9-beam element with axial force e-mail: dr. ahmet zafer Şenalp e-mail:...

9-Beam Element with Axial Force

Dr. Ahmet Zafer Şenalpe-mail: [email protected]

Mechanical Engineering DepartmentGebze Technical University

ME 520Fundamentals of Finite Element Analysis

mailto:[email protected]

Apply the formula,

To derive this, we introduce the shape functions:

Then, we can represent the deflection as:

which is a cubic function. Notice that,

Formal Approach

ME 520 Dr. Ahmet Zafer Şenalp 2Mechanical Engineering Department, GTU


which implies that the rigid body motion is represented by the assumed deformed shape of the beam. Curvature of the beam is,

where the strain-displacement matrix B is given by,

Formal Approach



Strain energy stored in the beam element is:

We conclude that the stiffness matrix for the simple beam element is

Formal Approach



Combining the axial stiffness (bar element), we obtain the stiffness matrix of a general 2-D beam element:

Formal Approach




It is clear that the beam element has 6 degrees of freedom (3 at each node)

For a structure with n nodes, the global stiffness matrix K will be of size 3nx3n.

The global stiffness matrix K is obtained by making calls to the Matlab function BeamAssemble which is written for this purpose.

Once the global stiffness matrix; K is obtained we have the following structure equation;

where U is the global nodal displacement vector and F is the global nodal force vector. At this step boundary conditions are applied manually to the vectors U and F.Then the matrix equation is solved by partitioning and Gaussion elimination.

FUK

Solution procedure with matlab 9-Beam Element with Axial Force


Finally once the unknown displacements and and reactions are found, the force is obtained for each element as follows:

where f is the 6x1 nodal force vector in the element and u is the 6x1 element displacement vector.

The first, second and third elements in each vector are the axial displacement, transverse displacement and rotation, respectively, at the first node, while the fourth fifth and sixth elements in each vector are the axial displacement, transverse displacement and rotation, respectively, at the second node.

uKf

Solution procedure with matlab

u

u


Matlab functions used


The 5 Matlab functions used for the beam element are:

BeamElementStiffness_axial_az( E,A,I,L)This function returns the element stiffness matrix for a beam element with axial force with modulus of elasticity E, area A, moment of inertia I, and length L. The size of the element stiffness matrix is 6 x 6.Function contents:function y = BeamElementStiffness_axial_az(E,A,I,L)%BeamElementStiffness This function returns the element % stiffness matrix for a beam % element with axial force % with modulus of elasticity E,% with area A% moment of inertia I, and length L.% The size of the element stiffness % matrix is 6 x 6.c=E*A/L;c12=2*E*I/(L);c14=4*E*I/(L);c2=6*E*I/(L*L);c3=12*E*I/(L*L*L);y = [c 0 0 -c 0 0; 0 c3 c2 0 -c3 c2; 0 c2 c14 0 -c2 c12; -c 0 0 c 0 0; 0 -c3 -c2 0 c3 -c2; 0 c2 c12 0 -c2 c14];




BeamAssemble_axial(K,k,i,j)This function assembles the element stiffness matrix k of the beam element with nodes i and j into the global stiffness matrix K. This function returns the 3nx3n global stiffness matrix K after the element stiffness matrix k is assembled.Function contents:function y = BeamAssemble_axial(K,k,i,j)%BeamAssemble_axial This function assembles the element stiffness% matrix k of the beam element with nodes% i and j into the global stiffness matrix K.% This function returns the global stiffness % matrix K after the element stiffness matrix % k is assembled.K(3*i-2,3*i-2) = K(3*i-2,3*i-2) + k(1,1);K(3*i-2,3*i-1) = K(3*i-2,3*i-1) + k(1,2);K(3*i-2,3*i) = K(3*i-2,3*i) + k(1,3);K(3*i-2,3*j-2) = K(3*i-2,3*j-2) + k(1,4);K(3*i-2,3*j-1) = K(3*i-2,3*j-1) + k(1,5);K(3*i-2,3*j) = K(3*i-2,3*j) + k(1,6);K(3*i-1,3*i-2) = K(3*i-1,3*i-2) + k(2,1);K(3*i-1,3*i-1) = K(3*i-1,3*i-1) + k(2,2);K(3*i-1,3*i) = K(3*i-1,3*i) + k(2,3);K(3*i-1,3*j-2) = K(3*i-1,3*j-2) + k(2,4);K(3*i-1,3*j-1) = K(3*i-1,3*j-1) + k(2,5);K(3*i-1,3*j) = K(3*i-1,3*j) + k(2,6);




