(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

Size: px
Start display at page:

Download "(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"

Transcription

1 ) T D = T = D = kn ) F W = F = W/ = kn/ = 15 kn ) R = W 1 + W = = 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 W = 1.4.1) P cos 6 P 1 cos 45 = 1.4.) P = P 1 cos 45 / cos 6 ) = P 1 1/ )/1/) = P ) P1 / + P 1 1/ W = P )/ = W P 1 = /{ + 1})W P = P 1 = / + 1)W 1.5 R A + R B 4 = 1.5.1) B R A = R A = N 1.5.1) R B = N

2 1.6 T 1.6.1)/ 1.6.) F T sin θ = F = T sin θ 1.6.1) T cos θ W = W = T cos θ 1.6.) tan θ = F/W θ = tan 1 F/W a A a/) sin α W a cos α F = tan α = F/W α = tan 1 F/W. 1.7 A R BX.4 = R BX = 6.8 kn. R BX R AX = R AX = R BX = 6.8 kn. R AY 8 kn kn = R AY = 8 kn. 1.8 A R BY 6 = R BY = 11. kn. R AX + + = R AX = 5 kn. R AY R BY = R AY = 18 R BY = 6.7 kn.

3 .1 1) P = kgf = 164 N. P 4 = P 1 + P P = N) = 14 N) = 1.8 kgf) ) 1 P 1 N P 1 + P 41 N 4 P 1 + P P 14 N.. σ = P A, ε = σ E, λ = P l = εl EA 4 1 N) σ 1 = π.) m ) = Pa) = 18 MPa) 1 18 MPa) ε 1 = = GPa) λ 1 = mm) = 1.54 mm) 4 1 N) σ = π.) m ) = Pa) = 141 MPa) 141 MPa) ε = = GPa) λ = mm) =.1 mm) 4 1 N) σ = π.5) m ) = Pa) = 5.9 MPa) 5.9 MPa) ε = = GPa) λ = mm) = 5.9 mm) ) 1 N mm = 1 N 1 ) m = 1 16 N m = 1 MPa N mm MPa. P N) = M kg) g m/s ) = Mg N)..1).17) λ S = Mg N) l m) E S 1 9 Pa) A S 1 6 m ) = λ A = Mgl 1 m) = Mgl mm)..) E S A S E S A S Mgl E A A A mm)..)

4 4..)..) A A = E S E A A S..4)..4) E S A S = E A A A..5) ).17).4 ε =.5 mm) 5 mm) = 1 1. ε = νε =. 1. λ = ε b =. 1 1 mm) =. mm)..5.17) 1 λ 1 = P l 1 E 1 A, λ = P l E A λ = λ 1 + λ = l1 + l ) P E 1 E A Pa).5 m).1m) = 5 kn) 7 5/7 kn) S m ) 5 1 N) 7 = 4 16 N/m ) S m ) S m ) = 5 1 N) N/m ) = m ) 4 d = 68 mm) = 18.5 mm) π

5 5 in ) 4 in 1 in = 5.4 mm ) in 4 in 5 mm 1 mm.7 σ a = σ B S = 84 MPa) W = π 4 D p 1 F F = π 4 d σ a n n W F = D p d σ a = = ) ) M M mm M1 1 M1 1.8 mm 1 mm ) )

6 6.1 l m) W = ρgal N).1.1) σ = W A = ρgl N/m ).1.).1.) l = σ ρg specific strength). =.11).7) σ ) σ = mg A A) x=l = A, ) ρgl A) x=l = C exp = A.11) σ C = A exp ρgl ) σ A = A exp ) ρg x l) σ. P 1 P 1 P 1 λ P 1 + P = P..1) λ = P 1 E 1 A 1 l = P E A l..)

7 7..1)..) E 1 A 1 + E A ) λ l = P λ = P l E 1 A 1 + E A.7 parallel model).9 series model).4 α 1 > α ) P =.).5 λ = l P + α 1 T l = E 1 A E 1 A 1 E A α 1 α ) T = E 1 A 1 E A l E A P + α T l ) P = α 1 α ) T E 1 A 1 E A α 1 α ) T E 1 A 1 + E A l P + α 1 T l = E A α 1 α ) T l E 1 A 1 E 1 A 1 + E A = E 1A 1 α 1 + E A α E 1 A 1 + E A C A 1a) ) T l + α 1 T l P AC = 1 P, P BC = P ) a) b) A 1

8 8 A-1b) ) AC λ AC = P AC l EA BC λ BC = P BC l = EA P l = EA P l EA C δ V ) δ H δ V = λ AC cos θ + λ BC cos θ 1 = + 4 δ H = λ AC sin θ λ BC sin θ 1 = 4 P l EA P l EA.6 A AB, AC A-a) A A P P A AB, AC P A AB, AC P AB, P AC AB A AC A a) A b) B c) C A A b) B B P AB BC, BD BC BD A c) C AC BC P AC, P BC ) CD, CE A A P.7.45) K = 1 + = 7. 5 MPa) 7 = 71 MPa).

9 9 A.8 n 1 + n = 1 N 1 = 1 1 6, N = 5 1 6, n 1 = 1 5 n =.7. N 1 N N n = A 4 A 4

10 P = τ S A = π = 1884 N) I p = πd4 = π 5 1 ) 4 Z p = πd 16 = π 5 1 ) 16 = m 4 ) = m ) I p = πd4 d 4 1) Z p = πd4 d 4 1) 16d = π 54 4 ) 1 8 = m 4 ) = π 54 4 ) = m ) 4.4 1) L 1 M t. πgd 4 = 1 π ) 4 = rad) ) L M t. πgd 4 = π ) 4 = rad) ) 1 : 16 4) = rad) 4.5 τ max = 16M t πd τ max τ u 16Mt d πτ u ) 1 / = ) 1 / 16 8 π = ) 1 / =.878 m)

11 MPa 5.6) = 16M t π.5 [1 ) ] M t = N m.5 =.46 rad 5.4).46 = M t π.5 4 [1 ) ] M t = N m 5. AB τ a = 5 MPa 5.6) = [ ) ] 4 d1 π.14 [1.14 ) ] 4 d1 = d 1 d 1 =.119 m AB ψ AB ψ AB = π =.15 rad CD τ a = 5 MPa = 16 9 π d d =.19 m

12 1 CD ψ CD ψ CD = BC ψ BC ψ BC = ψ =.8 rad π =.44 rad π.144 ψ = ψ AB + ψ BC + ψ CD =.44 rad =.5 5. B ψ AB M t1 BC M t M t1 = Gπd4 1ψ l 1 M t = Gπd4 ψ l M t1 + M t = M t [ πψ ] =.5 ψ =.56 1 rad = =.6 rad.6 = 1.5 [ ) ] π d 4 1 d =.6 m 8 MPa = 16 [ ) ] 4 1 π d 1 d =.5 m d d =.6 m

13 b) π 18 = H H = a) τ max = = 4.9 MPa) a) 75 = d d = 9.6 mm a) 75 = [ 1 d d = 44 mm 1 ) ] 4 1 d 1 = 17 mm ).4 = P P =.565 N 5.8) τ max = π ) = 5.5 MPa)

14 δ = 6 mm 5.1) [ 64P na Ra G d 4 + n br ] b a d 4 = δ b 64P [ P = 1.7 N ] =.6 a τ a 5.8) ) 4.5 τ a = π.1 =. MPa) b τ b 5.8) ) 4.1 τ b = π.15 =.5 MPa)

