lecture 27 –introduction to finite elements...

Lecture 27 – Introduction to finite elements methods

Instructor: Prof. Marcial Gonzalez

Fall, 2021ME 323 – Mechanics of Materials

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Last modified: 8/16/21 9:23:37 AM

Introduction to finite element methods (FEM)

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Finite element methods - Finite element analysis (FEA)Every structure studied in ME323 and much more …


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Finite element methods - Finite element analysis (FEA)

- It is an energy method that can handle any geometryand any three-dimensional states of stress.

- It is not limited to elastic materials and small deformations.

- FEA is a skill of high-demand in the job market and it isalways considered a plus!


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Finite element methods – Three dimensional elements- Example: Four point bending

L=100 in.E=29500 ksiν=0.3P=4.5 kip 10 in.

20 in.

Beam cross-section


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Finite element methods – Three dimensional elements- Example: Four point bending

Normal stressesmax

0

min


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Determine the principal stresses and the maximum shear stress at point A (i.e., the point on top of the wrench handle). The diameter of the circular crosssection is 12.5 mm.

Finite element methods – Three dimensional elements- Example: Combined loads (Lecture 39)


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max

0

min

max

0

min

max

min

max

min


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von Mises equivalent stress


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Finite element methods – One-dimensional rod elements- Review of axial deformations and rods

+ Elongation of the rod

+ Equivalent stiffness

Good approximation when


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+ Elastic strain energy

+ Stiffness matrix


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+ Three nodes+ Three degrees of freedom + Two elements+ Each element has two nodes

and two degrees of freedom


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+ Three nodes+ Three degrees of freedom + Two elements+ Each element has two nodes

and two degrees of freedom



Strainenergy

+ (N+1) nodes+ (N) elements

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Global stiffness matrix(symmetric matrix)

(combination of elemental stiffness matrices)

+ (N+1) nodes+ (N) elements


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Finite element methods – One-dimensional rod elements- We obtain the equilibrium solution using an energy principle

Principle of minimum potential energy“For a given set of admissible displacement fields for a conservative system,an equilibrium state of the system will correspond to a state for which thetotal potential energy is stationary.”

+ An admissible displacement field for a rod is one that satisfies all of thedisplacement boundary conditions of the problem.

+ The total potential energy of the system is equal to the sum of the potential ofthe applied external forces and the strain energy in the rod.

+ Stationarity of the potential energy correspond to its minimization with respectto the displacement field.

for each node in the mesh


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+ An admissible displacement field for a rod is one that satisfies all of thedisplacement boundary conditions of the problem.

Some displacements in are going to be zero. We will enforce theseconditions after the minimization of the potential energy.

(similar to Castigliano’ssecond therorem)


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+ The total potential energy of the system is equal to the sum of the potential ofthe applied external forces and the strain energy in the rod.


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+ Stationarity of the potential energy correspond to its minimization with respectto the displacement field.

In general: recall …

for each node in the mesh

for each node in the mesh, is equivalent to: (a linear system of N+1 equations)


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Finite element methods – One-dimensional rod elements- Example 55:

Number of nodes: 4

Number of elements: 3

Boundary conditions:

Stiffness of each element:


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Finite element methods – One-dimensional rod elements- Example 55, solved in 5 steps+ Step #1: Identify the degrees of freedom

+ Step #2: Build the global stiffness matrix

Number of nodes: 4Number of elements: 3

-

-


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Finite element methods – One-dimensional rod elements- Example 55, solved in 5 steps+ Step #3: Enforce boundary conditions

+ Step #4: Solve the reduced system of linear equations

Number of nodes: 4Number of elements: 3


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Finite element methods – One-dimensional rod elements- Example 55, solved in 5 steps+ Step #5: Recover the reaction at the supports


Finite element methods – One-dimensional rod elements- Example 56, using MATLAB:

clear% set number of elementsN=3;%define elemental propertiesEA=[1/4;1;9/4];L=[1;1;1];%set up forcing vectorF=[0;2;1;0];%define boundary conditionsBC=[1;0;0;1];

%set up global stiffness matrixk=EA./L;K=zeros(N+1,N+1);for ii=1:N

K(ii,ii)=K(ii,ii)+k(ii);K(ii+1,ii)=K(ii+1,ii)-k(ii);K(ii,ii+1)=K(ii,ii+1)-k(ii);K(ii+1,ii+1)=K(ii+1,ii+1)+k(ii);

end

%enforce BC's on [K] and {F}K_reduced = K;F_reduced = F;for jj=N+1:-1:1

if BC(jj)==1K_reduced(jj,:)=[];K_reduced(:,jj)=[];F_reduced(jj)=[];

endend%solve reduced system of equationsu_reduced=inv(K_reduced)*F_reduced;%determine reaction at supportsr=1;for jj=1:N+1

if BC(jj)==1nodal_u(jj,1) = 0;

elsenodal_u(jj,1) = u_reduced(r);r=r+1;

endenddisp('Nodal displacement'); disp(nodal_u');disp('Nodal force'); disp((K*nodal_u)');


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Finite element methods – One-dimensional rod elements- Example 56, using MATLAB:

clear% set number of elementsN=3;%define elemental propertiesEA=[1/4;1;9/4];L=[1;1;1];%set up forcing vectorF=[0;2;1;0];%define boundary conditionsBC=[1;0;0;1];

Output:

Nodal displacement 0 2.4490 1.0612 0

Nodal force-0.6122 2.0000 1.0000 -2.3878


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Finite element methods – One-dimensional rod elements- Can we use the same strategy for any other geometry?

Yes! This is called a ‘discretization’ of the object into elements.


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Finite element methods – Three-dimensional elements- Can we use the same strategy for any other geometry?

Yes! This is called a ‘discretization’ of the object into elements.

+ 8 nodes per element+ 3 degrees of freedom per node+ 24 degrees of freedom per element


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Any questions?

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lecture 27 –introduction to finite elements...

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