Stiffness method

MSc Structural Analysis

Andrew Phillips MEng PhD

Lecturer in Structural Engineering and Structural Biomechanics

Imperial College London

Structural Biomechanics, Structures SectionDepartment of Civil and Environmental Engineering

www.imperial.ac.uk/structuralbiomechanics

Implementing the Matrix Stiffness Method


Previously it was stated that the matrix stiffness methodcould easily be implemented in a program such as Matlab.

Here we will look at the implementation for a beam and aframe example.

The approach can be used for any structure made up ofdistinct elements connected together at nodes or joints.


Beam Example


Node and Element Numbering


Matlab Code: Setting Up

% Setting up a member stiffness matrix (bending)k = [4 2; 2 4];

% Setting up a matrix of distributed loadsw=[40 20 20 60];

% Setting up a matrix of lengthsL=[4 6 6 4];

% Setting up a matrix of EI valuesEI = [1 1.5 1.5 2];% Assuming use of UB402x140x42EI = EI. * (2.05 * 10ˆ8 * 0.0001569); % kN/mˆ2 * mˆ4


Matlab Code: Moment Stiffness Matrix

% Setting up the combined structure stiffness matrix% Based on the degree of kinematic indeterminacy (5)ndof=5;K=zeros(ndof,ndof);for n=1:ndof-1

K(n:n+1,n:n+1)=K(n:n+1,n:n+1)+k. * EI(n)./L(n);end


Matlab Code: Fixed End Moments

% Setting up fixed end moments for each span (FE-M)FEMspan = zeros(ndof-1,2);for n=1:ndof-1

FEMspan(n,1) = +w(n). * L(n)ˆ2./12;FEMspan(n,2) = -w(n). * L(n)ˆ2./12;

end

% Setting up fixed end moments at each support (FE-M)FEMsupport=zeros(ndof,1);FEMsupport(1,1)=FEMspan(1,1);for n=2:ndof-1

FEMsupport(n,1)=FEMspan(n-1,2)+FEMspan(n,1);endFEMsupport(ndof,1)=FEMspan(ndof-1,2);


Matlab Code: Finding θ Values

∆˜= K

˜

−1(−R˜

FE )

%%%% Calculation to find values of theta %%%%theta=(Kˆ-1) * (-FEMsupport);fprintf( 'Theta Values (rad): \n' );fprintf( '%1.4e, ' ,theta); fprintf( '\n\n' );

Theta Values (rad):-1.8199e-003, 3.2354e-004, 1.1122e-004, -7.6841e-004, 1 .6278e-003,


Matlab Code: Vertical Reactions Stiffness Matrix

% Setting up a vertical stiffness matrix% These values can be taken from the 3D member stiffness matri xkvert = [6 6; -6 -6];Kvert = zeros(ndof,ndof);for n=1:ndof-1

Kvert(n:n+1,n:n+1)=Kvert(n:n+1,n:n+1)+kvert. * EI(n)./L(n).ˆ2;end


Matlab Code: Fixed End Vertical Reactions

% Setting up fixed end vertical reactions for each span (FE-R 2)FER2span = zeros(ndof-1,2);for n=1:ndof-1

FER2span(n,1) = w(n). * L(n)./2;FER2span(n,2) = w(n). * L(n)./2;

end

% Setting up fixed end vertical reactions for each support (F E-R2)FER2support=zeros(ndof,1);FER2support(1,1)=FER2span(1,1);for n=2:ndof-1

FER2support(n,1)=FER2span(n-1,2)+FER2span(n,1);endFER2support(ndof,1)=FER2span(ndof-1,2);


Matlab Code: Vertical Reactions

%%%% Calculation to find values for the vertical support rea ctions (R2)R2 = FER2support+Kvert * theta;fprintf( 'Vertical Support Reactions (kN): \n' );fprintf( '%1.4e, ' ,R2); fprintf( '\n\n' );

Vertical Support Reactions (kN):6.1951e+001, 1.6154e+002, 1.1122e+002, 2.0602e+002, 9.9 268e+001,


Conclusions

The script can be generalised for any continuous beam.

