Chapter 3 Flow - stability Introduction • focus on stability of loose non-cohesive grains • rock: important material for protection • grains may vary in size from μc (sand) to m (rock) Uniform flow – Horizontal bed Forces on a grain in flow 1 2 1 2 1 2 2 DD D w 222 SS F ww 2 LL L w Drag force : = Cu FA Shear force : = F Cu ud FA Lift force : = Cu FA ρ ρρ ρ ⎫ ⎪ ⎪ ⎪ ≈ ⎬ ⎪ ⎪ ⎪ ⎭ Balance equations d g K = u d g = d g - u 2 c w ws 2 c Δ→Δ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ∝ ρ ρρ d g ) - ( d u dO W = dO F : 0 = M W = F : 0 = V FW x f = F : 0 = H 3 ws 22 c w SD, L F SD, ρρρ ∝ ⎪ ⎭ ⎪ ⎬ ⎫ ⋅⋅Σ Σ =Σ )()( Relation between load and strength Isbash (1930) g 2 u 0.7 = d or 1.7 = d g u or d g 2 1.2 = u 2 cc c Δ Δ Δ used for first approximation when: • relation between velocity and waterdepth not clear (e.g a jet entering a body of water) () () 2 ** * Re cc c c sw uud ff gd gd τ ψ ρ ρυ ⎛⎞ ==== ⎜⎟ −Δ ⎝⎠ Shields (1936) * * Re c ud υ = : c ψ Shields parameter (stability parameter) Note: : ψ Mobility parameter (when actual u used) * Re Re≠ Critical shear stress Shields Van Rijn () () ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ == Δ = − = υρρ τ ψ du ff dg u dg cc ws c c * * 2 * Re Example 1500 1033.1 002.01 Re 6 * * = × × == − ν du c What is u *c for sand with d = 2 mm? • Ψ c = 0.055 • smgdu gd u cc c c /042.0002.081.965.1055.0 * * =×××=Δ=→ Δ = ψ ψ 04.063 1033.1 002.0042.0 Re 6 * * =→= × × == − c c du ψ ν smgdu cc /036.0002.081.965.104.0 * = × × × =Δ= ψ Guess: u *c = 1 m/s Shields Example (cont.) 42 )1033.1( 81.965.1 002.0 3 26 3 2 * = × × = Δ = ν g dd d = 2 mm 0.04 c Ψ= • smgdu cc /036.0002.081.965.104.0 * = × × × =Δ= ψ Van Rijn Relative protrusion [...]... Kv - factors for various structures Shape Rectangular Trapezoidal RectAngular Round K v0 b 0 *K vG /b G K vG 1 .3 - 1.7 K vM 1.1 - 1.2 b 0 *K v b G 1.2 1 b 0 *K v /b G 1 .3 - 1.7 1.2 b 0 *K v /b G 1.2 - 1 .3 1.2 b 0 *K v /b G 1 - 1.1 1 - 1.1 Pier Stream Lined Round Outflow RectAngular Abruptly Sill Stream Lined Top Structure Groyne Abutment b 0 *K v /b G 2*K v b 0 *K v /b G 2*K v ⊗ 1.2 - 1.4 ⊗ 1 - 1.1... Round Outflow RectAngular Abruptly Sill Stream Lined Top Structure Groyne Abutment b 0 *K v /b G 2*K v b 0 *K v /b G 2*K v ⊗ 1.2 - 1.4 ⊗ 1 - 1.1 ⊗ 1.4 - 1.6 ⊗ 1.2 - 1 .3 1 0.9 Section 3. 6.1 Fig 3. 13 Section 3. 6.1 Fig 3. 13 Section 3. 6.1 Fig 3. 