untitled

untitled

400 312 400 2,3 ()1 1 /( ) / ( ) 2 1830.J. 6 1. 2. 3. 4. 5. 6. -300-793 -1191-1333 -1602-1867 -1911-1925 -1988 1989- () 剗 2 2 8 10 20 ( 22 7 )( 24 9 )( 26 4 ) () () 艜 10 10 20 20 5490km 1 "The roads

untitled

C L C L 4.5m 3.0m 10% 25% 50% 90% 75% 50% N N N 90% 90% 10% 10% 100 100 10 10 10% 10% :49kN :17 :17kN CBR CBR CBR 5 3,000 / 3,000 /mm /mm 1.2mm 89dB 190dB 3,000 3,000 /mm 20% 20%

66 σ σ (8.1) σ = 0 0 σd = 0 (8.2) (8.2) (8.1) E ρ d = 0... d = 0 (8.3) d 1 NN K K 8.1 d σd σd M = σd = E 2 d (8.4) ρ 2 d = I M = EI ρ 1 ρ = M EI ρ EI

65 8. K 8 8 7 8 K 6 7 8 K 6 M Q σ (6.4) M O ρ dθ D N d N 1 P Q B C (1 + ε)d M N N h 2 h 1 ( ) B (+) M 8.1: σ = E ρ (E, 1/ρ ) (8.1) 66 σ σ (8.1) σ = 0 0 σd = 0 (8.2) (8.2) (8.1) E ρ d = 0... d = 0 (8.3)

untitled

9118 154 B-1 B-3 B- 5cm 3cm 5cm 3m18m5.4m.5m.66m1.3m 1.13m 1.134m 1.35m.665m 5 , 4 13 7 56 M 1586.1.18 7.77.9 599.5.8 7 1596.9.5 7.57.75 684.11.9 8.5 165..3 7.9 87.8.11 6.57. 166.6.16 7.57.6 856 6.6.5

NETES No.CG V

1 2006 6 NETES No.CG-050001-V 2007 5 2 1 2 1 19 5 1 2 19 8 2 i 1 1 1.1 1 1.2 2 1.3 2 2 3 2.1 3 2.2 8 3 9 3.1 9 3.2 10 3.3 13 3.3.1 13 3.3.2 14 3.3.3 14 3.3.4 16 3.3.5 17 3.3.6 18 3.3.7 21 3.3.8 22 3.4

(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

A (1) = 4 A( 1, 4) 1 A 4 () = tan A(0, 0) π A π

4 4.1 4.1.1 A = f() = f() = a f (a) = f() (a, f(a)) = f() (a, f(a)) f(a) = f 0 (a)( a) 4.1 (4, ) = f() = f () = 1 = f (4) = 1 4 4 (4, ) = 1 ( 4) 4 = 1 4 + 1 17 18 4 4.1 A (1) = 4 A( 1, 4) 1 A 4 () = tan

( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( )

n n (n) (n) (n) (n) n n ( n) n n n n n en1, en ( n) nen1 + nen nen1, nen ( ) e + e ( ) ( ) e + e () ( ) e e Τ ( ) e e ( ) ( ) () () ( ) ( ) ( ) ( ) ( n) Τ n n n ( n) n + n ( n) (n) n + n n n n n n n n

[Ver. 0.2] 1 2 3 4 5 6 7 1 1.1 1.2 1.3 1.4 1.5 1 1.1 1 1.2 1. (elasticity) 2. (plasticity) 3. (strength) 4. 5. (toughness) 6. 1 1.2 1. (elasticity) } 1 1.2 2. (plasticity), 1 1.2 3. (strength) a < b F

untitled

( ) c a sin b c b c a cos a c b c a tan b a b cos sin a c b c a ccos b csin (4) Ma k Mg a (Gal) g(98gal) (Gal) a max (K-E) kh Zck.85.6. 4 Ma g a k a g k D τ f c + σ tanφ σ 3 3 /A τ f3 S S τ A σ /A σ /A

16 6 12 1 16 6 23 23 11 16 START 1 Out Ok 1,2 Ok END Out 3 1 1/ H24.2 2 1 L2-1 L2-2 H14.3 3 H9.10 PHC SC 19 1 24 3 18N/mm 2 24N/mm 2 30N/mm 2 25 10 13 12 13 12 11 11 11 11 19 7 25 10 24N 8cm 25(20)mm 45

8 300 mm 2.50 m/s L/s ( ) 1.13 kg/m MPa 240 C 5.00mm 120 kpa ( ) kg/s c p = 1.02kJ/kgK, R = 287J/kgK kPa, 17.0 C 118 C 870m 3 R = 287J

26 1 22 10 1 2 3 4 5 6 30.0 cm 1.59 kg 110kPa, 42.1 C, 18.0m/s 107kPa c p =1.02kJ/kgK 278J/kgK 30.0 C, 250kPa (c p = 1.02kJ/kgK, R = 287J/kgK) 18.0 C m/s 16.9 C 320kPa 270 m/s C c p = 1.02kJ/kgK, R = 292J/kgK

