Computer Aided Engineering Design

FINITE ELEMENT METHOD 313

u

k

F

k

k

k F

p

j

p

q

= k

(^1) + + 1⎛
⎝
⎜
⎞
⎠
⎟
⎡
⎣
⎢
⎤
⎦
⎥ (11.5f)
The reaction force Fi from Eq. (11.5e) is –kpuj = – (Fj + Fk) as expected. The displacements can be
verified for the case when Fj = 0. The effective spring constant using Newtonian mechanics is then
keff =
kk
kk
pq
pq +
and hence uk = Fk/keff = F
k kkpq
⎛ (^1) + 1
⎝
⎜
⎞
⎠
⎟ while uj is
F
k
k
p

11.3 Truss Elements

Plane trusses are often used in construction, particularly for roofing of residential and commercial
buildings and in short span bridges (Figure 11.2). Trusses, whether two- or three-dimensional, belong
to the class of skeletal structures consisting of elongated components called members connected at
joints. A member or a truss element of elastic modulus E, cross-section area Aand length l is shown
in Figure 11.3. Like in a spring, two degrees of freedom, namely, ui and uj are permitted along the bar
axis under the action of external loads fi and fj. Let the extension be dl = (uj – ui) so that the strain is

( – )uu

l

ji and thus the stress is E uu

l

( – )ji

. The internal force EA

uu

l

( – )ji must balance the

external forces at the nodes. Thus, at node i,fi = –EA

uu

l

( – )ji while at node j,f

j = EA

uu

l

( – )ji.

Comparing with Eq. (11.1) yields kp = AE/l and hence the stiffness matrix kt for a truss element
becomes

kt

AE

l

AE

l

AE

l

AE

l

AE

l

=

• 
  


• 
  

=

1–1

–1 1

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎤

⎦

⎥ (11.6)

Figure 11.2 Schematic of a roof with truss members

Supports

Roof

Members

ui x,ξ uj

fi x fj

i,ξ= –1 P xj,ξ= 1

Figure 11.3 A truss element

Although the derivation of the stiffness matrix for a truss element is straightforward using the
spring analogy, the same is derived using a more formal finite element procedure. Let the initial
positions of nodes i and j be xi and xj. A local coordinate measure ξ can be introduced so that at x =
xi,ξ = –1 and at x = xj,ξ = 1. The displacement u(x) at any point P in the element can be expressed
in terms of the unknown nodal displacements. Since there are only two such displacements, ui and uj,
it behooves to use a linear interpolating relation, that is

u(x) = c 1 + c 2 x or u(ξ) = d 1 + d 2 ξ (11.7a)

We know that u(xi) = ui and u(xj) = uj (or u(ξ= –1) = ui and u(ξ= 1) = ujin terms of the local measure)
substituting which we can solve for the unknown constants c 1 and c 2 (or d 1 and d 2 ). Solving for these