15 R A + R B = P 1. P 1 l + d) = R A l R A = l + d)p 1 l, R B = l 1 d)p 1 l 6. x l/ l/ x l Q = P 1, M = P 1 x Q = P 1 + f x l/), M = P 1 x + f x l/) / SFD BMD Q P 1 +f l/ M P 1 l+ f l /8 P 1 P 1 l/ l/ l l/ l SFD BMD 6. M max M max = f l = 5 1 = 1 kn m) a σ max = M max / a = 1 mm σ max = 6M max a = a = Pa) 6.4 M max = P l/4 σ max = M max /z z z = πd4 d 4 1) d

16 16 σ a P MPa) = 8d lp πd 4 d4 1 ) P = 16 πd 4 d 4 1) 8d l 87 kg = 859 N) = 87 kgf) 6.5 M 1, M M M 1 = τ max Q dx = Q =.Q Q = 6 kn) τ max = Q bh = 6 1 = MPa) P = 15. = 18 kgf) = 1764 N) f = 1764/ = 88 N/m) 6.7 v = 5f l 4 84 = = m) P ρ P = C D v wa = π.6 4 = 11.4 kgf) = 1.1 kn) v max I = πd4 d 4 1) = m 4 ) 64 v max = P l = =.111 m) = 1.1 mm) x a M = P x dv 1 dx = P x + C 1 ), v 1 = P x 6 + C 1x + C )

17 17 a x l a M = P a dv dx = P ax + C ), v = P ax + C x + C 4 ) x = a x = l/ v 1 = v =, dv /dx =, dv 1 dx = dv dx, v 1 = P 6 {x al a)x + a l 4a)}, dv 1 dx = P {x al a)}, v = P a {x lx + al a)} A x = ) v 1 dv 1 /dx C x = l/) v v 1 ) A = P ) a l 4a) dv1 P al a) P al a), =, v ) C = 6 dx 8 A 6.9 B R B R B = f l BD Mx) = f x lx) dv dx = f v = f x 4 x x = l 6.9.) lx + l 1 l x + l x v) C = 5f l 4 4 B x = 6.9.1) A dv dx = f l ) ) 6.9.1) 6.9.) v) A = f l l) = f l 4

18 I h 1 = 54, h = 6, h = 6, b 1 = 4, b = 8, b = 5, e 1 = 9., e =.7, c = 14.7) I = 1 b e b 1 c + b e 1) = 65 mm 4 ) = m 4 ) C ) dv = P l dx 6 = rad), C A v) A = v) C +.5 C v) C = P l 8 = m) ) dv = = m) dx C

19 x = 1 l R A = R B = P, M A = M B < x < l/ x M M = M A 1 P x d v dx = 1 M A 1 ) P x dv dx = 1 14 ) P x + M A x + C 1 v = 1 11 P x + 1 ) M Ax + C 1 x + C x dv dx = v = C 1 = C = x = 1 dv l dx = 1 16 P l + 1 M Al =, M A = M B = 1 8 P l dv dx = v max v max = P l 19 1/4 7. R A + R B = P R A l + M B P l = a), b) a) A b) A v a = R Al v b = P l l )l 6 a) b) A v a + v b =, R A = P l l l ) l R B = P P l l l ) l = P l ll + l) l

20 M B = P l R A l = P l l ll + l) l = P l 1l l1) l 7. A, C M A = M C = M B l 1 + l ) = f l 1 4 8M B = P b l l b ) ) M B =.9 kn m AB B R A = R A = 5 kn BC B y R C = R C = ) 1 = 7.1 kn) R A + R B + R C = = 17 1 R B = ) 1 = 9.9 kn) M B =.9 kn m, R A = 5 kn, R B = 9.9 kn, R C = 7.1 kn 7.4 M M = f x b h σ σ = 6M bh = f x bh b x /h = h h x /b = b x

21 1 7.5 σ x = l σ l σ = σ l M = P x I = b h 1 σ = 6P x b h σ l = 6P l b h l h = h l x l h l I l I = b h 1 = b h l 1 ) x = I lx x l l l x = l dv/dx = v = d v dx = P xl l l x x = P l l x dv dx = P l ) x 1 + C1 l v = P l ) 4 l x + C1 x + C x = v max C 1 = l, C = l v = P l ) 4 l x lx + l v max = 1P l EB h l l 8P l = Eb h l b h l 4P l /Eb h l

22 7.6 h 1 E 1 y 1 A 1 + E y A = 7 GPa)1 mm )h 1 5 mm) + 1 GPa)1 mm )h 1 6 mm) = h 1 = 7.4 mm, h = 7.6 mm y 1 =.4 mm, y =.6 mm I 1 = bt A 1y 1 = 1 1 m)1 1 m) 1 = m m ).4 1 m) I = m 4, I Z = I 1 + I = m 4 E i I i = E 1 I 1 + E I = n=1 = 8.5 kn m ) σ i = E iy ρ = E i M E 1 I 1 + E I y σ 1 = 4 kn m)7 16 kn/m )7.4 1 m) 8.5 kn m = 5. 1 kn/m = 5. MPa σ = 4 kn m)1 16 kn/m ) m) 8.5 kn m = 1.9 MPa σ 1 = 5. MPa, σ = 1.9 MPa 7.7 a) AB G P AB A, B θ 1 θ 1 = P l 16 AB b) M AB A, B θ θ = Ml

23 AB A θ = θ 1 θ AC, BD M AC, BD A, B θ θ = Ml 16 c) θ = θ P l 16 Ml = Ml M = P l C y, z I y, I z I y = hb 1 = = cm 4 ), I z = hb 1 = = 7 cm 4 ) y, z I yz = Y, Z I Y, I Z, I Y Z I Y = 1 I y + I z ) + 1 I y I z ) cos θ = 1 + 7) + 1 7) cos ) = 4. cm 4 ) I Z = 1 I y + I z ) 1 I y I z ) cos θ = 1 + 7) 1 7) cos ) = 6 cm 4 ) I Y Z = 1 I y I z ) sin θ = 1 7) sin ) = 17. cm 4 ) 7.9 A A = l 1 t + l t)t = ) 1 = 17 cm ) Y, Z S Y, S Z S Y = t l 1 S Z = l 1 t) t C y, z + l t)t + t l = = 8 1) 1 1 1) y = S Z A = =.15 cm) z = S Y A = =.15 cm) = 6.5 cm ) = 5.5 cm )

24 4 C y, z I y, I z I yz I y = I Y z A = tl 1 + l t)t = ) 1 + I z = I Z y A = l 1 t)t z A = 17.1 cm 4 ) + tl y A 8 1) 1 = = 8.1 cm 4 ) I yz = I Y Z y za = l 1t + l t t y za = t 4 l 1 + l t ) y za = ) = 74. cm 4 ) z θ θ = 1 tan 1 I yz = I z I y tan =.9 θ 1 =.9, θ = 67.1 z θ 1 =.9 Z I 1 θ = 67.1 Y I 7.1 y, z I y, I z I y = hb 1, I z = bh 1 AB 7.18 M = P l M y = P l sin θ, M z = P l cos θ A C σ max 7.7) y = h, z = b σ max = M y z I y = 6P l bh M z y I z = sin θ b + cos θ h P l sin θ) b ) ) hb 1 ) h P l cos θ) bh 1

