.

14The Direct

StiffnessMethod II

14–1

Lecture 14: THE DIRECT STIFFNESS METHOD II 14–2

TABLE OF CONTENTS

Page§14.1. Breakdown 14–3

§14.1.1. Disconnection . . . . . . . . . . . . . . . . . 14–3§14.1.2. Localization . . . . . . . . . . . . . . . . . 14–3§14.1.3. Computation of Member Stiffness Equations . . . . . . . 14–3

§14.2. Assembly: Globalization 14–5§14.2.1. Coordinate Transformations . . . . . . . . . . . . 14–5§14.2.2. Transformation to Global System . . . . . . . . . . . 14–6

The following material, as well as that presented in Lectures 14 and 15, is selectivelyextracted from Chapters 2 and 3 of the first-year core graduate course ASEN 5007:Introduction to Finite Element Methods or IFEM, which is taught every Fall. The ASEN5007 course material is posted athttp://www.colorado.edu/engineering/CAS/courses.d/IFEM.d/Home.html

14–2

14–3 §14.1 BREAKDOWN

In this Lecture we continue with the exposition of the Direct Stiffness Method (DSM) steps forFEM analysis. These steps are illustrated through a specific plane truss structure, introduced inLecture 13.

The DSM steps, major and minor, are sum-marized in Figure 14.1 for the convenienceof the reader. The two major processingsteps are Breakdown, followed by Assembly& Solution. A postprocessing substep mayfollow, although this is not part of the DSMproper.

The first 3 DSM substeps are: (1) discon-nection, (2) localization, and (3) computationof member stiffness equations. Collectivelythese form the breakdown. The first twoare flagged as conceptual in Figure 14.1because they are not actually programmedas such: they are implicitly carried outthrough the user-provided problem definition.Processing actually begins at the member-stiffness-equation forming substep.

Disconnection

Localization

Member (Element) Formation

Globalization

Merge

Application of BCs

Solution

Recovery of Derived Quantities

Breakdown

Assembly & Solution

(Lecture 14)

(Lectures 14-15)

post-processingsteps

processingsteps

conceptualsteps

Figure 14.1. The Direct Stiffness Method steps.Reproduced from last Lecture for convenience.

§14.1. Breakdown

§14.1.1. Disconnection

To carry out the first breakdown step we proceed to disconnect or disassemble the structure intoits components, namely the three truss members. This task is illustrated in Figure 14.2. To eachmember e = 1, 2, 3 assign a Cartesian system {x e, ye}. Axis x e is aligned along the axis of the eth

member. Actually x e runs along the member longitudinal axis; it is shown offset in that Figure forclarity.

By convention the positive direction of x e runs from joint i to joint j , where i < j . The angleformed by x e and x is the orientation angle ϕe. The axes origin is arbitrary and may be placed atthe member midpoint or at one of the end joints for convenience.

Systems {x e, ye} are called local coordinate systems or member-attached coordinate systems. Inthe general finite element method they also receive the name element coordinate systems.

§14.1.2. Localization

Next we drop the member identifier e so we are effectively dealing with a generic truss member, asillustrated in Figure 14.3(a). The local coordinate system is {x, y}. The two end joints are i and j .

As shown in that figure, a generic plane truss member has four joint force components and four jointdisplacement components (the member degrees of freedom). The member properties are length L ,elastic modulus E and cross-section area A.

14–3

Lecture 14: THE DIRECT STIFFNESS METHOD II 14–4

12

3

y

x

(3)

(1)

(2)

y_ (1)

x_

(1)

y_

(2)

x_

(2)y_ (3) x

_ (3)

Figure 14.2. Breakdown of example truss intoindividual members (1), (2) and (3), and selection

of local coordinate systems.

i

i

j

j

dL

x

Equivalent spring stiffnesss

−F F

f , u xi xi

_ _ f , u xj xj

_ _f , u yj yj

_ _f , u yi yi

_

_y_

_

k = EA / L

(a)

(b)

Figure 14.3. Generic truss member referred to its localcoordinate system {x, y}: (a) idealization as bar element,

(b) interpretation as equivalent spring.