K(3*i,3*i-2) = K(3*i,3*i-2) + k(3,1);K(3*i,3*i-1) = K(3*i,3*i-1) + k(3,2);K(3*i,3*i) = K(3*i,3*i) + k(3,3);K(3*i,3*j-2) = K(3*i,3*j-2) + k(3,4);K(3*i,3*j-1) = K(3*i,3*j-1) + k(3,5);K(3*i,3*j) = K(3*i,3*j) + k(3,6);K(3*j-2,3*i-2) = K(3*j-2,3*i-2) + k(4,1);K(3*j-2,3*i-1) = K(3*j-2,3*i-1) + k(4,2);K(3*j-2,3*i) = K(3*j-2,3*i) + k(4,3);K(3*j-2,3*j-2) = K(3*j-2,3*j-2) + k(4,4);K(3*j-2,3*j-1) = K(3*j-2,3*j-1) + k(4,5);K(3*j-2,3*j) = K(3*j-2,3*j) + k(4,6);K(3*j-1,3*i-2) = K(3*j-1,3*i-2) + k(5,1);K(3*j-1,3*i-1) = K(3*j-1,3*i-1) + k(5,2);K(3*j-1,3*i) = K(3*j-1,3*i) + k(5,3);K(3*j-1,3*j-2) = K(3*j-1,3*j-2) + k(5,4);K(3*j-1,3*j-1) = K(3*j-1,3*j-1) + k(5,5);K(3*j-1,3*j) = K(3*j-1,3*j) + k(5,6);K(3*j,3*i-2) = K(3*j,3*i-2) + k(6,1);K(3*j,3*i-1) = K(3*j,3*i-1) + k(6,2);K(3*j,3*i) = K(3*j,3*i) + k(6,3);K(3*j,3*j-2) = K(3*j,3*j-2) + k(6,4);K(3*j,3*j-1) = K(3*j,3*j-1) + k(6,5);K(3*j,3*j) = K(3*j,3*j) + k(6,6);y = K;




BeamElementForces(k,u)This function calculates the element element force vector using the element stiffness matrix k and the element displacement vector u. It returns the 6x1 element force vector fFunction contents:function y = BeamElementForces(k,u)%BeamElementForces This function returns the element nodal force% vector given the element stiffness matrix k % and the element nodal displacement vector u.y = k * u;




BeamElementShearDiagram(f, L)This function plots the shear force diagram for the beam element with nodal force vector f and length L.Function contents:function y = BeamElementShearDiagram(f, L)%BeamElementShearDiagram This function plots the shear force % diagram for the beam element with nodal% force vector f and length L.x = [0 ; L];z = [f(1) ; -f(3)];hold on;title('Shear Force Diagram');plot(x,z);y1 = [0 ; 0];plot(x,y1,'k')




BeamElementMomentDiagram(f, L)This function plots the bending moment diagram for the beam element with nodalforce vector f and length L.Function contents:function y = BeamElementMomentDiagram(f, L)%BeamElementMomentDiagram This function plots the bending moment % diagram for the beam element with nodal% force vector f and length L.x = [0 ; L];z = [-f(2) ; f(4)];hold on;title('Bending Moment Diagram');plot(x,z);y1 = [0 ; 0];plot(x,y1,'k')



Consider the beam as shownGiven E=210 GPaA=0.025 m2

I=60x10-6 m4

P1=20 kNP2=30 kNL=2 mDetermine:a) the global stiffness matrix for the structureb) horizontal and vertical displacements at node 2c) rotations at nodes 2 and 3d) the reactions at nodes 1 and 3e) the forces (shears and moments) in each elementf) the shear force diagram for each elementg) the bending moment diagram for each element

Solution of Example 1 with Matlab 9-Beam Element with Axial Force


Solution of Example 1 with Matlab

Solution:Use the 7 steps to solve the problem using beam element.

Step 1-Discretizing the domain:We will put a node (node2) at the location of the concentrated force so that we may determine the required quantities (displacements, rotation, shear, moment) at that point. The domain is subdivided into two elements and three nodes. The units used in Matlab calculations are kN and meter. The element connectivity is:

E# N1 N2

1 1 2

2 2 3




Step 2-Copying relevant files and starting MatlabCreate a directory

Copy BeamElementStiffness_axial_az.mBeamAssemble_axial.mBeamElementForces.mBeamElementShearDiagram.mBeamElementMomentDiagram.mfiles under the created directory

Open Matlab;Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory.

