1Apr 20, 2023

Last course Lecture plan and policies What is FEM? Brief history of the FEM Example of applications Discretization Example of FEM softwares Why study the theory of finite elements? Computational steps Introduction to Matlab Homework 1

2Apr 20, 2023

Today Stiffness equations for the linear

spring element and the bar element Application of the direct stiffness

method for structural systems made up of springs and bars in 1D space

Interpretation and properties of the stiffness matrix

Discretize the structure (problem domain) Divide the structure or continuum into finite elements

Once the structure has been discretized, the computational steps faithfully follow the steps in the direct stiffness method.

The direct stiffness method (DSM): The global stiffness matrix of the discrete structure are obtained

by superimposing (assembling) the stiffness matrices of the element in a direct manner.

Computational steps of the FEM- the direct stiffness method

Idealization Process

Source: C. Felippa (2012)

DSM: Breakdown Steps

Source: C. Felippa (2012)

DSM: Assembly & Solution Steps

Source: C. Felippa (2012)

72023.04.20

Any question before we discuss the simplest element, i.e. spring element?

Linear spring

Linear spring: Obeys Hooke’s law:

F : applied axial loadδ : elongation or contraction of the spring (deformation)

ks: spring constant (stiffness)

Resists forces only in the direction of the spring

F

F = ksδ

Linear spring as a finite element (1)

Points 1 and 2 are the finite element nodes. x: element (local) coordinate system f1, f2: element nodal forces, shown in the positive sense d1, d2: element nodal displacement, shown in the

positive sense Now let’s find the relationship between the nodal forces

and the nodal displacement when the spring is in equilibrium state.

f1, d1 f2, d2

x1 2

ks

Linear spring as a finite element (2)

First, let d1>0 and d2=0 (compression)

Nodal forces consistent with static equilibrium and Hooke’s law are

1 s 1f k d 2 s 1f k d

f1 f21 2

d1

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Linear spring as a finite element (3)

Next, let d1=0 and d2>0 (tension)

1 s 2f k d 2 s 2f k d

f1 f21 2

d2

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If both d1>0 and d2>0

In matrix format

1 s 1 2( )f k d d

2 s 1 2( )f k d d

1 1s

2 2

1 1

1 1

d fk

d f

orkd f

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This is equation is called the (discrete) element equilibrium equation or the element stiffness equation, refer to the element (local) coordinate system

kd f

s

1 1

1 1k

k

Spring element stiffness matrix

Linear spring as a finite element (4)

Example of a spring assemblage (1)

1, 2, 3: structural (global) node numbers (as opposed to element or local nodes)

X: global coordinate system F2 and F3 are the given applied forces, while F1 is

the unknown reaction force. D2 and D3 are the unknown nodal displacement,

while D1 is the given nodal displacement (in this case D1=0).

1 2 31 2

X

F1, D1

k1 k2F2, D2 F3, D3

Example of a spring assemblage (2)

The number of degree of freedom (DOF)= 3. The number of active DOF= 2.

We will establish the stiffness equation for the spring assemblage by using the element stiffness equations 1 and 2.

In this problem the local axes coincide with the global axis.

1 2 31 2

X

F1, D1

k1 k2F2, D2 F3, D3

Example of a spring assemblage (3)

Equilibrium condition for element 1:

Equilibrium condition for element 2:

12f

x1 2

k111f

11d

12d

1 11 1 1 1

1 11 1 2 2

k k d f

k k d f

22f

x1 2

k221f

21d

22d

2 22 2 1 1

2 22 2 2 2

k k d f

k k d f

Example of a spring assemblage (4)

Continuity or displacement compatibility condition:

1 2 31 2

X

F1, D1

k1 k2F2, D2 F3, D3

12f

x1 2

k111f

11d

12d

1 22 1 2d d D

11 1d D

22 3d D

22f

x1 2

k221f

21d

22d

The equilibrium equations expressed in terms of the global nodal displacements:

Example of a spring assemblage (5)

1 11 1 1 1

1 11 1 2 2

k k d f

k k d f

22 2 2 1

22 2 3 2

k k D f

k k D f

11 1 1 1

11 1 2 2

k k D f

k k D f

2 22 2 1 1

2 22 2 2 2

k k d f

k k d f

Expanding the matrix equations to the size of the DOF:

11 1 1 1

11 1 2 2

3

0

0

0 0 0 0

k k D f

k k D f

D

12

2 2 2 12

2 2 3 2

0 0 0 0

0

0

D

k k D f

k k D f

22 2 2 1

22 2 3 2

k k D f

k k D f

11 1 1 1

11 1 2 2

k k D f

k k D f

1 2 31 2

D1 D2 D3

1 2 31 2

D1 D2 D3

The addition of the two equations yields

11 1 1 1

1 21 1 2 2 2 2 1

22 2 3 2

0

0

k k D f

k k k k D f f

k k D f

Example of a spring assemblage (7)

The equilibrium conditions for nodes 1, 2, and 3:

12f

x1 2

k111f

11d

12d

22f

x1 2

k221f

21d

22d

Example of a spring assemblage (8)

11f1F 1

1 22 1f f

2F222f 3F3

11 1f F 1 2

2 1 2f f F 22 3f F

Example of a spring assemblage (9)

Substituting the last equations, we obtain:

or

1 1 1 1

1 1 2 2 2 2

2 2 3 3

0

0

k k D F

k k k k D F

k k D F

KD F

Structural or global nodal force matrix

Structural or global nodal displacement matrix

Structural or global stiffness matrix

To facilitate the assembling of the global stiffness equation, let us define a matrix in which each column contains the numerical labels of the global nodal displacement associated with each element.

