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 + dz + w + dw) P Qのように近接する２点間 の伸び 回転から固体の変形 を論じることができる ひずみ成分 点 P の変位 (u, v, w) 点 Q の変位 (u + du, v + dv, w + dw) u u u u + du = u + dx + dy + dz y z v v v v + dv = v + dx + dy + dz y z w w w w + dw = w + dx + dy + dz y z 37
x x x + x x x + u(x) x + x + u(x) + u(x) 38
x x x + x x x x + u(x) x + x + u(x) + u(x) ε(x) = x x x = ( x + u(x) ) x x = u(x) x ε = du dx 39
D D B A C B A C B x + dx y z B x + dx + u + ( u/)dx y + v + ( v/)dx z + w + ( w/)dx ds x = ( 1 + u ) 2 + ( v ) 2 + ( ) w 2 dx ε x = ds x dx dx = ( 1 + u ) 2 + ( v ) 2 + ( w ) 2 1 u 40
D D B A C B A C ε x = u ε y = v y ε z = w z 41
dy dx θ 1 tan θ 1 = θ 2 tan θ 2 = ( v + v ) dx v dx ( u + u ) y dy u dy = v = u y γ xy = θ 1 + θ 2 = u y + v 42
z ω z = 1 2 ( v u ) v x y ω x = 1 ( w 2 y v ) z ω y = 1 ( u 2 z w ) ω z = 1 ( v 2 u ) y 43
du dv dw du = u u u dx + dy + y z dz = u { ( 1 u dx + 2 y v ) + 1 ( u 2 y + v )} dy { ( 1 u + 2 z w ) + 1 ( u 2 z + w )} dz ( = ε x dx + ω z + 1 ) ( 2 γ xy dy + ω y + 1 ) 2 γ zx dz dv dw du dv dw = ε x γ xy /2 γ zx /2 γ xy /2 ε y γ yz /2 γ zx /2 γ yz /2 ε z dx dy dz + 0 ω z ω y ω z 0 ω x ω y ω x 0 dx dy dz 44
ε ij = 1 2 (u i, j + u j,i ) = = ε x γ xy /2 γ zx /2 γ xy /2 ε y γ yz /2 γ zx /2 γ yz /2 ε z 1 2 1 2 u ( v + u y ( w + u z ) ) ε xx ε xy ε xz ε yx ε yy ε yz ε zx ε zy ε zz ε x γ xy γ zx γ xy ε y γ yz γ zx γ yz ε z 1 2 1 2 ( u y + v v y ( w y + v z = ) ) 1 2 1 2 ( u z + w ( v z + w y w z ε xx ε yx ε zx ε xy ε yy ε zy ε xz ε yz ε zz ) ) 45
D D V = dxdydz B A C B A C V (1 + ε x ) dx (1 + ε y ) dy (1 + ε z ) dz = (1 + ε x + ε y + ε z + ε x ε y + ε y ε z + ε z ε x + ε x ε y ε z ) dxdydz e = V V V = ε x + ε y + ε z 46
ε = 1 3 (ε x + ε y + ε z ) = 1 3 e ε x γ xy γ zx γ xy ε y γ yz γ zx γ yz ε z = ε x 1 3 e γ xy γ zx γ xy ε y 1 3 e γ yz γ zx γ yz ε z 1 3 e + 1 3 e 0 0 0 e 0 0 0 e 47
48
σ x σ x x σ y = σ z = τ xy = τ yz = τ zx = 0 σ x = ε x y z σ x > 0 ε y 0, ε z 0 ε y, ε z ε x ε x ε y = νε x, ε z = νε x ν 0 50
(B) x y z (A) (A) x? σ x σ x x (A) σ x = ε x σ x σ x x (B) σ x = ε x 51
γ C τ δ C τ D D τ l γ γ tan γ = CC AC = δ l τ A τ A τ = Gγ G G ν 2G = 1 + ν 52
ε x = σ x ν σ y ν σ z = 1 + ν σ x ν (σ x + σ y + σ z ) ε y = σ y ν σ x ν σ z = 1 + ν σ y ν (σ x + σ y + σ z ) ε z = σ z ν σ x ν σ y = 1 + ν σ z ν (σ x + σ y + σ z ) γ xy = τ xy G γ yz = τ yz G γ zx = τ zx G = 2(1 + ν) = 2(1 + ν) = 2(1 + ν) τ xy τ yz τ zx 53