§14.1.3. Computation of Member Stiffness Equations

The force and displacement components of the generic truss member shown in Figure 14.3(a) arelinked by the member stiffness relations

f = K u, (14.1)

which written out in full is

fxi

f yi

fx j

f y j

=

Kxi xi Kxiyi Kxi x j Kxiy j

K yi xi K yiyi K yi x j K yiy j

Kx j xi Kx j yi Kx j x j Kx j y j

K y j xi K y j yi K y j x j K y j y j

uxi

u yi

ux j

u y j

. (14.2)

Vectors f and u are called the member joint forces and member joint displacements, respectively,whereas K is the member stiffness matrix or local stiffness matrix. When these relations areinterpreted from the standpoint of the general FEM, “member” is replaced by “element” and “joint”by ”node.”

There are several ways to construct the stiffness matrix K in terms of L , E and A. The moststraightforward technique relies on the Mechanics of Materials approach covered in undergraduatecourses. Think of the truss member in Figure 14.3(a) as a linear spring of equivalent stiffness ks , aninterpretation illustrated in Figure 14.3(b). If the member properties are uniform along its length,Mechanics of Materials bar theory tells us that1

ks = E A

L, (14.3)

1 See any introductory book on Mechanics of Materials.

14–4

14–5 §14.2 ASSEMBLY: GLOBALIZATION

Consequently the force-displacement equation is

F = ksd = E A

Ld, (14.4)

where F is the internal axial force and d the relative axial displacement, which physically is thebar elongation. The axial force and elongation can be immediately expressed in terms of the jointforces and displacements as

F = fx j = − fxi , d = ux j − uxi , (14.5)

which express force equilibrium2 and kinematic compatibility, respectively. Combining (14.4) and(14.5) we obtain the matrix relation3

f =

fxi

f yi

fx j

f y j

= E A

L

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

uxi

u yi

ux j

u y j

= K u, (14.6)

Hence

K = E A

L

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

. (14.7)

This is the truss stiffness matrix in local coordinates.

Two other methods for obtaining the local force-displacement relation (14.4) are covered in Exer-cises 2.6 and 2.7 of Chapter 2 of the IFEM Notes.

§14.2. Assembly: Globalization

The first substep in the assembly & solution major step, as shown in Figure 14.1, is globalization.This operation is done member by member. It refers the member stiffness equations to the globalsystem {x, y} so it can be merged into the master stiffness. Before entering into details we mustestablish relations that connect joint displacements and forces in the global and local coordinatesystems. These are given in terms of transformation matrices.

§14.2.1. Coordinate Transformations

The necessary transformations are easily obtained by inspection of Figure 14.4. For the displace-ments

uxi = uxi c + uyi s, u yi = −uxi s + uyi c,

ux j = ux j c + uyj s, u y j = −ux j s + uyj c,. (14.8)

2 Equations F = fx j = − fxi follow by considering the free body diagram (FBD) of each joint. For example, take joint ias a FBD. Equilibrium along x requires −F − fxi = 0 whence F = − fxi . Doing the same on joint j yields F = fx j .

3 The matrix derivation of (14.6) is the subject of Exercise 2.3 in Chapter 2 of the IFEM course

14–5

Lecture 14: THE DIRECT STIFFNESS METHOD II 14–6

x

y

(a) Displacementtransformation

(b) Forcetransformation

i

j

ϕ

fxi

fyi

fxj

fyj

fyi

_

fxi

_

fxj

_fyj

_

i

xy j

ϕ

uxi

uyi

uxj

uyj

uyi_

_ _

uxi_

uxj_

uyj_

Figure 14.4. The transformation of node displacement and forcecomponents from the local system {x, y} to the global system {x, y}.

where c = cos ϕ, s = sin ϕ and ϕ is the angle formed by x and x , measured positive counterclock-wise from x . The matrix form of this relation is

uxiu yiux ju y j

=

c s 0 0−s c 0 00 0 c s0 0 −s c

uxiuyiux juyj

. (14.9)

