E,I,L

ME 520

Fundamentals of Finite Element Analysis

8-Beam Element

Dr. Ahmet Zafer Şenalp

e-mail: azsenalp@gmail.com

Mechanical Engineering Department

Gebze Technical University

Simple Plane Beam Element 8-Beam Element

L: Length

I: Moment of inertia of the cross-sectional area

E: Elastic modulus v=v(x): Deflection (lateral displacement) of the neutral axis

  dv dx

: Rotation about the z-axis

F=F(x): Shear force

M=M(x): Moment about z-axis

ME 520


Mechanical Engineering Department, GTU 2

Simple Plane Beam Element 8-Beam Element

Elementary Beam Theory:

Direct Method

Using the results from elementary beam theory to compute each column of the stiffness matrix. Element stiffness equation (local node: i, j or 1, 2):

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Example 1 8-Beam Element

Find;

(a) The deflection and rotation at the center node

(b) the reaction forces and moments at the two ends

Solution:

Connectivity table:

E# N1 N2

1

2

1

2

2

3

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Example 1

Boundary conditions:

 Displacement boundary conditions: v

1



0 ,



1



0 , v

2



0 ,



2



0 , v

3



0 ,



3



0

 Force boundary conditions:

F

1 y



0 , M

1



0 , F

2 y

 

P , M

2



M , F

3 y



0 , M

3



0 a) Element Stiffness Matrices:

8-Beam Element

ME 520



Example 1

Global FE equation is:

8-Beam Element

Applying BC’s:

Reaction Forces:

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Stresses in the beam at the two ends can be calculated using the formula,

Note that the FE solution is exact according to the simple beam theory, since no distributed load is present between the nodes.

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Equivalent Nodal Loads of

Distributed Transverse Load

8-Beam Element

2 element case:

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Find ;

(a) The deflection and rotation at the right end

(b) The reaction force and moment at the left end.

Solution:

Connectivity table:

Equivalent system:

E# N1 N2

1 1 2

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Example 2



1



0 ,



1



0 , v

2



0 ,



2



0

 Force boundary conditions:

F

1 y



0 , M

1



0 , F

2 y

  f , M

2

 m

The structure FE equation:

8-Beam Element

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Example 2

Reaction forces:

8-Beam Element

This force vector gives the total effective nodal forces which include the equivalent

nodal forces for the distributed lateral load p given by :

The correct reaction forces can be obtained as follows,

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Example 3

Given;

Find ;

(a) Deflections, rotations

(b) reaction forces

Solution:

Connectivity table:

ME 520


E# N1 N2

1

2

3

1

2

3

2

3

4

Mechanical Engineering Department, GTU

8-Beam Element

12

Example 3



1



0 ,



1



0 , v

2



0 ,



2



0 , v

3



0 ,



3



0 , v

4



0

 Force boundary conditions :

F

1 y



0 , M

1



0 , F

2 y



0 , M

2



0 , F

3 y

 

P , M

3



0 , F

4 y



0

The spring stiffness matrix :

Adding this stiffness matrix to the global

FE equation:

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8-Beam Element

13

Example 3

Aplying BC’s:

8-Beam Element

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Example 3

Reaction Forces:

Checking the results: Draw free body diagram of the beam:

8-Beam Element

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Solution procedure with matlab 8-Beam Element

It is clear that the beam element has 4 degrees of freedom (2 at each node)

The sign convension used is that the displacement is positive if it points upwards and the rotation is positive if it is counterclockwise.

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

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.

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Solution procedure with matlab 8-Beam Element

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

 



    where f is the 4x1 nodal force vector in the element and u is the 4x1 element displacement vector.

  displacement and rotation, respectively, at the first node, while the third and

  rotation, respectively, at the second node.

ME 520



Matlab functions used 8-Beam Element

The 5 Matlab functions used for the beam element are:

BeamElementStiffness(E,I,L)

This function returns the element stiffness matrix for a beam element with modulus of elasticity E, moment of inertia I, and length L. The size of the element stiffness matrix is 4 x 4.

Function contents: function y = BeamElementStiffness(E,I,L)

%BeamElementStiffness This function returns the element

% stiffness matrix for a beam

% element with modulus of elasticity E,

% moment of inertia I, and length L.

% The size of the element stiffness

% matrix is 4 x 4.

y = E*I/(L*L*L)*[12 6*L -12 6*L ; 6*L 4*L*L -6*L 2*L*L ;

-12 -6*L 12 -6*L ; 6*L 2*L*L -6*L 4*L*L];

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BeamAssemble(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 2nx2n global stiffness matrix K after the element stiffness matrix k is assembled.

