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Permeability 滲透性 = Vi h. ++ 在土裡流很慢 U pressure hydraulic gradient 水力坡降; 單位高度下降多少水頭 L:平行V的長度 : oh head loss i sh Darcy's Law v=ki Vi 孔係大大 discharge velocity k coefficient of permeability 滲透係數 Vs. Seepage velocity V₁ = V (1+2) : : k = 1 k k absolute pormenbility 刂: dynamic viscosity 動力黏淨係數
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k20°c Hydralic conductivity = T°C 20°c 可查表 Constant-head test 定水頭 Table 7.2 Variation of rere Temperature, TCC) McMore Temperature, 7(°C) rcre 15 1.135 23 0.931 16 1.106 24 0.910 17. 1.077 25 0.889 18 1.051 26 0.869 19 1.025 27 0,850 20 1,000 28 0.832 21 0.976 29 0.814 22 0.953 30 0.797 總水頭B=Lz 總水頭A = L+ L h A A-B = h B k = QL Aht : Q volume of water A+V ·A· Soil Cross-sectional area 1 : t time Falling - head test 土 Stand pipe th m ( 71 ) 9 = k =—=—A = (學)(m1) adt. 9 flow rate a pipe cross-sectional area 細管截面積 Porous stone Soil specimen
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Stratified soil s horizontal" = Same q = keq ieq A = keq i/eq (1x H) H₁ H₂ = kv₁ H3. vertical 1 → V = A Same Avot ki i i H₁ + k ₂ i 2 Hz +... + kninH₂ i, H. kz iz ki = ki H, + kx H + + kn Hni H 2 : = bc Darcy's Law V-ki→ same H = L: 平行V 的長度 = →i V-kii-ki, k = = h = h₁ + h₂ + h 3 + ... + h n V = Hii + Hz iz +.... v = keq 1/4 keq = V + H = hi Hi •+ Hnin tx. H (kiii) (hith₂...thn) +.. .) kv H H.. Haiz kip ·H₁ + + H + ... +
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EX76 k = 0.08 9 = ? cm flow rate 50 Impervious layer Permeable layer 3 cos a (m) S cos a q=ki A = (b) Direction of flow Ah S tan a k h Cosc (3059×1)
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EX. 7.15 q= ? · k v Water supply hA Constant-head difference 300 mm 150 mm 150 mm 150 mm Figure 7.21 Three layers of soil in a tube 100 mm x 100 mm in cross section Hi H 450 + + 150 10-2 150 + + 150 3x103 4.9x q=ki A, i h = L = 12 x 10-3 cm = (12/× 10³) (-320) (10:10) (60.60) =291 cm³ hr s
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Ex. 7 9 0.0809 q ( cm ) = ? cm 3 Sec Water supply B C Constant-head difference 300 mm Soil A B C k (cm/sec) 10-2 3 x 10-3 4.9 × 10-4 150 mm 150 mm 150 mm → Figure 7.21 Three layers of soil in a tube 100 mm x 100 mm in cross section = q=ki A oh=24 KA koh A 0.08.09.15 102 (10·10) = = 12135 cm 12.135 mm 300-12.135 = 287.865 mm hA = 300 - oha = 94B KBA (3×10) (10.10) = 4.045 cm = 40. 45 mm. 0.0809.15 hB = (300-ohA) - ohB 247.415mm
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Permeability test 滲透試驗 抽水 觀測 q= = dh qk (½) (zh) Jr = 2 łuk Shind h dr = L: 平行V的長度
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seepage 滲流 When x. 方向的滲透係數一樣(kx=kz) Laplace's equation J³h dx² 十 flow nets 用采其地下水的流量 等势能線 equipotential line 有相同煦水頭 Jh 0 Jz2 流線:水流方向 flow line 1 梯 2 (水頭差) 厦 Nd: number of potential drops H₂ flow channel k Nf number of H 壓力 水頭 flow line hy 2 3 flow channels 高度 透水土 水頭 2 5 hz h3 3 4 hb h.4 不透水土 h-h₂ = H 告 = H NJ
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09 flow line h. h₂ h3 = 2 09-09 09... equipotential line * 1 2 14 = oh, q= ki A = hi-hz oq k (^_^) (1x1) = koh 流網總流量 q = k Nd 如果流網不是 l 而是 Nd h Nf 2) b l
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EX. 8.1 k = 5x 10's cm mih Water level Ground surface Scale 5 m 5.6 m 2.2 m Water table Datum 2-0 Flow channel I b Flow channel 2 1 I Flow channel 3 b 0.38 Impervious layer a Figure 8.6 Flow net for seepage around a single row of sheet piles 靜壓管(高度水頭+壓力水頭) How high (above the ground surface) the water will rise if piezometers are placed at points a and b H = 5.6-2.2 = 3.4m h= 3.4 3.4 = = Nd 0.567m a: 高度水頭 + 壓力水頭 總水頭 (-y) + (x+y) x= 5.6-h=5.6-0.567 b: 5.6-5h = 2.765m = 5.033 m
