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2 確率的手法による構造安全性の解析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 刷発行当時のものです.

4 i 25 7

5 ii Benjamin &Cornell Ang & Tang Schuëller Ang Mathematica Matlab Mathcad Excel(Microsoft-Office)

6 iii Mathcad Excel Excel (

8 v [1] ( ) [2] [3] [4]

9 vi [5] [6] A B C D

10 (1) (2) (3) (4) (5)

11 2 1 (6) (1) (3) (4) (5) (6) (randomness) (uncertainty) (probabilistic phenomena) (random phenomena) (probabilistic uncertainty) (random uncertainty) (statistical uncertainty) (deterministic model) (deterministic analysis) (probabilistic model) (probabilistic analysis) (non-deterministic analysis) (model uncertainty)

12 1.2 3 (random variable) (probability distribution) (expectation) 1.2 S (sample space) (trial) (experiment) s(sample point) s S s S (1.1) (discrete sample space) (continuous sample space) ( 1.1(a)) S = { 1, 2, 3, 4, 5, 6 } (1.2) S = { n : n = } (1.3)

13 4 1 S S a b x (a) (b) S =[, ] = { x : x } (1.4) ( 1.1(b)) S =[a, b] = { x : a x b } (1.5) (subset) (event) E S E S (1.6) (elementary event) (compound event) (certain event) (impossible event) 1.3 {2 4 6} 2 { } 5 { } 4 {4}

14 ( 1.2) ( ) 36 S = { (m, n) :m =1 6, n=1 6 } (1.7) n S m ( 1.3) x y y y max S y min x min x max x 1.3

15 6 1 ( ) ( ) (x, y) S = { } (x, y) :x min 5 x 5 x max,y min 5 y 5 y max (1.8) (probability) (a-priori probability) (experimental probability) 1 6 1/6 1/6 ( ) (statistical regularity) N A N A r A r A = N A N (1.9) N r A A p (Bernoulli) A P [A] lim r A = p = P [A], (N ) (1.1)

17 A A P [A] p A P [A] q p + q =1 (trial) (independent) (repeated) (Bernoulli trial) ( 3.1) p E s (success) f (failure) A A k n n 3.1 n (a 1 a 2 a n ) a i p k q n k (k = ) n k p(k)

18 p(k) = n C k p k q n k = n C k p k (1 p) n k (3.1) nc k = n! k!(n k)! (3.1) (binominal distribution) (3.1) (p + q) n n A K p(k) K k 1 2 n (n +1) K = {, 1, 2,,n} (3.2) K E[K] Var [K] ((p + q) n 1 1 ) n E[K] = k n C k p k (1 p) n k = k= n k k=1 n 1 = (u +1) u= n 1 = np u= = np(p + q) n 1 n! k!(n k)! pk (1 p) n k n! (u +1)!(n 1 u)! pu+1 (1 p) n 1 u, (u = k 1) (n 1)! u!(n 1 u)! pu (1 p) n 1 u = np (3.3) ( (3.3) u = k 1 u = u 1 (p + q) n 2 ) n E[K 2 ]= k 2 nc k p k (1 p) n k = k= n k= k 2 n! k!(n k)! pk (1 p) n k = n(n 1)p 2 + np

19 56 3 = n 2 p 2 np 2 + np (3.4) Var[K] =E[K 2 ] (E[K]) 2 = np(1 p) =npq (3.5) K p K K p() = (1 p) 3 p(1) = 3p(1 p) 2 p(2) = 3p 2 (1 p) p(3) = p 3 p =1/3 p =1/2 3.2 p(k) np=1 p(k) np=3/2.4 n=3.4 n=3 p=1/3 p=1/ k k (a) (b) 3.2 n n = 1 p =1/2 3.3

20 p(k).1 np = k X (geometric distribution) X 1 {1, 2,, } 1 p x X ( 3.4) p(x).2 p =1/5 1-p 1-p 1-p p x 1 x (a) (b) x 3.4 P [X = x] =p(x) =(1 p) x 1 p (3.6) ( ) x P [X 5 x] =F (x) = p(j) =1 (1 p) x (3.7) j=1 X (first occurrence time)