Function contents: function y = BeamAssemble(K,k,i,j)

%BeamAssemble 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.

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Matlab functions used

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;

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8-Beam Element

20


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 4x1 element force vecor f

Function 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;

ME 520




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' )

ME 520




BeamElementMomentDiagram(f, L)

This function plots the bending moment diagram for the beam element with nodal force 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' )

ME 520



Solution of Example 4 with Matlab

Consider the beam as shown

Given

E=210 GPa

I=60x10 -6 m 4

P=20 kN

L=2 m

Determine: a) the global stiffness matrix for the structure b) vertical displacement at node 2 c) rotations at nodes 2 and 3 d) the reactions at nodes 1 and 3 e) the forces (shears and moments) in each element f) the shear force diagram for each element g) the bending moment diagram for each element

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8-Beam Element

24

Solution of Example 4 with Matlab 8-Beam Element

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

ME 520




Step 2-Copying relevant files and starting Matlab

Create a directory

Copy

BeamElementStiffness.m

BeamAssemble.m

BeamElementForces.m

BeamElementShearDiagram.m

BeamElementMomentDiagram.m

files 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

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Step 3-Writing the element stiffness matrices:

The two element stiffness matrices k

1 and k

2 are obtained by making calls to the

Matlab function BeamElementStiffness. Each matrix has size 4x4.

Enter the data

>>E=210e6

>>I=60e-6

>>L=2

>>k1=BeamElementStiffness(E,I,L) k1 =

18900 18900 -18900 18900

18900 25200 -18900 12600

-18900 -18900 18900 -18900

18900 12600 -18900 25200

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>>k2=BeamElementStiffness(E,I,L) k2 =

18900 18900 -18900 18900

18900 25200 -18900 12600

-18900 -18900 18900 -18900

18900 12600 -18900 25200

Step 4-Assembling the global stiffness matrix:

Since the structure has 3 nodes, the size of the global stiffness matrix is 6x6.

>>K=zeros(6,6)

>>K=BeamAssemble(K,k1,1,2)


K =

18900 18900 -18900 18900 0 0

18900 25200 -18900 12600 0 0

-18900 -18900 37800 0 -18900 18900

18900 12600 0 50400 -18900 12600

0 0 -18900 -18900 18900 -18900

0 0 18900 12600 -18900 25200

ME 520




Step 5-Applying the boundary conditions:

Finite element equation for the problem is;

 







 v

 v

1

1

2 





 v

2







3

3





































F

1 y

M

F

M

F

2

3

 M

1 y

2 y

3



















The boundary conditions for the problem are; v

1



0 ,



1



0 , v

2



0 ,



2



0 , v

3



0 ,



3



0

F

1 y



0 , M

1



0 , F

2y

 

20 , M

2



0 , F

3y



0 , M

3



0

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8-Beam Element

29


Inserting the above conditions into finite element equation

10

3





















18

18

18

18

0

0

.

.

.

9

9

.

9

9

18 .

9

25 .

2



18 .

9

12 .

6

0

0



18 .

9



18 .

9

37 .

8

0



18 .

9

18 .

9

18 .

9

12 .

6

0

50 .

4



18 .

9

12 .

6

0

0



18 .

9



18 .

9

18 .

9



18 .

9



18

12

18

25

0

0

.

.

9

6

.

.

9

2

























 v

0

0

2 











2



0

3



























 









 0

F

3

F

1 y

M

1

20









0 y











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 3 to 4 and column 6, row 6 and columns 3 to 4, and row 6 and column 6. Therefore we obtain:

10

3









37

18

0

.

8

.

9

0

50 .

4

12 .

6

18

12

25

.

.

.

9

6

2









 u









2

2

3















20

0

0







ME 520




The solution of the above system is obtained using Matlab as follows.

Note that the ‘\’ operator is used for Gaussian elimination.

>>k=[K(3:4,3:4) K(3:4,6) ; K(6,3:4) K(6,6)] k =

37800 0 18900

0 50400 12600

18900 12600 25200

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

-20

0

0

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8-Beam Element

31


>>u=k\f u =

1.0e-03 *

-0.9259

-0.1984

0.7937

It is now clear that the vertical displacement at node 2=0.0009259 m (downward) rotation at node 2 =0.0001984 rad (clockwise) rotation at node 3 =0.0007937 rad (counterclockwise)

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.

ME 520




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

U =

1.0e-03 *

0

0

-0.9259

-0.1984

0

0.7937

>>F=K*U

F =

13.7500

15.0000

-20.0000

0

6.2500

-0.0000

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8-Beam Element

33

Solution of Example 4 with Matlab 8-Beam Element thus the recations are;

Force at node 1=13.75 kN

Moment at node 1=15 kNm (countereclockwise)


Next we set up the element nodal displacement vectors u

1 calculate the element force vectors f

1 and f

2 and u

2 then we by making calls to the Matlab function

BeamElementForces.

