6-Bar Elements in 2-D Space e-mail: Dr. Ahmet Zafer Şenalp e-mail: [email protected]@gmail.com Mechanical Engineering Department Gebze Technical.

6-Bar Elements in 2-D Space

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

Mechanical Engineering DepartmentGebze Technical University

ME 520Fundamentals of Finite Element Analysis

mailto:[email protected]

Bar (truss) structures:

Bar Element

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


Cross section examples forbar structures

2-D Case



Transformation:

In matrix form:

2-D Case



Transformation matrix:

For the two nodes of the bar element, we have:

The nodal forces are transformed in the same way:

2-D Case



In the local coordinate system, we have:

Using transformations:

Multiplying both sides by TT and noticing that TTT = I, we obtain:

Thus, the element stiffness matrix k in the global coordinate system is:

which is a 4´4 symmetric matrix.

Stiffness Matrix in the 2-D Space



Explicit form:

Calculation of the directional cosines l and m:

The structure stiffness matrix is assembled by using the element stiffness matrices in the usual way as in the 1-D case.




Element Stress:




Example 1



Given;Element 1: E,A,LElement 2: E,A,L

Find ; (a) displacement of node 2(b) stress in each bar

Solution:

Connectivity table: E# N1 N2 Angle

1 1 2 45

2 2 3 135

Example 1



Boundary conditions: Displacement boundary conditions:

Force boundary conditions:

a) Element Stiffness Matrices:

In local coordinate systems, we have;

These two matrices cannot be assembled together, because they are in different coordinate systems. We need to convert them to global coordinate system OXY.

0v ,0u ,0v ,0u ,0v,0u 332211

0F ,0F ,PF ,PF ,0F,0F y3x32y21x2y1x1

Example 1



Element 1:

Element 2:

Example 1



Assemble the structure FE equation:

Applying the BC’s:

Example 1



Stresses in the two bars:

Check the results:Look for the equilibrium conditions, symmetry, antisymmetry, etc.

Solution procedure with matlab


As plane truss element has 4 degrees of freedom (2 at each node) for a structure with n nodes, the global stiffness matrix K will be of size 2nx2n.

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

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

At this step boundary conditions are applied manually to the vectors U and F.

Then the matrix equation is solved by partioning and Gaussion elimination.

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

FUK



where f is the force in this element (a scalar) and u is the 4x1 element displacement vector.

The element stress is obtained by dividing the element force by the cross-sectional area A.

If there is an inclined support at one of the nodes of the truss then the global stiffness matrix needs to be modified using the following equation:

where [T] is a 2nx2n transformation matrix that is obtained by makinga call to the function PlaneTrussInclinedSupport.

usincossincosL

EAf

Toldnew TKTK

Solution procedure with matlab 6-Bar Elements in 2-D Space


The inclined supportis assumed to be at node i with an angle of inclination alpha as shownbelow

The new matrix K0 obtained is thus the global stiffness matrix for the structure.

Solution procedure with matlab 6-Bar Elements in 2-D Space

Matlab functions used


PlaneTrussElementLength(x1,y1,x2,y2)This function returns the length of the plane truss element whose first node has coordinates (x1,y1) and second node has coordinates (x2,y2). Function contents:function y = PlaneTrussElementLength(x1,y1,x2,y2)%PlaneTrussElementLength This function returns the length of the% plane truss element whose first node has % coordinates (x1,y1) and second node has % coordinates (x2,y2). y = sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1));




PlaneTrussElementStiffness(E,A,L,theta)This function returns the element stiffness matrix for a plane truss element with modulus of elasticity E, cross-sectional area A, length L, and angle theta (in degrees). The size of the element stiffness matrix is 4 x 4. Function contents:function y = PlaneTrussElementStiffness(E,A,L, theta)%PlaneTrussElementStiffness This function returns the element % stiffness matrix for a plane truss % element with modulus of elasticity E, % cross-sectional area A, length L, and% angle theta (in degrees).% The size of the element stiffness % matrix is 4 x 4.x = theta*pi/180;C = cos(x);S = sin(x);y = E*A/L*[C*C C*S -C*C -C*S ; C*S S*S -C*S -S*S ; -C*C -C*S C*C C*S ; -C*S -S*S C*S S*S];




