1.47 Stiffness matrices

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A stiffness matrix K is quadratic and symmetric. The number of rows and

columns is equal to the number of degrees of freedom of the element. The

elements of a stiffness matrix, as for example of a bar element or a beam

K = EA

l

_

1 −1

−1 1

_

, K = EI

l3

⎡

⎢⎢⎣ 12 −

6l −

12 −

6l

−6l 4l2 6l 2l2

−12 6l 12 6l

−6l 2l2 6l 4l2

⎤

⎥⎥⎦

,

(1.621)

are the strain energy products between the nodal unit displacements:

k ij = a(ϕi, ϕj) =

_ l

0

EAϕ

_

i ϕ

_

j dx =

_ l

0

σi εj Adx bar (1.622)

k ij = a(ϕi, ϕj) =

_ l

0

EI ϕ

__

i ϕ

__

j dx =

_ l

0

mi κj dx beam. (1.623)

The same holds for plates and slabs (Kirchhoff):

elements have the same displacements at the common nodes then in genelement,

see Fig. 1.162,

222 1 What are finite elements?

Fig. 1.162. A bar and a beam element, and the nodal unit displacements

k ij = a(ϕi,ϕj) =

_

Ω

Si •Ej dΩ =

_

Ω

σi • εj dΩ (1.624)

k ij = a(ϕi, ϕj) =

_

Ω

Mi •Kj dΩ =

_

Ω

mi •κj dΩ , (1.625)

where the corresponding terms are the scalar product between the stress tensor

of the field ϕi and the strain tensor of the field ϕj , or the moment tensor Mi

of ϕi and the curvature tensor Kj of ϕj .

If the stress vectors σi and strain vectors εi of the element nodal unit

displacement fields ϕi are written as row vectors, then

B(n×3) = [εT1

, εT2

, εT3

, . . . , εT

n ]T S(n×3) = [σT1

,σT2

,σT3

, . . . ,σT

n ]T ,

(1.626)

and the stiffness matrix becomes

K(n×n) =

_

Ω

B(n×3) ST

(3×n) dΩ =

_

Ω

B(n×3) D(3×3) BT

(3×n) dΩ ,

(1.627)

whereD is a 3×3-matrix which transforms the strains into stresses, σi = Dεi:

⎡

⎣

σxx

σyy

σxy

⎤

⎦ = E

1 − ν2

⎡

⎣

1 ν 0

ν 1 0

0 0(1− ν)/2

⎤

⎦

⎡

⎣

εxx

εyy

2 εxy

⎤

⎦ . (1.628)

This matrix D is for plane stress problems. For plane strain problems it has

the form

1.47 Stiffness matrices 223

D = E

(1 + ν)(1 − 2 ν)

⎡

⎣

(1 − ν) ν 0

ν (1 − ν) 0

0 0 (1− 2 ν)/2

⎤

⎦ . (1.629)

In plate bending the procedure is analogous.

The symmetric triple product in (1.627) has its origin in the structure of

the differential equation17

G(u, ˆu) =

_ l

0

− d

dx

EA

d

dx

u · ˆudx + [EA

d

dx

u · ˆu]l

0

−

_ l

0

d

dx

u · EA · d

dx

ˆudx = 0. (1.630)

Virtual work

The setup of a stiffness matrix can also be explained as follows. First all nodes

are kept fixed and the first degree of freedom, u1 = 1, (only) is activated.

The forces necessary to force the structure into this particular shape, u =

e1, constitute the unit load case p1. Next we let these forces consecutively

act through the nodal unit displacements ϕ1,ϕ2, . . . ,ϕn, and each time the

virtual work is calculated. These n numbers

δWe(p1,ϕi) = work done by p1 on acting through ϕi i = 1, 2, . . . n

(1.631)

form the first column of the stiffness matrix. Then the next degree of freedom,

u2, is activated (all other ui are zero) and the procedure is repeated. The result

is the stiffness matrix K.

Because the product Ku = u1 c1+u2 c2+. . .+un cn is the weighted (ui)

sum of the columns ci of K and because

Kei = ci ei = i-th unit vector, (1.632)

the columns ci of K are just the equivalent nodal forces that belong to the

single nodal unit displacements.

Three properties

Each stiffness matrix has the following three properties:

ˆu

T Ku = uT K ˆu symmetry

uT0Ku = 0 equilibrium

Ku0 = 0 u0 = rigid-body motion.

17 Gilbert Strang has written nearly a whole book—and very readable book—about

this subject, [232].

224 1 What are finite elements?

Because of the latter property the stiffness matrix of an unconstrained structure

or element is singular. If the rows and columns in K which correspond

to fixed degrees of freedom are deleted, then a regular matrix, the so-called

reduced stiffness matrix K, is obtained.

Energy

The elastic energy stored in the FE structure is

Wi =

1

2 uTKu (1.633)

and if a virtual displacement ϕi is applied then the virtual work is the scalar

product of row ri of the stiffness matrix and the vector u

δWe(ph, ϕi) = ri u =

_n

j=1

k ij uj . (1.634)

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