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みいかとどろき
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2 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. 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.

3 FAX

4 i

5 ii VCAD

6 iii

7 iv SI [m] [ ] = π/180 [rad] [N] [MPa] (= 10 6 [Pa]) =[N/mm 2 ] [N m] m 10 3 k 10 3 M 10 6 G 10 9

8 A A B OB AB OB 0.2% % % 1.1

9 2 1 1% mm (a) l 0 = 50 [mm] (b) l 1 = 56 [mm] (c) l 2 = 66 [mm] (d) l 3 = 70 [mm] (d) s e 1.3 P A 0 (1.1)

10 l 0 dl/l 0 l 0 l (1.2) s = P A 0 (1.1) e = l l 0 [ ] l dl l = = l l 0 (1.2) l 0 l 0 l 0 l (b) P =28.4[kN] = [N] A 0 A 0 =(π/4) (12.5 [mm]) 2 = 122.7[mm 2 ] = [m 2 ] (1.1) (1.2) s = P A 0 = [N] [m 2 ] = [Pa] = 264 [MPa] e = l l [mm] 50.0 [mm] = =0.12 l [mm] MPa 1.2 kn mm 2 Pa Pa = N/m 2 M 10 6 k 10 3 m 10 3 MPa (c)

11 O A (1.3) s = Ee (1.3) E A A Y 0.2% 0.2% 1.4 E E B B B C C 1.4

12 A A 0 (1.4) A σ = P A (1.4) (1.5) l dl/l l 0 l ε = l l 0 dl l = [ln l] l l 0 =lnl ln l 0 =ln l l 0 (1.5) Al = A 0 l 0 A (1.6) (1.6) (1.4) (1.1) (1.2) (1.7) A = A 0l 0 l (1.6) σ = P l = P l ( = s l0 + l l 0 = s 1+ l l ) 0 = s(1 + e) A 0 l 0 A 0 l 0 l 0 l 0 (1.7) (1.5) (1.2) (1.8) ε =ln l ( ) ( l0 + l l 0 =ln =ln 1+ l l ) 0 =ln(1+e) (1.8) l 0 l 0 l (1.7) B

13 6 1 (1.8) 1.2 e 1 =0.01, e 2 =0.001 e 3 = 0.01 ε 1, ε 2 ε 3 (1.8) ε 1 =ln(1+e 1 )=ln(1+0.01) = ε 2 =ln(1+e 2 )=ln( ) = ε 3 =ln(1+e 3 )=ln(1 0.01) = e 1 =0.10, e 2 =1.00, e 3 = 0.10 e 4 = 1.00 ε 1, ε 2, ε 3 ε (a) l 0 l 1 l 2 l 0 l 1 ε 0 1 (1.5) (1.9) l 1 l 2 ε 1 2 (1.10) (1.9) (1.10) (1.11) ε 0 1 ε 1 2 l 0 l 2 ε

14 ε 0 1 =ln ε 1 2 =ln ( l1 l 0 ( l2 l 1 ) ) ε ε 1 2 =ln =ln ( l1 l 0 ( l2 l (1.9) (1.10) ) ( ) l2 +ln =lnl 1 ln l 0 +lnl 2 ln l 1 l 1 ) = ε 0 2 (1.11) (1.2) l 0 l 1 e 0 1 l 1 l 2 e 1 2 (1.12) l 0 l 2 e 0 2 e e 1 2 = l 1 l 0 l 0 + l 2 l 1 l 1 l 2 l 0 l 0 = e 0 2 (1.12) 1.3 l 0 = 50 [mm] l 1 = 60 [mm] l 2 = 80 [mm] e e 1 2, e 0 2 ε ε 1 2, ε 0 2 e e 1 2 e 0 2 ε ε 1 2 = ε 0 2 (1.2) e 0 1 = l 1 l 0 l 0 e 1 2 = l 2 l 1 l 1 = = 60 [mm] 50 [mm] 50 [mm] 80 [mm] 60 [mm] 60 [mm] =0.20 =0.33 e 0 2 = l 2 l 0 80 [mm] 50 [mm] = =0.60 l 0 50 [mm] e e 1 2 = = 0.53 e 0 2 =0.60 (1.5) ( ) ( ) l1 60 [mm] ε 0 1 =ln =ln =0.18 l 0 50 [mm] ( ) ( ) l2 80 [mm] ε 1 2 =ln =ln =0.29 l 1 60 [mm] ( ) ( ) l2 80 [mm] ε 0 2 =ln =ln =0.47 l 0 50 [mm] ε ε 1 2 = = 0.47 = ε 0 2 =0.47