5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 = 4. () = 8 () = 4

... A F F l F l F(p, 0) = p p > 0 l p 0 P(, ) H P(, ) P l PH F PF = PH PF = PH p O p ( p) + = { ( p)} = 4p l = 4p (p 0) F(p, 0) = p O 3 5 5. F(, 0) = = 4 = 4 O = 4 =. ( = = 4 ) = 4 ( 4 ), 0 = 4 4 O 4 =

さくらの個別指導 ( さくら教育研究所 ) A a 1 a 2 a 3 a n {a n } a 1 a n n n 1 n n 0 a n = 1 n 1 n n O n {a n } n a n α {a n } α {a

... A a a a 3 a n {a n } a a n n 3 n n n 0 a n = n n n O 3 4 5 6 n {a n } n a n α {a n } α {a n } α α {a n } a n n a n α a n = α n n 0 n = 0 3 4. ()..0.00 + (0.) n () 0. 0.0 0.00 ( 0.) n 0 0 c c c c c

A(6, 13) B(1, 1) 65 y C 2 A(2, 1) B( 3, 2) C 66 x + 2y 1 = 0 2 A(1, 1) B(3, 0) P 67 3 A(3, 3) B(1, 2) C(4, 0) (1) ABC G (2) 3 A B C P 6

1 1 1.1 64 A6, 1) B1, 1) 65 C A, 1) B, ) C 66 + 1 = 0 A1, 1) B, 0) P 67 A, ) B1, ) C4, 0) 1) ABC G ) A B C P 64 A 1, 1) B, ) AB AB = 1) + 1) A 1, 1) 1 B, ) 1 65 66 65 C0, k) 66 1 p, p) 1 1 A B AB A 67

1

GL (a) (b) Ph l P N P h l l Ph Ph Ph Ph l l l l P Ph l P N h l P l .9 αl B βlt D E. 5.5 L r..8 e g s e,e l l W l s l g W W s g l l W W e s g e s g r e l ( s ) l ( l s ) r e l ( s ) l ( l s ) e R e r

12

12 1 2 3 4 5 6 1.2 AFRP (3.4.1)(3.4.3) ht M = 1.2M By0 M Ty0 h A n MBy0 h B AF p = 1000 = t AF AAF b 7 / 8 AF B M σ AFb h (tf m) (m) M T y0 (tf m) h T (m) M (tf m) AAF AFRPcm 2 σafb AFRPkgf/cm 2 σ AFb

dvipsj.8449.dvi

9 1 9 9.1 9 2 (1) 9.1 9.2 σ a = σ Y FS σ a : σ Y : σ b = M I c = M W FS : M : I : c : = σ b

1 12 ( )150 ( ( ) ) x M x 0 1 M 2 5x 2 + 4x + 3 x 2 1 M x M 2 1 M x (x + 1) 2 (1) x 2 + x + 1 M (2) 1 3 M (3) x 4 +

( )5 ( ( ) ) 4 6 7 9 M M 5 + 4 + M + M M + ( + ) () + + M () M () 4 + + M a b y = a + b a > () a b () y V a () V a b V n f() = n k= k k () < f() = log( ) t dt log () n+ (i) dt t (n + ) (ii) < t dt n+ n

sekibun.dvi

a d = a + a+ (a ), e d = e, sin d = cos, (af() + bg())d = a d = log, cosd = sin, f()d + b g()d d 3 d d d d d d d ( + 3 + )d ( + )d ( 3 )d (e )d ( sin 3 cos)d g ()f (g())d = f(g()) e d e d ( )e d cos d

<4D F736F F D B B BB2D834A836F815B82D082C88C60202D B2E646F63>

例題で学ぶはじめての塑性力学 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/066721 このサンプルページの内容は, 初版 1 刷発行当時のものです. http://www.morikita.co.jp/support/ 03 3817 5670 FAX 03 3815 8199 i 1

橡博論表紙.PDF

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

Part () () Γ Part ,

Contents a 6 6 6 6 6 6 6 7 7. 8.. 8.. 8.3. 8 Part. 9. 9.. 9.. 3. 3.. 3.. 3 4. 5 4.. 5 4.. 9 4.3. 3 Part. 6 5. () 6 5.. () 7 5.. 9 5.3. Γ 3 6. 3 6.. 3 6.. 3 6.3. 33 Part 3. 34 7. 34 7.. 34 7.. 34 8. 35

1 1 sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω 1 ω α V T m T m 1 100Hz m 2 36km 500Hz. 36km 1

sin cos P (primary) S (secondly) 2 P S A sin(ω2πt + α) A ω ω α 3 3 2 2V 3 33+.6T m T 5 34m Hz. 34 3.4m 2 36km 5Hz. 36km m 34 m 5 34 + m 5 33 5 =.66m 34m 34 x =.66 55Hz, 35 5 =.7 485.7Hz 2 V 5Hz.5V.5V V

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,

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,

Microsoft Word - 海岸紹介new.doc

H8.6 P.4-202-

9 5 ( α+ ) = (α + ) α (log ) = α d = α C d = log + C C 5. () d = 4 d = C = C = 3 + C 3 () d = d = C = C = 3 + C 3 =

5 5. 5.. A II f() f() F () f() F () = f() C (F () + C) = F () = f() F () + C f() F () G() f() G () = F () 39 G() = F () + C C f() F () f() F () + C C f() f() d f() f() C f() f() F () = f() f() f() d =

微分積分 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, 初版 1 刷発行時のものです.

