Tower Design and Manufacture - Small Wind Turbines: Analysis, Design, and Application - page 222

flanges used to join the sections. The turbine is a replacement for one that was
dropped during lowering when the guy wires tangled and snapped. The tractor
raising the turbine up can be seen on the skyline. At the end of the gin pole is a
block and tackle connected to the guy-wire anchor. Guys are often used for raising
and lowering a hinged tower. Hinging has a number of advantages, such as low-
ering tower stress levels by accommodating settling of the foundations, which
cannot always be made properly at remote sites. However, it has one major dis-
advantage: the guy foundations must be designed to withstand the base overturning
moment for the worst case of the extreme wind blowing in either direction along
the line joining the tractor and the turbine.
With the provision of turnbuckles, it is easy to adjust guy tension for any
movement in the foundation and, in principle, to tune for natural frequency var-
iation. Initial guy wire tensions should be around 10% of the allowable stress in the
wire [ 19 ].
Some useful guidelines for the design of guyed towers are given by Veers
et al. [ 20 ]. 1 They point out that guy-wires must be designed to avoid resonance at
blade passing frequencies and showed that the usual guy-wire angle of 35 to the
horizontal, which maximises the lateral stiffness, can be relaxed to 45 which
reduces the cost and the land area required for the turbine and tower.
Structural analysis of guy-wires is reasonably straightforward, but the tower
design usually requires FEA to deal with the concentrated loads imparted by the
guys at their attachment pints It can be seen in Fig. 10.10 that attachment in this
case occurs at the flange separating the top and middle tower section. This is a
common arrangement.

10.5.1 Exercises

1. Derive ( 10.17 ) for the approximate moment of inertia when t d 0 .
2. For a constant diameter octagonal tower with t d, and m tt = 170 kg, what
must the height be for the axial stress, r a , the second term on the right of
( 10.9 ), to exceed the 250 MPa when t = 5 mm and d = 0.15 m.
3. Show that the horizontal force, F wind , from a wind load given by Eq. 10.1 on a
tapered tower defined by Eq. 10.4 for a constant U is

:

F wind ¼ 1

2 qU 2 C d hd 0 þ d 1 h

2

4. Show that the formula in the previous exercise is dimensionally correct.

1 This reference is not generally available. The author is indebted to Dr. P. S. Veers for providing
copies of the slides used for the conference presentation—there was no formal paper—and for
permission to post them with the online materials.