25 5 7.76) tan β = I z I y tan θ = bh 1 hb 1 tan θ = h b tan θ

26 ) 15 / P = 1 π cos 6 = 54 N) σ = 1 cos 6 = 5 MPa) τ = 1 sin 6 = 87 MPa) 8. x = OC = OE + EC y = OD = AF AE OE = AO cos θ = x cos θ EC = AG = AP sin θ = BO sin θ = y sin θ AF = PG = AP cos θ = BO cos θ = y cos θ AE = AO sin θ = x sin θ x = x cos θ + y sin θ y = x sin θ + y cos θ ) σ 1 = 1 σ x + ) + 1 σ x ) + 4τxy = 1 σ x + 1 σx + 4τxy 8.) τ max = 1 σ x ) + 4τxy = 1 σx + 4τxy ) ε x = 1 ε x + ε y ) + 1 ε x ε y ) cos θ + 1 γ xy sin θ ε y = 1 ε x + ε y ) 1 ε x ε y ) cos θ 1 γ xy sin θ γ xy = ε y ε x ) sin θ + γ xy cos θ ε x = 1, ε y = 1, γ xy = 1 4

27 7 ε x = 1 cos θ sin θ ε y = 1 cos θ sin θ γ xy = 1 sin θ cos θ 8.5 1) σ x, σ y, τ xy ) = 5, 5, 5) σ 1, σ = 1,, θ = 45 ) σ x, σ y, τ xy ) = 5, 5, 5) σ 1, σ = 1,, θ = 45 ) σ x, σ y, τ xy ) = 5, 5, 5) σ 1, σ = 5, 5, θ =.5 4) σ x, σ y, τ xy ) = 5, 5, ) σ 1, σ =,, θ = 8.6 G = E 1 + ν) ν = E G 1 ν = ) 1 = ) ε x = σ x νσ y E, ε y = νσ x + σ y E 1 + ε x 1 + ε y = E + σ x νσ y E E νσ x + σ y E E + σ x νσ y E νσ x + σ y = E + σ x νσ y E νσ x + σ y = = γ + γ 1 γ 1 + γ γ)e νσ x + σ y ) = + γ)e + σ x νσ y ) E νσ x + σ y ) γe νσ x + σ y ) = E + σ x νσ y ) + γe + σ x νσ y ) γe νσ x + σ y + E + σ x νσ y ) = σ x + νσ x σ y νσ y ) γ{e νσ x + σ y ) + σ x + σ y } = {1 + ν)σ x 1 + ν)σ y } γ{e νσ x + σ y ) + σ x + σ y )} = {1 + ν)σ x σ y )} σ x + σ y =, σ x σ y = σ

28 8 Eγ = 1 + ν) σ) γ = 1 + ν)σ E = 1 + ν)τ E

29 ) S = σ t σ a 5 = σ t 9. σ t = 45 MPa. 9. 1) 1) ) 1) ) )) ) ) 1) σ ult = P A = P max πd 4

30 T τ τ = T ) d = T ) d I p πd 4 σ 1 σ 1 = τ = 16T πd = 16T πd = σ ult = 4P max πd T = d 4 P max. ) P σ x T τ 8 σ 1 = σ x + 1 σx + 4τ, θ 1 = 1 ) τ tan 1 σ x σ 1 = σ ult = 4P max πd 9. 1) σ θ = pr t θ 1 = p 1 = p, σ z = pr t = p 1 = 15p σ r = σ θ = σ 1, σ z = σ, σ r = σ 9.1) σ 1 σ ) + σ σ ) + σ σ 1 ) = σ Y σ θ σ z ) + σ z ) + σ θ ) = σ Y 15p) + 15p) + p) = ), p = = 11.5 MPa) ) 9.1) σ 1 σ = σ Y σ θ σ r = σ Y p = MPa), p = 1 MPa)

31 ) 1) ) 45 1) ) ) ) ) σ max = τ, σ min = τ 4 τ = T Z p = 16DT πd 4 d 4 ) τ = RT πr 4 r 4 ) 9.4.1) 8 Hooke ε x = 1 E {σ x νσ y + σ z )} ε x = ε, σ z =, σ x = σ max, σ y = σ min ε = 1 E {σ max νσ min } = τ {1 + ν}. E

32 9.4.1) G = ε = 1 E {σ max νσ min } = 1 + ν) E RT πr 4 r 4 ) = RT GπR 4 r 4 ) E 1 + ν) 9.5 M t 9.5) C τ = 16M t πd M 9.7) C N σ N = σ M = M πd N πd 4 = 4N πd M t M N τ = 16M t πd σ = σ M + σ N = M πd + 4N πd = M πd + 4Nd πd 4 = πd M + Nd ) 8 9.7) M M + Nd 8 9.8) M M + Nd ) ε a = 1 ε 1 + ε ) + 1 ε 1 ε ) cos θ ε b = 1 ε 1 + ε ) + 1 ε 1 ε ) cos θ + α) ε c = 1 ε 1 + ε ) + 1 ε 1 ε ) cos θ + α) α = 6 ε 1 + ε = A, ε 1 ε = B ε a = A + B cos θ ε b = A + B cos θ + 6 ) ε c = A + B cos θ + 1 )

33 ε a = A + B cos θ ε b = A + B cos θ cos 1 B sin θ sin 1 ε c = A + B cos θ cos 4 B sin θ sin ) 9.6.) 9.6.) 9.6.) 9.6.) 9.6.) 9.6.) ε a = A + B cos θ 9.6.1) cos θ sin θ ε b = A B B 9.6.) cos θ sin θ ε c = A B + B 9.6.) ε a + ε b + ε c = A ε b + ε c = A B cos θ ε b + ε c A = B cos θ 9.6.4) ε a A = B cos θ ε c ε b = B sin θ ε c ε b = B sin θ 9.6.5) 9.6.4) 9.6.5) ) ε a A) εc ε b + = B B A = ε a + ε b + ε c B = tan θ = ε a A) + sin θ cos θ = εb ε c ) = ε b ε c ε a A εa ε b ) + ε b ε c ) + ε c ε a )

34 4 ε 1 + ε = 1 ε a + ε b + ε c ) ε 1 ε = εa ε b ) + ε b ε c ) + ε c ε a ) tan θ = ε c ε b ε a ε b ε c 9.7 1) 8 ε x = ε x cos θ + ε y sin θ + γ xy sin θ cos θ = 1 ε x + ε y ) + 1 ε x ε y ) cos θ + 1 γ xy sin θ ε a = ε x cos θ 1 + ε y sin θ 1 + γ xy sin θ 1 cos θ 1 ε b = ε x cos θ + ε y sin θ + γ xy sin θ cos θ 9.7.1) ε c = ε x cos θ + ε y sin θ + γ xy sin θ cos θ ) 9.7.1) θ 1 =, θ = 6, θ = 1 ε a = ε x cos + ε y sin + γ xy sin cos ε b = ε x cos 6 + ε y sin 6 + γ xy sin 6 cos 6 ε c = ε x cos 1 + ε y sin 1 + γ xy sin 1 cos 1 ε a = ε x ε b = ε x 4 + ε y 4 + ε c = ε x 4 + ε y 4 γxy 4 γxy 4 ε x = ε a ε y = 1 ε b + ε c ε a ) γ xy = ε b ε c )