微分積分 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. ttp://www.morikita.co.jp/books/mid/00571 このサンプルページの内容は, 初版 1 刷発行時のものです. i ii 014 10 iii [note] 1 3 iv 4 5 3 6 4 x 0 sin x x 1 5 6 z = f(x, y) 1 y = f(x)

名古屋工業大の数学 2000 年 ~2015 年 大学入試数学動画解説サイト

名古屋工業大の数学 年 ~5 年 大学入試数学動画解説サイト http://mathroom.jugem.jp/ 68 i 4 3 III III 3 5 3 ii 5 6 45 99 5 4 3. () r \= S n = r + r + 3r 3 + + nr n () x > f n (x) = e x + e x + 3e 3x + + ne nx f(x) = lim f n(x) lim

untitled

( 9:: 3:6: (k 3 45 k F m tan 45 k 45 k F m tan S S F m tan( 6.8k tan k F m ( + k tan 373 S S + Σ Σ 3 + Σ os( sin( + Σ sin( os( + sin( os( p z ( γ z + K pzdz γ + K γ K + γ + 9 ( 9 (+ sin( sin { 9 ( } 4

9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x

2009 9 6 16 7 1 7.1 1 1 1 9 2 1 f(x, y) = xy sin x cos y x y cos y y x sin x d (x, y) = y cos y (x sin x) = y cos y(sin x + x cos x) x dx d (x, y) = x sin x (y cos y) = x sin x(cos y y sin y) y dy 1 sin

高校生の就職への数学II

II O Tped b L A TEX ε . II. 3. 4. 5. http://www.ocn.ne.jp/ oboetene/plan/ 7 9 i .......................................................................................... 3..3...............................

29

9 .,,, 3 () C k k C k C + C + C + + C 8 + C 9 + C k C + C + C + C 3 + C 4 + C 5 + + 45 + + + 5 + + 9 + 4 + 4 + 5 4 C k k k ( + ) 4 C k k ( k) 3 n( ) n n n ( ) n ( ) n 3 ( ) 3 3 3 n 4 ( ) 4 4 4 ( ) n n

() n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (5) (6 ) n C + nc + 3 nc n nc n (7 ) n C + nc + 3 nc n nc n (

3 n nc k+ k + 3 () n C r n C n r nc r C r + C r ( r n ) () n C + n C + n C + + n C n n (3) n C + n C + n C 4 + n C + n C 3 + n C 5 + (4) n C n n C + n C + n C + + n C n (5) k k n C k n C k (6) n C + nc

1. z dr er r sinθ dϕ eϕ r dθ eθ dr θ dr dθ r x 0 ϕ r sinθ dϕ r sinθ dϕ y dr dr er r dθ eθ r sinθ dϕ eϕ 2. (r, θ, φ) 2 dr 1 h r dr 1 e r h θ dθ 1 e θ h

IB IIA 1 1 r, θ, φ 1 (r, θ, φ)., r, θ, φ 0 r

[ ] 0.1 lim x 0 e 3x 1 x IC ( 11) ( s114901) 0.2 (1) y = e 2x (x 2 + 1) (2) y = x/(x 2 + 1) 0.3 dx (1) 1 4x 2 (2) e x sin 2xdx (3) sin 2 xdx ( 11) ( s

[ ]. lim e 3 IC ) s49). y = e + ) ) y = / + ).3 d 4 ) e sin d 3) sin d ) s49) s493).4 z = y z z y s494).5 + y = 4 =.6 s495) dy = 3e ) d dy d = y s496).7 lim ) lim e s49).8 y = e sin ) y = sin e 3) y =

2016 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 1 16 2 1 () X O 3 (O1) X O, O (O2) O O (O3) O O O X (X, O) O X X (O1), (O2), (O3) (O2) (O3) n (O2) U 1,..., U n O U k O k=1 (O3) U λ O( λ Λ) λ Λ U λ O 0 X 0 (O2) n =

(1) θ a = 5(cm) θ c = 4(cm) b = 3(cm) (2) ABC A A BC AD 10cm BC B D C 99 (1) A B 10m O AOB 37 sin 37 = cos 37 = tan 37

