(Part 1)(Part 2)

## Part 2

Niall McMahon © 2012 -- 2013

### Context

You can find out more about me here. Have a look here to find out about lecture times and about contacting me.

### These Notes

Part 2 of these notes introduces natural frequency and an overview of a blade flapping model. See also Mechanics and Dynamics, Part 1.

### Spring and Damper Systems

#### Spring Systems

A spring resists an applied force with a force equal to some constant times the straight-line deflection of the spring. This force can be written as -kx. The "-" indicates that the force acts to oppose the deflection, the k is the spring constant. x is the straight-line deflection. The spring constant depends on the spring material and geometry.

A system consists of a spring that is anchored at one end and attached a mass m at the other. The mass and spring are constrained to move along one axis only, the X axis. If x represents the deflection of the mass along this axis away from its rest, or equilibrium, position, the equation that describes the motion of the mass is:

m x'' = - kx

Where x'' is the acceleration of the mass. It's assumed that there is no other force acting on the mass other than the spring force. This would be true if the mass was first displaced and then released to move freely.

For this system, the general solution to the equation of motion, at a time t after release, takes the form,

x = x0cos(ωnt) + x0'/ωn sin(ωnt)

Where x0 and x0' are the initial position and velocity of the mass at the start time t0 respectively, and ωn is the ratio (k/m)½. This quantity is called the natural frequency; it is a measure of the ratio of the elastic forces to the inertial forces acting on the system.

Evidently, if x0 = 0, if the mass is displaced and held before release, the general solution becomes,

x = x0cos(ωnt)

#### Spring and Damper Systems

A dashpot or damper resists an applied force with a force equal to a constant times the speed of deflection, x'. This is written as -cx', where, as before, the "-" indicates that the force acts to oppose the applied force, c is the constant of proportionality, or the damping coefficient, and x' is the speed of deflection. A damper slows motion by absorbing and dissipating energy.

Consider the system described in the last section, now with a damper positioned in parallel with the spring. Both spring and damper are connected to the anchor point and to the mass. The mass and spring and damper are constrained to move along a fixed axis, denoted as the x axis. If x represents the deflection of the mass along this axis away from its rest or equilibrium position, the equation that describes the motion of the mass is:

m x'' = - kx - cx'

Where, as before, k is the spring constant and x'' is the acceleration of the mass. c is the damping coefficient and x' is the speed of the mass. It's assumed that there is no other force acting on the mass other than the spring force and the damping force. This would be the case if, as before, the mass was displaced, held and then released to move freely.

In this case, the general solution to the equation of motion, at a time t after release, takes the form,

x = C e- ζ ωn t sin(ωdt + φ)

C is a constant, the amplitude, that depends on the particular system's initial conditions, ζ is the ratio c/ccritical. ccritical is the value of the damping coefficient below which the system behaves increasingly as if only the spring force existed, and above which the system behaves increasingly like a system that is only damped, i.e. where there is no spring. These situations are referred to as underdamped and overdamped respectively. ωd is a term that includes ζ and ωn. φ is the phase angle; this is determined by the initial conditions and is relevant only when x'0 > 0.

### Vibration in Beams

Beams are useful models for: (i) wind turbine blades and (ii) wind turbine towers. Blades and towers can be idealised as cantilevered beams, i.e. beams that are firmly anchored at one end but free to move at the other.

The vibration in a beam depends on the natural frequencies, the material density, the second moment of area of the beam (J) and the modulus of elasicity of the material (E).

One method of determining how a beam will respond to an input vibration, i.e. how it will move and what its important natural frequencies are, is to consider the beam as consisting of a large number of point masses connected by massless rods. The overall behaviour of the beam is calculated by summing the behaviour of the point masses as they respond to the initial deflection and the motion of their neighbours. Simplified equations are used to describe the interaction between neighbouring point masses.

### Rotor Loads and a Model of Rotor Deflection

The lift and drag forces that act on a rotor can be resolved into axial (or thrust) and tangential components.

The thrust component leads to flapwise bending of the blades. The tangential component leads to edgewise bending. In general, flapwise bending is of most concern.

#### Flapwise Forces, Moments and Stresses

With a little bit of work, by summing the thrust force contributions along the blade, from blade element/momentum theory, the maximum flapwise bending moment in a blade, which occurs at the root, can be written as,

MB = 2 ⁄ 3 (T/B) R [Nm]

Where R is the blade radius, T is the total thrust force acting on the rotor and B is the number of blades.