21 58 3 (recurrence time) X X E[X] =X = x(1 p) x 1 p = 1 p x=1 (3.8) X ( ) (average return period) (3.8) E[X] =p{1+2(1 p)+3(1 p) 2 + } (1 p)e[x] =p{(1 p)+2(1 p) 2 +3(1 p) 3 + } pe[x] =p{1+(1 p)+(1 p) 2 + }= p 1 1 (1 p) =1 E[X 2 ]= x 2 (1 p) x 1 p = 2 p p 2 (3.9) x=1 (3.9) (3.8) E[X 2 ] (1 p)e[x 2 ] pe[x 2 ] pe[x 2 ] (1 p)pe[x 2 ] p 2 E[X 2 ] (1 p) Var [X] =E[X 2 ] (E[X]) 2 = 2 p p 2 1 p 2 = 1 p p 2 (3.1) X p p =(1 p) X = Å 1 1 X ã X (3.11) X 1 p Å p =1 p =1 1 1 X ã X (3.12) p X p e 1 =.368, p 1 e 1 =.632 (3.13)

22 ( lim 1 1 x ) x = e 1, (x ) X Å 1 1 ã 5 =.8 5 = Å 1 1 ã 1 =.9 1 = Å 1 1 ã 5 =.98 5 = p 1/1 p T R T R = 1 p = 1 (3.14) T R 1 1 ( ) p p = 1 T R = 1 1 (3.15) T R = 1 T =5 p f ((1 p) T T ) Å p f =1 (1 p) T =1 1 1 ã 5 = =.395 (3.16) 1 1 (= ) Å p f =1 (1 p) T =1 1 1 ã 1 = =.634 (3.17) 1 2 Å p f =1 (1 p) T =1 1 1 ã =.866 (3.18) 1 T R T r

23 ( ) ( ) R S R S f R (r) f S (s) ( 5.1) f S, f R f S (s) f R (r) s, r 5.1 R 5 S R >S R 5 S R >S

24 R 5 S p f (probability of failure) R >S p s (probability of survival) p f = P [R 5 S] (5.1) p s = P [R >S]=1 p f (5.2) (theory of structural reliability) (theory of structural safety) Freudenthal [1] (1947 ) Benjamin Cornell Ang Shinozuka Wen (load-resistance factor design, LRFD) ( Euro-code ) [8] RC [9,1] ( ) [11] [ 1] R S R S R S f R (r) f S (s) 5.2 S s s + ds P [s <S5 s + ds] =f S (s)ds (5.3) s R s P [R 5 s] = s f R (r)dr = F R (s) (5.4)

25 142 5 f S, f R f S (s) f S (s)ds=p[s<s s ds] f R (r) P[R s]=f R (s) s 5.2 ( ) s, r F R R S s s + ds R S P [R 5 s s <S5 s+ds] =f S (s)ds s f R (r)dr = f S (s)f R (s)ds (5.5) S p f ï s ò p f = f S (s) f R (r)dr ds = f S (s)f R (s)ds (5.6) R r r + dr f R (r)dr r S r 1 F S (r) ( 5.3 ) f S, f R f S (s) f R (r)dr =P[r<R r dr f R (r) P[S>r]=1-F S (r) r 5.3 ( ) s, r ï ò [ ] p f = f R (r) f S (s)ds dr = f R (r) 1 F S (r) dr (5.7) r (5.6) (5.7) p f 5.4

26 f S, f R 1. f S (s) F R (s) f S, f R 1. 1-F S (r) F S (r).5 p f.5 pf f R (r) s ( a ) (b ) 5.4 r 1 p f f S (s) = f S(s)F R (s) p f (5.8) f R (r) = f R(r)(1 F S (r)) p f (5.9) f S (s) f R (r) S R R S [ 2] R S f R (r)f S (s) f RS (r, s)drds = P [r <R5 r + dr s<s5 s + ds] = f R (r)f S (s)drds (5.1) f RS (r, s) r s 5.5 p f r 5 s f RS (r, s) r 5 s D F f RS p f = f RS (r, s)da = f R (r)f S (s)drds D f D f ï s ò = f S (s) f R (r)dr ds = f S (s)f R (s)ds (5.11) p f (5.6) (5.7) ds dr R 5 S