4. 98 () θ a = 5(cm) θ c = 4(cm) b = (cm) () D 0cm 0 60 D 99 () 0m O O 7 sin 7 = 0.60 cos 7 = 0.799 tan 7 = 0.754 () xkm km R km 00 () θ cos θ = sin θ = () θ sin θ = 4 tan θ = () 0 < x < 90 tan x = 4 sin

lim lim lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d

lim 5. 0 A B 5-5- A B lim 0 A B A 5. 5- 0 5-5- 0 0 lim lim 0 0 0 lim lim 0 0 d lim 5. d 0 d d d d d d 0 0 lim lim 0 d 0 0 5- 5-3 0 5-3 5-3b 5-3c lim lim d 0 0 5-3b 5-3c lim lim lim d 0 0 0 3 3 3 3 3 3

6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m f 4

35-8585 7 8 1 I I 1 1.1 6kg 1m P σ σ P 1 l l λ λ l 1.m 1 6kg 1.1m 1.m.1m.1 l λ ϵ λ l + λ l l l dl dl + dλ ϵ dλ dl dl + dλ dl dl 3 1. JIS 1 6kg 1% 66kg 1 13 σ a1 σ m σ a1 σ m σ m σ a1 f f σ a1 σ a1 σ m

1.500 m X Y m m m m m m m m m m m m N/ N/ ( ) qa N/ N/ 2 2

1.500 m X Y 0.200 m 0.200 m 0.200 m 0.200 m 0.200 m 0.000 m 1.200 m m 0.150 m 0.150 m m m 2 24.5 N/ 3 18.0 N/ 3 30.0 0.60 ( ) qa 50.79 N/ 2 0.0 N/ 2 20.000 20.000 15.000 15.000 X(m) Y(m) (kn/m 2 ) 10.000

xyz,, uvw,, Bernoulli-Euler u c c c v, w θ x c c c dv ( x) dw uxyz (,, ) = u( x) y z + ω( yz, ) φ dx dx c vxyz (,, ) = v( x) zθ x ( x) c wxyz (,, ) =

,, uvw,, Bernoull-Euler u v, w θ dv ( ) dw u (,, ) u( ) ω(, ) φ d d v (,, ) v( ) θ ( ) w (,, ) w( ) θ ( ) (11.1) ω φ φ dθ / dφ v v θ u w u w 11.1 θ θ θ 11. vw, (11.1) u du d v d w ε d d d u v ω γ φ w u

Taro11-aマニュアル.jtd

L A m ton m kn t t kn t kn t m m kn ton ton m m m kn/ CK CK = N/mm ca sa a cm kn/ kn/ kn/ kn/ kn/ kn/ kn/ - - kn/m WL % /m - - A c sin cos kn/m kn/m kn/m / - / A A H V H A cos V A sin - - = N/mm P P m

SO(2)

TOP URL http://amonphys.web.fc2.com/ 1 12 3 12.1.................................. 3 12.2.......................... 4 12.3............................. 5 12.4 SO(2).................................. 6

36 th IChO : - 3 ( ) , G O O D L U C K final 1

36 th ICh - - 5 - - : - 3 ( ) - 169 - -, - - - - - - - G D L U C K final 1 1 1.01 2 e 4.00 3 Li 6.94 4 Be 9.01 5 B 10.81 6 C 12.01 7 N 14.01 8 16.00 9 F 19.00 10 Ne 20.18 11 Na 22.99 12 Mg 24.31 Periodic

limit&derivative

- - 7 )................................................................................ 5.................................. 7.. e ).......................... 9 )..........................................

untitled

22112 Ss Ss 1 Ss S S BC 2 S1/2 S 1 1 3 S2/2 S 2 2 4 Ss Ss 1 1 Ss Ss 5 JEAG46011984JEAG4601 1987JEAG46011991 JSME S NC12005 JEAG4601-1991 6 NO NO YES YES 7 Ss Ss A1 A2 Ss Ss 8 B1 Ss 9 10 4.87 4.87 0.29

ρ ( ) sgv + ρwgv γ sv + γ wv γ s + γ w e e γ ρ g s s γ s ( ) + γ w( ) Vs + V Vs + V + e + e + e γ γ sa γ e e n( ) + e γ γ s ( n) + γ wn γ s, γ w γ γ +

σ P σ () n σ () n σ P ) σ ( σ P σ σ σ + u V e m w ρ w gv V V s m s ρ s gv s ρ ( ) sgv + ρwgv γ sv + γ wv γ s + γ w e e γ ρ g s s γ s ( ) + γ w( ) Vs + V Vs + V + e + e + e γ γ sa γ e e n( ) + e γ γ s (

II Karel Švadlenka * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* u = au + bv v = cu + dv v u a, b, c, d R

II Karel Švadlenka 2018 5 26 * [1] 1.1* 5 23 m d2 x dt 2 = cdx kx + mg dt. c, g, k, m 1.2* 5 23 1 u = au + bv v = cu + dv v u a, b, c, d R 1.3 14 14 60% 1.4 5 23 a, b R a 2 4b < 0 λ 2 + aλ + b = 0 λ =

土木工事共通仕様書（その２）

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 1 2 3 22 578 582 27 4 5 6 567 7 21 8 9 (9), 10 11 12 13 14 (1) (2) 16 532 35 15 (3) (4) (1) (1) 16 (4) () () () 17 () 18 170 19 20 21 10 22 23 24 25 26 27

29 4 ... 1... 1... 1... 2... 3... 4.... 4... 4... 7... 8... 8... 8... 8...12...14...14...14...16...18...18...19...21... 42...42...42....42....46....49...51....51....51... 52...52...52...53 I. I. I. I.

