MODELING MECHANICAL AND ELECTRICAL
2.6 VARIATIONAL PRINCIPLE OF DYNAMIC SYSTEMS
Applying the variation to the energy and work terms produces δVT =kll12sinψcosψδψ+ktψδψ δWM = f l2cosψδψ.
Combining these two terms according to the variational principle, equation (2.173), yields
fl2 cosψ−kll21sinψcosψ−ktψ
δψ=0.
For this expression to be valid, the term in parentheses must be equal to zero for arbitrary variational displacements. This leads to the equilibrium expression
kll12sinψcosψ+ktψ=fl2 cosψ.
For small angles, sinψ≈ψand cosψ≈1, leading to the expressions kll21+kt
ψ=fl2.
This equation represents the equilibrium expression for small angles. Note that the stiffness of the system is a combination of the stiffness due to the torsional spring and the stiffness due to the liner spring. The stiffness due to the linear spring is modified by the square of the distance due to the lever.
where we must explicitly define the fact that the force and coordinates are time- dependent functions. Also, Newton’s laws state that the sum of the forces in a dynamic system are equal to the time derivative of the linear momentum. The linear momentum is denotedp(t) in equation (2.184). To proceed with the analysis without too much confusion regarding notation, let us assume that all functions in equation (2.184) are functions of time and simply write the expression as
f= dp
dt, (2.185)
for clarity. Equation (2.185) is rewritten in a manner similar to a system in static equilibrium, by subtracting the time derivative of the linear momentum from both sides of the expression:
f−dp
dt =0. (2.186)
Now we can interpret the momentum term as simply a force due to the motion of the particle.
Before proceeding with the derivation, we must reexamine our definition of the variational displacement. For systems in static equilibrium we defined the variational displacement as a differential motion that is consistent with the geometric boundary conditions. For dynamic systems we need to augment this definition with certain assumptions regarding the path that the variation displacement takes as a function of time. Consider a system that begins at the pointu(t1) at timet1 and has the final positionu(t2) at timet2. One interpretation of the solution of the dynamic equations of motion is that we are trying to find the path that the system takes as it moves from u(t1) tou(t2).
Consider a system that begins atu(t1) and travels tou(t2) through the solid path shown in Figure 2.20. Notice that there are an infinite number of paths that connect these two points in the coordinate space; the question arises: Why does the solid path represent the actual motion of the dynamic system? Let’s consider applying a variational displacement to the actual path at any function of time. As stated, this
u(t1)
u(t2)
Figure 2.20 Paths associated with the variational displacements for a dynamic problem.
variational displacement is consistent with the geometric boundary conditions of the problem. Furthermore, we assume thatthe variational displacement at time t1and t2
is equal to zero. Effectively, we are assuming that the initial and final positions of the system are fixed and cannot be varied. With this assumption we can state that
δu(t1)=δu(t2)=0. (2.187)
Continuing with the derivation, next we take the dot product of equation (2.186) with the variational displacement,
f·δu−dp
dt ·δu=0. (2.188)
We recognize that the first term of equation (2.188) can be written as the varia- tion of the total external work,δWM+δWE, and the variation of the total potential energy,−δVT = −δVM−δVE; thus, we can write
δWM+δWE −δVT−dp
dt ·δu=0. (2.189)
The question is how to eliminate the momentum term in equation (2.189). Under the assumption of Newtonian mechanics, the momentum can be written as
p(t)=mdu
dt, (2.190)
wheremis the mass. Substituting the momentum expression, equation (2.190), into equation (2.189), and integrating between t1and t2yields
t2 t1
(δWM+δWE −δVT)dt− t2
t1
mu˙·δudt=0, (2.191) where the overdot represents differentiation with respect to time. Applying integration by parts to the momentum term produces
t2
t1
mu˙·δudt =mu˙·δut2
t1− t2
t1
mu˙· d
dtδudt. (2.192)
The terms evaluated att1andt2are equal to zero by our definition of the variational displacement, and the term in the integral can be rewritten
− t2
t1
m ˙u· d
dtδudt= − t2
t1
m ˙u·δu˙dt (2.193)
by virtue of the fact that variation and differentiation are interchangeable operations.
Now we must realize that the term in the integral can be written as a variation of δm
2 ˙u·˙u
=m˙u·δ˙u. (2.194)
Denoting the term in parentheses on the left-hand side of the expression as a variation of thekinetic energy, T, we can write
δT=m˙u·δ˙u. (2.195)
Substituting equation (2.195) into equation (2.193) and then incorporating into the work and energy expression, equation (2.191), yields
t2
t1
(δWM+δWE −δVT+δT)dt=0. (2.196) The term T−VTis called theLagrangianand is given the symbolL. The variational operator is additive; therefore, we can write equation (2.196) as
t2
t1
{δL+δWM+δWE}dt=0. (2.197)
Equation (2.196) is the variational principle for dynamic systems and is often called Hamilton’s principle.
The variational principle for dynamic systems is equally powerful as the principle applied to systems in static equilibrium. Once again it transforms the problem of solving for the equations of motion from one involving vector terms to one that only involves scalar quantities. The additional terms that are required for dynamic analysis are incorporated in the kinetic energy of the Lagrangian. The kinetic energy can be visualized as the energy associated with the motion of the system, whereas the potential energy is the energy stored in the system.
