Computer Aided Engineering Design

FINITE ELEMENT METHOD 323

ui vi θi uj vj θj

ktruss =

00– 00

000000

000000

–00 00

000000

000000

AE

l

AE

l

AE

l

AE

l

u

u

i i i j j j

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

v

v

θ

θ

and for beam element
ui vi θi uj vj θj

kbeam

32 32

22

32 3 2

22

=

00 000 0

0 12 6 0 – 12 6

06 4 0–6 2

00 000 0

0 – 12 – 6 0 12 – 6

06 2 0–6 4

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

EI

l

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

⎥⎥

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

u

u

i i i j j j

v

v

θ

θ

Adding the two matrices, we get the frame element stiffness matrix as

Figure 11.9 A frame element

y

l

νi, Fvi

θi, Mi θj, Mj

vj, Fvi

ui, Fui uj, Fuj

and rotation perpendicular to the plane of the element (Figure 11.9). For u = [ui,vi,θi, uj,vj,θj]T
chosen as the displacement vector for the element, the stiffness matrices of the truss and beam
elements in Eqs. (11.7l) and (11.9g) can be combined. Expressing the stiffness in all six degrees of
freedom, for the truss element, we have

x