Results are found to match those obtained from Oasys GSAexcluding shear deformations.


Frame Example


Node and Element Numbering



% Defining the number of elementsnumelem=13;

% Defining the number of nodesnumnodes=11;

% Defining pinned supportssupports=[1 1 1 0; 2 1 1 0; 3 1 1 0; 4 1 1 0];



% Taking a Steel UB406x140x46E = 2.05 * 10ˆ8; %(kN/mˆ2)I = 15685 * 10ˆ-8; %(mˆ4)A = 58.6 * 10ˆ-4; %(mˆ2)

% See the labelling on the diagramL = zeros(numelem,1); %number of elementsL(1:12,1)=4; %refer to diagramL(13,1)=4 * sqrt(2); %refer to diagram


Matlab Code: Member Stiffness Matrices

% Setting up a reference member stiffness matrix for each mem berk=zeros(6,6,numelem);for n=1:numelemk(1:6,1:6,n) = [E * A/L(n) 0 0 -E * A/L(n) 0 0; ...

0 12* E* I/L(n)ˆ3 6 * E* I/L(n)ˆ2 0 -12 * E* I/L(n)ˆ3 6 * E* I/L(n)ˆ2; ...0 6* E* I/L(n)ˆ2 4 * E* I/L(n) 0 -6 * E* I/L(n)ˆ2 2 * E* I/L(n); ...-E * A/L(n) 0 0 E * A/L(n) 0 0; ...0 -12 * E* I/L(n)ˆ3 -6 * E* I/L(n)ˆ2 0 12 * E* I/L(n)ˆ3 -6 * E* I/L(n)ˆ2; ...0 6* E* I/L(n)ˆ2 2 * E* I/L(n) 0 -6 * E* I/L(n)ˆ2 4 * E* I/L(n)];

end



% Rotating the member stiffness matices into the global axesalpha=[90 90 90 90 90 90 90 0 0 0 0 0 45];alpha=alpha * pi./180;

N=zeros(3,3,numelem);kdash=zeros(6,6,numelem);

for n=1:numelem;% Setting up a transformation matrix for each of the four quad rantsN(1:3,1:3,n)=[cos(alpha(n)) +sin(alpha(n)) 0; ...

-sin(alpha(n)) cos(alpha(n)) 0; ...0 0 1];

%kdash(1:3,1:3,n)=N(:,:,n)' * k(1:3,1:3,n) * N(:,:,n);kdash(1:3,4:6,n)=N(:,:,n)' * k(1:3,4:6,n) * N(:,:,n);kdash(4:6,1:3,n)=N(:,:,n)' * k(4:6,1:3,n) * N(:,:,n);kdash(4:6,4:6,n)=N(:,:,n)' * k(4:6,4:6,n) * N(:,:,n);

end



Alternative code to transform k to kdash for each element

%%%% The same transformation can be achieved using this code blockN=zeros(6,6,numelem);kdash=zeros(6,6,numelem);for n=1:numelem;

% setting up a transformation matrix for all four quadrants% top-left and bottom-right quadrants of N contain trigonom etric terms% top-right and bottom-left quadrants of N contain zero term sN(1:3,1:3,n)=[cos(alpha(n)) +sin(alpha(n)) 0; ...

-sin(alpha(n)) cos(alpha(n)) 0; ...0 0 1];

N(4:6,4:6,n)=N(1:3,1:3,n);%kdash(1:6,1:6,n)=N(:,:,n)' * k(:,:,n) * N(:,:,n);

end


Assembling the Structure Stiffness Matrix

% Setting up connectivity tables for the elements% Refer to the diagram% This will be used to shape the global stiffness matrixconnelem=[1 5; 2 6; 3 7; 4 8; 6 9; 7 10; 8 11; ...