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b. The total rate of seepage through the permeable layer per unit length q = k H Nf = Nd m³ (5 × 10³) 3.4 2.38 6.74 × 10's in s = C. The approximate average hydraulic gradient at c head loss length of flow 0.567 4.1 = 0.138
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EX.8.2 k = 1.5× 103 cm 5 m Nd=10 Figure 8.7 50m Retaining wall 56 n do 8 9 Sand Impervious layer q = k Hj Nf 5 = (15×105) 03 Q = 09x50 10 :3 2.25×105 m m (2.25 x 10³) x 50 x (60x60x24) = 97.2 day
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EX.8.3 Nf = 2.9 NJ-10 H=215+2 2.5 m 2.0 m * 3.0 m Sheet pile 4.0 m 2 3 45 / 10.9 21 Figure 8.9 Impervious stratum Ground Level Clayey sand 10-40 k = 2×10 cm H q = k 1 Np Nd Nf = (2×106) 21572 -2.9 = 2.61×106 (1) 10 左半+右半 (2-b1×10 ) x2 x (60x60x24) = 5.22 x 10° (box box24) 0.452
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Anisotrophic soil kx + kz 異向 q = √kx ky HJ Nf uplift pressure Ad 7m 10 m ↓ 8m 2m 壓力水设 14m abcd ef 2 3 45 (a) Datum k = k = k 2=0 Impermeable layer A 總水頭 = 2+7=9 NJ=7,壓力水頭每格降7 : 1m a 的總水頭= 9-1 = 8 m Pa = 8rw 14m a b C d 7ye kN/m² 6y.0 kN/m² 8ye kN/m² (b) 3ye kN/m² 4ye kN/m² Syue kN/m²
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Earth dam on an imprevious base. 土霸 不透水基底 Water level (觀察到) d B 0.3A a' a H B b dz dx Impervious layer Figure 8.15 Flow through an earth dam constructed over an impervious base 0 H tanp J d² L = Cosa q= kl tan & sin a H Sin
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·EX. 8.6 3:45°, α-30°, B=3m, H = 6m 0 = H = 6 m min height of dam - 7.6m, k = 61×106 q ( day ) = ? tanß = 6 tan 45° = 6 Water level H B Impervious layer Figure 8.15 Flow through an earth dam constructed over an impervious base 2=0.30+ 1.6 tang + B 7.6. + B + = 19.56 tand L= J Cosa 19.56 C05300 - d³ H Cos²α Sil (12.5625)=3.45 q = kl tang sing 61x 106 (60x24) x 3.45 tan 30° Sin 30° 0.0876
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Stress in Saturated soil. total stress 總應力 ·σ = HYW + (HA-H) I sat HA. effective stress TAZZ J' 1 X X l X 2 P] 17 D neutral stress 孔隙水壓力 (pore water pressure) u = HAYW σ = σ + u q. = - = HYw + (HA-H) I sat - HA Tw = (HA-H) (rsat-rw) 土壤高度.. ☑
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EX.9.1 H = 6 m HA-H-13m J-165 kN m³ rsat = 19.25 kN m³ A soil profile is shown in Figure 9.3. Calculate the total stress, pore water pres sure, and effective stress at points A, B, and C. 6m 13m Dry sand Yary 16.5 kN/m³ Groundwater table Saturated sand Yat 19.25 kN/m³ A Dry sand Saturated sand Clay Figure 9.3 Soil profile 總應力 孔隙水壓力u 有效應力 0 0 0 6x 16.5 0 B = 99 N m² 99 C 6 x 16.5 + 13 x 19.25 = 349.25KN m² = 13×9.8 12753N m² = 349.25-127.53 22172 KN m²
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EX.9.2 如果他會下陷,C=190N m² 水位要上升多少h? 總應力 孔隙水壓力u 有效應力5. (6-h)×16.5 C (h+13) 9.8 190 new +(h+13)×19.25 (6-h)×16.5+ (h+13)×19.25-(h+13) 9.8 = 190 -7.05h+221.85= 190 7.05h=31.85 h = 4.52 m
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Stress in Saturated soil with upward seepage. Inflow (a) Valve (open) h+ H+H₂ 總應力厂 孔倞水壓力有效應力 A H. Yw Hi Yw 0 B Hirw+HY (H. THz + h) rw H₂r-hrw C Hirw+zrsat (H₁ + 2 + 12) tw = (H₁+2) rw + Z H₂ hrw H₂
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boiling (quick condition) when Jeritical 0 = Zr- 1 cr ZV w 1 cr= riv Seepage force Seepage force unit volume = 1 Yw (+)
