5-Bar Element e-mail: Dr. Ahmet Zafer Şenalp e-mail: [email protected]@gmail.com Mechanical Engineering Department Gebze Technical University.

5-Bar Element

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

5-Bar Element

Cross section examples forbar structures

Consider a uniform prismatic bar:

L: lengthA: cross-sectional areaE: elastic modulusu=u(x): displacemente=e(x): strains=s(x): stress

Strain-displacement relation:

Stress-strain relation:

Bar Element


5-Bar Element

i j

FE Model

Assuming that the displacement u is varying linearly along the axis of the bar, i.e.,

The bar is acting like a spring in this case and we conclude that element stiffness matrix is:

Stiffness Matrix - Direct Method


5-Bar Element

D: elongation, F: force in bar

k: stiffness of the bar

We derive the same stiffness matrix for the bar using a formal approach which can be applied to many other more complicated situations:

Define two linear shape functions as follows:

B: Element strain-displacement matrix

Stiffness Matrix - A Formal Approach


5-Bar Element

Stress can be written as:

Consider the strain energy stored in the bar:

The work done by the two nodal forces is:

For conservative system, we state that:

which gives:



5-Bar Element

k is the element stiffness matrix. The above expression is a general result which can be used for the construction of other types of elements. This expression can also be derived using other more rigorous approaches, such as the Principle of Minimum Potential Energy, or the Galerkin’s Method.Now, we evaluate for the bar element

which is the same as we derived using the direct method. Strain energy in the element can be written as:



5-Bar Element

Number of components of the displacement vector at a node.For 1-D bar element: one dof at each node.

Physical Meaning of the Coefficients in k:The jth column of k (here j = 1 or 2) represents the forcesapplied to the bar to maintain a deformed shape with unitdisplacement at node j and zero displacement at the other node.

Degree of Freedom (dof)


5-Bar Element

Example 1


5-Bar Element

Find; (a) the global stiffness matrix(b) displacements of nodes (c) the reaction forces (d) stresses at the elements

Solution :

Connectivity table: E# N1 N2

1 1 2

2 2 3

1 3

FE Model

2

21

Example 1


5-Bar Element

Boundary conditions : Displacement boundary conditions :

Force boundary conditions :

a) Element Stiffness Matrices :

0u ,0u ,0u 321

0F ,PF ,0F 321

Example 1


5-Bar Element

Construction of global stiffness matrix :

Equilibrium (FE) equation for the whole system is;

Example 1


5-Bar Element

b) Applying boundary conditions; 0u ,0u ,0u 321 0F ,PF ,0F 321

Example 1


5-Bar Element

For the data given;

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244

mm

0

0.5208

0

u

u

u

3

2

1

Example 1


5-Bar Element

c) From the 1st and 3rd row of the finite element equation, the reaction forces can be calculated as;

d) Stresses in the elements:

N -16666.7F

N -33333.3F

2

1

MPa 33.83

MPa 33.83

2

1

· In this case, the calculated stresses in elements 1 and 2 are exact within the linear theory for 1-D bar structures. It will not help if we further divide element 1 or 2 intosmaller finite elements.· For tapered bars, averaged values of the cross-sectional areas should be used for the elements. · We need to find the displacements first in order to find the stresses, since we are using the displacement based FEM.

Notes


5-Bar Element


5-Bar Element

As linear bar element has 2 degrees of freedom (1 at each node) for a structure with n nodes, the global stiffness matrix K will be of size nxn.

The global stiffness matrix K is obtained by making calls to the Matlab function LinearBarAssemble 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 portioning and Gaussion elimination.

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

FUK

Solution procedure with matlab


5-Bar Element

where f is the 2x1 element force vector and u is the 2x1 element displacement vector.

The element stress is obtained by using local stiffness matrix and local displacement vector.

ukf

Solution procedure with matlab

Matlab functions used


5-Bar Element

LinearBarElementStiffness(E,A,L)This function returns the element stiffness matrix for a linear bar with modulus of elasticity E, cross-sectional area A, and length L. The size of the element stiffness matrix is 2 x 2. Function contents:function y = LinearBarElementStiffness(E,A,L) %LinearBarElementStiffness This function returns the element % stiffness matrix for a linear bar with % modulus of elasticity E, cross-sectional % area A, and length L. The size of the % element stiffness matrix is 2 x 2. y = [E*A/L -E*A/L ; -E*A/L E*A/L];



5-Bar Element

LinearBarAssemble(K,k,i,j)This function assembles the element stiffness matrix k of the linear bar 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 = LinearBarAssemble(K,k,i,j) %LinearBarAssemble This function assembles the element stiffness % matrix k of the linear bar 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(i,i) = K(i,i) + k(1,1); K(i,j) = K(i,j) + k(1,2); K(j,i) = K(j,i) + k(2,1); K(j,j) = K(j,j) + k(2,2); y = K;