5 6; 6 7; 7 8; 9 10; 10 11; ...5 9];

% As mentioned the k matrices can be broken into quadrants 3x3 in size% Deriving an initial global stiffness matrix, K% Note the final matrix must be symmetrical about the diagona l% Taking each of the nodes in turn and working out where the dif ferent% quadrant of the original kdash matrices should be placed in K

K=zeros(numnodes * 3,numnodes * 3);Klogic=zeros(numnodes,numnodes);



% placing the top-left quadrantfor n=1:numnodes

clear aa=find(connelem(:,1)==n);if isempty(a) 6=1;

for m=1:length(a);K(1+(n-1) * 3:n * 3,1+(n-1) * 3:n * 3)= ...

K(1+(n-1) * 3:n * 3,1+(n-1) * 3:n * 3)+kdash(1:3,1:3,a(m));%Klogic(n,n)=Klogic(n,n)+1;

endend

end



% placing the bottom-right quadrantfor n=1:numnodes

clear aa=find(connelem(:,2)==n);if isempty(a) 6=1;

for m=1:length(a);K(1+(n-1) * 3:n * 3,1+(n-1) * 3:n * 3)= ...

K(1+(n-1) * 3:n * 3,1+(n-1) * 3:n * 3)+kdash(4:6,4:6,a(m));%Klogic(n,n)=Klogic(n,n)+1;

endend

end



% placing the top-right and bottom-left quadrantfor n=1:numnodes

clear a ba=find(connelem(:,1)==n);if isempty(a) 6=1;

b=connelem(a,2);for m=1:length(a);

K(1+(n-1) * 3:n * 3,1+(b(m)-1) * 3:b(m) * 3)= ...K(1+(n-1) * 3:n * 3,1+(b(m)-1) * 3:b(m) * 3)+kdash(1:3,4:6,a(m));

K(1+(b(m)-1) * 3:b(m) * 3,1+(n-1) * 3:n * 3)= ...K(1+(b(m)-1) * 3:b(m) * 3,1+(n-1) * 3:n * 3)+kdash(4:6,1:3,a(m));

%Klogic(n,b(m))=Klogic(n,b(m))+1;Klogic(b(m),n)=Klogic(b(m),n)+1;

endend

end



% Addressing rows and columns in the stiffness matrix associ ated with% applied support constraints

a=size(supports);b=0;Ksupportrefs=zeros(1,sum(sum(supports(:,2:4) 6=0)));for n=1:a(1)

for m=2:a(2)if supports(n,m)==1;

b=b+1;Ksupportrefs(b)=(supports(n,1)-1) * 3+(m-1);

endend

endclear a b

K(Ksupportrefs(1,:),:)=0;K(:,Ksupportrefs(1,:))=0;for n=1:length(Ksupportrefs)

K(Ksupportrefs(1,n),Ksupportrefs(1,n))=1;end


Assembling the Fixed End Reactions Matrix

%Setting up matrix of distributed loadsw=[0 0 0 0 0 0 0 10 10 10 10 10 0];

%Setting up fixed end moments for each element (FE-M)FEMelem = zeros(numelem,2);for n=1:numelem

FEMelem(n,1) = +w(n). * L(n)ˆ2./12;FEMelem(n,2) = -w(n). * L(n)ˆ2./12;

end

%Setting up fixed end x2 reactions for each element (FE-R2)FER2elem = zeros(numelem,2);for n=1:numelem