15 (a) 1.2 σ Y 3.1(b) τ τ k σ = Y, τ = k (3.1) 3.1

16 F C F (σ x,σ y,σ z,τ xy, τ yz,τ zx )=C (3.2) C (3.2) F (σ 1,σ 2,σ 3 )=C (3.3) (3.3) 2.5 J 1, J 2 J 3 (3.3) F (J 1,J 2,J 3 )=C (3.4) 1.1 J 2 J 3 F (J 2,J 3 )=C (3.5) 2.5 J 1 =0 (3.5) τ max

17 46 3 σ 1 σ 2 2, σ 2 σ 3 2, σ 3 σ 1 2 (3.6) σ 1 σ 2 σ 3 τ max τ max = σ 1 σ 3 2 (3.7) τ max = σ 1 σ 3 2 = C (3.8) C 3.1(a) (b) C σ 1 σ σ 2 σ 3 0 (3.1) σ = Y (3.8) C C = σ 1 σ 3 2 = σ 0 2 = σ 2 = Y 2, C = Y 2 (3.9) 3.1(b) σ 1 = τ, σ 2 =0, σ 3 = τ (3.10) (3.1) τ = k (3.8) C C = σ 1 σ 3 2 = τ ( τ) 2 = τ = k, C = k (3.11) 3.2

18 (3.9) (3.11) Y =2k (3.12) (3.5) J 2 J 3 (3.8) (3.8) σ F (J 2,J 3)=J 2 = 1 {(σ 1 σ 2 ) 2 +(σ 2 σ 3 ) 2 +(σ 3 σ 1 ) 2} 6 = 1 { σ 2 x + σ 2 y + σ 2 z +2 ( τxy 2 + τyz 2 + τ 2 ) } zx 2 = 1 { (σ x σ y ) 2 +(σ y σ z ) 2 +(σ z σ x ) ( τ 2 xy + τ 2 yz + τ 2 zx) } = C (3.13) C (3.13) (3.5) (3.13) J 2 U s U s = 1 { (σ 1 σ 2 ) 2 +(σ 2 σ 3 ) 2 +(σ 3 σ 1 ) 2} = J 2 12G 2G (3.14) (3.9) (3.11) (3.14) C (3.9) (3.13) C

19 48 3 C = J 2 = 1 6 { (σ 0) 2 +(0 0) 2 +(0 σ) 2} = 1 6 2σ2 = 1 3 Y 2 C = 1 3 Y 2 (3.15) (3.10) τ = k (3.13) C C = J 2 = 1 { (τ 0) 2 + {0 ( τ)} 2 +( τ τ) 2} = τ 2 =k 2 C = k 2 (3.16) (3.15) (3.16) C = 1 3 Y 2 = k 2, Y = 3k (3.17) (3.13) Y = 250 [MPa] k [MPa] (3.12) k = [MPa] Y = = 125 [MPa] 2 2 (3.17) k = Y 3 = 250 [MPa] 3 = 144 [MPa] 3.1 k = 100 [MPa] Y [MPa] (a) x y z

