7 1 7 : 7.1 3.5 (b)
7 2 7.1 7.2 7.3
7 3 7.2 7.4
7 4 x M = Pw (7.3) ρ M (EI : ) M = EI ρ = w EId2 (7.4) dx 2 ( (7.3) (7.4) ) EI d2 w + Pw =0 (7.5) dx2 P/EI = α 2 (7.5) w = A sin αx + B cos αx 7.5 7.6 : x =0,l w =0 B =0, Asin αl + B cos αl =0 A sin αl =0 (7.7) w = A sin αx 7.7 A =0 A 0 sin αl =0 αl = nπ (n =1, 2,,n) P = (nπ)2 EI l 2 n =1 P = π2 EI l 2 = P E
7 5 7.3 σ E = P E EI = π2 A Al = π2 E 2 (l/r) = π2 E 2 λ 2 r = I/A : λ = l/r : λ c = λ = Y ( λ Y σ E λ Y = π E/σ Y, λ = π E/σ E ) 7.8 7.9
7 6 dx 7.4 ( ) x P P dw dx P dw dx + d2 w dx dx 2 = P d2 w dx 2 q = P d2 w dx 2 w q(x) EI d4 w dx 4 = q(x) EI d4 w dx 4 + P d2 w dx 2 =0 P E = π2 EI (βl) 2 = π2 EI (l ef ) 2 l ef : β : x =0 x = l w(0) = 0, M(0) = EI d2 w dx 2 =0 w(l) =0, M(l) = EId2 w dx 2 =0 w(0) = 0, θ(0) = dw dx =0 w(l) =0, θ(l) =dw dx =0
7 7 7.1 ( : ) l : (cm)
7 8 7.10 a
7 9 7.10 b
7 10 7.5 (1) EI d2 w +(w + e)p =0 dx2 α 2 = P/EI w = A sin αx + B cos αx e : x =0,l w =0 e(1 cos αl) B = e, A = sin αl (1 cos αl) sin αx + cos αx sin αl w = e 1 sin αl { } sin αl cos αx cos αl sin αx + sin αx = e 1 sin αl sin(αl αx) + sin αx = e 1 sin αl 1 w c = e cos(αl/2) 1 M c = P (w c + e) = Pe cos(αl/2) 7.11
7 11 7.12 7.13 (e=0.01k)
7 12 (2) ( ) EI d2 w dx + P (w + w 0)=0 2 w 0 w 0 = A 0 sin πx l α 2 = P/EI α 2 w = A sin αx + B cos αx α 2 (π/l) 2w 0 : x =0,l w =0 x =0 B =0 x = l A sin αl =0 A =0 (sin αl =0 ) w = A 0 α 2 sin(πx/l) (π/l) 2 α 2 w c = M c = P (w c + A 0 )=A 0 P A 0 α 2 (π/l) 2 α 2 = 1 1 P/P E A 0P P E P 7.14
7 13 7.15 7.16 A 0 σ cr
7 14 7.6 : 7.17
7 15 (tangent modulus theoty) E E t σ = P/A > σ cr = π2 E t (l/r) 2 E t : σ ( ) l = π E t (7.33) r σ cr cr 1 2 3 σ ε σ E t (7.33) 7.18
7 16 (reduce modulus theoty) 1 σ cr = π2 E r (l/r) 2 E r : 2 3 : : > 7.19
7 17 : M ext = Pw : ( ) ( ) ( ) σ = Eε = E y R = Eφy φ =1/R (R : ) ) ) M int = 1 2 (φe td 1 )d 1 b ( 2 3 d 1 + 1 2 (φed 2)d 2 b ( 2 3 d 2 ( ) 1 b = R 3 (E td 3 1 + Ed3 2 )=E ri R ( )( 1 b E r = (E t d I 3) 3 1 + Ed3 2 ) (7.39) 1 2 φe td 1 d 1 b = 1 2 φed 2 d 2 b d 2 1 = E E t d 2 2, d 1 d 2 = E E t (7.41) 7.20 ( ) (7.41) I = 1 12 b(d 1 + d 2 ) 3 (7.39) E r = 4EE t ( E + E t ) 2 (7.42)
7 18 Shanley 7.22 7.21 Shanley
7 19 7.23 ( )
7 20 σ Y 1/3 7.24 H 7.25
