Superelement approach

In the superelement approach, a (part of a) flexible body is approximated with a number of so-called superelements. Each symmetric superelement consists of 3 rigid bodies connected by joints (marked o in Fig. 3.15) containing ideal torsional springs that model the elastic properties in bending direction. The attractive feature of modeling the flexibility by joint springs and dampers is that the spring and damper forces are readily incorporated into the standard (rigid) multibody body packages (e.g. SD/FAST® [109]).

It should be noted that the centre body of a superelement can be divided in two parts of equal length to include axial deflection, and torsion deformation as well. In this thesis we limit ourselves (initially) to bending since the first torsional mode and the first two (non-rotating) bending modes of the Lagerwey LW-50/750 wind turbine are sufficient apart (see Table A.2 on page 223). That is, for the Lagerwey LW-50/750 wind turbine it is not necessary to take the torsional mode of vibration into account.

Figure 3.15: Deflections and slopes of a superelement with bending stifness EI, and length L. Each (symmetric) superelement consists of three rigid bodies (with lengths kL, (1 — 2k)L, and kL) connected by joints (O).

The main question is "What should the values of the spring constants be in order to produce a comprehensive and accurate dynamic model of a flexible body?"

Figure 3.15: Deflections and slopes of a superelement with bending stifness EI, and length L. Each (symmetric) superelement consists of three rigid bodies (with lengths kL, (1 — 2k)L, and kL) connected by joints (O).

Accurate in the sense that i) the elastic deformations of the superelement under a static load should be equal to those of a flexible beam, ii) the superelement should have the same mass and inertia properties as a rigid beam with identical dimensions, and iii) the eigenfrequencies of the superelement should be as close as possible to those of a continuous beam. Next, we will derive the spring constants in bending direction required in the superelement approach. The interested reader is referred to Molenaar [189] for the determination of the spring constants representing the axial deflection and the torsion deformation.

Continuous bending expressions

If it is assumed that both the shear deformation and rotational inertia of the flexible body cross-sections are negligible if compared with bending deformation and trans-lational inertia, the spring constants can be derived from the differential equation of the deflection curve of a prismatic beam (i.e. beam with constant cross section throughout its length). This equation is given in Gere & Timoshenko [76] as d2v _ M



Transverse displacement (v ± y)



Distance from the origin



Bending moment



Modulus of elasticity



Area moment of inertia


It should be noted that Eq. (3.57) is valid only when Hooke's law applies for the material, and when the slope of the deflection curve is very small. Also, since effects of shear deformations are disregarded, the equation describes only deformations due to pure bending.

For a tapered beam, the presented relationship gives satisfactory results provided that the angle of taper is small (i.e. < 10°). In that case, Eq. (3.57) has to be written in the following form d2v M

in which I(y) is the area moment of inertia of the cross-section at distance y from the origin.

Determination parameters superelement

The superelement parameters (i.e. the torsional spring constants cz1, cz2, and cz3 see Fig. 3.15) are found by comparing the deflection and the angle of rotation at the free end of a Euler-Bernoulli beam (i.e. a prismatic beam with length L, cross-section area A, constant flexural rigidity EIz, and uniformly distributed mass per unit length p = m/L, where m is the total mass of the beam) subjected to a load F and couple M at the free end of the beam (see Fig. 3.16). Since this is a case of pure bending, we may use Eq. (3.57) to determine the total deflection S and the total angle of rotation 0 at the free end [76].

Substituting the expression for the bending moment, the differential equation becomes

EIzv" = -M = FL - Fy with v" = d—v. The first integration of this equation gives

The constant of integration C\ can be found from the condition that the slope of the beam is zero at the support; thus v'(0) = 0, which results in Ci = 0. Therefore

Integration of this equation yields

EIz v

FLy2 Fy3

The boundary condition on the deflection at the support is v(0) = 0, which shows that C2 =0. Thus, the equation of the deflection curve is

The angle of rotation 0F and the deflection SF at the free end of the beam loaded by a force F are readily found by substituting y = L into Eqs. (3.59) and (3.60) respectively.

The equation of the deflection curve for an Euler-Bernoulli beam loaded by a couple M at the end of the beam (see Fig. 3.16) can be determined analogously. The results for both cases are summarized in the following equation

6 EIz

Inversion of this equation results in

2L3 3L2 3L2 6L





' S '






From Fig. 3.15 it can be easily derived that

A7i A72

E,I= constant

E,I= constant

Figure 3.16: Deflections and slopes of an Euler-Bernoulli beam with bending stiffness EI, and length L.

and that the following relation holds

■ Ô '

' L(1 — k) kL


Czl + Cz2


" A71 "


Cz2 + Cz3

. AY2


12k2 —

12k +

4 —12k2

+ 12k — 2 "

" A71 "


— 12k2 -

f 12k -

- 2 12k2 —

12k + 4


with k the partitioning coefficient. Substituting Eq. (3.64) in Eq. (3.62) and back-substituting the result in Eq. (3.63) gives

The spring coefficients cz1, cz2, and cz3 are found by comparing the elements in the above equation. The resulting spring coefficients are


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  • kifle
    What the relationship 0f moment of inertia and beam deflection?
    6 years ago

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