7-bar elements in 3-d space e-mail: dr. ahmet zafer Şenalp e-mail:...

7-Bar Elements in 3-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



Element stiffness matrices are calculated in the localcoordinate systems and then transformed into the globalcoordinate system (X, Y, Z) where they are assembled.

FEA software packages will do this transformation automatically.

Input data for bar elements:· (X, Y, Z) for each node· E and A for each element

The space truss element is characterized by linear shape functions.



Each truss element has two nodes and is inclined with anglesmeasured from the global X, Y and Z axes respectively to the local x axis as shown below;

zyx and ,

Stiffness matrix



LetIn this case the element stiffness matrix is as folllows;

zzyyxx cosC and cosC ,cosC



As space truss 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 SpaceTrussAssemble 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

Solution procedure with matlab



where f is the force in this element (a scalar) and u is the 6x1 element displacement vector.

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

uCCCCCCL

EAf zyxzyx

Solution procedure with matlab

Matlab functions used



SpaceTrussElementLength(x1,y1,z1,x2,y2,z2)This function returns the length of the space truss element whose first node has coordinates (x1,y1,z1) and second node has coordinates (x2,y2,z2)Function contents:function y = SpaceTrussElementLength(x1,y1,z1,x2,y2,z2)%SpaceTrussElementLength This function returns the length of the% space truss element whose first node has % coordinates (x1,y1,z1) and second node has % coordinates (x2,y2,z2). y = sqrt((x2-x1)*(x2-x1) + (y2-y1)*(y2-y1) + (z2-z1)*(z2-z1));




SpaceTrussElementStiffness(E,A,L,thetax,thetay,thetaz)This function returns the element stiffness matrix for a space truss element with modulus of elasticity E, cross-sectional area A, length L, and angles thetax, thetay, thetaz (in degrees). The size of the element stiffness matrix is 6 x 6.Function contents:function y = SpaceTrussElementStiffness(E,A,L,thetax,thetay,thetaz)%SpaceTrussElementStiffness This function returns the element % stiffness matrix for a space truss % element with modulus of elasticity E, % cross-sectional area A, length L, and% angles thetax, thetay, thetaz % (in degrees). The size of the element % stiffness matrix is 6 x 6.x = thetax*pi/180;u = thetay*pi/180;v = thetaz*pi/180;Cx = cos(x);Cy = cos(u);Cz = cos(v);w = [Cx*Cx Cx*Cy Cx*Cz ; Cy*Cx Cy*Cy Cy*Cz ; Cz*Cx Cz*Cy Cz*Cz];y = E*A/L*[w -w ; -w w];




SpaceTrussAssemble(K,k,i,j)This function assembles the element stiffness matrix k of the space truss 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 = SpaceTrussAssemble(K,k,i,j)%SpaceTrussAssemble This function assembles the element stiffness% matrix k of the space 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(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;




SpaceTrussElementForce(E,A,L,thetax,thetay,thetaz,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 = SpaceTrussElementForce(E,A,L,thetax,thetay,thetaz,u)%SpaceTrussElementForce This function returns the element force% given the modulus of elasticity E, the % cross-sectional area A, the length L, % the angles thetax, thetay, thetaz% (in degrees), and the element nodal % displacement vector u.x = thetax * pi/180;w = thetay * pi/180;v = thetaz * pi/180;Cx = cos(x);Cy = cos(w);Cz = cos(v);y = E*A/L*[-Cx -Cy -Cz Cx Cy Cz]*u;




SpaceTrussElementStress(E,L,thetax,thetay,thetaz,u)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.Function contents:function y = SpaceTrussElementStress(E,L,thetax,thetay,thetaz,u)%SpaceTrussElementStress 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.x = thetax * pi/180;w = thetay * pi/180;v = thetaz * pi/180;Cx = cos(x);Cy = cos(w);Cz = cos(v);y = E/L*[-Cx -Cy -Cz Cx Cy Cz]*u;



Consider the space truss shown below.The supports at nodes 1, 2, and 3 are ball-and-socket joints allowing rotation but no translation.Given E=200 GPaA14=0.001 m2