35 ) q N/mm ) x σ = qax A = qx N/mm ) x dx 1.6) du = σ E Adx = q x E Adx q = mg Al U = du = q A E U = mg/al) Al 6E x dx = q Al 6E = m g l 6AE 1. a)b)c) U a, U b, U c σ e U a = σ E Al = σ e E π 4 d) l = σ eπd l E b) d P b / π 4 d ) σ e d P b / π 4 d) ) σ e /4 U b = σ E Al = σ e E = σ eπd l 8E U c = σ E Al = σ e E U a : U b : U c = σ eπd l E σe ) π 4 d) l + 4 π E 4 d) l + σ eπd l E = 5σ eπd l E π 4 d) l = σ eπd l 4E : 5σ eπd l E : σ eπd l 4E = : 5 : 8 1. x d A σ A = π 4 d, σ = P π 4 d

36 6 dx du ) 4P πd du = σ E Adx = E π 4 d dx = P dx πed d = d 1 + d d 1 )x/l U 1.4 U = = P πe = P πe du = P πe dx d = P dx πe d 1 + d d 1 )x l l l 1 { d d 1 d 1 + d d 1 ) x } l [ l 1 1 ] = P l d 1 d d d 1 πe d 1 d A x M x = R A x + M A + f x U R A =, U = U R A M x M x = 1 R A U M A =, U = = 1 = 1 = 1 [ RA M x dx M x M x dx R A R A x + M A + f x x M A x f RA l M A l f 8 l4 8 x4 ] l ) ) = ) x dx 8R A l 1M A l f l 4 = 1M A + 8R A l f l = 1.4.1) U = U M A M x M x = 1 M A = 1 = 1 M x M x dx M A R A x + M A + f x [ RA x + M A x + f 6 x ) dx ] l = 1 6 R Al + 6M A + f l ) =

37 ) 1.4.) 6M A R A l + f l = 1.4.) R A = f l = R B, M A = f l 1 = M B 1.5 x C M M M = M = x l ) x l/) l/ x l) M = x l/) M = 1 l/ x l) δ = = i = = / / / / 1.6 U U = A δ A MM P x MM P x dx + P x) dx + l/ dx + dx + δ A = U P = dx + l/ l/ MM dx l/ P x MM x l ) P x 1 dx dx = dx = P l 8 5P l 48 ) l P l) dx = P + l = 4 P l U = P P l P l ) = AB M AB = P x T AB = P b BC M CB = P x 1 U U = a M AB dx + a = P a 6 + P ab + P b GI p 6 TAB b dx + GI p P l M CB dx 1

38 8 C δ C 1.8 δ C = U P = P a + b ) + P ab GI p AB T 1 BC T T 1 cos θ + T =, T 1 sin θ = P T 1, T T 1 = P sin θ, T = P tan θ T 1, T δ B = T 1 l 1 P AE + T ) l = T 1L 1 AE AE T 1 P + T l AE T P l 1 = l cos θ, l = l, T 1 P = 1 sin θ, T P = 1 tan θ 1.9 δ B = 1 AE = P l AE P sin θ 1 + cos θ sin θ cos θ l cos θ 1 sin θ + 1 AE P tan θ ) l 1 tan θ AC CB M 1, M, U 1, U U = U 1 + U M 1 = R A x + 1 f x, U 1 = 1 / A R A δ A δ A = M = R A x + f l M 1 dx, U = 1 l/ x l ) 6 M dx U = U 1 + U = U 1 M 1 + U M R A R A R A M 1 R A M R A = 1 { / M 1 M 1 R A dx + l/ M M R A dx M 1, M M 1 = M = x A x = δ A = R A R A / R A x 1 ) { f x x dx + R A x f l x l )} x dx = l/ 6 R A = f l } )

39 ) M = P x M t = P s ) U B = 1 U T = 1 GI p M dx = P l 6 M t dx = P s l GI p A M = 1 x A M t = 1 δ = 1 ψ = 1 GI p MM dx = 1 M t M t dx = 1 GI p P x dx = P l P s dx = P s GI p l

40 I = = 1184 cm 4 ) = m 4 ) P cr = π 4 l = π = 59π = N) = 19.5 kn) 11. I = π cm 4 = π m 4 P cr = 4π l = 4π σ cr = 4π Al = 4π π π = π 1 4 = N) = 6 kn) π = 8π 1 6 = Pa) = 79 MPa) 11. v = C 1 sin α x l d4 v dx 4 + P d v dx = + C cos α x l + C x l + C 4 C + C + C 4 = α l C l C = α l C 1 cos α α l C 1 sin α + C = l α ) α ) C1 cos α + C sin α = l l C + C 4 = C = C 4 C 1 =, C = C sin α = n = 1 α = nπ P cr = π l

41 vx) x M = d v dx vx) v max M P Mx) = P v max v) d v dx + P v = P v max ) v h = v max ) v = C 1 sin α x l + C cos α x l + v max x = v =, k dv dx = P v max α k l vl) = v max C + v max = ) C 1 = P v max C 1 sin α + C cos α = α tan α = kl k α = π/ k α = kl/ P cr = k/l A = = cm ) = 1 4 m ) I = σ cr = π l A = 14 cm 4 ) = m 4 ) = π E = π E

42 4 σ cr σ cr = σ cr /G 1 = π E/E/1 +.5))) 1 = π =.961 4% 11.6 σ cr = 1974 MPa > σ Y / σ cr = 4 [ r = 1 I πd 4 A = π 8 4 πd = d 4, l e = l ) ] [ 1 = π 8 = = 19 MPa) ]

43 ) { σ θ ) max = ρr ω + ν) ν + ν r1 r ) } = {4 4 π 6 )} +.) { ) } = Pa) = 8.9 MPa) ) 8σ θ ) max v = ωr = + ν)ρ = =.4 m/s) +.) σ a = σ θ ) max 1.47) 1.4 p 1 = r r 1)σ θ ) max r + r 1 =.1.1 ) = Pa) = 1.8 MPa) σ θ ) max σ θ ) max = p 1r 1 r r 1 = p 1 n 1 σ θ ) max 1.47) ) + 1} σ θ ) max = r 1 + r )p 1 r r 1 = { r r 1 r ) 1 p 1 = n + 1)p 1 n 1 r 1 σ θ ) max σ θ ) max = n + 1 n ) δ = 4p m E r r r 1) r r )r r 1 ) = ).7.6 ).6.5 ) = m) =.141 mm)

44 ) 1.61) σ θ = σ θ + σ θ = r + r r p m + r = = Pa) = 479 MPa) ) 1.6) r r 1 r r 1 r r 1 r ) p ) σ θ1 = σ θ1 + σ θ1 = r r p m + r + r1 r 1 r p 1 r 1 = = Pa) = 46.8 MPa) ) 1.78) σ θ ) max = r 1 + r )p 1 r r 1 = 1 + n 1 n p 1 σ max = r + r 1 r r 1 ) p 1 = 1 + n 1 n ) p 1 σ θ ) max σ max ) 1.9 p 1 = r r 1) r + r 1 = 1 + n 1 n 1 n ) 1 + n = ) =.8 σ max 1 =.14.8 ) = Pa) = 118 MPa) 1.84) + ν) t = p a 8 σ max +.) = =. m) = mm) ) w w = 1 D p 64 r4 + C 1 r + C + C r log r + C 4 log r)