Let call this matrix “code matrix”, denote Mcode

The size of Mcode is Ndofe x Nelem, where Ndofe: number of degrees of freedom per element Nelem: total number of elements in the structure

Assembling process (1)

Assembling process (2)

1 2 31 2

X

F1, D1

k1 k2F2, D2 F3, D3

1 2

d1 d2

Assembling process (3)

The rows and columns of each stiffness matrix are labeled according to the DOFs associated with them; the labels are copied from Mcode.

1 1

1 1

k k

k k

1 2

1

2

2 2

2 2

k k

k k

2 32

3

1 1

1 1 2 2

2 2

0

0

k k

k k k k

k k

1 2 3

1

2

3

Boundary Conditions (1)

Given F1, F2 and F3, can you now solve this equation to obtain D1, D2, and D3?

The support or boundary conditions must be specified, otherwise K will be singular.

Types of boundary conditions: Homogeneous b.c.’s: zero values of nodal displacement are

specified. Non-homogeneous b.c.’s: nonzero values are specified.

1 1 1 1

1 1 2 2 2 2

2 2 3 3

0

0

k k D F

k k k k D F

k k D F

Example of homogeneous b.c.’s: D1=0 The global stiffness equation can be written as

The unknown values: F1, D2, D3

1 1 1

1 1 2 2 2 2

2 2 3 3

0 0

0

k k F

k k k k D F

k k D F

Boundary Conditions (2)

Boundary Conditions (3)

Considering the 2nd and the 3rd equations, we can write the matrix equation as

This equation is called reduced stiffness equation. For all homogeneous b.c.’s, the reduced stiffness equation can be

obtained by simply deleting the rows and columns corresponding to the zero-displacement DOF’s from the original stiffness equation.

2 21 2 2

3 32 2

D Fk k k

D Fk k

Boundary Conditions (4)

Solving the reduced stiffness equation, we can obtain

Substituting D2 and D3 into the original stiffness equation, we obtain

2 32

1

F FD

k

2 3 3

31 2

F F FD

k k

1 2 3( )F F F

Example of nonhomogeneous b.c.’s: D1=δ The global stiffness equation can be written as

The unknown values: F1, D2, D3

1 1 1

1 1 2 2 2 2

2 2 3 3

0

0

k k F

k k k k D F

k k D F

Boundary Conditions (5)

Boundary Conditions (6)

Considering the 2nd and the 3rd equations,

or

When dealing with non-homogeneous b.c.’s, we must transform the terms associated with the known displacements to the right-side force matrix.

2 21 2 21

3 32 20

D Fk k kk

D Fk k

2 21 2 2 1

3 32 2 0

D Fk k k k

D Fk k

Boundary Conditions (7)

In general, specified boundary conditions are treated by partitioning the global stiffness equation as follows:

Subscript a: associated with active (free, unconstrained) DOF’s Subscript p: associated with passive (locked, constrained) DOF’s Unknown vectors: Da, Fp

aa ap a a

pa pp p p

K K D F

K K D F

Boundary Conditions (8)

From this equation, we have

aa ap a a

pa pp p p

K K D F

K K D F

aa a a ap p K D F K D

p pa a pp p F K D K D

Example

Given a system of three springs supporting three equal weights W.

W=1 kN, k= 0.2 kN/mm Treating the springs as finite

elements, determine the vertical displacement of each weight and the support reaction.

Repeat the problem if the second weight is “forced” to move 5 mm upward.

3k

2k

W

W

k

W

Homework 2 (due date next class)

The spring stiffness k1=1 kN/mm, k2=2 kN/mm, k3=3 kN/mm, k4=4 kN/mm, k5=5 kN/mm.

Applying a force of 2 kN at node 2 in the positive x direction, determine the nodal displacements, the reactions, and the forces in each element. Use the direct stiffness method, implemented in Matlab.

Instead of applying the force, now at node 2 given a fixed, known displacement – 2 mm. Repeat the calculations.

Any questions before we proceed to 1D bar element?

Linear-elastic Bar-- Assumptions

The bar is geometrically straight. The material obeys Hooke’s law, i.e.

A: sectional area of the bar E: modulus of elasticity of the material σ: axial stress in the bar, σx = F/A

εx: axial strain of the bar

x xE FF

A, E

Linear-elastic Bar-- Assumptions

Forces are applied only at the ends of the bar. The bar supports axial loading only.

