o_19jdnovcg1eh21852116govv1lgra.pdf

Critical load of a tower – a buckling problem
Slender structural elements under pressure can fail due to
buckling. This means the system is not able to undo
unintentional deflections from a certain pressure load, but
the deflection is continuously increasing which means that
the system is failing. The load because of which this critical
state can occur is called critical load or limit load.
The load state of a tower under critical load can be
visualized by the following analogy:
If the existing load is lower than the critical load
the tower behaves like a ball in a bowl. If the ball
is moved it always rolls back to the middle of the
bowl by itself. This is called a stable system.
If the existing load is higher than the critical load
the tower behaves like a ball on a bowl that is turned
upside down. As soon as it is moved only
some millimeters the ball drops. This is called an
instable system.
In this context the tower can be regarded as free standing
column with bending resistant connection to the basement
(exception: inclined or braced towers) and the critical load
can be determined with Euler’s formula:
Fk = π x EI / (2 x H)²
Decisive for the size of the critical load are the stiffness of
the trusses (EI) and the height of tower (H). The stiffer and
lower the tower is the higher is the critical load and thus the
higher the tower can be loaded. Referring to 4-point-trusses
it can be said further:
The permissible pay load declines quadratically with the
height (twice the height => 1/4 pay load)
The pay load rises quadratically with the axial distance of
the mainchords (twice the axial distance => quadrupled
pay load)
The pay load rises linearly with the cross section area of the
mainchords (twice the area => twice the pay load)
The following chart shows the critical loads for different
tower heights for some common trusses (calculation see
example): the critical load applies to towers with central loading
without horizontal loads; for the calculations a safety
of 2.5 is taken into account. Eccentric loads and wind loads
lead to further reduction of the permissible load.
Please note again, this is a simplified calculation.
For a more precise verification further points need to be
taken into account:
- Bending resistant connection to a weak basement
(in this case the critical load declines even more, which
means the permissible pay load also declines)
- If the load is not suspended directly at the tower but
lifted with a pulley the pressure load on the tower is
doubled which means the permissible pay load is halved.
- A dynamic increase of lifting or lowering the pay load
should be taken into account by a load increase of
20 – 40%.
- For inclined towers or towers with cantilevered loads the
critical load is diminished due to the additional bending
stress. In this case a more detailed calculation is
necessary.
Example: Determination of the critical
load according to Euler
Tower H = 10 m of square aluminium trusses
with main chords 50 x 4 center distance 47 cm
critical load according
to Euler:
Fk = \pi x E x I / (2 x H)²
with \pi = circuit number = 3,14
E = E = E modulus of aluminium = 7.000 kN/cm²
I = A chord x e² = 5,78 cm² x 47² = 12768 cm 4
H= height tower = 10 m = 1000 cm
permissible load F with safety = 2,5
I = area moment of inertia 2. degree
for square trusses:
I = cross section area main chord
multiplied by the square of center distance
F < Fk / 2,5 = 3,14 x 7.000 x 12768 / (2,5 x 2000²)
= 28,1kN = 2,81 t
PA-TOWER
main chords center distance I buckling load in [kN] bucklingd with safety of 2.5
main chords
at H in [m]
[cm] [cm 2 ] 5,0 7,5 10,0 12,5
50 x 2,0 24 1737 15,3 6,8 3,8 2,4
48 x 4,5 30 5535 48,7 21,6 12,2 7,8
50 x 4,0 47 12769 112,3 49,9 28,1 18,0
48 x 4,5 57 19980 175,8 78,1 43,9 28,1
Further informationen of the autors:
www.krasenbrink-bastians.de, www.vom-felde.de
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