The maximum stress can be written,

σmax = MB c / J [Nm-2]

Where c in this context is the distance from the outside surface of the blade to the primary span-wise axis of the blade (strictly the neutral axis, where the net bending moment is zero) at the root. If the root is circular in cross-sectional shape, you can think of c as being the radius of the cross-section.

The shear force at the root is simply the total rotor thrust divided by the number of blades, i.e. T/B [N].

#### Edgewise Forces, Moments and Stresses

The maximum edgewise bending moment in a blade, i.e. in the plane of the rotor, occurs at the root and is equal to the total aerodynamic torque divided by the number of blades, i.e. Q / B.

#### Linearised Hinge-Spring Model

##### Overview

The linearised hinge-spring model is a simplification of the material properties of a wind turbine blade. Springs and dampers are used to model the response of the blade to the various forces that act during normal operation. Its primary function is to determine how wind turbine blades will deflect and vibrate in response to these loads.

While the spring-hinge model allows for flap-wise, edge-wise and rotational motion about the span-wise blade axis, the flap-wise motion is the most important and the forces involved are the most severe, in general. In the course textbook, Wind Energy Explained, the authors detail the development of a flap-wise hinge-spring model.

An equation that describes flap-wise blade deflection in the most general case can be more easily deduced by first considering the more specific situation where the rotor turns freely, i.e. there are no wind or yaw loads. Gravity and rotation dominate. This situation could arise when the wind drops suddenly and the rotor is still turning.

##### Assumptions

The spring-hinge model is built with assumptions that:

• Blades are rigid and attached to the rigid hub by a sprung hinge.
• A simplified, linear, representation of aerodynamic forces is good enough.
• Non-linear effects, e.g. turbulence and vortices, can be resprented well enough as finite deflections of the blade.
• The solutions can be represented by sinusoidal functions; this approximation makes the mathematics neater.
• Blade cross-sections are uniform, i.e. the profile remains the same along the length of each blade.
• The hinged joint can be offset from the root of the blade by a small distance.
• Rate of rotation is constant.

The loads captured by the model include those that arise because of:

• Rotation.
• Gravity.
• Rotor misalignment relative to the free-stream wind direction, i.e. yaw error.
• Wind shear across the rotor, i.e. from top to bottom, assumed to be linear.
##### Overview of Derivation

The most general case, when the rotor is driven by the wind and subjected to forces associated with steady yaw or yaw error, follows naturally from the free model.

The equations used to describe blade deflection take the form:

Angle of deflection from rotor plane (β) = (forces that tend to cause flap-wise deflection, e.g. thrust) - (forces that tend to restore blade to the rotor plane, i.e. spring force).

(See Figures 4.11 and 4.12 in Wind Energy Explained.)

One of the principal concerns of the model is to determine how blades will deflect and vibrate in response to normal disturbances. The equations are constructed to yield information about the frequency and amplitude of the vibrations.

Specific problems are useful stepping-stones on the way to a general model:

(i) Spring; (ii) no rotation; (iii) no offset from root. This is similar to simple vibration.
(i) No spring; (ii) rotation; (iii) no offset from root.
(i) Spring; (ii) rotation; (iii) no offset from root.
(i) Spring; (ii) rotation; (iii) offset.

The equations developed for the simplified situations can be adapted to include,

• Centrifugal effects.
• Gravity.
• Spring force.
• Rotor acceleration.

By summing the moments associated with each force along the blade, a more general solution for a freely turning rotor is obtained.

General Solution for Forced Rotor

From the solution for a freely turning rotor, the general solution can be derived by considering the additional force and moment contributions from,

• Steady winds; a linear aerodynamics model is used, i.e. the following assumptions are made:
• Rotational speed >> wind speed.
• CL depends linearly on angle of attack (α).
• α is small.
• No wake rotation.
• Constant chord and twist.
• Yaw error.
• Wind shear.

The result is a general model for blade deflection and vibration. The general model equation(s) are difficult to solve for the most general set-up but solutions for specific situations fall out relatively easily.

#### Other Solution Methods

More faithful models of blade deflection can be constructed using computational methods such as finite elements. These numerical approaches require computer hardware to work through the solutions and, for this reason, they are unlike the pen and paper solution described in overview here.