7 21 7.7 : = / : = P cr = π2 E t (βl/r) 2A g = σ cr A g E t : A g : βl/r : β : l : r : : 7.26
7 22 H (da ) dm =(θe t y)(da)(y) ( ) M = θe ty 2 da = θ A R = 1 θ = 1 θ R = M E I = M E I θ = E ty 2 da E = 1 A I E ty 2 da A : I e E = E I A:elastic y2 da = E I e I P cr = π2 E y 2 da (βl/r) 2 A g = π2 E(I e /I) I (βl/r) 2 A g A E ty 2 da 7.18 7.27
7 23 A H B k = 2x 0 = A e b A f E I e I = E t f(2x 0 ) 3 12 12 t f b 3 = Ek 3 E t = = dp/a = A ee dp/a e A E E t A = A e E =(A w +2kA f )E (7.54) A w : A f : A : (7.54) k k = E ta A w 2EA f 2A f σ cr = π2 Ek 3 (βl/r) 2 = π2 E (βl/r) 2 AE t 2A f E A w 2A f 3 E I e I = E 2A e(d/2) 2 2A f (d/2) = Ek 2 σ cr = π2 Ek (βl/r) 2 (7.61) E I e I = E 2kA f(d 2 /4) + t w d 3 /12 2A f (d 2 /4) + t w d 3 /12 = E 2kA f + A w /3 = E ta/e 2A w /3 2A f + A w /3 2A f + A w /3 ( (7.54) 2kA f = E t A/E A w ) σ cr = π2 E E (βl/r) 2 ta/e 2A w /3 2A f + A w /3 (7.62) 7.28 H
7 24 [ 1] H (σ cr βl) (a) P = A σda = σa P =(A A e )σ Y + A e σda σ cr = P/A (2/3)σ Y E t = E, E = EI e /I, I e = I σ cr = 2 3 σ Y = π2 E (βl/r) 2 βl r = π2 (200000) =65.4 2/3(690) σ cr = P/A > (2/3)σ Y : 7.30 [ -1] I e /I =(b/2) 3 /b 3 =1/8 σ cr = 2 3 σ Y = π2 E(I e /I) βl r =23.2 σ cr = P/A = σ Y σ cr = σ Y π 2 E 8(βl/r) 2 σ cr = σ Y = π2 E (βl/r) 2 8(βl/r) 2 βl r =18.9 βl r =53.5
7 25 [ 2] H 7.31 2
7 26 7.32 H σ cr = P/A (2/3)σ Y E t = E, σ cr = 2 3 σ Y = π2 E (βl/r) 2 σ cr = P/A > (2/3)σ Y : σ cr = π2 EI e /I (βl/r) 2 I e I = 2(1/12)(2z 0) 3 t 2(1/12)b 3 t = 8(z 0) 3 b 3 σ cr = 8π2 E(z 0 /b) 3 (βl/r) 2 [ ( )( 1 P cr =2 σbt 2 σ 2 2 3 σ Y σ 2 3 σ Y ( 1 2 z 0 b ) = {( P cr =2bt 1 z 0 b = A f σ Y σ cr = P cr A f = σ Y 2 3 σ Y b 2 ) 4 3 σ Y 1 4 3 ( z0 b σ = [( )( 1 2 z 0 b [ 1 z 0 b ) ] bt 1 z ) 0 4 b 3 σ Y 2 3 σ Y ) 3 1 4 3 ( z0 b ) 3 ] 4 3 σ Y ]( 1 2 z )} 0 b
7 27 7.33 σ cr = σ Y σ cr σ Y 7.34 SSRC 1 =1 λ2 c 4 σ Y 4π 2 E ( βl r (λ c 2), )2, λ c = βl σy r π 2 E 1 λ 2 c (λ c 2)
7 28 σ cr =1.0 (λ λ 0 ) σ Y σ cr = 1 1+α(λ σ Y 2λ 2 λ 0 )+λ 2 λ 0 : : ECCS Eurocode 3 {1+α(λ λ 0 )+λ 22 4λ 2 } (λ >λ 0 ) 7.35 ECCS
7 29 7.36
7 30 σ =1.0 (λ c 0.2) σ =1.109 0.545λ c (0.2 <λ c 1.0) σ =1.0/(0.773 + λ 2 c) (1.0 <λ c ) 7.37 ( )
7 31 ( 7.37), 1.7 σ ca = σ cag σ cal /σ cao σ ca : σ cag : σ cal : σ cao : 7.2 ( )