27 144 5 f RS (r, s) f R (r) r r D S (r>s, ) p S ( ) r=s f R (r) f S (s) p F ( ) D F (r s, ) f RS (r, s) s (a) D S (r>s, ) r=s dr (5 7) r ds s D F (r s, ) dr s (5.6) f S (s) ds (b) 5.5 [ 3] 3.5 Z = R S Z ( + ) f Z (z) = f R (z + s)f S (s)ds (5.12) (3.261) a =1 b = 1 p f Z p f = P [Z 5 ] = f Z (z)dz = f R (r)f S (s)dsdr = ï s ò f S (s) f R (r)dr ds = f R (z + s)f S (s)dsdz (5.13) z + s = r z = r = s z r (r = z + s) R S ( ) s s p f = f S (s)[ f R (r)dr]ds = f S (s)f R (s)ds (5.14) R S p f ( 1 ) R S µ R µ S σr 2 σ2 S ( 5.6)

28 f S, f R m S f S (s) m R f R (r) s, r 5.6 R = N(µ R,σR), 2 S = N(µ S,σS) 2 (5.15) R, S + R S R S Z (safety margin) Z = R S (5.16) Z (3.3 ) Z = N(µ Z,σ 2 Z)=N(µ R µ S,σ 2 R + σ 2 S) (5.17) p f = P [R S] =P [Z ] = = f Z (z)dz 1 2πσZ e (z µz)2 /2σ 2 Z dz (5.18) µ Z = µ R µ S, σ 2 Z = σ 2 R + σ 2 S (5.19) y = z µ Z σ Z p f = 1 2π µz/σ Z =1 Φ Ñ µ R µ S» σ 2 R + σ2 S Å e y2 /2 dy = Φ µ Z é σ Z ã Å µz =1 Φ σ Z ã (5.2) (5.21) Φ ( )

29 146 5 β β = µ Z = µ R µ» S σ Z σr 2 + σ2 S (5.22) p f p f =1 Φ(β) =Φ( β) (5.23) β (safety index) (reliability index) p f β 5.7 β Z v Z f(z) m Z = bs Z p f m Z z 5.7 β = 1 v Z, v Z = σ Z µ Z (5.24) (µ R /µ S ) (µ S /µ R ) v R v S p f =1 Φ (µ R /µ S ) 1» vr 2 (µ =1 Φ 1 (µ S /µ R )» R/µ S ) 2 + vs 2 vr 2 + v2 S (µ (5.25) S/µ R ) 2 (µ R /µ S ) (µ S /µ R ) v R v S β p f Φ (y) ( 5.8) β p f 1 1 β = Φ 1 (1 p f ) (5.26) p f β 5.1 β p f.5

30 F(y) f(y).5 p f =1-F(b) p f =F(-b) p f p f -b b y -b b y (a) (b) p f β β p f E E E E E E E-5 (E- n =1 n ) β 5.1 (µ S /µ R ) p f v R v S 5 % ( 5.9) p f 1. v R =v S =.3.5 v R =v S = m S /m R

31 148 5 µ S = p f ( 2 ) R S R, S ln R ln S ln R = N(λ R,ξR), 2 ln S = N(λ S,ξS) 2 (5.27) λ R ξr 2 ln R λ S ξs 2 ln S R S µ R µ S σr 2 σ2 S R S λ R λ S ξr 2 ξ2 S 3.1 ( 5.1) f R, f S f S (s) f R (r) m S m R r, s 5.1 Z = R S (5.28) ln Z =lnr ln S (5.29) ln R ln S ln Z ln Z = N(λ Z,ξZ) 2 (5.3) λ Z = λ R λ S, ξz 2 = ξr 2 + ξs 2 (5.31) Z

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35 25 Printed in Japan ISBN

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. 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