PlaneTrussAssemble(K,k,i,j)This function assembles the element stiffness matrix k of the plane truss 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.Function contents:function y = PlaneTrussAssemble(K,k,i,j)%PlaneTrussAssemble This function assembles the element stiffness% matrix k of the plane truss 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(2*i-1,2*i-1) = K(2*i-1,2*i-1) + k(1,1);K(2*i-1,2*i) = K(2*i-1,2*i) + k(1,2);K(2*i-1,2*j-1) = K(2*i-1,2*j-1) + k(1,3);K(2*i-1,2*j) = K(2*i-1,2*j) + k(1,4);K(2*i,2*i-1) = K(2*i,2*i-1) + k(2,1);K(2*i,2*i) = K(2*i,2*i) + k(2,2);K(2*i,2*j-1) = K(2*i,2*j-1) + k(2,3);K(2*i,2*j) = K(2*i,2*j) + k(2,4);




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




PlaneTrussElementForce(E,A,L,theta,u)This function returns the element force given the modulus of elasticity E, the cross-sectional area A, the length L, the angle theta (in degrees), and the element nodal displacement vector u.Function contents:function y = PlaneTrussElementForce(E,A,L,theta,u)%PlaneTrussElementForce This function returns the element force% given the modulus of elasticity E, the % cross-sectional area A, the length L, % the angle theta (in degrees), and the % element nodal displacement vector u.x = theta * pi/180;C = cos(x);S = sin(x);y = E*A/L*[-C -S C S]* u;




PlaneTrussElementStress(E,L,thetax,thetay,thetaz,u)This function returns the element stress given the modulus of elasticity E, the length L, the angles thetax, thetay, thetaz (in degrees), and the element nodal displacement vector u. It returns the element stress as a scalar.Function contents:function y = PlaneTrussElementStress(E,L,theta,u)%PlaneTrussElementStress This function returns the element stress% given the modulus of elasticity E, the % the length L, the angle theta (in % degrees), and the element nodal % displacement vector u.x = theta * pi/180;C = cos(x);S = sin(x);y = E/L*[-C -S C S]* u;




PlaneTrussInclinedSupport(T,i,alpha)This function calculates the transformation matrix T of the inclined support at node i with angle of inclination alpha (in degrees).Function contents:function y = PlaneTrussInclinedSupport(T,i,alpha)%PlaneTrussInclinedSupport This function calculates the% tranformation matrix T of the inclined% support at node i with angle of% inclination alpha (in degrees).x = alpha*pi/180;T(2*i-1,2*i-1) = cos(x);T(2*i-1,2*i) = sin(x);T(2*i,2*i-1) = -sin(x) ;T(2*i,2*i) = cos(x);y = T;


Solution of Example 2with Matlab



Given;E=210 GPaA=1x10-4 m2

Find ; (a) global stiffness matrix of the structure(b) horizontal displacement at node 2(c) horizontal and vertical displacementsat node 3(d) reactions at nodes 1 and 2(b) stress in each element

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




Step 1-Discretizing the domain:This problem is already discretized. The domain is subdivided into 3 elements and 3 nodes. The units used in Matlab calculations are kN and meter. The element connectivity is:

E# N1 N2

1 1 2

2 1 3

3 2 3




Step 2-Copying relevant files and starting MatlabCreate a directory

Copy PlaneTrussAssemble.mPlaneTrussElementForce.mPlaneTrussElementLength.mPlaneTrussElementStiffness.mPlaneTrussElementStress.mPlaneTrussInclinedSupport.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

E# N1 N2

1 1 2

2 1 3

3 2 3





Enter the data>>E=210e6>>A=1e-4>>L1=4>>L2=PlaneTrussElementLength(0,0,2,3)>>L3= PlaneTrussElementLength(4,0,2,3)