Start solving the problem in Command Window:>>clearvars>>clc




Step 3-Writing the element stiffness matrices:The two element stiffness matrices k1 and k2 are obtained by making calls to the Matlab function BeamElementStiffness_axial_az. Each matrix has size 6x6.Enter the data>>E=210e6>>A=0.025>>I=60e-6>>L=2>>k1=BeamElementStiffness_axial_az(E,A,I,L)k1 =

2625000 0 0 -2625000 0 0 0 18900 18900 0 -18900 18900 0 18900 25200 0 -18900 12600 -2625000 0 0 2625000 0 0 0 -18900 -18900 0 18900 -18900 0 18900 12600 0 -18900 25200




>>k2=BeamElementStiffness_axial_az(E,A,I,L)k2 =

2625000 0 0 -2625000 0 0 0 18900 18900 0 -18900 18900 0 18900 25200 0 -18900 12600 -2625000 0 0 2625000 0 0 0 -18900 -18900 0 18900 -18900 0 18900 12600 0 -18900 25200Step 4-Assembling the global stiffness matrix:Since the structure has 3 nodes, the size of the global stiffness matrix is 9x9.>>K=zeros(9,9)>>K=BeamAssemble_axial(K,k1,1,2)>>K=BeamAssemble_axial(K,k2,2,3)K =

2625000 0 0 -2625000 0 0 0 0 0 0 18900 18900 0 -18900 18900 0 0 0 0 18900 25200 0 -18900 12600 0 0 0 -2625000 0 0 5250000 0 0 -2625000 0 0 0 -18900 -18900 0 37800 0 0 -18900 18900 0 18900 12600 0 0 50400 0 -18900 12600 0 0 0 -2625000 0 0 2625000 0 0 0 0 0 0 -18900 -18900 0 18900 -18900 0 0 0 0 18900 12600 0 -18900 25200




Step 5-Applying the boundary conditions:Finite element equation for the problem is;

The boundary conditions for the problem are;

3

y3

x3

2

y2

x2

1

y1

x1

3

3

3

2

2

2

1

1

1

M

F

F

M

F

F

M

F

F

v

u

v

u

v

u

K

0 ,0 v,0u ,0 ,0 v,0u ,0 ,0v,0u 333222111

0M ,0F ,0F ,0M ,20F,30F ,0M ,0F,0F 33y3x22y2x1y1x1





Inserting the above conditions into finite element equation

Step 6-Solving the equations:Solving the above system of equations will be performed by partitioning (manually) and Gaussian elimination (with Matlab)First we partition the above equation by extracting the submatrices in rows 4 to 6 and columns 4 to 6 and row 9 column 9. Therefore we obtain:

0

0

0

0

20

30

M

F

F

0

0

v

u

0

0

0

K

1

y1

x1

3

2

2

2




The solution of the above system is obtained using Matlab as follows.Note that the ‘\’ operator is used for Gaussian elimination.>>k=[K(4:6,4:6) K(4:6,9) ; K(9,4:6) K(9,9)]

k =

5250000 0 0 0 0 37800 0 18900 0 0 50400 12600 0 18900 12600 25200

>>f=[30; -20 ; 0 ; 0]

f =

-20 0 0




>>u=k\fu =

1.0e-03 *

0.0057 -0.9259 -0.1984 0.7937

Step 7-Post-processing:In this step we obtain the reactions at nodes 1 and 3 and the forces (shears and moments) in each beam element using Matlab as follows.First we set up the global nodal displacement vector U, then we calculate the nodal force vector F.




>>U=[0 ; 0 ; 0; u(1) ; u(2) ; u(3); 0; 0; u(4)]U =

1.0e-03 *

0 0 0 0.0057 -0.9259 -0.1984 0 0 0.7937

>>F=K*UF =

-15.0000 13.7500 15.0000 30.0000 -20.0000 0 -15.0000 6.2500 -0.0000

0

0

0

0

20

30

M

F

F

0

0

v

u

0

0

0

K

1

y1

x1

3

2

2

2




Next we set up the element nodal displacement vectors u1 and u2 then we calculate the element force vectors f1 and f2 by making calls to the Matlab function BeamElementForces.>> u1=[U(1) ; U(2) ; U(3) ; U(4) ; U(5) ; U(6)]u1 =

1.0e-03 *

0 0 0 0.0057 -0.9259 -0.1984




>> u2=[U(4) ; U(5) ; U(6) ; U(7) ; U(8) ; U(9)]u2 =

1.0e-03 *

0.0057 -0.9259 -0.1984 0 0 0.7937>>f1 =BeamElementForces(k1,u1)f1 =

-15.0000 13.7500 15.0000 15.0000 -13.7500 12.5000




>>f2 =BeamElementForces(k2,u2)f2 =

15.0000 -6.2500 -12.5000 -15.0000 6.2500 -0.0000

Finally we call the Matlab functions BeamElementShearDiagram and BeamElementMomentDiagram, respectively for each element.




>>BeamElementShearDiagram(f1,L)




>>BeamElementShearDiagram(f2,L)




>>BeamElementMomentDiagram(f1, L)




>>BeamElementMomentDiagram(f2, L)

9-beam element with axial force e-mail: dr. ahmet zafer Şenalp e-mail:...

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