5-Bar Element

LinearBarElementForces(k,u)This function returns the element nodal force vector given the element stiffness matrix k and the element nodal displacement vector u. Function contents:function y = LinearBarElementForces(k,u) %LinearBarElementForces This function returns the element nodal % force vector given the element stiffness % matrix k and the element nodal displacement % vector u. y = k * u;



5-Bar Element

LinearBarElementStressess_azLinearBarElementStresses_az(k, u, L, E)This function returns the element nodal stress vector given the element stiffness matrix k, the element nodal displacement vector u, the length of the element and EFunction contents:function y = LinearBarElementStresses_az(k, u, L, E) %LinearBarElementStresses This function returns the element nodal % stress vector given the element stiffness % matrix k, the element nodal displacement % vector u, the length of the element and EL_m=[-1/L 1/L];y = E*L_m*u;


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

Step 1-Discretizing the domain:This problem is already discretized. The domain is subdivided into 2 elements and 3 nodes. The units used in Matlab calculations are N and milimeter. The element connectivity is:

Solution of Example 1with Matlab

E# N1 N2

1 1 2

2 2 3

5-Bar Element



5-Bar Element

Step 2-Copying relevant files and starting MatlabCreate a directoryCopy LinearBarElementStiffness.m LinearBarAssemble.mLinearBarElementForces.mLinearBarElementStresses_az.m files under the created directoryOpen Matlab;Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory.

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244



5-Bar Element

Start solving the problem in Command Window:>>clearvars>>clcEnter the data>>E=2e4>>A(1)=400>>A(2)=200>>L=125

Step 3-Writing the element stiffness matrices:>>k1=LinearBarElementStiffness(E,A(1),L)k1 =

64000 -64000 -64000 64000>>k2=LinearBarElementStiffness(E,A(2),L)

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244



5-Bar Element

>>k2=LinearBarElementStiffness(E,A(2),L)k2 =

32000 -32000 -32000 32000

Step 4-Assembling the global stiffness matrix:

Since the structure has 3 nodes, the size of the global stiffness matrix is 3x3.Therefore to obtain K we first set up a zero matrix of size 3x3 then make 2 calls to the Matlab function LinearBarAssemble since we have 2 bar elements in the system.Each call to the function will assemble one element. The following are the Matlab commands:

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244



5-Bar Element

>>K=zeros(3,3)K =

0 0 0 0 0 0 0 0 0

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

64000 -64000 0 -64000 64000 0 0 0 0>>K=LinearBarAssemble(K,k2,2,3)K =

64000 -64000 0 -64000 96000 -32000 0 -32000 32000

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244



5-Bar Element

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

The boundary conditions for the problem are;

Inserting the above conditions into finite element equation

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244

3

2

1

3

2

1

F

F

F

u

u

u

3200032000-0

32000-9600064000-

064000-64000

3

1

2

F

50000

F

0

u

0

3200032000-0

32000-9600064000-

064000-64000

0u ,0u ,0u 321 0F ,PF ,0F 321



5-Bar Element

3

1

2

F

50000

F

0

u

0

3200032000-0

32000-9600064000-

064000-64000

50000u96000 2

0.5208u 2

>>u2=50000/96000

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;

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



5-Bar Element

>>U=[0; u2; 0]

U =

0 0.5208 0

>>F=K*U

F =

1.0e+04 *

-3.3333 5.0000 -1.6667



5-Bar Element

mm 125L ,mm 200A

mm/N 2x10E ,N 10x5P

2

244

>>u_el_1=[U(1); U(2)]>>u_el_2=[U(2); U(3)]>>stress1=LinearBarElementStresses_az(k1, u_el_1, L, E)>>stress2=LinearBarElementStresses_az(k2, u_el_2, L, E)

stress1 =

83.3333

stress2 =

-83.3333

:element stresses



5-Bar Element

m 0.002by

right the todisplayed is 3 node

m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

Create a directoryCopy LinearBarElementStiffness.m LinearBarAssemble.mLinearBarElementForces.mLinearBarElementStresses_az.m files under the created directoryOpen Matlab;Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory.



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

Start solving the problem in Command Window:>>clearvars>>clcEnter the data>>E=210e6>>A=0.003>>L1=1.5>>L2=1

Calculate stiffness matrices>>k1=LinearBarElementStiffness(E,A,L1)>>k2=LinearBarElementStiffness(E,A,L2)



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

Assemble the global stiffness matrix;>>K=zeros(3,3)>>K=LinearBarAssemble(K,k1,1,2)>>K=LinearBarAssemble(K,k2,2,3)K is calculated as:

K =

420000 -420000 0 -420000 1050000 -630000 0 -630000 630000



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

FE Equation is:

Apply BCs:

3

2

1

3

2

1

F

F

F

u

u

u

630000630000-0

630000-1050000420000-

0420000-420000

3

1

2

F

10000

F

002.0

u

0

3200032000-0

32000-9600064000-

064000-64000

002.0u ,0u ,0u 321 0F ,PF ,0F 321



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

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 row 2 and column 2 which turns to be a 1x1 matrix.