Bending, torsion, and shear are not transmitted to the element

FFx xE

Bar element (1)

Now let’s find the relationship between the nodal forces and the nodal displacement when the spring is in equilibrium state.

f1, d1 f2, d2

x1 2

EA

L

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Bar element (2)

First, let d1>0 and d2=0 (compression)

Nodal forces consistent with static equilibrium and Hooke’s law are

1 1

EAf d

L 2 1

EAf d

L

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Bar element (3)

Next, let d1=0 and d2>0 (tension)

1 2

EAf d

L 2 2

EAf d

L

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If both d1>0 and d2>0

In matrix format

1 1 2( )EA

f d dL

2 1 2( )EA

f d dL

1 1

2 2

1 1

1 1

d fEA

d fL

orkd f

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Example – A bar structure (1)

The area of a cross section at a distance x:

A=A(x)

0x Tx L

03A A0A A

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Example – A bar structure (2) Discretization

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Example – A bar structure (3) Generate element stiffness matrices

Apply the element stiffness matrix formula to the bar problem

1 0

T

1 15

1 1

EA

L

k 2 0

T

1 13

1 1

EA

L

k

Element 1 Element 2

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Example – A bar structure (4) Element 1

D1 D2 D3

1 0

T

1 1 05

1 1 0

0 0 0

EA

L

K

D1 D2 D3

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Example – A bar structure (4) Element 2

D1 D2 D3

D1 D2 D3

2 0

T

0 0 03

0 1 1

0 1 1

EA

L

K

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Example – A bar structure (5) All elements are present

1 2 0

T

5 5 0

5 8 3

0 3 3

EA

L

K K K

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Example – A bar structure (6) Applying the loads

Nodal force vector

0

H

F

F

Unknown reaction force

D1

D2

D3

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Example – A bar structure (7) Imposing the displacement boundary

conditions D1=0

In consequence, now there are only two unknown parameters D2 and D3

Thus the “active” DOF=2

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Solving the global equation,

The unknowns are D2, D3, and H

Discarding row 1 and column 1 from K, we obtain

02

T3

5 5 0 0

5 8 3 0

0 3 3

HEA

DL

D F

20

3T

8 3 0

3 3

DEA

D FL

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Solution

Support reaction

Example – A bar structure (9)

2 T

3 0

3/15

8 /15

D FL

D EA

02

T

5EA

D HL

T2

05

FLD

EA

Thus H F

Agree with the global equilibrium

F F

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Example – A bar structure (10) “Disassembly” of the structural nodal

displacement

1

T2

03

0

3/15

8 /15

DFL

DEA

D

D

111 1 T

12 02

0

3/15

Dd FL

D EAd

d

222 1 T

23 02

3/15

8 /15

Dd FLD EAd

d

Element 2

Element 1

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Example – A bar structure (11) Calculating stresses

Element 1

Element 2

1 11 1 2 1

T 0

2

0.5 5x x

d d FE E

L A

2 22 2 2 1

T 0

2

0.5 3x x

d d FE E

L A

1 T

0

0

3/15

FL

EA

d

2 T

0

3/15

8 /15

FL

EA

d

Any questions before we discuss the interpretation and properties of the stiffness matrix?

Element stiffness matrix

Recall the bar element

The stiffness equation for the element

The stiffness matrix

56Apr 20, 2023

1 1

2 2

1 1

1 1

d fEA

d fL

1 1

1 1

EA

L

k

From this eq., can you give a definition of stiffness matrix?

A stiffness matrix is a matrix relating the nodal displacement vector to nodal force vector

How did we derive this matrix?

By directly applying the basic physical laws, i.e. the Hooke’s and equilibrium laws

This direct approach is limited to simple elements

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kd f

Consider the stiffness equation

If d1=1 and d2=0, then

On the other hand, if d1=0 and d2=1

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1 1

2 2

1 1

1 1

d fEA

d fL

1

2

1

1

f EA

f L

The 1st column of k

1

2

1

1

f EA

f L

The 2nd column of k

Thus, a column of k is the vector of loads that must be applied to an element at its nodes to maintain a deformation state in which the corresponding nodal d.o.f. has unit value while all other nodal d.o.f are zero

This concept is general; it is applicable to all stiffness matrices, including the structural stiffness matrix

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Properties of stiffness matrices

Nonnegative main diagonal values

This is because it is physically unreasonable that a single load in a given direction would produce a displacement component in the opposite direction

Symmetry

k (or K) is symmetric if loads are linearly related to displacements

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0iik 0iiK

Properties of stiffness matrices (cont’d) Singularity

This is because the element or structure can under go “rigid body motion”

Sparsity (for the global or structural stiffness matrix) A matrix is called “sparse” if it contains many zeros A practical FE structure may have thousands of d.o.f, and more than 99% of

coefficients in its stiffness matrix may be zero http://web-ext.u-aizu.ac.jp/~niki/javaappl/jassem/jassem.html

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