20 b w h z z ε z z xy xy b 5.1

21 x = x σ x x = x + dx σ x + dσ x dσ x y σ x x p p μp μ x =0 y x b x (σ x + dσ x )h b x = x + dx σ x h b x = x 2μp dx b (σ x + dσ x ) h b σ x h b 2μp dx b =0 dσ x h b =2μp dx b, dσ x dx = 2μp h (5.1) (3.61)

22 σ = { } (σ x σ y ) 2 +6τxy (5.2) (5.2) τ xy =0 μp τ xy 0 { } 1 3 σ = 2 2 (σ x σ y ) 2 3 = 2 σ x σ y (5.3) 3 Y = 2 σ x σ y k = 1 2 σ x σ y (5.4) σ x >σ y σ x <σ y y x σ x >σ y 2 Y = σ x σ y 2k = σ x σ y (5.5) 3 σ y p p = σ y 2 3 Y = σ x + p 2k = σ x + p (5.6) (5.1) (5.6) x dσ x dx + dp =0 (5.7) dx (5.1) (5.7) p 2μp h + dp dp =0 dx p = 2μ dx (5.8) h

23 104 5 (5.8) p C, C exp C C dp p = 2μ dx ln p = 2μ ( h h x + C p =exp 2μ ) h x+c p =expc exp ( 2μh ) x = C exp ( 2μh ) x (5.9) C (x =+w/2) σ x =0 (5.6) p =2k ( p =2k x =+ w ) (5.10) 2 (5.10) (5.9) C ( 2k = C exp 2μ h w ) ( μw ), C =2kexp (5.11) 2 h (5.11) (5.9) p ( μw ) p =2kexp exp ( 2μh ) h x = 2 ( μw ) Y exp exp ( 2μh ) 3 h x (5.12) (5.12) (3.17) Y k Y = 3k [mm] 20 [mm] 400 [MPa] 0.1 (5.12) (x =0) (5.12) p = 2 ( μw Y exp 3 h = [MPa] exp = 762 [MPa] ) exp ( 2μ ) h x ( ) ( [mm] exp 20 [mm] [mm] 0 [mm] ) [mm] 10 [mm] 200 [MPa] [mm]

24 p 2k p/2k (x =0) 5.3 (μ =0.1,w/h=5) 5.4 μ =0 p =2k μ (x = ±w/2) p p (x =0) 5.4 (w/h =1) p P p x =0 (5.12) p 0 x +w/2 0 x +w/2 P

25 106 5 P =2b +w/2 0 ( μw =4kb exp h ( μw =4kb exp h = 4kbh 2μ = 2kbh μ = 2Ybh 3μ {exp pdx =2b +w/2 0 ( μw 2k exp h ) exp ( 2μh ) x 2μ h 0 ) ( h ) { ( exp μw 2μ h ){ exp ( μw exp h { ( μw 1 exp h ( μw h )} ) } 1 ( μw h = 2kbh μ +w/2 ) } 1 ) exp ( 2μh ) x dx ) } 1 { ( μw exp h ) } 1 (5.13) (5.13) P bw p p= P bw = 2kh { ( μw ) } exp 1 = 2Yh ( μw ) } {exp 1 (5.14) μw h 3μw h μ =0 p (5.14) (5.12) x μ =0 p =2k = ( 2/ 3 ) Y p p =2k = 2 3 Y (μ = 0) (5.15) [mm] 20 [mm] 600 [mm] 400 [MPa] 0.1 P p (5.13) P = 2Ybh ( μw {exp 3μ h = ) } [MPa] 600 [mm] 20 [mm] { ( ) } [mm] exp 1 20 [mm]

26 168 n 11 n , ,

27 , , , 40, 68 40, ,

28 The Japan Society for Technology of Plasticity Y S K 4F FAX C FAX / / / Printed in Japan ISBN

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63>

<4D F736F F D B B BB2D834A836F815B82D082C88C602E646F63> 入門モーター工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/074351 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. 10 kw 21 20 50 2 20 IGBT IGBT IGBT 21 (1) 1 2 (2) (3) ii 20 2013 2 iii iv...