ε x 1 ν ν 0 0 0 ε y ε z = 1 ν 1 ν 0 0 0 ν ν 1 0 0 0 γ xy 0 0 0 2(1 + ν) 0 0 γ yz 0 0 0 0 2(1 + ν) 0 γ zx 0 0 0 0 0 2(1 + ν) σ x σ y σ z τ xy τ yz τ zx {ε} =[C] {σ } {ε} {σ } [C] 54
σ x σ y σ z τ xy τ yz τ zx = (1+ν)(1 2ν) 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 ν ν 1 ν 0 0 0 0 0 0 1 2ν 2 0 0 0 0 0 0 0 0 0 0 0 1 2ν 2 0 1 2ν 2 ε x ε y ε z γ xy γ yz γ zx {σ }=[D] {ε} [D] 55
σ x = (1 + ν)(1 2ν) σ y = (1 + ν)(1 2ν) σ z = (1 + ν)(1 2ν) τ xy = 2(1 + ν) γ xy τ yz = 2(1 + ν) γ yz τ yz = 2(1 + ν) γ yz { (1 ν)εx + νε y + νε z } { νεx + (1 ν)ε y + νε z } { νεx + νε y + (1 ν)ε z } 56
σ x σ y σ z τ xy τ yz τ zx = = (1+ν)(1 2ν) (1+ν)(1 2ν) 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 ν ν 1 ν 0 0 0 0 0 0 1 2ν 2 0 0 0 0 0 0 0 0 0 0 0 1 2ν 2 0 1 2ν 2 1 ν ν ν 0 0 0 ν 1 ν ν 0 0 0 ν ν 1 ν 0 0 0 1 2ν 0 0 0 2 0 0 1 2ν 0 0 0 0 2 0 0 0 0 0 0 1 2ν 2 ε x αt ε y αt ε z αt γ xy γ yz ε x ε y ε z γ xy γ yz γ zx γ zx α 1 2ν T T T 0 0 0 57
ε x = 1 + ν σ x ν (σ x + σ y + σ z ) ε y = 1 + ν σ y ν (σ x + σ y + σ z ) ε z = 1 + ν σ z ν (σ x + σ y + σ z ) ε xy = 1 2 γ xy = 1 + ν σ xy ε yz = 1 2 γ yz = 1 + ν ε zx = 1 2 γ zx = 1 + ν σ yz σ zx ε ij = 1 + ν σ ij ν δ ijσ kk 58
ε ij = 1 + ν ε mm = 1 + ν = 1 + ν σ ij ν δ ijσ kk i = j = m σ kk = 1 2ν ε kk σ mm ν δ mmσ kk = 1 + ν σ kk 3ν σ kk = 1 2ν ε ij = 1 + ν σ ij ν 1 2ν δ ijε kk σ mm 3 ν σ kk σ kk = ε kk λ = σ ij = ν (1 2ν)(1 + ν) δ ijε kk + 1 + ν ε ij ν (1 2ν)(1+ν), µ = G = 2(1+ν) (Lamé) 59
σ x x γ xy = γ yz = γ zx = 0 τ xy = τ yz = τ zx = 0 y z 0 ε y = σ y ν σ x ν σ z = 0, σ y = σ z = ν 1 ν σ x ε x = σ x ν σ y ν σ z σ x = (1 ν) (1 + ν)(1 2ν) ε x ε z = σ z ν σ x ν σ y = 0 ε x = (1 + ν)(1 2ν) (1 ν) σ x (1 ν) 0 ν < 0.5 (1 + ν)(1 2ν) 60
{ } σ x = (1 ν)εx + νε y + νε z (1 + ν)(1 2ν) { } σ y = νεx + (1 ν)ε y + νε z (1 + ν)(1 2ν) { } σ z = νεx + νε y + (1 ν)ε z (1 + ν)(1 2ν) τ xy = 2(1 + ν) γ xy, τ yz = 2(1 + ν) γ yz, τ yz = 2(1 + ν) γ yz ε z = γ yz = γ zx = 0 σ x = (1 + ν)(1 2ν) σ y = (1 + ν)(1 2ν) τ xy = 2(1 + ν) γ xy { (1 ν)εx + νε y } { νεx + (1 ν)ε y } σ z 61
σ z = τ yz = τ zx = 0 ε z = ν 1 ν ( εx + ε y ) σ x = σ y = τ xy = (1 + ν)(1 ν) (ε x + νε y ) (1 + ν)(1 ν) (νε x + ε y ) 2(1 + ν) γ xy 62
: σ x = (1 + ν)(1 ν) (ε x + νε y ) σ y = (1 + ν)(1 ν) (νε x + ε y ) τ xy = 2(1 + ν) γ xy ν = ν 1 + ν, Ē = (1 ν 2 ) σ x = σ y = τ xy = Ē (1 + ν)(1 2 ν) Ē (1 + ν)(1 2 ν) Ē 2(1 + ν) γ xy { (1 ν)εx + νε y } { νεx + (1 ν)ε y } 63
σ = 1 3 (σ x + σ y + σ z ) e = ε x + ε y + ε z K = σ x + σ y + σ z = σ = 3(1 2ν) 1 2ν (ε x + ε y + ε z ) 3(1 2ν) e 64
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