Because of the applied displacement of 0.002 m at node 3, we need to extract the submatrix in row 2 and column 3 which also turns out to be a 1x1 matrix.

Therefore we obtain;

10002.0630000u1050000 2



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

The solution of the above system is obtained using Matlab as follows. Note that the backslash operator ‘\’ is used for Gaussian elimination.>>k=K(2,2)>>k0=K(2,3)>>u0=0.002>>f=[-10]>>f0=f-k0*u0>>u=k\f0result is;u =

0.0012



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

In this step, we obtain the reactions at nodes 1 and 3, and the stress in each bar using Matlab as follows.

First we set up the global displacement vector; U, then we calculate the golbal force vector F.>>U=[0; u; u0]>>F=K*Uresults;F =

-500.0000 -10.0000 510.0000



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

Next step is to set up element nodal displacement vectors u1 and u2.Then we calculate the element force vectors f1 and f2 by making calls to the Matlab function LinearBarElementForces.>>u1=[0; U(2)]>>f1=LinearBarElementForces(k1,u1)result is;f1 =

-500.0000 500.0000



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

>>u2=[U(2); U(3)]>>f2=LinearBarElementForces(k2,u2)result is;f2 =

-510.0000 510.0000Finally, we call LinearBarElementStresses_az to calculate stresses.>>sigma1=LinearBarElementStresses_az(k1, u1, L1, E) results;sigma1 =

1.6667e+05



5-Bar Element

m 0.002by


m 003.0A

GPa 012E ,Nk 10P2

002.0u ,0u ,0u 321

0F ,PF ,0F 321

>>sigma2=LinearBarElementStresses_az(k2, u2, L2, E) results;sigma2 =

1.7000e+05

Example 3


5-Bar Element

Given;

Find; The support reaction forces at the two ends of the bar

Solution :

Connectivity table:E# N1 N2

1 1 2

2 2 3

Example 3


5-Bar Element

We first check to see if or not the contact of the bar with the wall on the right will occur. To do this, we imagine the wall on the right is removed and calculate the displacement at the right end,

BCs;

contact occurs

The global FE equation is found to be,

321 u ,0u ,0u 0F ,PF ,0F 321

Example 3


5-Bar Element

Applying boundary conditions; 321 u ,0u ,0u 0F ,PF ,0F 321

Example: 3


5-Bar Element

Reaction forces:



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321

contact occurs

BCs;



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321

Create a directoryCopy LinearBarElementStiffness.m LinearBarAssemble.mLinearBarElementForces.mLinearBarElementStresses_az.m files under the created directoryOpen Matlab;Open ‘Set Path’ command and by using ‘Add Folder’ command add the current directory.



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321 Start solving the problem in Command Window:>>clearvars>>clcEnter the data>>P=6e4>>E=2e4>>A=250>>L=150>>DELTA=1.2

Calculate stiffness matrices>>k1=LinearBarElementStiffness(E,A,L)>>k2=LinearBarElementStiffness(E,A,L)



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321 Assemble the global stiffness matrix;>>K=zeros(3,3)>>K=LinearBarAssemble(K,k1,1,2)>>K=LinearBarAssemble(K,k2,2,3)K is calculated as:

K =

1.0e+04 *

3.3333 -3.3333 0 -3.3333 6.6667 -3.3333 0 -3.3333 3.3333



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321

FE Equation is:

Apply BCs:

3

2

1

3

2

1

F

F

F

u

u

u

3333333333-0

33333-6666733333-

033333-33333

321 u ,0u ,0u 0F ,PF ,0F 321

3

1

2

F

60000

F

2.1

u

0

3333333333-0

33333-6666733333-

033333-33333



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321

600002.133333u66667 2

>>k=K(2,2)>>k0=K(2,3)>>u0=1.2>>f=[60000]>>f0=f-k0*u0>>u=k\f0result is;u =

1.5000



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321

>>U=[0; u; u0]>>F=K*Uresults;F =

-50000 60000 -10000



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321 >>u1=[0; U(2)]>>f1=LinearBarElementForces(k1,u1)results;f1 =

-50000 50000>>u2=[U(2); U(3)]>>f2=LinearBarElementForces(k2,u2)results;f2 =

10000 -10000



5-Bar Element

E# N1 N2

1 1 2

2 2 3

321 u ,0u ,0u

0F ,PF ,0F 321 >>sigma1=LinearBarElementStresses_az(k1, u1, L, E) results;sigma1 =

200>>sigma2=LinearBarElementStresses_az(k2, u2, L, E) results;sigma2 =

-40

Uniformly distributed axial load q (N/mm, N/m, lb/in) can be converted to two equivalent nodal forces of magnitude qL/2.We verify this by considering the work done by the load q,

Distributed Load


5-Bar Element

Thus, from the U=W concept for the element, we have:

The new nodal force vector is:

Distributed Load


5-Bar Element

Distributed Load


5-Bar Element

5-Bar Element e-mail: Dr. Ahmet Zafer Şenalp e-mail: [email protected]@gmail.com Mechanical Engineering Department Gebze Technical University.

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