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#求解线性刚度阵2,文献1的公式10d # [1] Finite-element method for large-amplitude two-dimensional panel flutter at hypersonic speeds.pdf w1, s1, w2, s2, ae, x = var('w1, s1, w2, s2, ae, x ') #形函数 N1=1-3*x*x+2*x*x*x N2=(x*(1-2*x+x*x))*ae N3=3*x*x-2*x*x*x N4=(x*x*x-x*x)*ae #形函数1阶导数 N1d1=6*x*x-6*x N2d1=(3*x*x-4*x+1)*ae N3d1=6*x-6*x*x N4d1=(3*x*x-2*x)*ae #形函数2阶导数 N1d2=12*x-6 N2d2=(6*x-4)*ae N3d2=6-12*x N4d2=(6*x-2)*ae f1=N1d1 f2=N2d1 f3=N3d1 f4=N4d1 kk11=f1*f1 kk12=f1*f2 kk13=f1*f3 kk14=f1*f4 kk21=f2*f1 kk22=f2*f2 kk23=f2*f3 kk24=f2*f4 kk31=f3*f1 kk32=f3*f2 kk33=f3*f3 kk34=f3*f4 kk41=f4*f1 kk42=f4*f2 kk43=f4*f3 kk44=f4*f4 k11=integral(kk11, x, 0, 1) k12=integral(kk12, x, 0, 1) k13=integral(kk13, x, 0, 1) k14=integral(kk14, x, 0, 1) k21=integral(kk21, x, 0, 1) k22=integral(kk22, x, 0, 1) k23=integral(kk23, x, 0, 1) k24=integral(kk24, x, 0, 1) k31=integral(kk31, x, 0, 1) k32=integral(kk32, x, 0, 1) k33=integral(kk33, x, 0, 1) k34=integral(kk34, x, 0, 1) k41=integral(kk41, x, 0, 1) k42=integral(kk42, x, 0, 1) k43=integral(kk43, x, 0, 1) k44=integral(kk44, x, 0, 1) print k11;print k12;print k13;print k14; print k21;print k22;print k23;print k24; print k31;print k32;print k33;print k34; print k41;print k42;print k43;print k44; show(k11);show(k12);show(k13);show(k14); show(k21);show(k22);show(k23);show(k24); show(k31);show(k32);show(k33);show(k34); show(k41);show(k42);show(k43);show(k44);
6/5 1/10*ae -6/5 1/10*ae 1/10*ae 2/15*ae^2 -1/10*ae -1/30*ae^2 -6/5 -1/10*ae 6/5 -1/10*ae 1/10*ae -1/30*ae^2 -1/10*ae 2/15*ae^2
65\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{6}{5}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10} \, \mbox{ae}
65\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{6}{5}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10} \, \mbox{ae}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10} \, \mbox{ae}
215ae2\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{2}{15} \, \mbox{ae}^{2}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{10} \, \mbox{ae}
130ae2\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{30} \, \mbox{ae}^{2}
65\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{6}{5}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{10} \, \mbox{ae}
65\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{6}{5}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{10} \, \mbox{ae}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{10} \, \mbox{ae}
130ae2\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{30} \, \mbox{ae}^{2}
110ae\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{1}{10} \, \mbox{ae}
215ae2\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{2}{15} \, \mbox{ae}^{2}