13 Down Stream ⊗ For many piers in a river the first expression for K v is appropriate The second is valid for a detached pier in an infinitely wide flow,... non-cohesive materials 1:1 1:1.5 1:2 1 :3 Influence of slope on stability φ = 40ο Case b: slope parallel to flow Case c: slope perpendicular to flow Slope parallel to current K( α // ) = F( α // ) W cos α tan φ - W sin α = = F(0) W tan φ α sin φ cos α − cos φ sin α sin ( φ - α ) = = sin φ sin φ Slope perpendicular to current 2 2 2 F( α ) cos α tan φ - sin α cos α K( α ) = = = 2 F(0) tan φ tan α = 1 -. .. zero) * qs = 6.56 ⋅ 1018 ψ 16 * qs = 13 ψ 2.5 (for ψ < 0.05) ⎫ qs ⎪ * with qs = ⎬ (for ψ > 0.05) ⎪ Δ g d3 ⎭ Stone dimension Nominal diameter: dn = 3 V = 3 M / ρ d n50 ≠ d50 d n50 ≈ 0.84 d50 Influence of waterdepth Uniform flow: u* = u g C 2 u*c ψc = Δgd u Δ g d Isbash: = c n50 uic Δgd Attention: uc ≠ uic =1.7 C ψ c g 12 R C =18log kr roughness kr = 2*d50 or kr = 3* d50 Influence waterdepth on critical... =18log = 18log 2d n50 d n50 α = 30 o; φ = 40o Stability on head of dam Deceleration u c without structure u cu Kv = = u c with structure u cs ucu: vertically averaged critical velocity in uniform flow ucs: velocity in case with a structure Effect of flow field Relation between Kv and turbulence level b g (1 + 3rcu ) ucu = 1 + 3rcs ucs ucu 1 + 3rcs ⎯ → Kv = ⎯ = ucs 1 + 3rcu ucu : vertically averaged critical... (hu - hd ) = (0.5 + 0.04 2 1 discharge coefficient hd ) 2 g (hu - hd ) d n 50 Vertical constriction Stability with flow under weir Shields in horizontal constriction (horizontal closure with trucks) ψc u gap =C g Δgd n 50 General formula sin α 4 12 sin φ 2 Correction α slope of construction ϕ angle of repose (internal stability) ⎛ 3h ⎞ = 4.5log ⎜ ⎟ ψc ⎝ d n 50 ⎠ Damage at half depth: 12 × 1/ 2h 3h C... locations 2 frequent movement at some locations 3 frequent movement at several locations 4 frequent movement at many locations 5 frequent movement at all locations 6 continuous movement at all locations 7 general transport of the grains Shields Videos on stability of rock on a bed with current only u = 0.60 m/s, Ψ = 0. 03 u = 0.70 m/s, Ψ = 0.04 u = 0. 83 m/s, Ψ = 0.05 u = 0.92 m/s, Ψ = 0.06 u = 0.97 . 0.055 • smgdu gd u cc c c /042.0002.081.965.1055.0 * * =×××=Δ=→ Δ = ψ ψ 04.0 63 1 033 .1 002.0042.0 Re 6 * * =→= × × == − c c du ψ ν smgdu cc / 036 .0002.081.965.104.0 * = × × × =Δ= ψ Guess: u *c = 1 m/s Shields Example (cont.) 42 )1 033 .1( 81.965.1 002.0 3 26 3 2 * = × × = Δ = ν g dd d = 2 mm 0.04 c Ψ= • smgdu cc / 036 .0002.081.965.104.0 * = × × × =Δ= ψ Van. Rijn () () ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ == Δ = − = υρρ τ ψ du ff dg u dg cc ws c c * * 2 * Re Example 1500 1 033 .1 002.01 Re 6 * * = × × == − ν du c What is u *c for sand with d = 2 mm? • Ψ c = 0.055 • smgdu gd u cc c c /042.0002.081.965.1055.0 * * =×××=Δ=→ Δ = ψ ψ 04.0 63 1 033 .1 002.0042.0 Re 6 * * =→= × × == − c c du ψ ν smgdu cc / 036 .0002.081.965.104.0 * = × × × =Δ= ψ Guess:. Chapter 3 Flow - stability Introduction • focus on stability of loose non-cohesive grains • rock: important material for protection • grains may vary in size