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初级会员
注册日期: 2008-04-07
年龄: 42
帖子: 2
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方程A(r)w=0
根据DET(A(r))=0,求得{ri} i=1:12 然后代入原方程,求得{wi=(w1i,w2i,w3i,w4i,w5i,w6i)}i=1:12 又u1(x)=c1*w1i*exp(ri*x) u2(x)=c2*w2i*exp(ri*x) p1(x)=c3*w3i*exp(ri*x) p2(x)=c4*w4i*exp(ri*x) v1(x)=c5*w5i*exp(ri*x) v2(x)=c6*w6i*exp(ri*x) 边界条件为: u1'(0)=0 u1'(L)=0 u2'(0)=0 u2'(L)=0 p1'(0)=0 p1'(L)=0 p2'(0)=0 p2'(L)=0 v1'(0)-p1(0)=0 v1'(L)-p1(L)=0 v2'(0)-p2(0)=0 v2'(L)-p2(L)=0 将w,r代入上述方程组,可得Mc=0 求DET(M) 具体程序如下: m=2*pi*58.339 L=3.5 A1=3*10^-2 J1=9*10^-6 E1=4.539*10^10 b1=2600 G1=1.945*10^10 A2=1.64*10^-3 J2=5.41*10^-6 E2=2.1*10^11 b2=7850 G2=8.08*10^10 es=0.1 ec=0 EA=8.59*10^8 K=2.858*10^8 d=0.22 K1=1.2 K2=2.49 k=K/d p=EA/d b3=b1+b2 r=sym('r') A=[E1*A1*r^2-k+b1*A1*m^2,k,0,k*es,k*ec*r/2,k*ec*r/2; k,E2*A2*r^2-k+b2*A2*m^2,0,-k*es,-k*ec*r/2,-k*ec*r/2; 0,0,-r,0,r^2+b1*K1*m^2/G1-K1*p/(G1*A1),K1*p/(G1*A1); 0,0,0,-r,K2*p/(G2*A2),r^2+b2*K2*m^2/G2-K2*p/(G2*A2); k*ec/2,-k*ec/2,E1*J1*r^2+b1*J1*m^2-G1*A1/K1,-k*ec*es/2,-k*ec^2*r/2+G1*A1*r/K1,-k*ec^2*r/6; k*es+k*ec/2,-k*es-k*ec/2,0,E2*J2*r^2+b2*J2*m^2-G2*A2/K2-k*es^2,-k*ec*es*r/2-k*ec^2*r/6,-k*ec^2*r/3+G2*A2*r/K2-k*ec*ec*r/2] B=det(A) C=sym2poly(B) E=roots(C) Matrix=[] for i=1:12 Temp=subs(A,r,E(i)) H=null(Temp) Matrix=[Matrix,H] end w1=Matrix(1, ![]() w2=Matrix(2, ![]() w3=Matrix(3, ![]() w4=Matrix(4, ![]() w5=Matrix(5, ![]() w6=Matrix(6, ![]() syms c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12 x c=[c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 c11 c12]' for i=1:12 y1(i)=c(i)*w1(i)*exp(E(i)*x) y2(i)=c(i)*w2(i)*exp(E(i)*x) y3(i)=c(i)*w3(i)*exp(E(i)*x) y4(i)=c(i)*w4(i)*exp(E(i)*x) y5(i)=c(i)*w5(i)*exp(E(i)*x) y6(i)=c(i)*w6(i)*exp(E(i)*x) u1=sum(y1) u2=sum(y2) q1=sum(y3) q2=sum(y4) v1=sum(y5) v2=sum(y6) end d1u1=vpa(diff(u1,x,1),5) d1u2=vpa(diff(u2,x,1),5) d1q1=vpa(diff(q1,x,1),5) d1q2=vpa(diff(q2,x,1),5) d1v1=vpa(diff(v1,x,1),5) d1v2=vpa(diff(v2,x,1),5) d1u10=vpa(subs(d1u1,x,0),5) d1u1L=vpa(subs(d1u1,x,L),5) d1u20=vpa(subs(d1u2,x,0),5) d1u2L=vpa(subs(d1u2,x,L),5) d1q10=vpa(subs(d1q1,x,0),5) d1q1L=vpa(subs(d1q1,x,L),5) d1q20=vpa(subs(d1q2,x,0),5) d1q2L=vpa(subs(d1q2,x,L),5) d1v10=vpa(subs(d1v1,x,0),5) d1v1L=vpa(subs(d1v1,x,L),5) d1v20=vpa(subs(d1v2,x,0),5) d1v2L=vpa(subs(d1v2,x,L),5) q10=vpa(subs(q1,x,0),5) q1L=vpa(subs(q1,x,L),5) q20=vpa(subs(q2,x,0),5) q2L=vpa(subs(q2,x,L),5) T10=d1v10-q10 T1L=d1v1L-q1L T20=d1v20-q20 T2L=d1v2L-q2L SV=[d1u10,d1u1L,d1u20,d1u2L,d1q10,d1q1L,d1q20,d1q2L,T10,T1L,T20,T2L] M1=[] for i=1:12 B1=eye(12) Tem1=subs(SV(1),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B1(i, ![]() M1=[M1, Tem1] end M2=[] for i=1:12 B2=eye(12) Tem2=subs(SV(2),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B2(i, ![]() M2=[M2, Tem2] end M3=[] for i=1:12 B3=eye(12) Tem3=subs(SV(3),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B3(i, ![]() M3=[M3, Tem3] end M4=[] for i=1:12 B4=eye(12) Tem4=subs(SV(4),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B4(i, ![]() M4=[M4, Tem4] end M5=[] for i=1:12 B5=eye(12) Tem5=subs(SV(5),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B1(i, ![]() M5=[M5, Tem5] end M6=[] for i=1:12 B6=eye(12) Tem6=subs(SV(6),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B6(i, ![]() M6=[M6, Tem6] end M7=[] for i=1:12 B7=eye(12) Tem7=subs(SV(7),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B7(i, ![]() M7=[M7, Tem7] end M8=[] for i=1:12 B8=eye(12) Tem8=subs(SV(8),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B8(i, ![]() M8=[M8, Tem8] end M9=[] for i=1:12 B9=eye(12) Tem9=subs(SV(9),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B9(i, ![]() M9=[M9, Tem9] end M10=[] for i=1:12 B10=eye(12) Tem10=subs(SV(10),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B10(i, ![]() M10=[M10, Tem10] end M11=[] for i=1:12 B11=eye(12) Tem11=subs(SV(11),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B11(i, ![]() M11=[M11, Tem11] end M12=[] for i=1:12 B12=eye(12) Tem12=subs(SV(12),{c1,c2,c3,c4,c5,c6,c7,c8,c9,c10,c11,c12},B12(i, ![]() M12=[M12, Tem12] end M=[M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12] P=det(M) 偶计算的结果和书上的差很多,不知道问题出在哪?请高手指点~拜托了~~ |
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