45 45 p = w = 1 D C1 r + C + C r log r + C 4 log r ) C 4 = 1.79) 1.8) d w Q = D dr + 1 r w d w dr 1 ) dw r = D d dr dr Q = 4C r { 1 r r Q P P + πrq = d dr r dw )} dr C = P 8π w w = 1 D C 1 r + C + P ) 8π r log r r = a w = M θ = 1.8) C 1 = P 16 log a + + ν ), C = P a 1 + ν 16π 1.8) M θ M r M θ = P 4π 1 + ν) log a r, M r = P 4π + ν 1 + ν {1 + ν) log a r + 1 ν) } σ θ σ r 1.15) 1.16) σ θ = E 1 ν ε θ + νε r ) = Ez d w 1 ν dr + ν ) dw = 1M θ r dr t z σ r = E 1 ν ε r + νε θ ) = Ez ) 1 dw 1 ν r dr + ν d w dr = 1M r t z σ θ σ r σ θ = 1M θ t σ r = 1M r t z = 1z t P 4π 1 + ν) log a r = 1P z 1 + ν) 4πt z = 1z P {1 t + ν) log a } 4π r + 1 ν) = 1P z { 4πt 1 + ν) log a } r + 1 ν)

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

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

More information

dynamics-solution2.dvi

dynamics-solution2.dvi 1 1. (1) a + b = i +3i + k () a b =5i 5j +3k (3) a b =1 (4) a b = 7i j +1k. a = 14 l =/ 14, m=1/ 14, n=3/ 14 3. 4. 5. df (t) d [a(t)e(t)] =ti +9t j +4k, = d a(t) d[a(t)e(t)] e(t)+ da(t) d f (t) =i +18tj

More information

untitled

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 =

More information

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) (

( ) sin 1 x, cos 1 x, tan 1 x sin x, cos x, tan x, arcsin x, arccos x, arctan x. π 2 sin 1 x π 2, 0 cos 1 x π, π 2 < tan 1 x < π 2 1 (1) ( 6 20 ( ) sin, cos, tan sin, cos, tan, arcsin, arccos, arctan. π 2 sin π 2, 0 cos π, π 2 < tan < π 2 () ( 2 2 lim 2 ( 2 ) ) 2 = 3 sin (2) lim 5 0 = 2 2 0 0 2 2 3 3 4 5 5 2 5 6 3 5 7 4 5 8 4 9 3 4 a 3 b

More information

[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

More information

Part () () Γ Part ,

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

More information

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 +

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

More information

Gmech08.dvi

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

More information

() 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)

() 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 ()

More information

all.dvi

all.dvi 38 5 Cauchy.,,,,., σ.,, 3,,. 5.1 Cauchy (a) (b) (a) (b) 5.1: 5.1. Cauchy 39 F Q Newton F F F Q F Q 5.2: n n ds df n ( 5.1). df n n df(n) df n, t n. t n = df n (5.1) ds 40 5 Cauchy t l n mds df n 5.3: t

More information

( )

( ) 18 10 01 ( ) 1 2018 4 1.1 2018............................... 4 1.2 2018......................... 5 2 2017 7 2.1 2017............................... 7 2.2 2017......................... 8 3 2016 9 3.1 2016...............................

More information

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

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

More information

変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy,

変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, 変 位 変位とは 物体中のある点が変形後に 別の点に異動したときの位置の変化で あり ベクトル量である 変位には 物体の変形の他に剛体運動 剛体変位 が含まれている 剛体変位 P(x, y, z) 平行移動と回転 P! (x + u, y + v, z + w) Q(x + d x, y + dy, z + dz) Q! (x + d x + u + du, y + dy + v + dv, z +

More information

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y

(3) (2),,. ( 20) ( s200103) 0.7 x C,, x 2 + y 2 + ax = 0 a.. D,. D, y C, C (x, y) (y 0) C m. (2) D y = y(x) (x ± y 0), (x, y) D, m, m = 1., D. (x 2 y [ ] 7 0.1 2 2 + y = t sin t IC ( 9) ( s090101) 0.2 y = d2 y 2, y = x 3 y + y 2 = 0 (2) y + 2y 3y = e 2x 0.3 1 ( y ) = f x C u = y x ( 15) ( s150102) [ ] y/x du x = Cexp f(u) u (2) x y = xey/x ( 16) ( s160101)

More information

s s U s L e A = P A l l + dl dε = dl l l

s s U s L e A = P A l l + dl dε = dl l l P (ε) A o B s= P A s B o Y l o s Y l e = l l 0.% o 0. s e s B 1 s (e) s Y s s U s L e A = P A l l + dl dε = dl l l ε = dε = l dl o + l lo l = log l o + l =log(1+ e) l o Β F Α E YA C Ο D ε YF B YA A YA

More information

熊本県数学問題正解

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... (

More information

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x

x () g(x) = f(t) dt f(x), F (x) 3x () g(x) g (x) f(x), F (x) (3) h(x) = x 3x tf(t) dt.9 = {(x, y) ; x, y, x + y } f(x, y) = xy( x y). h (x) f(x), F (x [ ] IC. f(x) = e x () f(x) f (x) () lim f(x) lim f(x) x + x (3) lim f(x) lim f(x) x + x (4) y = f(x) ( ) ( s46). < a < () a () lim a log xdx a log xdx ( ) n (3) lim log k log n n n k=.3 z = log(x + y ),

More information

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

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 2005 9/8-11 2 2.2 ( 2-5) γ ( ) γ cos θ 2πr πρhr 2 g h = 2γ cos θ ρgr (2.1) γ = ρgrh (2.2) 2 cos θ θ cos θ = 1 (2.2) γ = 1 ρgrh (2.) 2 2. p p ρgh p ( ) p p = p ρgh (2.) h p p = 2γ r 1 1 (Berry,1975) 2-6

More information

.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 =,

.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

More information

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

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

More information

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

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

More information

[ ] 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

[ ] 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 =

More information

pdf

pdf http://www.ns.kogakuin.ac.jp/~ft13389/lecture/physics1a2b/ pdf I 1 1 1.1 ( ) 1. 30 m µm 2. 20 cm km 3. 10 m 2 cm 2 4. 5 cm 3 km 3 5. 1 6. 1 7. 1 1.2 ( ) 1. 1 m + 10 cm 2. 1 hr + 6400 sec 3. 3.0 10 5 kg

More information

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P

6 2 2 x y x y t P P = P t P = I P P P ( ) ( ) ,, ( ) ( ) cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ y x θ x θ P 6 x x 6.1 t P P = P t P = I P P P 1 0 1 0,, 0 1 0 1 cos θ sin θ cos θ sin θ, sin θ cos θ sin θ cos θ x θ x θ P x P x, P ) = t P x)p ) = t x t P P ) = t x = x, ) 6.1) x = Figure 6.1 Px = x, P=, θ = θ P

More information

untitled

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

More information

29

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

More information

) 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 = 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) 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

More information

1

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

More information

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C

1990 IMO 1990/1/15 1:00-4:00 1 N N N 1, N 1 N 2, N 2 N 3 N 3 2 x x + 52 = 3 x x , A, B, C 3,, A B, C 2,,,, 7, A, B, C 0 9 (1990 1999 ) 10 (2000 ) 1900 1994 1995 1999 2 SAT ACT 1 1990 IMO 1990/1/15 1:00-4:00 1 N 1990 9 N N 1, N 1 N 2, N 2 N 3 N 3 2 x 2 + 25x + 52 = 3 x 2 + 25x + 80 3 2, 3 0 4 A, B, C 3,, A B, C 2,,,, 7,