Another benefit of the variational approach is that application of the method is almost identical to application of the method for systems in static equilibrium:
1. Choose a complete set of independent generalized coordinates and generalized velocities.
2. Write the potential and kinetic energy functions in terms of the generalized coordinates and find the variationδVT andδT.
3. Determine the work expression for each external force on the system and find the variationδWM+δWE.
4. Apply equation (2.196) and collect the terms associated with each independent variational displacement. Due to the fact that the variational displacements are independent, the coefficients that multiply the variational displacements must all be equal to zero. These coefficients are the equilibrium expressions for the system.
The concept of generalized coordinates applies to the variational principle for dynamic systems in the same manner as for the variational principle for systems in equilibrium. The displacement vector is written as a function of a set of generalized coordinate that form a complete set and are independent. Once this has been done, the Lagrangian can be written as
L( ˙r1, . . . ,r˙Nr,r1, . . . ,rNr,q1, . . . ,qNq)
=T( ˙r1, . . . ,r˙Nr)−VT(r1, . . . ,rNr,q1, . . . ,qNq). (2.198) The variation of the Lagrangian is
δL=
Nr
i=1
∂T
∂r˙iδr˙i+
Nr
i=1
∂VT
∂ri δri+
Nq
j=1
∂VT
∂qj δqj. (2.199)
Substituting the variation of the Lagrangian into equation (2.197) and combining it with the expressions for the generalized mechanical forces and the electrical work yields
t2 t1
N r
i=1
Fi−∂VT
∂ri
δri+ ∂T
∂r˙i
δr˙i+
Nq
j=1
vj−∂VT
∂qj
δqj
dt=0. (2.200)
The integrand of equation (2.200) contains variations of both the generalized coordi- nates and the time derivatives of the generalized coordinates, also calledgeneralized velocities. Before the governing equations can be derived, all variations of the gen- eralized velocities must be eliminated from the integrand. The generalized velocities are eliminated by applying integration by parts to the terms associated withδr˙i:
t2
t1
∂T
∂r˙i
δr˙idt= ∂T
∂r˙i
δri
t2
t1
− t2
t1
d dt
∂T
∂ri
δridt. (2.201)
The first term on the right-hand side of equation (2.201) is equal to zero since the generalized coordinates are defined to be equal to zero att1andt2. Setting this term equal to zero, we can combine equations (2.200) and (2.201) and write the variational principle as
t2 t1
N r
i=1
Fi−∂VT
∂r˙i
− d dt
∂T
∂r˙i
δri+
Nq
j=1
vj−∂VT
∂qj
δqj
dt=0.
(2.202) Eliminating the generalized velocities allows us to obtain the governing equations from the variational principle. Since the variational coordinates δri and δqj are
independent, the integral is equal to zero if and only if the set of equations Fi−∂VT
∂ri
− d dt
∂T
∂r˙i
=0 i =1, . . . ,Nr
(2.203) vj−∂VT
∂qj
=0 j=1, . . . ,Nq
are satisfied. These are the set of governing equations for the dynamic system. Rewrite the governing equations as
Fi = ∂VT
∂ri
− d dt
∂T
∂r˙i
i =1,· · ·,Nr
(2.204) vj = ∂VT
∂qj
j =1,· · ·,Nq.
The governing equations for a dynamic system are similar in form to the governing equations for a system in static equilibrium, equation (2.183), except for the ad- ditional time derivative of the kinetic energy term. This term represents the forces associated with the motion of the system. There are no time derivative terms in the governing equations for the electrical system since the kinetic energy does not have any functional dependence on the time derivative of the charge coordinates.
Example 2.10 Derive the equations of motion for the system shown in Figure 2.18 when a massmis placed at each node.
Solution The coordinates of the system are defined asu1andu2. The kinetic energy of the system is the summation of the kinetic energy terms for the two masses:
T= 1
2mu˙21+1 2mu˙22. The potential energy term is identical to that in Example 2.9:
VT = 1
2ku21+1
2αk(u2−u1)2+1 2ku22.
To apply equation (2.205), it is necessary to find the generalized mechanical forces and partial derivatives of the potential and kinetic energy functions. As in Example 2.9, the electrical work and energy terms are zero since there are no electrical elements.
Finding the partial derivatives results in
∂T
∂u˙1
=mu˙1
∂T
∂u˙2
=mu˙2
∂VT
∂u1 =ku1+αk(u2−u1)(−1)
∂VT
∂u2 =αk(u2−u1)+ku2. Taking the time derivative of the kinetic energy terms yields
d dt
∂T
∂u˙1 =mu¨1
d dt
∂T
∂u˙2
=mu¨2.
The generalized mechanical forces are F1= f1 F2= f2.
Combining the terms according to equation (2.205) produces the governing equations f1 =ku1−αk(u2−u1)+mu¨1
f2 =αk(u2−u1)+ku2+mu¨2.
The two equations can be rewritten with the forcing terms on the right-hand side as mu¨1+k(1+α)u1−αku2= f1
mu¨2−αku1+k(1+α)u2= f2
and the two equations can also be rewritten in matrix form as
m 1 0
0 1
¨ u1
¨ u2
+k
1+α −α
−α 1+α u1 u2
= f1
f2
,
which is a standard second-order form for vibrating systems.