Q = va = kia (1.2) 1.2 ( ) 2 ( 1.2) 1.2(a) (1.2) k = Q/iA = Q L/h A (1.3) 1.2(b) t 1 t 2 h 1 h 2 a

1 1 1.1 (Darcy) v(cm/s) (1.1) v = ki (1.1) v k i 1.1 h ( )L i = h/l 1.1 t 1 h(cm) (t 2 t 1 ) 1.1 A Q(cm 3 /s)

2 1 1.1 Q = va = kia (1.2) 1.2 ( ) 2 ( 1.2) 1.2(a) (1.2) k = Q/iA = Q L/h A (1.3) 1.2(b) t 1 t 2 h 1 h 2 a

1.2 3 (a) (b) 1.2 a dt dh dq dq = adh dq dq = kaidt adh = kaidt = ka(h/l)dt h2 1 1 dh = ka h al dt h 1 1 h 1 t2 dh = ka dt al t 1

4 1 [log e h] h 2 h 1 = ka 1 al [t]t 2 t 1 (log e h 2 log e h 1 ) = ka 1 al (t 2 t 1 ) (log e h 1 log e h 2 ) = ka 1 al (t 2 t 1 ) k = k = h 1 log e = ka (t 2 t 1 ) h 2 al al A(t 2 t 1 ) log h 1 e (1.4) h 2 2.30aL A(t 2 t 1 ) log h 1 10 (1.5) h 2 1.3 ( 10 cm 12.7 cm) 5 B ( θ S r k r ) B 0.95 ( 0.9 ) B B

1.3 5 (1) 1.3 110 cm ( 28 cm 16 cm 9850 cm 3 ) ( 23,000 cm 3 ) 1.0 cm 12.0 cm 26.2 cm 50 kpa 1.3 (1.6) k s = QL Ath = 0.26 Q th (1.6)

6 第 1 章 飽和土の透水 ここに Q は流量 A は供試体の断面積 (19.6 cm2 ) h は水位差 L は供試体の高さ (5.1 cm) t は計測時間である 変水位透水試験の飽和透水係数は 次の式 (1.7) より求めることができる ks = 2.3aL h1 16.25 h1 log10 = log10 At h2 t h2 (1.7) ここで a はアクリル管の断面積 (13.85 cm2 ) A は供試体の断面積 (19.6 cm2 ) L は 供試体の高さ (5.1 cm) h1 は計測開始時の水位差 h2 は計測終了時の水位差 t は計測 時間である チャンバー内の供試体寸法は 式 (1.6) 式 (1.7) にも出てきたように 直径 5.0 cm 高さ 5.1 cm で体積は 100 cm3 である (写真-1.1) 供試体の組立は カラムにゴム製の O リングを取り付け回転させるだけで 試料受器に固定できるように工夫してあり 通水断 面を容易に確保できる 写真 1.1 チャンバー内の供試体 測定手順は 供試体をセットし底面から浸透させ チャンバー内の所定の水位まで給水 する この水位と二重管ビューレット内の水位を一致させ 真空脱気を 1 時間行い 背圧 P (圧力が低いと計測が難しいので 196.2 kpa) をかけて供試体内に流入した水量 V を 計測し 式 (1.8) から飽和度を換算する Sr = 1 P0 V 1 H P Vv ここで H はヘンリー係数 P0 は大気圧 Vv は供試体の間隙部分の体積である (1.8)

1.3 7 B (29.4 kpa 9.81 kpa B ) 1 3 3 B (2) 1.4 B B B B B 1.4 B 1.5 B ( ) 1.0 cm

8 1 1.5 1.4 0.5 1.0 Re (Re < 1) 1.6 1 1 B B 3 B B B

1.3 9 1.6 B 0.98 (3) a) Re (laminar flow) (turbulent flow) (Reynolds) (Reynold s number) (Critical Reynold s number) R e = vd ν (1.9)

10 1 v (m/s) D (m) ν (m 2 /s) R e 2000 R e 1 R e = ρ wvd µ (1.10) v (cm/s) D (D 50 D 60 )(cm) µ (g/s cm) (1.1) m 0.5 < m < 1 v = ki m (1.11) b) B (A.W. Skempton) u = B { σ 3 + A ( σ 1 + σ 3 )} (1.12) A B A A f > 0 A f < 0 A( σ 1 σ 3 ) B B < 1 B = 1 K 0 V 0 e 0 n 0 V 0 n 0 σ v0 ( σ = hγ w + σ v0 h γ w ) u