More information

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1

1 I 1.1 ± e = = - = C C MKSA [m], [Kg] [s] [A] 1C 1A 1 MKSA 1C 1C +q q +q q 1 1 I 1.1 ± e = = - =1.602 10 19 C C MKA [m], [Kg] [s] [A] 1C 1A 1 MKA 1C 1C +q q +q q 1 1.1 r 1,2 q 1, q 2 r 12 2 q 1, q 2 2 F 12 = k q 1q 2 r 12 2 (1.1) k 2 k 2 ( r 1 r 2 ) ( r 2 r 1 ) q 1 q 2 (q 1 q 2

More information

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α

18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 18 I ( ) (1) I-1,I-2,I-3 (2) (3) I-1 ( ) (100 ) θ ϕ θ ϕ m m l l θ ϕ θ ϕ 2 g (1) (2) 0 (3) θ ϕ (4) (3) θ(t) = A 1 cos(ω 1 t + α 1 ) + A 2 cos(ω 2 t + α 2 ), ϕ(t) = B 1 cos(ω 1 t + α 1 ) + B 2 cos(ω 2 t

More information

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2

II No.01 [n/2] [1]H n (x) H n (x) = ( 1) r n! r!(n 2r)! (2x)n 2r. r=0 [2]H n (x) n,, H n ( x) = ( 1) n H n (x). [3] H n (x) = ( 1) n dn x2 e dx n e x2 II No.1 [n/] [1]H n x) H n x) = 1) r n! r!n r)! x)n r r= []H n x) n,, H n x) = 1) n H n x) [3] H n x) = 1) n dn x e dx n e x [4] H n+1 x) = xh n x) nh n 1 x) ) d dx x H n x) = H n+1 x) d dx H nx) = nh

More information

Note.tex 2008/09/19( )

Note.tex 2008/09/19( ) 1 20 9 19 2 1 5 1.1........................ 5 1.2............................. 8 2 9 2.1............................. 9 2.2.............................. 10 3 13 3.1.............................. 13 3.2..................................

More information

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

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

More information

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H

V(x) m e V 0 cos x π x π V(x) = x < π, x > π V 0 (i) x = 0 (V(x) V 0 (1 x 2 /2)) n n d 2 f dξ 2ξ d f 2 dξ + 2n f = 0 H n (ξ) (ii) H 199 1 1 199 1 1. Vx) m e V cos x π x π Vx) = x < π, x > π V i) x = Vx) V 1 x /)) n n d f dξ ξ d f dξ + n f = H n ξ) ii) H n ξ) = 1) n expξ ) dn dξ n exp ξ )) H n ξ)h m ξ) exp ξ )dξ = π n n!δ n,m x = Vx)

More information

K E N Z U 01 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.................................... 4 1..1..................................... 4 1...................................... 5................................

More information

4 5.............................................. 5............................................ 6.............................................. 7......................................... 8.3.................................................4.........................................4..............................................4................................................4.3...............................................

More information

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 * [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 =

More information

21 2 26 i 1 1 1.1............................ 1 1.2............................ 3 2 9 2.1................... 9 2.2.......... 9 2.3................... 11 2.4....................... 12 3 15 3.1..........

More information

K E N Z U 2012 7 16 HP M. 1 1 4 1.1 3.......................... 4 1.2................................... 4 1.2.1..................................... 4 1.2.2.................................... 5................................

More information

untitled

untitled 1 17 () BAC9ABC6ACB3 1 tan 6 = 3, cos 6 = AB=1 BC=2, AC= 3 2 A BC D 2 BDBD=BA 1 2 ABD BADBDA ABC6 BAD = (18 6 ) / 2 = 6 θ = 18 BAD = 12 () AD AD=BADCAD9 ABD ACD A 1 1 1 1 dsinαsinα = d 3 sin β 3 sin β

More information

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e

Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e 7 -a 7 -a February 4, 2007 1. 2. 3. 4. 1. 2. 3. 1 Gauss Gauss ɛ 0 E ds = Q (1) xy σ (x, y, z) (2) a ρ(x, y, z) = x 2 + y 2 (r, θ, φ) (1) xy A Gauss ɛ 0 E ds = ɛ 0 EA Q = ρa ɛ 0 EA = ρea E = (ρ/ɛ 0 )e z

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc.com/ 3.............................. 3.............................. 4.3 4................... 5.4........................ 6.5........................ 8.6...........................7

More information

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

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 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

More information

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5.

4. ϵ(ν, T ) = c 4 u(ν, T ) ϵ(ν, T ) T ν π4 Planck dx = 0 e x 1 15 U(T ) x 3 U(T ) = σt 4 Stefan-Boltzmann σ 2π5 k 4 15c 2 h 3 = W m 2 K 4 5. A 1. Boltzmann Planck u(ν, T )dν = 8πh ν 3 c 3 kt 1 dν h 6.63 10 34 J s Planck k 1.38 10 23 J K 1 Boltzmann u(ν, T ) T ν e hν c = 3 10 8 m s 1 2. Planck λ = c/ν Rayleigh-Jeans u(ν, T )dν = 8πν2 kt dν c

More information

all.dvi

all.dvi 5,, Euclid.,..,... Euclid,.,.,, e i (i =,, ). 6 x a x e e e x.:,,. a,,. a a = a e + a e + a e = {e, e, e } a (.) = a i e i = a i e i (.) i= {a,a,a } T ( T ),.,,,,. (.),.,...,,. a 0 0 a = a 0 + a + a 0

More information

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 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

More information

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

( ; ) 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 =

More information

I

I I 6 4 10 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

More information

2011de.dvi

2011de.dvi 211 ( 4 2 1. 3 1.1............................... 3 1.2 1- -......................... 13 1.3 2-1 -................... 19 1.4 3- -......................... 29 2. 37 2.1................................ 37

More information

I ( ) 2019

I ( ) 2019 I ( ) 2019 i 1 I,, III,, 1,,,, III,,,, (1 ) (,,, ), :...,, : NHK... NHK, (YouTube ),!!, manaba http://pen.envr.tsukuba.ac.jp/lec/physics/,, Richard Feynman Lectures on Physics Addison-Wesley,,,, x χ,

More information

20 4 20 i 1 1 1.1............................ 1 1.2............................ 4 2 11 2.1................... 11 2.2......................... 11 2.3....................... 19 3 25 3.1.............................

More information

006 11 8 0 3 1 5 1.1..................... 5 1......................... 6 1.3.................... 6 1.4.................. 8 1.5................... 8 1.6................... 10 1.6.1......................

More information

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x (

II (10 4 ) 1. p (x, y) (a, b) ε(x, y; a, b) 0 f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a 2 x a 0 A = f x ( II (1 4 ) 1. p.13 1 (x, y) (a, b) ε(x, y; a, b) f (x, y) f (a, b) A, B (6.5) y = b f (x, b) f (a, b) x a = A + ε(x, b; a, b) x a x a A = f x (a, b) y x 3 3y 3 (x, y) (, ) f (x, y) = x + y (x, y) = (, )

More information

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

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

More information

http://www.ike-dyn.ritsumei.ac.jp/ hyoo/wave.html 1 1, 5 3 1.1 1..................................... 3 1.2 5.1................................... 4 1.3.......................... 5 1.4 5.2, 5.3....................