Step 3-Writing the element stiffness matrices:>>k1=PlaneTrussElementStiffness(E,A,L1, 0)>>theta2=atan(3/2)*180/pi>>theta3=180-theta2>>k2=PlaneTrussElementStiffness(E,A,L2, theta2)>>k3=PlaneTrussElementStiffness(E,A,L3, theta3)

E# N1 N2

1 1 2

2 1 3

3 2 3





yields;k2 =

1.0e+03 *

1.7921 2.6882 -1.7921 -2.6882 2.6882 4.0322 -2.6882 -4.0322 -1.7921 -2.6882 1.7921 2.6882 -2.6882 -4.0322 2.6882 4.0322

k3 =

1.0e+03 *

1.7921 -2.6882 -1.7921 2.6882 -2.6882 4.0322 2.6882 -4.0322 -1.7921 2.6882 1.7921 -2.6882 2.6882 -4.0322 -2.6882 4.0322

E# N1 N2

1 1 2

2 1 3

3 2 3





Step 4-Assembling the global stiffness matrix:Since the structure has 3 nodes the size of the global stiffness matrixis 6x6. So to find global stiffness matrix;>>K=zeros(6,6)>>K=PlaneTrussAssemble(K,k1,1,2)>>K=PlaneTrussAssemble(K,k2,1,3)>>K=PlaneTrussAssemble(K,k3,2,3)yields;K =

1.0e+03 *

7.0421 2.6882 -5.2500 0 -1.7921 -2.6882 2.6882 4.0322 0 0 -2.6882 -4.0322 -5.2500 0 7.0421 -2.6882 -1.7921 2.6882 0 0 -2.6882 4.0322 2.6882 -4.0322 -1.7921 -2.6882 -1.7921 2.6882 3.5842 0.0000 -2.6882 -4.0322 2.6882 -4.0322 0.0000 8.0645

E# N1 N2

1 1 2

2 1 3

3 2 3





Step 5-Applying the boundary conditions:BCs are:

E# N1 N2

1 1 2

2 1 3

3 2 3


Y3

X3

Y2

X2

Y1

X1

3

3

2

2

1

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

F

F

F

F

F

F

v

u

v

u

v

u

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

,0v,0u,0v,0u,0v,0u 332211 10F,5F,0F,0F,0F,0F Y3X3Y2X2Y1X1

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;




E# N1 N2

1 1 2

2 1 3

3 2 3


10

5

0

0

0

0

v

u

0

u

0

0

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

KKKKKK

3

3

2

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

10

5

0

v

u

u

KKK

KKK

KKK

3

3

2

666563

565553

363533

>>k=[K(3,3) K(3,5) K(3,6); K(5,3) K(5,5) K(5,6); K(6,3) K(6,5) K(6,6)] >>f=[0; 5 ; -10]>>u=k\fu =

0.0011 0.0020 -0.0016




Step 7-Post-processing:In this step we obtain the reactions and and the force in each 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 ; u(1) ; 0; u(2:3)]yields;U = 0 0 0.0011 0 0.0020 -0.0016

E# N1 N2

1 1 2

2 1 3

3 2 3





>>F=K*Uyields;F = -5.0000 1.2500 -0.0000 8.7500 5.0000 -10.0000

E# N1 N2

1 1 2

2 1 3

3 2 3





Form element displacement vector>>u1=[U(1) ; U(2) ; U(3) ; U(4)]>>u2=[U(1) ; U(2) ; U(5) ; U(6)]>>u3=[U(3) ; U(4) ; U(5) ; U(6)]To find stresses recall the below function subroutine>>sigma1=PlaneTrussElementStress(E,L1,0,u1)>>sigma2=PlaneTrussElementStress(E,L2,theta2,u2)>>sigma3=PlaneTrussElementStress(E,L3,theta3,u3)yields;sigma1 =

5.8333e+04

sigma2 = -1.5023e+04

sigma3 = -1.0516e+05

E# N1 N2

1 1 2

2 1 3

3 2 3


ME 520 35GYTE-Makine Mühendisliği Bölümü


Given;

Element 1 and 2:Element 3:

Find; (a) displacements(b) reaction forces

Solution:

Connectivity table: E# N1 N2 Açı

1 1 2 90

2 2 3 0

3 1 3 45

Example 3-Multipoint constraint






a) Element Stiffness Matrices (In Global coordinate system):

Element 1:

0v ,0u ,0v ,0u ,0v,0u 332211

0F ,0F ,0F ,PF ,0F,0F y3x3y2x2y1x1 E# N1 N2

1 1 2

2 2 3

3 1 3






a) Element Stiffness Matrices (In Global coordinate system):

Element 1:

0v ,0u ,0v ,0u ,0v,0u 332211

0F ,0F ,0F ,PF ,0F,0F y3x3y2x2y1x1 E# N1 N2

1 1 2

2 2 3

3 1 3




Element 2:

Element 3:




The global FE equation is:

BC’s: 0v ,0u ,0v ,0u ,0v,0u 332211 0F ,0F ,0F ,PF ,0F,0F y3x3y2x2y1x1




Applying BC’s:

Solving for unknowns:




From the global FE equation, we can calculate the reaction forces:

Check the results!A general multipoint constraint (MPC) can be described as,

where Aj’s are constants and uj’s are nodal displacement components. In the FE software, such as MSC/NASTRAN, users only need to specify this relation to the software. The software will take care of the solution.

Solution of Example 4(Multipoint constraint)with Matlab



Given:E=70 GPaA=0.004 m2

Find:a) global stiffness matrixb) displacements at nodes 2, 3, and 4c) reactions at nodes 1 and 4d) stress in each element

Solution:

Connectivity table;

E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Create a directory

Copy PlaneTrussAssemble.mPlaneTrussElementForce.mPlaneTrussElementLength.mPlaneTrussElementStiffness.mPlaneTrussElementStress.mPlaneTrussInclinedSupport.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




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Enter the data>>E=70e6>>A=0.004>>L1=3.5>>theta1=90>>L2=4>>theta2=0>>L3=PlaneTrussElementLength(0,0,4,3.5)>>theta3=atan(3.5/4)*180/pi>>L4= L3>>theta4=360-theta3>>L5=4>>theta5=0>>L6=3.5>>theta6=270




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Calculate stiffness matrices;>>k1=PlaneTrussElementStiffness(E,A,L1, theta1)>>k2=PlaneTrussElementStiffness(E,A,L2, theta2)>>k3=PlaneTrussElementStiffness(E,A,L3, theta3)>>k4=PlaneTrussElementStiffness(E,A,L4, theta4)>>k5=PlaneTrussElementStiffness(E,A,L5, theta5)>>k6=PlaneTrussElementStiffness(E,A,L6, theta6)Assemble the global stiffness matrix;

Since the structure has 4 nodes the size of the global stiffness matrixis 8x8. So to find global stiffness matrix;>>K=zeros(8,8)>>K=PlaneTrussAssemble(K,k1,1,2)>>K=PlaneTrussAssemble(K,k2,1,4)>>K=PlaneTrussAssemble(K,k3,1,3)




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

>>K=PlaneTrussAssemble(K,k4,2,4)>>K=PlaneTrussAssemble(K,k5,2,3)>>K=PlaneTrussAssemble(K,k6,3,4)yields;K =

1.0e+05 *

0.9984 0.2611 -0.0000 -0.0000 -0.2984 -0.2611 -0.7000 0 0.2611 1.0284 -0.0000 -0.8000 -0.2611 -0.2284 0 0 -0.0000 -0.0000 0.9984 -0.2611 -0.7000 0 -0.2984 0.2611 -0.0000 -0.8000 -0.2611 1.0284 0 0 0.2611 -0.2284 -0.2984 -0.2611 -0.7000 0 0.9984 0.2611 -0.0000 -0.0000 -0.2611 -0.2284 0 0 0.2611 1.0284 -0.0000 -0.8000 -0.7000 0 -0.2984 0.2611 -0.0000 -0.0000 0.9984 -0.2611 0 0 0.2611 -0.2284 -0.0000 -0.8000 -0.2611 1.0284

Next we need to modify the global stiffness matrix obtainedabove to take the effect of the inclined support at node 4.