1.3 11 V w K 0 C w V w = n 0 V 0 C w u (1.13) ( σ = hγ w + σ v0 ) σ v ε v ε v = C s σ v (1.14) C s K 0 V s V s = V 0 C s σ v (1.15) (1.13) (1.15) n 0 V 0 C w u = V 0 C s σ v (1.16) σ v = σ v + u (1.16) σ v u = 1 1 + n 0 C w /C s σ v (1.17) (1.17) C w /C s C w /C s (1.17) (1.18) u = σ v (1.18) (1.18) B B = u σ v = 1 (1.19) (1.20) B = u σ v < 1 (1.20)

12 1 (back pressure) c) 1.1 C K( ) H 0 273 0.0288 5 278 0.0260 10 283 0.0235 15 288 0.0216 20 293 0.0201 25 298 0.0188 30 303 0.0176 35 308 0.0165 (1.21) S r S r = 1 [ ] 1 H P 0 1 P 0 P 1 H S r0 (1.21) H 1.1 P 0 (98.1 kn/m 2 ) P S r0 P 0 (V a0 + HV w0 ) = (P 0 + P )(V a1 + HV w1 ) (1.22) V a0 V a1 P V w0 V w1 P

1.4 13 V V a0 V a1 = V V w1 V w0 = V (1.23) (1.22) (1.23) V a1 V w1 V (H 1)(P 0 + P ) + P (V a0 + HV w0 ) = 0 (1.24) S r0 S r0 = 1 H V (P 0 + P ) V v P (1.25) (1.25) V v V P (1.21) (1.25) ( (1.8)) S r = 1 1 H P 0 V P V v (1.26) 1.4 (v = ki) k v k h 1.7 x v x ab (1.27) (1.28) ( v x x, z + z ) = v x (x, z) + 1 2 2 ( v x (x, z) + 1 ) v x 2 z z v x z (1.27) z z (1.28)

14 1 1.7 cd x (1.29) (1.30) ( v x x + x, z + z ) = v x (x, z) + v x 2 x x + 1 v x z (1.29) 2 z ( v x (x, z) + v x x x + 1 ) v x 2 z z z (1.30) 1.8(a) x q x (1.28) (1.30) ( q x = v x (x, z) + 1 ) ( v x 2 z z z v x (x, z) + v x x x + 1 ) v x 2 z z z = v x x z (1.31) x 1.8(b) q z = v z x z (1.32) z (1.31) (1.32) 0 q x + q z = v x x x z + v z z x z = v x x + v z z = 0 (1.33)

1.4 15 (a) (b) 1.8 v x k h ( ) h v x = k h i = k h x i h x i = h x, i = h z, v x = k h h x, v z = k v h z (1.34) (1.34) (1.33) k h = k v 2 h k h x 2 + k 2 h v z 2 = 0 (1.35) 2 h x 2 + 2 h z 2 = 0 (1.36) (1.36)

16 1 1.5 1 2 3 ( ) 1.9 1.9 H v ( α+ β = 90 ) H h H =

1.5 17 0 H x z(x z ) h h x = z = 0 (1.37) x z 1.9 (1.38) (1.39) v x H = cos α, z h v x = k h x, v x v = cos β, v z v H = sin α (1.38) v h z = k v z (1.39) = sin β (1.40) (1.39) (1.40) cos β = k h v h x, sin β = k v v h z (1.41) (1.38) (1.41) cos(α + β) = cos α cos β sin α sin β = x H = k ( ) h h x + Hv x z z ( k v ) ( h z ) ( k x H v ) h z (1.42) (1.42) (1.37) 0 y = cos x 0 π/2 α + β = 90 1.10 ( ) 1 Q Q = kiat(cm 3 ) ( 1.11) a h 8 1 8( 1 8 ) N d

18 1 1.10 1.11 N f ( 4(1 4 ) ) N d N f 1 q q = kia = k h N d 1 a a 1 = k h N d (1.43) h/n d (1/a) i = h L = h 1 a 1 a N d a 1 A Q (1.43)