>> u1=[U(1) ; U(2) ; U(3) ; U(4)] u1 =

1.0e-03 *

0

0

-0.9259

-0.1984

ME 520




>> u2=[U(3) ; U(4) ; U(5) ; U(6)] u2 =

1.0e-03 *

-0.9259

-0.1984

0

0.7937

>>f1 =BeamElementForces(k1,u1) f1 =

13.7500

15.0000

-13.7500

12.5000

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8-Beam Element

35



-6.2500

-12.5000

6.2500

-0.0000

Shear force at centilever region=13.75 kN

Bending moment at centilever region=15 kNm

Shear force at pin joint=6.25 kN

Finally we call the Matlab functions BeamElementShearDiagram and

BeamElementMomentDiagram, respectively for each element.

8-Beam Element

ME 520




>>BeamElementShearDiagram(f1,L)

8-Beam Element

ME 520




>>BeamElementShearDiagram(f2,L)

8-Beam Element

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>>BeamElementMomentDiagram(f1, L)

8-Beam Element

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>>BeamElementMomentDiagram(f2, L)

8-Beam Element

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Consider the beam as shown

Given

E=210 GPa

I=5x10 -6 m 4 w=7 kN/m

Determine: a) the global stiffness matrix for the structure b) rotations at nodes 1, 2 and 3 c) the reactions at nodes 1, 2, 3 and 4 d) the forces (shears and moments) in each element e) the shear force diagram for each element f) the bending moment diagram for each element

ME 520



8-Beam Element

41


Solution:

Step 1-Discretizing the domain:

We need first to replace the distributed loading on element 2 by equivalent nodal loads. This is performed as follows for element 2 with a uniformly distributed load.

The resulting beam with eqivalent nodal load is shown below:

 

2

 

















  wL

2 wL

12

 wL

2 wL

2

2

12































-





14

14 kN

9.333

kNm kN

9.333

kNm









(**)

ME 520




The units used in Matlab calculations are kN and meter.

The element connectivity is:

2

3

E# N1 N2

1 1 2

2

3

3

4

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Step 2-Copying relevant files and starting Matlab

Create a directory

Copy

BeamElementStiffness

BeamAssemble

BeamElementForces

BeamElementShearDiagram

BeamElementMomentDiagram files 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

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Step 3-Writing the element stiffness matrices:

The two element stiffness matrices k

1 and k

2 are obtained by making calls to the

Matlab function BeamElementStiffness. Each matrix has size 4x4.

Enter the data

>>E=210e6

>>I=5e-6

>>L1=3

>>L2=4

>>L3=2

>>k1=BeamElementStiffness(E,I,L1) k1 =

1.0e+03 *

0.4667 0.7000 -0.4667 0.7000

0.7000 1.4000 -0.7000 0.7000

-0.4667 -0.7000 0.4667 -0.7000

0.7000 0.7000 -0.7000 1.4000

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45



1.0e+03 *

0.1969 0.3937 -0.1969 0.3937

0.3937 1.0500 -0.3937 0.5250

-0.1969 -0.3937 0.1969 -0.3937

0.3937 0.5250 -0.3937 1.0500


1575 1575 -1575 1575

1575 2100 -1575 1050

-1575 -1575 1575 -1575

1575 1050 -1575 2100

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8-Beam Element

46


Step 4-Assembling the global stiffness matrix:

Since the structure has 4 nodes, the size of the global stiffness matrix is 8x8.

>>K=zeros(8,8)




K =

1.0e+03 *

0.4667 0.7000 -0.4667 0.7000 0 0 0 0

0.7000 1.4000 -0.7000 0.7000 0 0 0 0

-0.4667 -0.7000 0.6635 -0.3063 -0.1969 0.3937 0 0

0.7000 0.7000 -0.3063 2.4500 -0.3937 0.5250 0 0

0 0 -0.1969 -0.3937 1.7719 1.1812 -1.5750 1.5750

0 0 0.3937 0.5250 1.1812 3.1500 -1.5750 1.0500

0 0 0 0 -1.5750 -1.5750 1.5750 -1.5750

0 0 0 0 1.5750 1.0500 -1.5750 2.1000

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47


Step 5-Applying the boundary conditions:

Finite element equation for the problem is;

 





 v





 v



2

1

1

2







 v



3

3



 v



4

4































F

1 y

M





F

2

M y

1













F

3

M

F

4

M

2 y

3 y

4























The boundary conditions for the problem are; v

1



0 ,



1



0 , v

2



0 ,



2



0 , v

3



0 ,



3



0 , v

4



0 ,



4



0

F

1 y



0 , M

1



0 , F

2y

 

14 , M

2

 

9 .