FER2elem(n,1) = w(n). * L(n)./2;FER2elem(n,2) = w(n). * L(n)./2;

end



%Setting up fixed end moment for each nodeFEMnode=zeros(numnodes,1);for n=1:numnodes

clear a ba = find(connelem(:,1)==n);if isempty(a) 6=1;

for m=1:length(a);FEMnode(n,1)=FEMnode(n,1)+FEMelem(a(m),1);

endendb = find(connelem(:,2)==n);if isempty(b) 6=1;

for m=1:length(b);FEMnode(n,1)=FEMnode(n,1)+FEMelem(b(m),2);

endend

end



%Setting up fixed end x2 reactions for each nodeFER2node=zeros(numnodes,1);for n=1:numnodes

clear a ba = find(connelem(:,1)==n);if isempty(a) 6=1;

for m=1:length(a);FER2node(n,1)=FER2node(n,1)+FER2elem(a(m),1);

endendb = find(connelem(:,2)==n);if isempty(b) 6=1;

for m=1:length(b);FER2node(n,1)=FER2node(n,1)+FER2elem(b(m),2);

endend

end



%Setting up matrix of FE reactionsFERnode=zeros(3 * numnodes,1);FERnode(3:3:3 * numnodes,1)=FEMnode(1:1:numnodes,1);FERnode(2:3:3 * numnodes,1)=FER2node(1:1:numnodes,1);FERnode(Ksupportrefs,1)=0;


Finding the Joint Displacements

∆˜= K

˜

−1(−R˜

FE )

%Finding the joint displacementsdisplacements=(Kˆ-1) * (-FERnode);reshapedisplacements=reshape(displacements,3,[]);fprintf( '\nDisplacements at the nodes (m,m,rad):\n\n' );disp(reshapedisplacements);


Results: Joint Displacements

Displacements at the nodes (m,m,rad):

1.0e-003 *

Columns 1 through 6

0 0 0 0 0.0672 0.07600 0 0 0 -0.0661 -0.1966

0.0689 -0.0355 -0.0198 -0.0971 -0.1882 0.0140

Columns 7 through 11

0.0893 0.1013 0.2331 0.2170 0.2021-0.2742 -0.1290 -0.2546 -0.4155 -0.1922-0.0274 0.1182 -0.1916 -0.0053 0.2021


Find the Support Reactions

R˜= k

˜

′∆˜+ R

˜

FE

%Finding the support reactions%Isolating the elements that have restrained ends%This relies on supports being defined for single elementsa=size(supports);for n=1:a(1)

b=find(connelem(:,1)==n);EndAdisp=displacements(1+(connelem(b,1)-1) * 3:3+(connelem(b,1)-1) * 3,1);EndBdisp=displacements(1+(connelem(b,2)-1) * 3:3+(connelem(b,2)-1) * 3,1);Endsdisp(1:3,1)=EndAdisp;Endsdisp(4:6,1)=EndBdisp;reactions(1+(n-1) * 3:3+(n-1) * 3,1)=kdash(1:3,1:6,b) * Endsdisp+ ...

FERnode(1+(connelem(b,1)-1) * 3:3+(connelem(b,1)-1) * 3,1);endreshapereactions=reshape(reactions,3,[]);fprintf( '\nReactions (kN,kN,kNm):\n\n' );disp(reshapereactions);


Results: Support Reactions

Reactions (kN,kN,kNm):

1.0333 -0.1987 0.0305 -0.865119.8530 59.0480 82.3450 38.7540

0.0000 0.0000 -0.0000 0.0000


Check: Assembled Structure Stiffness Matrix

Assembly logic of the K matrix:

1 0 0 0 1 0 0 0 0 0 00 1 0 0 0 1 0 0 0 0 00 0 1 0 0 0 1 0 0 0 00 0 0 1 0 0 0 1 0 0 01 0 0 0 3 1 0 0 1 0 00 1 0 0 1 4 1 0 1 0 00 0 1 0 0 1 4 1 0 1 00 0 0 1 0 0 1 3 0 0 10 0 0 0 1 1 0 0 3 1 00 0 0 0 0 0 1 0 1 3 10 0 0 0 0 0 0 1 0 1 2


Conclusions

The script can be generalised for any plane frame.

The script can be adapted to include different cross sections.

Results are found to match those obtained from Oasys GSAexcluding shear deformations.


Ask not what you can do for the matrix stiffness method...

Ask what the matrix stiffness method can do for you.


Stiffness method

Education

Transcript of Stiffness method