sehimini bulunuz. Ç.Ü. İnş.Müh.Böl.

27 Kemer Barajlar (Hesap Esasları) n Bağımsız Halkalar Yöntemi t h t/2 t/2 t r m r m = r d t/2 O r d Ç.Ü. İnş.Müh.Böl.

28 Çözüm 2 2 σh = σb em = = 3 rd = rm + t 2 t = p σ h r d 200 t / m² t = p ( γ h) r σ rd h m t + 2 = p ( γ h) ( r t) m σ h t = γ h r m + γ h 0.5 t σ γ = 1 t/m³, σ h = 200 t/m² için: h t t = γ h r ( σ γ h 0.5) h h r m = bulunur. ( 200 h 0.5) m Sehim: δ = σ h r m E Ç.Ü. İnş.Müh.Böl.

29 t = h r m ( 200 h 0.5) δ = σ h r m E Sıra No Baraj tepesinden itibaren derinlik h (m) Eksen yarıçapı rm (m) Kemer kalınlığı t (m) Sehim δ (cm) ,00 0, ,67 0, ,00 0, ,77 0, ,33 0,50

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