More information

meiji_resume_1.PDF

meiji_resume_1.PDF β β β (q 1,q,..., q n ; p 1, p,..., p n ) H(q 1,q,..., q n ; p 1, p,..., p n ) Hψ = εψ ε k = k +1/ ε k = k(k 1) (x, y, z; p x, p y, p z ) (r; p r ), (θ; p θ ), (ϕ; p ϕ ) ε k = 1/ k p i dq i E total = E

More information

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2

No δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i x j δx j (5) δs 2 No.2 1 2 2 δs δs = r + δr r = δr (3) δs δs = r r = δr + u(r + δr, t) u(r, t) (4) δr = (δx, δy, δz) u i (r + δr, t) u i (r, t) = u i δx j (5) δs 2 = δx i δx i + 2 u i δx i δx j = δs 2 + 2s ij δx i δx j

More information

TOP URL 1

TOP URL   1 TOP URL http://amonphys.web.fc2.com/ 1 6 3 6.1................................ 3 6.2.............................. 4 6.3................................ 5 6.4.......................... 6 6.5......................

More information

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 % 0.2% 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

More information

入試の軌跡

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

More information

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

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

More information

数学の基礎訓練I

数学の基礎訓練I I 9 6 13 1 1 1.1............... 1 1................ 1 1.3.................... 1.4............... 1.4.1.............. 1.4................. 3 1.4.3........... 3 1.4.4.. 3 1.5.......... 3 1.5.1..............

More information

II ( ) (7/31) II ( [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re

II ( ) (7/31) II (  [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Re II 29 7 29-7-27 ( ) (7/31) II (http://www.damp.tottori-u.ac.jp/~ooshida/edu/fluid/) [ (3.4)] Navier Stokes [ (6/29)] Navier Stokes 3 [ (6/19)] Reynolds [ (4.6), (45.8)] [ p.186] Navier Stokes I Euler Navier

More information

/02/18

/02/18 3 09/0/8 i III,,,, III,?,,,,,,,,,,,,,,,,,,,,?,?,,,,,,,,,,,,,,!!!,? 3,,,, ii,,,!,,,, OK! :!,,,, :!,,,,,, 3:!,, 4:!,,,, 5:!,,! 7:!,,,,, 8:!,! 9:!,,,,,,,,, ( ),, :, ( ), ( ), 6:!,,, :... : 3 ( )... iii,,

More information

構造と連続体の力学基礎

構造と連続体の力学基礎 II 37 Wabash Avenue Bridge, Illinois 州 Winnipeg にある歩道橋 Esplanade Riel 橋6 6 斜張橋である必要は多分無いと思われる すぐ横に道路用桁橋有り しかも塔基部のレストランは 8 年には営業していなかった 9 9. 9.. () 97 [3] [5] k 9. m w(t) f (t) = f (t) + mg k w(t) Newton

More information

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka )

II A A441 : October 02, 2014 Version : Kawahira, Tomoki TA (Kondo, Hirotaka ) II 214-1 : October 2, 214 Version : 1.1 Kawahira, Tomoki TA (Kondo, Hirotaka ) http://www.math.nagoya-u.ac.jp/~kawahira/courses/14w-biseki.html pdf 1 2 1 9 1 16 1 23 1 3 11 6 11 13 11 2 11 27 12 4 12 11

More information

The Physics of Atmospheres CAPTER :

The Physics of Atmospheres CAPTER : The Physics of Atmospheres CAPTER 4 1 4 2 41 : 2 42 14 43 17 44 25 45 27 46 3 47 31 48 32 49 34 41 35 411 36 maintex 23/11/28 The Physics of Atmospheres CAPTER 4 2 4 41 : 2 1 σ 2 (21) (22) k I = I exp(

More information

i I II I II II IC IIC I II ii 5 8 5 3 7 8 iii I 3........................... 5......................... 7........................... 4........................ 8.3......................... 33.4...................

More information

untitled

untitled (a) (b) (c) (d) Wunderlich 2.5.1 = = =90 2 1 (hkl) {hkl} [hkl] L tan 2θ = r L nλ = 2dsinθ dhkl ( ) = 1 2 2 2 h k l + + a b c c l=2 l=1 l=0 Polanyi nλ = I sinφ I: B A a 110 B c 110 b b 110 µ a 110

More information

8 i, III,,,, III,, :!,,,, :!,,,,, 4:!,,,,,,!,,,, OK! 5:!,,,,,,,,,, OK 6:!, 0, 3:!,,,,! 7:!,,,,,, ii,,,,,, ( ),, :, ( ), ( ), :... : 3 ( )...,, () : ( )..., :,,, ( ), (,,, ),, (ϵ δ ), ( ), (ˆ ˆ;),,,,,,!,,,,.,,

More information

Gmech08.dvi

Gmech08.dvi 145 13 13.1 13.1.1 0 m mg S 13.1 F 13.1 F /m S F F 13.1 F mg S F F mg 13.1: m d2 r 2 = F + F = 0 (13.1) 146 13 F = F (13.2) S S S S S P r S P r r = r 0 + r (13.3) r 0 S S m d2 r 2 = F (13.4) (13.3) d 2

More information

ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,.

ii 3.,. 4. F. ( ), ,,. 8.,. 1. (75% ) (25% ) =7 24, =7 25, =7 26 (. ). 1.,, ( ). 3.,...,.,.,.,.,. ( ) (1 2 )., ( ), 0., 1., 0,. (1 C205) 4 10 (2 C206) 4 11 (2 B202) 4 12 25(2013) http://www.math.is.tohoku.ac.jp/~obata,.,,,..,,. 1. 2. 3. 4. 5. 6. 7. 8. 1., 2007 ( ).,. 2. P. G., 1995. 3. J. C., 1988. 1... 2.,,. ii 3.,. 4. F. ( ),..

More information

n ξ n,i, i = 1,, n S n ξ n,i n 0 R 1,.. σ 1 σ i .10.14.15 0 1 0 1 1 3.14 3.18 3.19 3.14 3.14,. ii 1 1 1.1..................................... 1 1............................... 3 1.3.........................

More information

K E N Z OU

K E N Z OU K E N Z OU 11 1 1 1.1..................................... 1.1.1............................ 1.1..................................................................................... 4 1.........................................

