 |
 |
|
Fig. 199 - Horse barn with |
open Queenpost truss roof |
|
(scale visualization model) |
click image to enlarge |
|
. |
The
sloped plane of the roof must also be stabilized with some sort of bracing
to keep the |
trusses
from tipping over (braces are indicated by blue pinges). |
|
|
◄
Fig. 200 - Bracing roof |
trusses to keep them |
from tipping over |
(static demonstration model) |
|
|
Combining all of these planes together gives an |
|
idealized model of a completely stabilized |
house framework complete with bracing. |
|
Fig. 201 - Stable skeletal framework of a house ► |
(static demonstration model) |
|
|
Lattice action |
. |
Thus far
our stability analysis of a home's framework has treated the structure as
though it |
was
comprised of discrete triangulated planes in order to simplify the
stability analysis. |
This
approach is similar to treating polyhedra as though they are assemblies of
discrete |
planar
faces, which of course they are. However, we can also do a stability
analysis of the |
entire
three-dimensional structure by using another equation Euler devised for
polyhedra: |
. |
M = 3 ( J ) - 6 where M = members |
J = joints |
. |
Lets
apply it to the house framework you just studied. For example, the
stabilized cube has |
a total
of 18 members (12 edges plus 6 diagonal braces) and 8 joints.
Substituting these |
values
into Euler's equation gives 18 = 3 ( 8 ) - 6, confirming that it is
stable. |
|
|
|
a) stabilized cube |
b) stabilized roof |
c) stabilized house |
18 = 3 ( 8 ) - 6 |
12 = 3 ( 6 ) - 6 |
24 = 3 ( 10 ) - 6 |
stable |
stable |
stable |
click image to enlarge |
Fig. 202 - Analyzing the stability of
idealized lattice frameworks with M = 3 ( J ) - 6 |
|
. |
Back
to Knowhere |
 |
Page 119
- Building stability - Lattice action |
 |
|