how to simplify symbolic expressions

3 次查看（过去 30 天）
rakesh kumar
rakesh kumar 2022-1-12
评论： rakesh kumar 2022-1-12
how can i simplify the given expession .i am getting this result when I am using symbolic math
(4.2e-41*((0.46*(1.3*(7.5e+123*(-1.0*EI^12)^(1/2) + 3.6e+124*EI^6)^(2/3) + 2.6e+42*EI^2*(7.5e+123*(-1.0*EI^12)^(1/2)
  2 个评论
KSSV
KSSV 2022-1-12
编辑：KSSV 2022-1-12
Show us the full code. Read about the function simplify.
rakesh kumar
rakesh kumar 2022-1-12
syms EI w
nel=2; % number of elements
nnel=2; % number of nodes per element
ndof=2; % number of dofs per node
nnode=(nnel-1)*nel+1; % total number of nodes in system
sdof=nnode*ndof; % total system dofs
% elastic modulus
%xi=1; % moment of inertia of cross-section
tleng=1; % length of a half of the beam
leng=tleng/nel; % element length of equal size
area=1; % cross-sectional area of the beam
rho=1; % mass density (arbitrary value for this problem because
% it is not used for the static problem)
ipt=1; % option for mass matrix (arbitrary value and not used here)
bcdof(1)=1; % first dof (deflection at left end) is constrained
bcval(1)=0; % whose described value is 0
bcdof(2)=2; % 12th dof (slope at the symmetric end) is constrained
bcval(2)=0; % whose described value is 0
displacement=zeros(sdof/2,1);
ff=zeros(sdof,1); % initialization of system force vector
kk=sym(zeros(sdof,sdof)); % initialization of system matrix
index=zeros(nnel*ndof,1); % initialization of index vector
ff(2*nel+1)=-1; % because a half of the load is applied due to symmetry
mm=zeros(sdof,sdof);
force=zeros(sdof, 1);
for iel=1:nel % loop for the total number of elements
edof = nnel*ndof;
start = (iel-1)*(nnel-1)*ndof;
%
for i=1:edof
index(i)=start+i;
end
c=EI/(leng^3);
k=c*[12 6*leng -12 6*leng; 6*leng 4*leng^2 -6*leng 2*leng^2; -12 -6*leng 12 -6*leng;6*leng 2*leng^2 -6*leng 4*leng^2];
edof = length (index);
for i=1:edof
ii=index(i) ;
for j=1:edof
jj=index(j) ;
kk(ii,jj)=kk(ii,jj)+k(i,j);
end
end
if ipt==1
%
c2=rho*area*leng/420 ;
m=c2*[156 22*leng 54 -13*leng;...
22*leng 4*leng^2 13*leng -3*leng^2;...
54 13*leng 156 -22*leng;...
-13*leng -3*leng^2 -22*leng 4*leng^2];
%
% lumped mass matrix
%
elseif ipt==2
m=zeros(4,4);
mass=rho*area*leng;
m=diag([mass/2 0 mass/2 0]);
%
% diagonal mass matrix
%
else
%
m=zeros(4,4);
mass=rho*area*leng;
m=mass*diag([l/2 leng^2/78 1/2 leng^2/78]);
%
end
edof = length (index)
for i=1:edof
ii=index(i) ;
for j=1:edof
jj=index(j) ;
mm(ii,jj)=mm(ii,jj)+m(i,j);
end
end
n=length(bcdof);
sdof=size(kk);
for i=1:n
c=bcdof(i);
for j=1:sdof
kk(c,j)=0;
kk(j,c)=0;
mm(c,j)=0;
mm(j,c)=0;
end
%
mm(c,c)=1;
end
end
% solve the matrix equation and print
% w^2 M * X = K*X
% w^2*X = Minv * K * X
%(Minv*K – w^2*I)X = 0
%This is in the same standard eigenvalue form as above with
%A = Minv*K
%Lambda = w^2
A=(inv(mm)*kk)
A1=sym(A)
[V D]=eig(A1);
a=simplify(V)
b=simplify(D)
a1=vpa(a,2)
b1=vpa(b,2)

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

KSSV
KSSV 2022-1-12
pretty(vpa(b1(3,3),8))
It looks that equation has lot of numbers and fractional powers of EI. FEw numbers are very small and few are very big. You need to check your calculations. Looks like this is an FEM problem. Why you want to use syms in the first place? Rethink on your problem.
  1 个评论
rakesh kumar
rakesh kumar 2022-1-12
i want to find the eigen vectors of two noded beam. i want to verify that eigen vectors are independent of EI .For one node I have got the result. Also I wnat to use it to check wether eigen vectors are dependent on EI if we use different values of EI in the two element.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Applications 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by