A24=0.002 m2

A34=0.001 m2

P=12 kNDetermine:a) The global stiffness matrix for the structureb) the displacement at node 4c) the reactions at nodes 1, 2, and 3d) the stress in each element

Solution of Example 1 with Matlab



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 three elements and four nodes. The units used in Matlab calculations are KN and meter. The element connectivity is:

E# N1 N2

1 1 4

2 2 4

3 3 4




Step 2-Copying relevant files and starting MatlabCreate a directory

Copy SpaceTrussElementLength.mSpaceTrussElementStiffness.mSpaceTrussAssemble.mSpaceTrussElementForce.mSpaceTrussElementStress.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:Enter the data>>E=200e6>>A1=0.001>>A2=0.002>>A3=0.001>>L1=SpaceTrussElementLength(0,0,-4,0,5,0)L1 =

6.4031

>>L2=SpaceTrussElementLength(-3,0,0,0,5,0)L2 =

5.8310>>L3=SpaceTrussElementLength(0,0,4,0,5,0)L3 =

6.4031




>>theta1x=acos(0/L1)*180/pitheta1x =

90>>theta1y=acos(5/L1)*180/pitheta1y =

38.6598>>theta1z=acos(4/L1)*180/pitheta1z =

51.3402>>theta2x=acos(3/L2)*180/pitheta2x =

59.0362




>>theta2y=acos(5/L2)*180/pitheta2y =

30.9638>>theta2z=acos(0/L2)*180/pitheta2z =

90>>theta3x=acos(0/L3)*180/pitheta3x =

90>>theta3y=acos(5/L3)*180/pitheta3y =

38.6598>>theta3z=acos(-4/L3)*180/pitheta3z =

128.6598




>>k1=SpaceTrussElementStiffness(E,A1,L1,theta1x,theta1y,theta1z)k1 = 1.0e+04 *

0.0000 0.0000 0.0000 -0.0000 -0.0000 -0.0000 0.0000 1.9046 1.5236 -0.0000 -1.9046 -1.5236 0.0000 1.5236 1.2189 -0.0000 -1.5236 -1.2189 -0.0000 -0.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000 -1.9046 -1.5236 0.0000 1.9046 1.5236 -0.0000 -1.5236 -1.2189 0.0000 1.5236 1.2189>>k2=SpaceTrussElementStiffness(E,A2,L2,theta2x,theta2y,theta2z)k2 = 1.0e+04 *

1.8159 3.0264 0.0000 -1.8159 -3.0264 -0.0000 3.0264 5.0441 0.0000 -3.0264 -5.0441 -0.0000 0.0000 0.0000 0.0000 -0.0000 -0.0000 -0.0000 -1.8159 -3.0264 -0.0000 1.8159 3.0264 0.0000 -3.0264 -5.0441 -0.0000 3.0264 5.0441 0.0000 -0.0000 -0.0000 -0.0000 0.0000 0.0000 0.0000




>>k3=SpaceTrussElementStiffness(E,A3,L3,theta3x,theta3y,theta3z)k3 =

1.0e+04 *

0.0000 0.0000 -0.0000 -0.0000 -0.0000 0.0000 0.0000 1.9046 -1.5236 -0.0000 -1.9046 1.5236 -0.0000 -1.5236 1.2189 0.0000 1.5236 -1.2189 -0.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000 -0.0000 -1.9046 1.5236 0.0000 1.9046 -1.5236 0.0000 1.5236 -1.2189 -0.0000 -1.5236 1.2189Step 4-Assembling the global stiffness matrix:Since the structure has 4 nodes, the size of the golbal stiffness matrix is 12x12.>>K=zeros(12,12)K = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0




>>K=SpaceTrussAssemble(K,k1,1,4)>>K=SpaceTrussAssemble(K,k2,2,4)>>K=SpaceTrussAssemble(K,k3,3,4)yields;K =