20 9 19 1 3 11 1 3 111 3 112 1 4 12 6 121 6 122 7 13 7 131 8 132 10 133 10 134 12 14 13 141 13 142 13 143 15 144 16 145 17 15 19 151 1 19 152 20 2 21 21 21 211 21 212 1 23 213 1 23 214 25 215 31 22 33

19 σ = P/A o σ B Maximum tensile strength σ % 0.2% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional

19 σ = P/A o σ B Maximum tensile strength σ 0. 0.% 0.% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional limit ε p = 0.% ε e = σ 0. /E plastic strain ε = ε e

85 4

85 4 86 Copright c 005 Kumanekosha 4.1 ( ) ( t ) t, t 4.1.1 t Step! (Step 1) (, 0) (Step ) ±V t (, t) I Check! P P V t π 54 t = 0 + V (, t) π θ : = θ : π ) θ = π ± sin ± cos t = 0 (, 0) = sin π V + t +V

基礎数学I

I & II ii ii........... 22................. 25 12............... 28.................. 28.................... 31............. 32.................. 34 3 1 9.................... 1....................... 1............

さくらの個別指導 ( さくら教育研究所 ) A 2 2 Q ABC 2 1 BC AB, AC AB, BC AC 1 B BC AB = QR PQ = 1 2 AC AB = PR 3 PQ = 2 BC AC = QR PR = 1

... 0 60 Q,, = QR PQ = = PR PQ = = QR PR = P 0 0 R 5 6 θ r xy r y y r, x r, y x θ x θ θ (sine) (cosine) (tangent) sin θ, cos θ, tan θ. θ sin θ = = 5 cos θ = = 4 5 tan θ = = 4 θ 5 4 sin θ = y r cos θ =

1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3

1. 1 A : l l : (1) l m (m 3) (2) m (3) n (n 3) (4) A 2 1 2 1 2 3 α, β γ α β + γ = 2 m l lm n nα nα = lm. α = lm n. m lm 2β 2β = lm β = lm 2. γ l 2. 3 4 P, Q R n = {(x 1, x 2,, x n ) ; x 1, x 2,, x n R}

, 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p, p 3,..., p n p, p,..., p n N, 3,,,,

6,,3,4,, 3 4 8 6 6................................. 6.................................. , 3, 6 = 3, 3,,,, 3,, 9, 3, 9, 3, 3, 4, 43, 4, 3, 9, 6, 6,, 0 p, p, p 3,..., p n N = p p p 3 p n + N p n N p p p,

untitled

.m 5m :.45.4m.m 3.m.6m (N/mm ).8.6 σ.4 h.m. h.68m h(m) b.35m θ4..5.5.5 -. σ ta.n/mm c 3kN/m 3 w 9.8kN/m 3 -.4 ck 6N/mm -.6 σ -.8 3 () :. 4 5 3.75m :. 7.m :. 874mm 4 865mm mm/ :. 7.m 4.m 4.m 6 7 4. 3.5

4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P = 90, = ( ) = X

4 4. 4.. 5 5 0 A P P P X X X X +45 45 0 45 60 70 X 60 X 0 P P 4 4 θ X θ P θ 4. 0, 405 P 0 X 405 X P 4. () 60 () 45 () 40 (4) 765 (5) 40 B 60 0 P 0 0 + 60 = 90, 0 + 60 = 750 0 + 60 ( ) = 0 90 750 0 90 0

3-1. 1) 1-1) =10.92m =18.20m m 2 6,480 3, =30 30kN/m 2 Z=0.9

3-1. 3-2. 3-3. 3-1. 1) 1-1) =10.92m =18.20m 198.74m 2 6,480 3,800 4.5 =30 30kN/m 2 Z=0.9 1-2) G1 G2 G3 G4 1-3) G1 G2 H3 1-4) t = 12 2.5 2) 2-1) No ( ) 1 120 120 2 120 120 3 120 180 360 4 120 150 210 5

1 1.1 Excel Excel Excel log 1, log 2, log 3,, log 10 e = ln 10 log cm 1mm 1 10 =0.1mm = f(x) f(x) = n

1 1.1 Excel Excel Excel log 1, log, log,, log e.7188188 ln log 1. 5cm 1mm 1 0.1mm 0.1 4 4 1 4.1 fx) fx) n0 f n) 0) x n n! n + 1 R n+1 x) fx) f0) + f 0) 1! x + f 0)! x + + f n) 0) x n + R n+1 x) n! 1 .

