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
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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 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
3 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 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 aL A(t 2 t 1 ) log h 1 10 (1.5) h ( 10 cm 12.7 cm) 5 B ( θ S r k r ) B 0.95 ( 0.9 ) B B
5 1.3 5 (1) 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 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 h 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 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)
7 1.3 7 B (29.4 kpa 9.81 kpa B ) B (2) 1.4 B B B B B 1.4 B 1.5 B ( ) 1.0 cm
8 Re (Re < 1) B B 3 B B B
9 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 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
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 = 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 12 1 (back pressure) c) 1.1 C K( ) H (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
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) ( v x (x, z) + 1 ) v x 2 z z v x z (1.27) z z (1.28)
14 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)
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 h z 2 = 0 (1.36) (1.36)
16 ( ) H v ( α+ β = 90 ) H h H =
17 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 = ( ) 1 Q Q = kiat(cm 3 ) ( 1.11) a h 8 1 8( 1 8 ) N d
18 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)
19 N f Q = k h N f N d (1.44) k = cm/s 1 m Q = k h N f N d = = 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 = =7.5 m El.18.0 m A 3.0 m B
20 m + = A p + z = H h ρ w N d p + ( 3.0) = ρ w 8 1 p = m (1.48) ρ w H z (7.5/8) 1 ( 1) B ( 6) p + ( 6.0) = ρ w 8 6 p = 9.38 m ρ w El.0.0 m A H = 27.0 m z = 15.0 m p = ρ w 8 1 p = m ρ w
21 N d N f ( ) 1 % Lambe N f = 4 N d = 12 (N f /N d = 0.333) k = m/s Q m 3 /s Lambe N f N d N f /N d Q( 10 6 m 3 /s) (%) (1.48) ( ) 1.14(a) (b) (a) (b) N f = 4 N d = m II 1 ( ) p + ( 1.5) = ρ w p = 4.71 m p = 4.71 t/m 2 = 46.2 kn/m 2 ρ w
22
23 (kn/m 2 ) II , III III 1 ( ) p + ( 1.5) = ρ w 14 2 p = 6.64 m p = 6.64 t/m 2 = 65.2 kn/m 2 ρ w
24 24 1 II x z R ( ) 1/R 1.15 k h k v 9
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 = cm/s 1 m 1.16 Muskat Q = k h N f N d = = 0.75 cm3 /s
26 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 = Q/kh = 0.5 Q Q = 0.5 kh = = 0.75 cm 3 /s Q/kh N f /N d d 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
27 (Dupuit) (a) (b) (c) (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 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 nr q = πrv = πrk dh dr (1.54)
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
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1 (Berry,1975) 2-6 p (S πr 2 )p πr 2 p 2πRγ p p = 2γ R (2.5).1-1 : : : : ( ).2 α, β α, β () X S = X X α X β (.1) 1 2
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(1.2) T D = 0 T = D = 30 kn 1.2 (1.4) 2F W = 0 F = W/2 = 300 kn/2 = 150 kn 1.3 (1.9) R = W 1 + W 2 = = 1100 N. (1.9) W 2 b W 1 a = 0
1 1 1.1 1.) T D = T = D = kn 1. 1.4) F W = F = W/ = kn/ = 15 kn 1. 1.9) R = W 1 + W = 6 + 5 = 11 N. 1.9) W b W 1 a = a = W /W 1 )b = 5/6) = 5 cm 1.4 AB AC P 1, P x, y x, y y x 1.4.) P sin 6 + P 1 sin 45
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II 14 14-7-8 8/4 II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ 6/ ] Navier Stokes 3 [ ] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I 1 balance law t (ρv i )+ j
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知能科学：ニューラルネットワーク
