|
Analyzing the stability of an entire three-dimensional framework of
members is referred to |
|
as
lattice stability analysis, as opposed to the planar stability analysis we
employed for |
|
polygons
and trusses previously. If the lattice framework is completely
triangulated and all |
|
of its
joints, or hubs, are flexible, it is said to exhibit pure lattice action
and is statically |
|
determinate. That is, static analysis techniques, like those we
employed to analyze the |
|
forces
acting on truss bridges, can be used to quantify and qualify those forces.
However, |
|
doing a
static analysis of a lattice involves significantly more work than doing
one for a |
|
truss. |
|
. |
|
Referring back to the lattice calculations just presented in
Fig. 202, notice what happens |
|
when the
roof structure, b), is added on top of the room structure, a), to build
the entire |
|
house
structure, c). The resulting structure needs only 24 members to be
stable with its 10 |
|
joints.
One might have guessed that the house needs 25 members, presuming that the
top |
|
face of
the cube shares its diagonal bracing strut with the one in the base of the
roof. |
|
Actually
this strut is altogether unnecessary, or redundant, since the room
structure and the |
|
roof
structure combine to stabilize each other's top and base respectively
without the need |
|
for any
shared brace between them. The stabilizing elements of these lattice
structures |
|
combine
to achieve a synergistic effect whereby the whole is not equal to the sum
of its |
|
parts.
Rather it exhibits a unique behavior of its own. One has to be
careful when doing |
|
a
stability analysis of three-dimensional structures to be sure that a
common sense notion |
|
of what
a structure requires for stability agrees with reality. |
|
. |
|
Exercise: 1)Construct a model of the completely triangulated house
structure shown in |
|
|
Fig. 202 c). Examine its stability
with and without the extra diagonal brace in |
|
|
question to confirm it is redundant. |
|
|
. |
|
Plate
action |
|
. |
|
Thus far
we have modeled the stability of three-dimensional polyhedra as though
they are |
|
lattices
comprised of rigid members, or struts, connected together end to end by
flexible |
|
joints,
or hubs. We have also briefly mentioned that the faces of a cube's
framework could |
|
click image to enlarge |
be
stabilized by using a rigid panel, or shear plate. Such a |
|
 |
plate can be imagined as being a four-sided polygon braced |
|
by
an infinite number if diagonal struts acting together to |
|
create the stable plate. That is, it is infinitely
triangulated |
|
like the blue colored plate in the image shown on the left. In |
|
this respect a plate does not differ from a strut in how it |
|
stabilizes a structure. However, plates do differ in the way |
|
12 = 3 ( 6 ) - 6 |
that they are joined together as an assembly to build stable |
|
stable plate structure |
structures. They are joined edge to edge around their entire |
|
Fig. 203 - Pure plate action |
perimeter but not at the vertices where their edges converge. |
|
As
a result plate structures differ in how they react to outside |
|
|
. |
|
Back
to Knowhere |
|
 |
Page 120 -
Building stability - Plate action |
 |
|