1.5 19 N f Q = k h N f N d (1.44) k = 2.0 10 5 cm/s 1 m Q = k h N f N d = 20 10 5 750 4 8 100 = 0.75 cm3 /s (1.45) h 27.0 m 19.5 m 7.5 m=750 cm cm 1 m 100 cm 2 v = ki = k( h/l) ( ) ( ) 3 (1.46) (p/ρ w ) z v 2 /2g p ρ w + z + v2 2g = const. (1.46) p ρ w + z = const. (1.47) (1.47) 1.12 A B h = 27.0 19.5 =7.5 m El.18.0 m A 3.0 m B

20 1 1.12 6.0 m + = A p + z = H h ρ w N d p + ( 3.0) = 9.0 7.5 ρ w 8 1 p = 11.06 m (1.48) ρ w H z (7.5/8) 1 ( 1) B ( 6) p + ( 6.0) = 9.0 7.5 ρ w 8 6 p = 9.38 m ρ w El.0.0 m A H = 27.0 m z = 15.0 m p + 15.0 = 27.0 7.5 ρ w 8 1 p = 11.06 m ρ w

1.5 21 1.13 N d N f ( ) 1 % Lambe N f = 4 N d = 12 (N f /N d = 0.333) k = 5.145 10 6 m/s Q 10.29 10 6 m 3 /s 1.2 1.13 Lambe 1.2 1.13 N f N d N f /N d Q( 10 6 m 3 /s) (%) 3.20 10 0.320 9.88 4.0 4.03 12 0.336 10.37 0.8 5.39 16 0.337 10.40 1.0 1.14 (1.48) ( ) 1.14(a) (b) 1 7 1.14(a) (b) N f = 4 N d = 14 1.5 m II 1 ( ) p + ( 1.5) = 6.0 6.0 ρ w 14 6.5 p = 4.71 m p = 4.71 t/m 2 = 46.2 kn/m 2 ρ w

22 1 1.13

1.5 23 1.14 1.3 (kn/m 2 ) 1 2 3 4 5 6 7 II 46.2 44.1 39,9 35.7 31.5 27.3 23.1 III 65.2 61.0 56.8 52.6 48.3 44.1 42.0 III 1 ( ) p + ( 1.5) = 6.0 6.0 ρ w 14 2 p = 6.64 m p = 6.64 t/m 2 = 65.2 kn/m 2 ρ w

24 1 II 1 7 1.3 1.6 x z R ( ) 1/R 1.15 k h k v 9

1.7 25 R = kv 1 = k h 9 = 1 3 (1.49) x ( ) R 1.15 III (a) (b) 1.14(b) 1.7 (Muskat) 1.16 Muskat D d 1 d 2 Q k h 1.10 (1.45) 1.10 k = 2.0 10 5 cm/s 1 m 1.16 Muskat Q = k h N f N d = 20 10 5 750 4 8 100 = 0.75 cm3 /s

26 1 1.16 D = 18.0 m d 1 = 9.0 m d 2 = 0 m d 1 /D = 9.0/18.0 = 0.5 d 2 /d 1 = 0 1.16 Q/kh = 0.5 Q Q = 0.5 kh = 0.5 2.0 10 5 750 100 = 0.75 cm 3 /s Q/kh N f /N d d 2 1.17 1.17 α α (1.44) α = N d /N f k = Q h N d N f = Q h α (1.50) N d N f α 3.5 4.5

1.8 27 1.8 (Dupuit) (a) (b) (c) 1.18 1 1.18(a) o R abc 1.18(b) 1.18(b) (c) n q = kd dh dx (1.51) D = 2 tan θx θ = 45 D = 2x x = nr h = H x = R h = H/2 (1.52)

28 1 (1.51) (1.52) dh = q 2xk dx h = q 2k ln x + C C = q ln nr + H 2k H 2 = q 2k ln R q ln nr + H 2k H 2 = q 2k ln R nr k = q H ln 1 n (1.53) (1.50) (1.53) α ln n 1 1.19 2 1.19 nr q = πrv = πrk dh dr (1.54)

1.8 29 (1.54) (1.52) dh = q πrk dr h = q πk ln r + C C = q ln nr + H πk H 2 = q πk ln R + q ln nr πk H 2 = q πk ln R nr k = q H ln 1 2 n π (1.55) α ln n 1 2/π (1.53) (1.55) α n 1.20 α 1.20