how to simplify symbolic expressions

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)

[KSSV]

Show us the full code. Read about the function simplify.

[rakesh kumar]

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)

Answer 1

[KSSV]

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.

[rakesh kumar]

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.