Inertial Loads

Deformable bodies undergoing rigid body motions are subject to inertial loads. To model these phenomena correctly it is convenient to separate rigid displacements due to rigid motion from non-rigid displacements due to the deformation of the solid.

In Solid Mechanics, the total motion of the body can be described in a moving reference frame.

Kinematics of Deformable Bodies Subject to Rigid Motions

Considering the body without the effects of prescribed rigid body motion, the position vector for the deformed configuration is given as:

\hat{x} = X + \hat{u}

(4465)

where

X

is the undeformed position vector and

\hat{u}

is the displacement, defined in the moving reference frame. The current position vector in the laboratory coordinate system can be expressed as:

x_{n} = R \hat{x} + T = R (X + \hat{u}) + T

(4466)

where $R$ is the rotation matrix and $T$ is the translation vector.

To determine the material velocity and acceleration in the presence of a rigid body motion, the current position vector is expressed in the time-dependent form:

x_{n} (X, t) = R (t) (X + \hat{u} (X, t)) + T (t)

(4467)

From this, the total derivative with respect to time provides the material velocity at point $x_{n}$ :

v_{d} (X, t) = {\dot{x}}_{n} = \frac{d x_{n}}{d t} = \dot{R} (t) (X + \hat{u} (X, t)) + R (t) \hat{v} (X, t) + \dot{T} (t)

(4468)

where

\hat{v}

is the velocity associated with the deformations

\hat{u}

\hat{v} = \dot{\hat{u}}

(4469)

The time derivative of the rotation matrix is often expressed as:

\dot{R} = Ω R

(4470)

where

Ω

is the angular velocity tensor:

Ω = [\begin{matrix} 0 & {\begin{matrix} - ω \end{matrix}}_{z} & ω_{y} \\ ω_{z} & 0 & {- ω}_{x} \\ {- ω}_{y} & ω_{x} & 0 \end{matrix}]

(4471)

The angular velocity tensor can also be defined as the cross product between the angular velocity vector (

ω

) and the current position vector:

Ω x_{n} = ω \times x_{n}

(4472)

Inserting Eqn. (4470) into Eqn. (4468), and removing the explicit time dependence of the different variables, leads to an alternative expression for the velocity:

v_{d} = \dot{x_{n}} = \frac{d x_{n}}{d t} = Ω R (X + \hat{u}) + R \hat{v} + \dot{T}

(4473)

The acceleration of the position vector follows as the total time derivative of the velocity expression:

a_{d} = \ddot{x} = \frac{d v_{d}}{d t} = [\dot{Ω} R + Ω Ω R] (X + \hat{u}) + 2 Ω R \hat{v} + R \hat{a} + \ddot{T}

(4474)

To identify the different contributions to the acceleration, the terms in Eqn. (4474) can be regrouped as follows:

\begin{array}{l} a_{d} = & \dot{Ω} R (X + \hat{u}) & E u l e r A c c e l e r a t i o n \\ + Ω Ω R (X + \hat{u}) & C e n t r i f u g a l A c c e l e r a t i o n \\ + 2 Ω R \hat{v} & C o r i o l i s A c c e l e r a t i o n \\ + R \hat{a} & D e f o r m a t i o n A c c e l e r a t i o n (\hat{a} = \frac{d \hat{v}}{d t}) \\ + \ddot{T} & T r a n s l a t i o n a l R i g i d B o d y A c c e l e r a t i o n \end{array}

(4475)

Virtual work due to Inertial Forces

Following Eqn. (4475), the virtual work is expressed with respect to the initial configuration using the material density per unit of undeformed volume (