1.0e+04 *

0.0000 0.0000 0.0000 0 0 0 0 0 0 -0.0000 -0.0000 -0.0000 0.0000 1.9046 1.5236 0 0 0 0 0 0 -0.0000 -1.9046 -1.5236 0.0000 1.5236 1.2189 0 0 0 0 0 0 -0.0000 -1.5236 -1.2189 0 0 0 1.8159 3.0264 0.0000 0 0 0 -1.8159 -3.0264 -0.0000 0 0 0 3.0264 5.0441 0.0000 0 0 0 -3.0264 -5.0441 -0.0000 0 0 0 0.0000 0.0000 0.0000 0 0 0 -0.0000 -0.0000 -0.0000 0 0 0 0 0 0 0.0000 0.0000 -0.0000 -0.0000 -0.0000 0.0000 0 0 0 0 0 0 0.0000 1.9046 -1.5236 -0.0000 -1.9046 1.5236 0 0 0 0 0 0 -0.0000 -1.5236 1.2189 0.0000 1.5236 -1.2189 -0.0000 -0.0000 -0.0000 -1.8159 -3.0264 -0.0000 -0.0000 -0.0000 0.0000 1.8159 3.0264 0.0000 -0.0000 -1.9046 -1.5236 -3.0264 -5.0441 -0.0000 -0.0000 -1.9046 1.5236 3.0264 8.8532 0 -0.0000 -1.5236 -1.2189 -0.0000 -0.0000 -0.0000 0.0000 1.5236 -1.2189 0.0000 0 2.4378




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

The boundary conditions for the problem are;

z4

y4

x4

z3

y3

x3

z2

y2

x2

z1

y1

x1

4

4

4

3

3

3

2

2

2

1

1

1

F

F

F

F

F

F

F

F

F

F

F

F

w

v

u

w

v

u

w

v

u

w

v

u

K

0wvuwvuwvu 333222111

0F,0F,12F

0F,0F,0F,0F,0F,0F,0F,0F,0F

z4y4x4

z3y3x3z2y2x2z1y1x1




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 submatrix in rows 10 to 12and columns 10 to 12.

0

0

12

F

F

F

F

F

F

F

F

F

w

v

u

0

0

0

0

0

0

0

0

0

K

z3

y3

x3

z2

y2

x2

z1

y1

x1

4

4

4




Therefore we obtain;

The solution of the above system is obtained using Matlab as follows.Note that the ‘\’ operator is used for Gaussian elimination.

>>k=K(10:12,10:12)k =

1.0e+04 *

1.8159 3.0264 0.0000 3.0264 8.8532 0 0.0000 0 2.4378>>f=[12; 0; 0]f = 12 0 0

0

0

12

w

v

u

4.200

09.80.3

0.00.38.1

10

4

4

44




u=k\fu =

0.0015 -0.0005 -0.0000Step 7-Post-processing:In this step we obtain the reactions at nodes 1, 2, and 3 and the stress 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 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; 0 ; u]U =

0 0 0 0 0 0 0 0 0 0.0015 -0.0005 -0.0000




>>F=K*UF =

0.0000 10.0000 8.0000 -12.0000 -20.0000 -0.0000 0.0000 10.0000 -8.0000 12.0000 -0.0000 -0.0000so the recation forces are;at node 1: (0,10,8) Nat node 2: (-12,-20,0) Nat node 3: (0,10,-8) NObviously force equilibrium is satisfied for this problem.




Next we set up the element nodal displacement vectors u1, u2 and u3 then we calculate the element stresses sigma1, sigma2 and sigma3 by making calls to the Matlab function SpaceTrussElementStress.>> u1=[U(1) ; U(2) ; U(3) ; U(10) ; U(11); U(12)]u1 =

0 0 0 0.0015 -0.0005 -0.0000>> u2=[U(4) ; U(5) ; U(6) ; U(10) ; U(11); U(12)]u2 =

0 0 0 0.0015 -0.0005 -0.0000




>> u3=[U(7) ; U(8) ; U(9) ; U(10) ; U(11); U(12)]u3 =

0 0 0 0.0015 -0.0005 -0.0000

>>sigma1=SpaceTrussElementStress(E,L1,theta1x,theta1y,theta1z,u1)sigma1 =

-1.2806e+04


1.1662e+04






-1.2806e+04

Thus it is clear that the stresses:In element 1=12.806 MPa (compressive)In element 2=11.662 MPa (tensile)In element 3=12.806 MPa (compressive)

The symmetry in the results regarding elements 1 and 3 is obvious.

7-bar elements in 3-d space e-mail: dr. ahmet zafer Şenalp e-mail:...

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