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Stress in Saturated soil with downward seepage Outflow Inflow Valve (open) (a) 總應力 孔倞水壓力 有效應力 A Hirw Hi Yw 0 B Hirw + H₂rsat (Hitto - h) rw H₂r' + hrw - (H₁+212) w C H₁rw + Zr sat = (H.+3)rw - Z Hz zr² + 1/2 hr w H₂
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EX. 93 JAZD H Max = ? 9 m 3 m Ysat 18 kN/m² 3.6 m Yat 16.5 kN/m³ ☐ Saturated clay Sand (和下面這層無關) Figure 9.5 JA = (9-H) sat : MA = 3.68w J₁₁ = JA -MA = 0 (9-H)· 18-36.981 20 126.684 = 18H H ≤ 1038 m = H Max
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EX. 9.4. JA 20 h = ? 7 m H=5m 2 m Figure 9.6 Saturated clay Sand Ysat = 19 kN/m³ 4.5 m Year = 18 kN/m³ (7-5)19 ·JA = h·981 + (1-5) 19 MA= 4.59.81 = JÁ JA -MAZO ·h. 981 2.194.5 9.81 20 h≤ 0626 m
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EX. 9.5 e: 0.52 = 2.67 G₁ = rsat = (Gste) w He (2.67+0.52)981 1+052 20. 6 kN m³ 0.7 m H₂ V 2 m Inflow 1 m Sand 1.5 m Valve (open) Figure 9.10 Upward flow of water through a layer of sand in a tank a. J M, at A. B 總應力 07.981+20.6 孔隙水壓力 有效應力 (Hi+ z) rw + 1/1/1 t A =27467KN KN (1+0.7) 9.81 + 1.5.981 27.467-24.0345 t EN = = :24.0345t/m 3.4325 m² m² 0.7.9.81 +2.20.6 (1.5+07+2) 9.81 B = = 48 kN 48-41.202 = m² 41.202 N = m² 6.798 KN m² b. upward seepage Seepage force unit volume force per unit volume = ? = = iɣw - w - 15981 = 73575 N =
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heaving around sheet piles. 隆起 板樁 factor of safety FS = 安全係數 D 2 'W' U 總應力(capacity) (demand) I av Yw FS= I critical 臨界 i exit 實際
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b D 2 2 EX. 9.6 tickness = 18m 8 Sat 17.7 = FS= ? oh = H Nd LN H₁-10 m Sheet pile H-15m Y-17.7 kN/m 3 T 3 m lav = 1.6+3 2 Texit = oh icr Fs = icr Texit = Heave zone Impermeable layer Figure 9.14 Flow net for seepage of water around sheet piles driven into permeable layer 10-1.5 = 1.42 b = 2.3 1.42 = = 2.3 0617 177-9.81 = = 10.804 9.81 0804 0617 = 1.3
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EX. 9.7 FS = ? Figure 9.16 H=4.2m -8- = 1.65m. 2 3 456 , H = 49 Nd i exit = sh 4.2 = 0.525m 0.525 = 5 m Granular soil e=0.55 G=2.68 Impervious layer & 1.65 Ter ☑' G5-1 0.318 2.68-1 =1.08 1+0.55 rw He Fs F₁ = 1 cr 1.08 ·3.14. Texit 0.318
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Mohr's circle 1. 圖解法 逆時針 →乙為正 N Zxy Jx 法向量 受力面 力的方向 的法向量 ㅍ公式法 法力向 .7 + t + (上圖) - + + + = principal stress 主力(when (0) On = ( Ty + Ox ) + ( σy ox)² + Zxy 2 半徑
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A N B Jx 轉90° A B zxy BC or DA 最小主應力 180° 最大主應力 t AB or CD Oy + Ox Jy-Jx On = Co≤20+ 2. z In ·Jy - Ox sinzo 2 - Txy cos20 Zxy sin20
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E.X. 10.1 J x = 120 m N 7=-40 KN m² = 300 KN m² a. oy 9 = 20° principal stress Jn On = = ? ( Ty + J x ) ± 2 (300 +120) Figure 10.4 Soil element with stresses acting on it (oy-ox)²+ Zxy 2 (302-120) + 40² ± + 士 N m² b. J & Z on AB ? = 2/0 ± 98.5 Jy + Jx 2 300+120 Jy - Jx + z 300-120 + 2 2 Cos 20+ 01 = 308.5 kN = 111.5 IN Zxy sin 20 m² ma cos 40° - 40 sin 40° = 253.23 In 04-0x Sin 20 - 2 ·Zxy cos 20 300-120 sin 40° + 40 sin to° = 88.5 IN 2 40° m² KN 2 m
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3 On = ( Ty+0x) : (oy-ox)²+ Zxy tan 20 = 2 (90+36) 士 63+34.2 2exy Oy -σx 0 = -19° 00=19+45=64' 2 (-90-36)² + (-21)² 2 (-21) = -0.78 90-36 最主魔力 乙Max=34.2 圓上90,實際45° 1534.2 广 最大主應力
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