:-0.+2iT 02b r-21 2 Oy- Okay 0.0 i|φ ax a 2 z 可得0,0x+21cx=20(z)+x(z) 或 yx+2ry=2[0"(2)+v(z) 26
26 22 4 i 2 2i 22 22 2i 2 zx y x y x y y x xy = − = − − − + = 2i 2[z ''( z ) ''( z)] 可得 y − x + xy = + 2i 2[z ''( z ) '( z)] 或 y − x + xy = +
Only given (2) andy(a), we can divide the right side of above equation into imaginary part and real part, from Imaginary part we get txy, from real part we get oy-ox x+,=2()+(x)=4Re(x) ando,-0x+2rx=2[2"()+v() is complex-variable representation of stress component. Of course by building equations, o τ can be represented by z) and y(z) respectively, but that will make equations become lengthiness and it's not convenient to use 27
27 Only given (z) andψ (z), we can divide the right side of above equation into imaginary part and real part, from imaginary part we get τxy, from real part we get σy-σx. 2[ '(z) '(z)] 4Re '(z) x + y = + = 2i 2[z ''(z) '(z)] and y − x + xy = + is complex-variable representation of stress component. Of course by building equations, σx、σy 、τxy can be represented by (z) and ψ (z) respectively, but that will make equations become lengthiness and it’s not convenient to use
只要已知(z)及v(z),就可以把上述公式右 边的虚部和实部分开,由虚部得出τx,由实部得 出0y0x x+,=2()+(x)=4Re(x) 和 0ya+2ix=210()+v(= 就是应力分量的复变函数表示。当然也可以建立 公式,把0x0,、τn三者分开用o(z)和v(z) 来表示,但那些公式将比较冗长,用起来很不方 便 28
28 只要已知(z)及ψ (z)，就可以把上述公式右 边的虚部和实部分开，由虚部得出τ xy，由实部得 出σ y-σ x。 2[ '(z) '(z)] 4Re '(z) x + y = + = 2i 2[z ''(z) '(z)] 和 y − x + xy = + 就是应力分量的复变函数表示。当然也可以建立 公式，把σ x、σ y 、τ xy三者分开用(z)和ψ (z) 来表示，但那些公式将比较冗长，用起来很不方 便
dextrane Methods for Pane bestial Two Complex-variable representation of displacement component Assuming plane stress problems, according to geometrical equation and physical equation E HO=(o+on)-(1+) E=o-Ho=o+o )-(1+u) x e av a +--)= 2(1+) OX yields E=29()+(2)-(1+2 =2[(z)+(=)]-(1+) ax 29
29 Two Complex-variable representation of displacement component Assuming plane stress problems, according to geometrical equation and physical equation yields x y x y y x u E = − = ( + ) − (1+ ) y x x y x y v E = − = ( + ) − (1+ ) xy y u x E v = + + ( ) 2(1 ) 2 2 2 2 ' ' 2 [ ( ) ( )] (1 ) 2[ ( ) ( )] (1 ) x z z x x z z x u E + − + = = + − +
二位移分量的复变函数表示 假定为平面应力问题。由几何方程及物理方程 E=0x=(0x+0,)-(1+1) E=o-Ho=o+o )-(1+u) x e av a +--)= 2(1+) OX 可得E=20()+9()-(1+)0 =2[(z)+(=)]-(1+) ax
30 二 位移分量的复变函数表示 假定为平面应力问题。由几何方程及物理方程 x y x y y x u E = − = ( + ) − (1+ ) y x x y x y v E = − = ( + ) − (1+ ) xy y u x E v = + + ( ) 2(1 ) 2 2 2 2 ' ' 2 [ ( ) ( )] (1 ) 2[ ( ) ( )] (1 ) x z z x x z z x u E + − + = = + − + 可得