64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () m/s : : a) b) kg/m kg/m k

63 3 Section 3.1 g 3.1 3.1: : 64 3 g=9.85 m/s 2 g=9.791 m/s 2 36, km ( ) 1 () 2 () 3 9.8 m/s 2 3.2 3.2: : a) b) 5 15 4 1 1. 1 3 14. 1 3 kg/m 3 2 3.3 1 3 5.8 1 3 kg/m 3 3 2.65 1 3 kg/m 3 4 6 m 3.1. 65 5

untitled

GeoFem 1 1.1 1 1.2 1 1.3 1 2 2.1 2 2.2 3 2.3 FEM 5 (1) 5 (2) 5 (3) 6 2.4 GeoFem 7 2.5 FEM 16 2.6 19 2.7 26 3.1 33 3.2 35 3.3 GeoFem 36 3.4 48 3.5 49 A A1 A2 A3 A4 A5 A6 A7 GeoFem GeoFem CRS GeoFem GeoFem

熊本県数学問題正解

00 y O x Typed by L A TEX ε ( ) (00 ) 5 4 4 ( ) http://www.ocn.ne.jp/ oboetene/plan/. ( ) (009 ) ( ).. http://www.ocn.ne.jp/ oboetene/plan/eng.html 8 i i..................................... ( )0... (

入試の軌跡

4 y O x 4 Typed by L A TEX ε ) ) ) 6 4 ) 4 75 ) http://kumamoto.s.xrea.com/plan/.. PDF) Ctrl +L) Ctrl +) Ctrl + Ctrl + ) ) Alt + ) Alt + ) ESC. http://kumamoto.s.xrea.com/nyusi/kumadai kiseki ri i.pdf

untitled

0 ( L CONTENTS 0 . sin(-x-sinx, (-x(x, sin(90-xx,(90-xsinx sin(80-xsinx,(80-x-x ( sin{90-(ωφ}(ωφ. :n :m.0 m.0 n tn. 0 n.0 tn ω m :n.0n tn n.0 tn.0 m c ω sinω c ω c tnω ecω sin ω ω sin c ω c ω tn c tn ω

main.dvi

6 FIR FIR FIR FIR 6.1 FIR 6.1.1 H(e jω ) H(e jω )= H(e jω ) e jθ(ω) = H(e jω ) (cos θ(ω)+jsin θ(ω)) (6.1) H(e jω ) θ(ω) θ(ω) = KωT, K > 0 (6.2) 6.1.2 6.1 6.1 FIR 123 6.1 H(e jω 1, ω

IMO 1 n, 21n n (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a

1 40 (1959 1999 ) (IMO) 41 (2000 ) WEB 1 1959 1 IMO 1 n, 21n + 4 13n + 3 2 (x + 2x 1) + (x 2x 1) = A, x, (a) A = 2, (b) A = 1, (c) A = 2?, 3 a, b, c cos x a cos 2 x + b cos x + c = 0 cos 2x a = 4, b =

(Compton Scattering) Beaming 1 exp [i (k x ωt)] k λ k = 2π/λ ω = 2πν k = ω/c k x ωt ( ω ) k α c, k k x ωt η αβ k α x β diag( + ++) x β = (ct, x) O O x

Compton Scattering Beaming exp [i k x ωt] k λ k π/λ ω πν k ω/c k x ωt ω k α c, k k x ωt η αβ k α x β diag + ++ x β ct, x O O x O O v k α k α β, γ k γ k βk, k γ k + βk k γ k k, k γ k + βk 3 k k 4 k 3 k

修士論文

SAW 14 2 M3622 i 1 1 1-1 1 1-2 2 1-3 2 2 3 2-1 3 2-2 5 2-3 7 2-3-1 7 2-3-2 2-3-3 SAW 12 3 13 3-1 13 3-2 14 4 SAW 19 4-1 19 4-2 21 4-2-1 21 4-2-2 22 4-3 24 4-4 35 5 SAW 36 5-1 Wedge 36 5-1-1 SAW 36 5-1-2

極限

si θ = ) θ 0 θ cos θ θ 0 θ = ) P T θ H A, 0) θ, 0 < θ < π ) AP, P H A P T PH < AP < AT si θ < θ < ta θ si θ < θ < si θ cos θ θ cos θ < si θ θ < θ < 0 θ = h θ 0 cos θ =, θ 0 si θ θ =. θ 0 cos θ θ θ 0 cos

1. (8) (1) (x + y) + (x + y) = 0 () (x + y ) 5xy = 0 (3) (x y + 3y 3 ) (x 3 + xy ) = 0 (4) x tan y x y + x = 0 (5) x = y + x + y (6) = x + y 1 x y 3 (

1 1.1 (1) (1 + x) + (1 + y) = 0 () x + y = 0 (3) xy = x (4) x(y + 3) + y(y + 3) = 0 (5) (a + y ) = x ax a (6) x y 1 + y x 1 = 0 (7) cos x + sin x cos y = 0 (8) = tan y tan x (9) = (y 1) tan x (10) (1 +