More information

73

73 73 74 ( u w + bw) d = Ɣ t tw dɣ u = N u + N u + N 3 u 3 + N 4 u 4 + [K ] {u = {F 75 u δu L σ (L) σ dx σ + dσ x δu b δu + d(δu) ALW W = L b δu dv + Aσ (L)δu(L) δu = (= ) W = A L b δu dx + Aσ (L)δu(L) Aσ

More information

LLG-R8.Nisus.pdf

LLG-R8.Nisus.pdf d M d t = γ M H + α M d M d t M γ [ 1/ ( Oe sec) ] α γ γ = gµ B h g g µ B h / π γ g = γ = 1.76 10 [ 7 1/ ( Oe sec) ] α α = λ γ λ λ λ α γ α α H α = γ H ω ω H α α H K K H K / M 1 1 > 0 α 1 M > 0 γ α γ =

More information

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r

l µ l µ l 0 (1, x r, y r, z r ) 1 r (1, x r, y r, z r ) l µ g µν η µν 2ml µ l ν 1 2m r 2mx r 2 2my r 2 2mz r 2 2mx r 2 1 2mx2 2mxy 2mxz 2my r 2mz 2 r 2 1 (7a)(7b) λ i( w w ) + [ w + w ] 1 + w w l 2 0 Re(γ) α (7a)(7b) 2 γ 0, ( w) 2 1, w 1 γ (1) l µ, λ j γ l 2 0 Re(γ) α, λ w + w i( w w ) 1 + w w γ γ 1 w 1 r [x2 + y 2 + z 2 ] 1/2 ( w) 2 x2 + y 2 + z 2

More information

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B

BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF : B 2000 8 3.4 p q θ = 80 B E a H F b θ/2 O θ/2 D A B E BD = a, EA = b, BH = a, BF = b 3 EF B, EOA, BOD EF B EOA BF : AO = BE : AE, b : = BE : b, AF = BF = b BE = bb. () EF = b AF = b b. (2) EF B BOD EF :

More information

(1) 1.1

(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

More information

<4D F736F F D B B BB2D834A836F815B82D082C88C60202D B2E646F63>

<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

More information

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0

1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 1 No.1 5 C 1 I III F 1 F 2 F 1 F 2 2 Φ 2 (t) = Φ 1 (t) Φ 1 (t t). = Φ 1(t) t = ( 1.5e 0.5t 2.4e 4t 2e 10t ) τ < 0 t > τ Φ 2 (t) < 0 lim t Φ 2 (t) = 0 0 < t < τ I II 0 No.2 2 C x y x y > 0 x 0 x > b a dx

More information

II 2 II

II 2 II II 2 II 2005 yugami@cc.utsunomiya-u.ac.jp 2005 4 1 1 2 5 2.1.................................... 5 2.2................................. 6 2.3............................. 6 2.4.................................

More information

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B

さくらの個別指導 ( さくら教育研究所 ) A AB A B A B A AB AB AB B 1 1.1 1.1.1 1 1 1 1 a a a a C a a = = CD CD a a a a a a = a = = D 1.1 CD D= C = DC C D 1.1 (1) 1 3 4 5 8 7 () 6 (3) 1.1. 3 1.1. a = C = C C C a a + a + + C = a C 1. a a + (1) () (3) b a a a b CD D = D

More information

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

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

More information

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT

I ( ) 1 de Broglie 1 (de Broglie) p λ k h Planck ( Js) p = h λ = k (1) h 2π : Dirac k B Boltzmann ( J/K) T U = 3 2 k BT I (008 4 0 de Broglie (de Broglie p λ k h Planck ( 6.63 0 34 Js p = h λ = k ( h π : Dirac k B Boltzmann (.38 0 3 J/K T U = 3 k BT ( = λ m k B T h m = 0.067m 0 m 0 = 9. 0 3 kg GaAs( a T = 300 K 3 fg 07345

More information

TOP URL 1

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.............................

More information

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,.

,. Black-Scholes u t t, x c u 0 t, x x u t t, x c u t, x x u t t, x + σ x u t, x + rx ut, x rux, t 0 x x,,.,. Step 3, 7,,, Step 6., Step 4,. Step 5,,. 9 α ν β Ξ ξ Γ γ o δ Π π ε ρ ζ Σ σ η τ Θ θ Υ υ ι Φ φ κ χ Λ λ Ψ ψ µ Ω ω Def, Prop, Th, Lem, Note, Remark, Ex,, Proof, R, N, Q, C [a, b {x R : a x b} : a, b {x R : a < x < b} : [a, b {x R : a x < b} : a,

More information

i

i 009 I 1 8 5 i 0 1 0.1..................................... 1 0.................................................. 1 0.3................................. 0.4........................................... 3

More information

sec13.dvi

sec13.dvi 13 13.1 O r F R = m d 2 r dt 2 m r m = F = m r M M d2 R dt 2 = m d 2 r dt 2 = F = F (13.1) F O L = r p = m r ṙ dl dt = m ṙ ṙ + m r r = r (m r ) = r F N. (13.2) N N = R F 13.2 O ˆn ω L O r u u = ω r 1 1:

More information

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt

S I. dy fx x fx y fx + C 3 C vt dy fx 4 x, y dy yt gt + Ct + C dt v e kt xt v e kt + C k x v k + C C xt v k 3 r r + dr e kt S Sr πr dt d v } dt k e kt S I. x yx y y, y,. F x, y, y, y,, y n http://ayapin.film.s.dendai.ac.jp/~matuda n /TeX/lecture.html PDF PS yx.................................... 3.3.................... 9.4................5..............

More information

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d

ma22-9 u ( v w) = u v w sin θê = v w sin θ u cos φ = = 2.3 ( a b) ( c d) = ( a c)( b d) ( a d)( b c) ( a b) ( c d) = (a 2 b 3 a 3 b 2 )(c 2 d 3 c 3 d A 2. x F (t) =f sin ωt x(0) = ẋ(0) = 0 ω θ sin θ θ 3! θ3 v = f mω cos ωt x = f mω (t sin ωt) ω t 0 = f ( cos ωt) mω x ma2-2 t ω x f (t mω ω (ωt ) 6 (ωt)3 = f 6m ωt3 2.2 u ( v w) = v ( w u) = w ( u v) ma22-9

More information

NETES No.CG V

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

More information

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d

S I. dy fx x fx y fx + C 3 C dy fx 4 x, y dy v C xt y C v e kt k > xt yt gt [ v dt dt v e kt xt v e kt + C k x v + C C k xt v k 3 r r + dr e kt S dt d S I.. http://ayapin.film.s.dendai.ac.jp/~matuda /TeX/lecture.html PDF PS.................................... 3.3.................... 9.4................5.............. 3 5. Laplace................. 5....

More information

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3

> > <., vs. > x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D > 0 x (2) D = 0 x (3 13 2 13.0 2 ( ) ( ) 2 13.1 ( ) ax 2 + bx + c > 0 ( a, b, c ) ( ) 275 > > 2 2 13.3 x 2 x y = ax 2 + bx + c y = 0 2 ax 2 + bx + c = 0 y = 0 x ( x ) y = ax 2 + bx + c D = b 2 4ac (1) D >

More information

Untitled

Untitled 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

More information

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

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)

More information

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b)

x = a 1 f (a r, a + r) f(a) r a f f(a) 2 2. (a, b) 2 f (a, b) r f(a, b) r (a, b) f f(a, b) 2011 I 2 II III 17, 18, 19 7 7 1 2 2 2 1 2 1 1 1.1.............................. 2 1.2 : 1.................... 4 1.2.1 2............................... 5 1.3 : 2.................... 5 1.3.1 2.....................................

More information

I 1

I 1 I 1 1 1.1 1. 3 m = 3 1 7 µm. cm = 1 4 km 3. 1 m = 1 1 5 cm 4. 5 cm 3 = 5 1 15 km 3 5. 1 = 36 6. 1 = 8.64 1 4 7. 1 = 3.15 1 7 1 =3 1 7 1 3 π 1. 1. 1 m + 1 cm = 1.1 m. 1 hr + 64 sec = 1 4 sec 3. 3. 1 5 kg

More information

( )

( ) 7..-8..8.......................................................................... 4.................................... 3...................................... 3..3.................................. 4.3....................................

More information