2 3 4 (Neural Network) (Deep Learning) (Deep Learning) ( x x = ax + b x x x ? x x x w σ b = σ(wx + b) x w b w b .2.8.6 σ(x) = + e x.4.2 -.2 - -5 5 x w x2 w2 σ x3 w3 b = σ(w x + w 2 x 2 + w 3 x 3 + b) x,
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x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ dt iωζ = ẍ + ω2 x (2.1) ζ ζ = Aωe iωt = Aω cos ωt + iaω sin
2 2.1 F (t) 2.1.1 mẍ + kx = F (t). m ẍ + ω 2 x = F (t)/m ω = k/m. 1 : (ẋ, x) x = A sin ωt, ẋ = Aω cos ωt 1 2-1 x A Aω ẋ ẋ 2 + ω 2 x 2 = ω 2 A 2. (ẋ, ωx) ζ ẋ + iωx ζ ζ dζ = ẍ + iωẋ = ẍ + iω(ζ iωx) dt dζ
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1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1
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9 7 A = A x x + A y y + A, B = B x x + B y y + B, C = C x x + C y y + C..6 x y A B C = A x x + A y y + A B x B y B C x C y C { B = A x x + A y y + A y B B x x B } B C y C y + x B y C x C C x C y B = A
ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 +
2.6 2.6.1 ω 0 m(ẍ + γẋ + ω0x) 2 = ee (2.118) e iωt x = e 1 m ω0 2 E(ω). (2.119) ω2 iωγ Z N P(ω) = χ(ω)e = exzn (2.120) ϵ = ϵ 0 (1 + χ) ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.121) Z ω ω j γ j f j
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3 3.1 ( 1 m d2 x(t dt 2 = kx(t k = (3.1 d 2 x dt 2 = ω2 x, ω = x(t = 0, ẋ(0 = v 0 k m (3.2 x = v 0 ω sin ωt (ẋ = v 0 cos ωt (3.3 E = 1 2 mẋ2 + 1 2 kx2 = 1 2 mv2 0 cos 2 ωt + 1 2 k v2 0 ω 2 sin2 ωt = 1
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4 4 ) a + b = i + 6 b c = 6i j ) a = 0 b = c = 0 ) â = i + j 0 ˆb = 4) a b = b c = j + ) cos α = cos β = 6) a ˆb = b ĉ = 0 7) a b = 6i j b c = i + 6j + 8) a b a b = 6i j 4 b c b c 9) a b = 4 a b) c = 7
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Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction 2003 3 Study on Retaining Wall Design For Circular Deep Shaft Undergoing Lateral Pressure During Construction
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5. [. ] z = f(, y) () z = 3 4 y + y + 3y () z = y (3) z = sin( y) (4) z = cos y (5) z = 4y (6) z = tan y (7) z = log( + y ) (8) z = tan y + + y ( ) () z = 3 8y + y z y = 4 + + 6y () z = y z y = (3) z =
m(ẍ + γẋ + ω 0 x) = ee (2.118) e iωt P(ω) = χ(ω)e = ex = e2 E(ω) m ω0 2 ω2 iωγ (2.119) Z N ϵ(ω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j (2.120)
2.6 2.6.1 mẍ + γẋ + ω 0 x) = ee 2.118) e iωt Pω) = χω)e = ex = e2 Eω) m ω0 2 ω2 iωγ 2.119) Z N ϵω) ϵ 0 = 1 + Ne2 m j f j ω 2 j ω2 iωγ j 2.120) Z ω ω j γ j f j f j f j sum j f j = Z 2.120 ω ω j, γ ϵω) ϵ
50 2 I SI MKSA r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq
49 2 I II 2.1 3 e e = 1.602 10 19 A s (2.1 50 2 I SI MKSA 2.1.1 r q r q F F = 1 qq 4πε 0 r r 2 r r r r (2.2 ε 0 = 1 c 2 µ 0 c = 3 10 8 m/s q 2.1 r q' F r = 0 µ 0 = 4π 10 7 N/A 2 k = 1/(4πε 0 qq F = k r
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液晶の物理１：連続体理論（弾性，粘性）
The Physics of Liquid Crystals P. G. de Gennes and J. Prost (Oxford University Press, 1993) Liquid crystals are beautiful and mysterious; I am fond of them for both reasons. My hope is that some readers
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医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. このサンプルページの内容は, 第 2 版 1 刷発行時のものです.
医系の統計入門第 2 版 サンプルページ この本の定価 判型などは, 以下の URL からご覧いただけます. http://www.morikita.co.jp/books/mid/009192 このサンプルページの内容は, 第 2 版 1 刷発行時のものです. i 2 t 1. 2. 3 2 3. 6 4. 7 5. n 2 ν 6. 2 7. 2003 ii 2 2013 10 iii 1987
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H 0 H = H 0 + V (t), V (t) = gµ B S α qb e e iωt i t Ψ(t) = [H 0 + V (t)]ψ(t) Φ(t) Ψ(t) = e ih0t Φ(t) H 0 e ih0t Φ(t) + ie ih0t t Φ(t) = [
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