Gmech08.dvi

51 5 5.1 5.1.1 P r P z θ P P P z e r e, z ) r, θ, ) 5.1 z r e θ,, z r, θ, = r sin θ cos = r sin θ sin 5.1) e θ e z = r cos θ r, θ, 5.1: 0 r

I A A441 : April 15, 2013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida )

I013 00-1 : April 15, 013 Version : 1.1 I Kawahira, Tomoki TA (Shigehiro, Yoshida) http://www.math.nagoya-u.ac.jp/~kawahira/courses/13s-tenbou.html pdf * 4 15 4 5 13 e πi = 1 5 0 5 7 3 4 6 3 6 10 6 17

f : R R f(x, y) = x + y axy f = 0, x + y axy = 0 y 直線 x+y+a=0 に漸近し 原点で交叉する美しい形をしている x +y axy=0 X+Y+a=0 o x t x = at 1 + t, y = at (a > 0) 1 + t f(x, y

017 8 10 f : R R f(x) = x n + x n 1 + 1, f(x) = sin 1, log x x n m :f : R n R m z = f(x, y) R R R R, R R R n R m R n R m R n R m f : R R f (x) = lim h 0 f(x + h) f(x) h f : R n R m m n M Jacobi( ) m n

untitled

- k k k = y. k = ky. y du dx = ε ux ( ) ux ( ) = ax+ b x u() = ; u( ) = AE u() = b= u () = a= ; a= d x du ε x = = = dx dx N = σ da = E ε da = EA ε A x A x x - σ x σ x = Eε x N = EAε x = EA = N = EA k =

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

(1) 1.1

1 1 1.1 1.1.1 1.1 ( ) ( ) ( ) { ( ) ( ) { ( ) ( ) { ( ) ( ) { ( ) ( ) { ( ) ( ) ( ) ( ) ( ) 2 1 1.1.2 (1) 1.1 1.1 3 (2) 1.2 4 1 (3) 1.3 ( ) ( ) (4) 1.1 5 (5) ( ) 1.4 6 1 (6) 1.5 (7) ( ) (8) 1.1 7 1.1.3

46 4 E E E E E 0 0 E E = E E E = ) E =0 2) φ = 3) ρ =0 1) 0 2) E φ E = grad φ E =0 P P φ = E ds 0

4 4.1 conductor E E E 4.1: 45 46 4 E E E E E 0 0 E E = E E E =0 4.1.1 1) E =0 2) φ = 3) ρ =0 1) 0 2) E φ E = grad φ E =0 P P φ = E ds 0 4.1 47 0 0 3) ε 0 div E = ρ E =0 ρ =0 0 0 a Q Q/4πa 2 ) r E r 0 Gauss

B

B YES NO 5 7 6 1 4 3 2 BB BB BB AA AA BB 510J B B A 510J B A A A A A A 510J B A 510J B A A A A A 510J M = σ Z Z = M σ AAA π T T = a ZP ZP = a AAA π B M + M 2 +T 2 M T Me = = 1 + 1 + 2 2 M σ Te = M 2 +T

卓球の試合への興味度に関する確率論的分析

17 i 1 1 1.1..................................... 1 1.2....................................... 1 1.3..................................... 2 2 5 2.1................................ 5 2.2 (1).........................

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F

III 1 (X, d) d U d X (X, d). 1. (X, d).. (i) d(x, y) d(z, y) d(x, z) (ii) d(x, y) d(z, w) d(x, z) + d(y, w) 2. (X, d). F X.. (1), X F, (2) F 1, F 2 F F 1 F 2 F, (3) F λ F λ F λ F. 3., A λ λ A λ. B λ λ

1 1 3 ABCD ABD AC BD E E BD 1 : 2 (1) AB = AD =, AB AD = (2) AE = AB + (3) A F AD AE 2 = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD 1 1

ABCD ABD AC BD E E BD : () AB = AD =, AB AD = () AE = AB + () A F AD AE = AF = AB + AD AF AE = t AC = t AE AC FC = t = (4) ABD ABCD AB + AD AB + 7 9 AD AB + AD AB + 9 7 4 9 AD () AB sin π = AB = ABD AD

(, Goo Ishikawa, Go-o Ishikawa) ( ) 1

(, Goo Ishikawa, Go-o Ishikawa) ( ) 1 ( ) ( ) ( ) G7( ) ( ) ( ) () ( ) BD = 1 DC CE EA AF FB 0 0 BD DC CE EA AF FB =1 ( ) 2 (geometry) ( ) ( ) 3 (?) (Topology) ( ) DNA ( ) 4 ( ) ( ) 5 ( ) H. 1 : 1+ 5 2

ORIGINAL TEXT I II A B 1 4 13 21 27 44 54 64 84 98 113 126 138 146 165 175 181 188 198 213 225 234 244 261 268 273 2 281 I II A B 292 3 I II A B c 1 1 (1) x 2 + 4xy + 4y 2 x 2y 2 (2) 8x 2 + 16xy + 6y 2

