4.9 Truss and frame models

Back

If a truss is used to approximate a plate (similar to Fig. 4.36), then in each

bar element the longitudinal displacement is a linear function

ue(x) = u1 ( l − x

l

) + u2

x

l

(4.73)

and the deformed shape of the truss is found by minimizing the potential

energy

Π(u) =

1

2

_

e

_ l

0

N2

e

EA

dx −

_

e

_ l

0

pue dx . (4.74)

In an FE analysis with plate elements, the expression for the potential energy

is instead

364 4 Plane problems

Fig. 4.36. Frame model of a shear wall

Π(u) =

1

2

_

Ω

S •E dΩ −

_

Ω

p •udΩ . (4.75)

These two equations are not that different. If the plate Ω were subdivided in

accordance with the direction of the principal forces into elements of constant

width, and if all elements with stresses below a certain threshold limit were

neglected, and if it were assumed that the stresses across the elements are

constant, then the result would be Eq. (4.74). Hence truss or frame models of

2-D and 3-D continua are simplified FE models. Or stated otherwise, a truss

model is not per se a better or more authentic engineering model than an

FE model. It even has severe defects because it does not register the stress

concentrations at the corner points, see Fig. 4.36. And if we ignore these

stresses and other finer details what is an FE analysis of a plate good for?

But the true value of a truss model is that it enables us to visualize easily

how the loads are carried by a structure. The more FE technology advances

and the more massively parallel computing power is employed, the more a

sound analysis tool is needed that is capable of filtering the essential and

important information from the vast output produced by a computer program.

Thus, it seems clear that we will soon see a revival of the old manual

methods—but under a different name. Experienced structural engineers have

always trusted these methods more than FE methods...

4.11 Multistory shear wall 365

Table 4.6. Support reactions (kN) of the wall in Fig. 4.37, Q4 = bilinear elements,

Q4+2 = Wilson’s element

support Q4 Q4+2 BE beam

A 87 87 86 75

B 281 282 284 302

C 132 131 130 123

sum 500 500 500 500

Навигация