The 4×4 matrix that appears above is called a displacement transformation matrix and is denoted4

by T. The node forces transform as fxi = fxi c − f yi s, etc., which in matrix form become

fxifyifx jfy j

=

c −s 0 0s c 0 00 0 c −s0 0 s c

fxif yi

fx j

f y j

. (14.10)

The 4×4 matrix that appears above is called a force transformation matrix. A comparison of (14.9)and (14.10) reveals that the force transformation matrix is the transpose TT of the displacementtransformation matrix T. This relation is not accidental and can be proved to hold generally.5

Remark 14.1. Note that in (14.9) the local system (barred) quantities appear on the left-hand side, whereas in(14.10) they show up on the right-hand side. The expressions (14.9) and and (14.10) are discrete counterpartsof what are called covariant and contravariant transformations, respectively, in continuum mechanics. Thecontinuum counterpart of the transposition relation is called adjointness. Colectively these relations, whetherdiscrete or continuous, pertain to the subject of duality.

Remark 14.2. For this particular structural element T is square and orthogonal, that is, TT = T−1. Butthis property does not extend to more general elements. Furthermore in the general case T is not even asquare matrix, and does not possess an ordinary inverse. However the congruent transformation relations(14.11)–(14.13) do hold generally.

4 This matrix will be called Td when its association with displacements is to be emphasized, as in Exercise 14.5.5 A simple proof that relies on the invariance of external work is given in Exercise 14.5. However this invariance was only

checked by explicit computation for a truss member in Exercise 14.4. The general proof relies on the Principle of VirtualWork, which is discussed later.

14–6

14–7 §14.2 ASSEMBLY: GLOBALIZATION

§14.2.2. Transformation to Global System

From now on we reintroduce the member (element) index, e. The member stiffness equations inglobal coordinates will be written

f e = Keue. (14.11)

The compact form of (14.9) and (14.10) for the eth member is

ue = Teue, fe = (Te)T fe. (14.12)

Inserting these matrix expressions into fe = K

eue and comparing with (14.11) we find that the

member stiffness in the global system {x, y} can be computed from the member stiffness Ke

in thelocal system {x, y} through the congruent transformation6

Ke = (Te)T KeTe. (14.13)

Carrying out the matrix multiplications in closed form (Exercise 2.8) we get

Ke = Ee Ae

Le

c2 sc −c2 −scsc s2 −sc −s2

−c2 −sc c2 sc−sc −s2 sc s2

, (14.14)

in which c = cos ϕe, s = sin ϕe, with e superscripts of c and s suppressed to reduce clutter. If theangle is zero we recover (14.10), as may be expected. Ke is called a member stiffness matrix inglobal coordinates. The proof of (14.13) and verification of (14.14) is left as Exercise 14.8.

The globalized member stiffness matrices for the example truss can now be easily obtained byinserting appropriate values into (14.14). For member (1), with end joints 1–2, angle ϕ(1) = 0◦ andthe member properties given in Figure 13.2 of the previous lecture, we get

f (1)x1

f (1)y1

f (1)x2

f (1)y2

= 10

1 0 −1 00 0 0 0

−1 0 1 00 0 0 0

u(1)x1

u(1)y1

u(1)x2

u(1)y2

. (14.15)

For member (2), with end joints 2–3, and angle ϕ(2) = 90◦:

f (2)x2

f (2)y2

f (2)x3

f (2)y3

= 5

0 0 0 00 1 0 −10 0 0 00 −1 0 1

u(2)x2

u(2)y2

u(2)x3

u(2)y3

. (14.16)

Finally, for member (3), with end joints 1–3, and angle ϕ(3) = 45◦:

f (3)x1

f (3)y1

f (3)x3

f (3)y3

= 20

0.5 0.5 −0.5 −0.50.5 0.5 −0.5 −0.5

−0.5 −0.5 0.5 0.5−0.5 −0.5 0.5 0.5

u(3)x1

u(3)y1

u(3)x3

u(3)y3

. (14.17)

To be continued in the next Lecture, which finishes the DSM exposition.

6 Also known as congruential transformation and congruence transformation in linear algebra books.

14–7