If there is an inclined support at one of the nodes of the truss (multipoint constraint) then the global stiffness matrix needs to be modified using the following equation:

where [T] is a 2nx2n transformation matrix that is obtained by makinga call to the function PlaneTrussInclinedSupport.




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Toldnew TKTK




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

>>T=eye(8,8)results;T =

1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1

>>T=PlaneTrussInclinedSupport(T,4,45)




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

yields;T =

1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 1.0000 0 0 0 0 0 0 0 0 0.7071 0.7071 0 0 0 0 0 0 -0.7071 0.7071

>>K0=T*K*T’

yields;




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

K0 =

1.0e+05 *

0.9984 0.2611 -0.0000 -0.0000 -0.2984 -0.2611 -0.4950 0.4950 0.2611 1.0284 -0.0000 -0.8000 -0.2611 -0.2284 0 0 -0.0000 -0.0000 0.9984 -0.2611 -0.7000 0 -0.0264 0.3956 -0.0000 -0.8000 -0.2611 1.0284 0 0 0.0231 -0.3461 -0.2984 -0.2611 -0.7000 0 0.9984 0.2611 -0.0000 -0.0000 -0.2611 -0.2284 0 0 0.2611 1.0284 -0.5657 -0.5657 -0.4950 0 -0.0264 0.0231 -0.0000 -0.5657 0.7523 0.0150 0.4950 0 0.3956 -0.3461 -0.0000 -0.5657 0.0150 1.2745

Applying the BCs:FE equation is;

FKU




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

BCs are:

Imposing BCs>>k=K0(3:7,3:7)yields;k =

1.0e+05 *

0.9984 -0.2611 -0.7000 0 -0.0264 -0.2611 1.0284 0 0 0.0231 -0.7000 0 0.9984 0.2611 -0.0000 0 0 0.2611 1.0284 -0.5657 -0.0264 0.0231 -0.0000 -0.5657 0.7523

0v,0u,0v ,0u ,0v ,0u ,0v,0u 44332211

0F ,0F,0F ,30F ,0F ,0F ,0F,0F y4x4y3x3y2x2y1x1




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

>>f=[0 ;0 ;30 ;0 ;0 ]>>u=k\fyields,u = 1.0e-03 * 0.6053 0.1590 0.8129 -0.3366 -0.2367

0

0

30

0

0

u

v

u

v

u

k

4

3

3

2

2




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Post-processing;>>U=[0 ; 0 ; u ; 0]>>F=K0*Uyields;F =

-3.7500 -26.2500 0.0000 0.0000 30.0000 -0.0000 0 37.1231




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

Stresses;>>u1=[U(1) ; U(2) ; U(3) ; U(4) ]>>sigma1=PlaneTrussElementStress(E,L1,theta1,u1) results;sigma1 = 3.1791e+03>>u2=[U(1) ; U(2) ; U(7) ; U(8) ]>>sigma2=PlaneTrussElementStress(E,L2,theta2,u2) results;sigma2 = -4.1425e+03>>u3=[U(1) ; U(2) ; U(5) ; U(6) ]>>sigma3=PlaneTrussElementStress(E,L3,theta3,u3)results;sigma3 = 5.1380e+03




E# N1 N2

1 1 2

2 1 4

3 1 3

4 2 4

5 2 3

6 3 4

>>u4=[U(3) ; U(4) ; U(7) ; U(8) ]>>sigma4=PlaneTrussElementStress(E,L4,theta4,u4) results;sigma4 = -6.9666e+03>>u5=[U(3) ; U(4) ; U(5) ; U(6) ]>>sigma5=PlaneTrussElementStress(E,L5,theta5,u5) results;sigma5 = 3.6333e+03>>u6=[U(5) ; U(6) ; U(7) ; U(8) ]>>sigma6=PlaneTrussElementStress(E,L6,theta6,u6)results;sigma6 = -6.7311e+03

6-Bar Elements in 2-D Space e-mail: Dr. Ahmet Zafer Şenalp e-mail: [email protected]@gmail.com Mechanical Engineering Department Gebze Technical.

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