ρ_{0}

). The first term of Eqn. (4463) can be rewritten as follows, substituting

a

for

\ddot{u}

\begin{array}{l} {δ Π}^{m} (δ \hat{u}, \hat{u}, \hat{v}, \hat{a}) = & \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} \dot{Ω} R (X + \hat{u}) d V & (= {δ Π}^{m, a} (δ \hat{u}, \hat{u})) \\ + \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} Ω Ω R (X + \hat{u}) d V & (= {δ Π}^{m, c f} (δ \hat{u}, \hat{u})) \\ + {2 \int}_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} Ω R \hat{v} d V & (= {δ Π}^{m, c o} (δ \hat{u}, \hat{v})) \\ + \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{T} R \hat{a} d V & (= {δ Π}^{m, \hat{a}} (δ \hat{u}, \hat{a})) \\ + \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} \ddot{T} d V & (= {δ Π}^{m, t} (δ \hat{u})) \end{array}

(4476)

The fourth term is the inertial load due to local acceleration. This term leads to the mass matrix contribution in the finite element discretization. The fifth term is the translational rigid body motion acting as an external load, independent of the unknowns $\hat{u}$ , and its derivatives $\hat{v}$ and $\hat{a}$ .

The centrifugal force (second term) and the Euler acceleration (first term), are loads which are linearly dependent on $\hat{u}$ . Similarly, the Coriolis load (third term) depends linearly on the velocities ( $\hat{v}$ ). The Coriolis load therefore leads to a linear damping contribution in the finite element problem but does not dissipate energy (unlike viscous damping).

Linearization of the Inertial Force Virtual Work

The linearization of the inertial work terms is straightforward. For completeness the centrifugal force and Euler acceleration terms are as follows:

Δ ({δ Π}^{m, a} (δ \hat{u}, \hat{u})) = \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} \dot{Ω} R Δ \hat{u} d V = \int_{V_{0}}^{} ρ_{o} {(R δ \hat{u})}^{⊤} \dot{Ω} (R Δ \hat{u}) d V

(4477)

Δ ({δ Π}^{m, c f} (δ \hat{u}, \hat{u})) = \int_{V_{0}}^{} ρ_{o} δ \hat{u} \cdot R^{⊤} Ω Ω R Δ \hat{u} d V = \int_{V_{0}}^{} ρ_{o} (Ω^{⊤} R δ \hat{u}) \cdot (Ω R Δ \hat{u}) d V = {- \int}_{V_{0}}^{} ρ_{o} (Ω R δ \hat{u}) \cdot (Ω R Δ \hat{u}) d V

(4478)

The linearization of the centrifugal force introduces an apparent stiffness in the system (centrifugal stiffness or spin softening effect). This stiffness is important for rotor applications and also influences the modal analysis of rotating structures. From Eqn. (4478) the importance of this term increases quadratically as the rotational speed increases.

To clarify the anti-symmetry of the linearizations the integrand terms have been rearranged. Eqn. (4477) shows the symmetric bilinear form of the centrifugal stiffness term, whilst the linearization of the Euler acceleration term is in a skew-symmetric bilinear form due to the skew-symmetric nature of the angular acceleration tensor ( $\dot{Ω}$ ). The Euler acceleration term is only relevant in transient simulations with a time-dependent rotational speed.

A similar rearrangement of the Coriolis term (

{δ Π}^{m, c o} (δ \hat{u}, \hat{v})

) leads to:

{δ Π}^{m, c o} (δ \hat{u}, \hat{v}) = 2 \int_{V_{0}} ρ_{0} δ \hat{u} \cdot R^{T} Ω R \hat{v} d V = 2 \int_{V_{0}} ρ_{0} {(R δ \hat{u})}^{T} Ω (R \hat{v}) d V

(4479)

Considering the skew-symmetry of the angular velocity, Eqn. (4476) shows the Coriolis term is in a skew-symmetric bilinear form. Owing to its linearity in the velocities (

\hat{v}

), the linearization of this term follows:

Δ ({δ Π}^{m, c o} (δ \hat{u}, \hat{v})) = 2 \int_{V_{0}} ρ_{0} {(R δ \hat{u})}^{T} Ω (R Δ \hat{v}) d V

(4480)