() x + y + y + x dy dx = 0 () dy + xy = x dx y + x y ( 5) ( s55906) 0.7. (). 5 (). ( 6) ( s6590) 0.8 m n. 0.9 n n A. ( 6) ( s6590) f A (λ) = det(a λi)

0. A A = 4 IC () det A () A () x + y + z = x y z X Y Z = A x y z ( 5) ( s5590) 0. a + b + c b c () a a + b + c c a b a + b + c 0 a b c () a 0 c b b c 0 a c b a 0 0. A A = 7 5 4 5 0 ( 5) ( s5590) () A ()

#A A A F, F d F P + F P = d P F, F y P F F x A.1 ( α, 0), (α, 0) α > 0) (x, y) (x + α) 2 + y 2, (x α) 2 + y 2 d (x + α)2 + y 2 + (x α) 2 + y 2 =

#A A A. F, F d F P + F P = d P F, F P F F A. α, 0, α, 0 α > 0, + α +, α + d + α + + α + = d d F, F 0 < α < d + α + = d α + + α + = d d α + + α + d α + = d 4 4d α + = d 4 8d + 6 http://mth.cs.kitmi-it.c.jp/

e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1, σ,..., σ N ) i σ i i n S n n = 1,,

01 10 18 ( ) 1 6 6 1 8 8 1 6 1 0 0 0 0 1 Table 1: 10 0 8 180 1 1 1. ( : 60 60 ) : 1. 1 e a b a b b a a a 1 a a 1 = a 1 a = e G G G : x ( x =, 8, 1 ) x 1,, 60 θ, ϕ ψ θ G G H H G x. n n 1 n 1 n σ = (σ 1,

x E E E e i ω = t + ikx 0 k λ λ 2π k 2π/λ k ω/v v n v c/n k = nω c c ω/2π λ k 2πn/λ 2π/(λ/n) κ n n κ N n iκ k = Nω c iωt + inωx c iωt + i( n+ iκ ) ωx

x E E E e i ω t + ikx k λ λ π k π/λ k ω/v v n v c/n k nω c c ω/π λ k πn/λ π/(λ/n) κ n n κ N n iκ k Nω c iωt + inωx c iωt + i( n+ iκ ) ωx c κω x c iω ( t nx c) E E e E e E e e κ e ωκx/c e iω(t nx/c) I I

Gmech08.dvi

63 6 6.1 6.1.1 v = v 0 =v 0x,v 0y, 0) t =0 x 0,y 0, 0) t x x 0 + v 0x t v x v 0x = y = y 0 + v 0y t, v = v y = v 0y 6.1) z 0 0 v z yv z zv y zv x xv z xv y yv x = 0 0 x 0 v 0y y 0 v 0x 6.) 6.) 6.1) 6.)

TOP URL 1

TOP URL http://amonphys.web.fc.com/ 1 19 3 19.1................... 3 19.............................. 4 19.3............................... 6 19.4.............................. 8 19.5.............................

.5 z = a + b + c n.6 = a sin t y = b cos t dy d a e e b e + e c e e e + e 3 s36 3 a + y = a, b > b 3 s363.7 y = + 3 y = + 3 s364.8 cos a 3 s365.9 y =,

[ ] IC. r, θ r, θ π, y y = 3 3 = r cos θ r sin θ D D = {, y ; y }, y D r, θ ep y yddy D D 9 s96. d y dt + 3dy + y = cos t dt t = y = e π + e π +. t = π y =.9 s6.3 d y d + dy d + y = y =, dy d = 3 a, b

JKR Point loading of an elastic half-space 2 3 Pressure applied to a circular region Boussinesq, n =

JKR 17 9 15 1 Point loading of an elastic half-space Pressure applied to a circular region 4.1 Boussinesq, n = 1.............................. 4. Hertz, n = 1.................................. 6 4 Hertz

() (, y) E(, y) () E(, y) (3) q ( ) () E(, y) = k q q (, y) () E(, y) = k r r (3).3 [.7 ] f y = f y () f(, y) = y () f(, y) = tan y y ( ) () f y = f y

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 =

( ; ) C. H. Scholz, The Mechanics of Earthquakes and Faulting : - ( ) σ = σ t sin 2π(r a) λ dσ d(r a) =

1 9 8 1 1 1 ; 1 11 16 C. H. Scholz, The Mechanics of Earthquakes and Faulting 1. 1.1 1.1.1 : - σ = σ t sin πr a λ dσ dr a = E a = π λ σ πr a t cos λ 1 r a/λ 1 cos 1 E: σ t = Eλ πa a λ E/π γ : λ/ 3 γ =

chap1.dvi

1 1 007 1 e iθ = cos θ + isin θ 1) θ = π e iπ + 1 = 0 1 ) 3 11 f 0 r 1 1 ) k f k = 1 + r) k f 0 f k k = 01) f k+1 = 1 + r)f k ) f k+1 f k = rf k 3) 1 ) ) ) 1+r/)f 0 1 1 + r/) f 0 = 1 + r + r /4)f 0 1 f