Arch Dams Translated from the slides of Prof. Dr. (Ç.Ü.) by his kind courtesy ercan kahya 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.
Arch Dams!
Ç.Ü. İnş.Müh.Böl.
Ç.Ü. İnş.Müh.Böl.
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.
Arch Dams Base width Arch Width Crest Upstream face Downstream face Height Base width radiousp Toe Axis Central angle Downstream Face Base
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
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 3-4 1-2 Ç.Ü. İnş.Müh.Böl.
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.
Arch Dam Types 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 q Independent ring method q Trial load method q Elasticity theory q Shell theory
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
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
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
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)]
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
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
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)
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
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
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 2.4 1.0 b h = 1 γʹ m b 1 = 120 2.4 0.8 b = 120 1 2.4 0.8 b= 95 m. Hacim= 120 95 = 5700 m³ 1 2 Ç.Ü. İnş.Müh.Böl.
n Kemer-Ağrlık baraj için: m = 0 b ʹ 1 = h γʹ 1 b ʹ = 120 = 2.4 1 2 Hacim = 120 78 = 4680 m³ 78 m 120 5700 4680 5700 Hacim azalması : = % 18 78 95 Ağırlık yerine kemer-ağırlık seçmeyle beton hacminde %18 azalma olacaktır. Bu seçim ancak yamaçlar yeterince sağlam olduğunda yapılabilir. Ç.Ü. İnş.Müh.Böl.
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 10 6 t/m² olduğuna göre tepede anahtar noktasında ortaya çıkacak sehimi bulunuz. Ç.Ü. İnş.Müh.Böl.
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 1 220 = rd = = = 55 m 4 ( γ h) 4 γ 4 1 t 20 rm = rd = 55 = 45 2 2 m δ = r σ m h E δ = 45 220 6 2 10 = 0.005 m 0.5 cm sehim oluşur. Ç.Ü. İnş.Müh.Böl.
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 10 6 t/m² dir. Her seviyede kemer halkasının kalınlık ve sehimini bulunuz. Ç.Ü. İnş.Müh.Böl.
Kemer Barajlar (Hesap Esasları) n Bağımsız Halkalar Yöntemi t h 25 25 25 25 0 25 50 75 100 t/2 t/2 t r m r m = r d t/2 O r d Ç.Ü. İnş.Müh.Böl.
Çözüm 2 2 σh = σb 300 3 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 + 0.5 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.
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) 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