33 , F

3y

 

14 , M

3



9 .

33 , F

4 y



0 , M

4



0

8-Beam Element

ME 520




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 2, 4 and 6 and columns2, 4 and 6. Therefore we obtain:

10 3









0

1 .

4

.

70

0

0 .

7

2 .

45

0 .

53

0 .

3 .

0

53





15













2



3

1













9

9 .

333

.

0

333





The solution of the above system is obtained using Matlab as follows.

Note that the ‘\’ operator is used for Gaussian elimination.

>>k=[K(2,2) K(2,4) K(2,6) ; K(4,2) K(4,4) K(4,6) ; K(6,2) K(6,4) K(6,6)] k =

1400 700 0

700 2450 525

0 525 3150

ME 520




>>f=[0 ; -9.333 ; 9.333] f =

0

-9.3330

9.3330

>>u=k\f u =

0.0027

-0.0054

0.0039

It is now clear that rotation at node 1 =0.0027 rad (counterclockwise) rotation at node 2 =0.0054 rad (clockwise) rotation at node 3 =0.0039 rad (counterclockwise)

ME 520



8-Beam Element

50


Step 7-Post-processing:

In this step we obtain the reactions at nodes 1, 2, 3 and 4 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 ;u(1) ;0 ; u(2) ; 0 ; u(3); 0 ; 0]

U =

0

0.0027

0

-0.0054

0

0.0039

0

0

ME 520




>>F=K*U

F =

-1.8937

-0.0000

1.2850

-9.3330

6.6954

9.3330

-6.0867

4.0578

thus the recations are;

Force at node 1=-1.8937 kN



Force at node 4=-6.0867 kN

Moment at node 4 (at fixed support)=4.0578 kNm (counterclockwise)

ME 520



8-Beam Element

52


Next we set up the element nodal displacement vectors u calculate the element force vectors f

1

, f

2 and f

3

1

, u

2 and u

3 then we by making calls to the Matlab function BeamElementForces.

>> u1=[U(1) ; U(2) ; U(3) ; U(4)] u1 =

0

0.0027

0

-0.0054

>> u2=[U(3) ; U(4) ; U(5) ; U(6)] u2 =

0

-0.0054

0

0.0039

>> u3=[U(5) ; U(6) ; U(7) ; U(8)] u3 =

0

0.0039

0

0

ME 520





-1.8937

-0.0000

1.8937

-5.6810


-0.6087

-3.6520

0.6087

1.2173


6.0867

8.1157

-6.0867

4.0578

ME 520



8-Beam Element

54


Note that the forces for element 2 need to be modified because of the distributed load. In order to obtain the correct forces for element 2 we need to subtract from f2 the vector of equivalent nodal loads given in equation (**). This is performed using Matlab as follows:

>>f2=f2-[-14 ; -9.333 ; -14 ; 9.333] f2 =

13.3913

5.6810

14.6087

-8.1157

Element 1 has a shear force of -1.8937 kN and a bending moment of 0 kNm at its left end while it has a shear force of 1.8937 kN and a bending moment of -5.6810 kNm at its right end.

Element 2 has a shear force of 13.3913 kN and a bending moment of 5.6810 kNm at its left end while it has a shear force of 14.6087 kN and a bending moment of -

8.1157 kNm at its right end.

ME 520




Element 3 has a shear force of 6.0867 kN and a bending moment of 8.1157 kNm at its left end while it has a shear force of -6.0867 kN and a bending moment of

4.0578 kNm at its right end.

Obviously the roller at the left end has zero moment.

Finally we call the Matlab functions BeamElementShearDiagram and

BeamElementMomentDiagram, respectively for each element.

ME 520




>>BeamElementShearDiagram(f1,L1)

8-Beam Element

ME 520





8-Beam Element

ME 520





8-Beam Element

ME 520




>>BeamElementMomentDiagram(f1, L1)

8-Beam Element

ME 520





8-Beam Element

ME 520





8-Beam Element

ME 520



E,I,L

ME 520

Fundamentals of Finite Element Analysis

8-Beam Element

Dr. Ahmet Zafer Şenalp

Mechanical Engineering Department

Gebze Technical University

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E,I,L

ME 520

Fundamentals of Finite Element Analysis

8-Beam Element

Dr. Ahmet Zafer Şenalp

Mechanical Engineering Department

Gebze Technical University

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