MODELING MECHANICAL AND ELECTRICAL
2.2 WORK AND ENERGY METHODS
2.2.1 Mechanical Work
which yields
D1=0
D2=0.0013 C/m2 D3=0.0159 C/m2.
This example illustrates that the electric field applied in the direction that is isotropic (in this case, the 1 direction) produces electric displacement in the same direction.
Applying the field in a direction that exhibits anisotropy (the 3 direction) produces an electric displacement vector that has components in orthogonal directions.
f 1
u 2
du
f
Figure 2.9 Concept of a differential unit of work applied to a moving particle.
path,dui, the force produces a differential amount of work,dW, defined by
dW=f·du. (2.79)
The total work on the particle fromu1tou2is defined by the integral W12=
u2
u1
f·du. (2.80)
The units of work are N·m, or joules, which is denoted by the symbol J.
An important feature of work, as compared to force and displacement, is thatwork is a scalar quantity, due to the fact that it is a dot product of the force vector with the dis- placement vector. Another important property of the dot product is that only the force that is in the direction of the motion performs work; force that is orthogonal to the di- rection of the motion does no work on the particle. Although work is a scalar quantity, the sign of the work is important. Positive work is work that is performed in the direc- tion of motion, whereas negative work is work that is in a direction opposite the motion.
The force vector shown in Figure 2.9 is an example of a prescribed force that is independent of the particle location. In many instances, though, the force is a function of the particle displacement, f=f(u), and the work performed on the particle is obtained from the expression
W12= u2
u1
f(u)·du. (2.81)
In this case the work expression can be integrated to yield a function ofu that we define as an energy function U,
U= u2
u1
f(u)·du. (2.82)
The integrand of equation (2.82) is defined as a differential amount of work,
dU=f·du. (2.83)
Expanding equation (2.83) in a Cartesian coordinate system defined in the 1, 2, and 3 directions results in the expressions
dU=(f1xˆ1+ f2xˆ2+ f3xˆ3)·(du1xˆ1+du2xˆ2+du3xˆ3)
= f1du1+ f2du2+ f3du3. (2.84)
The energy function U is a function of the displacement coordinatesui; therefore, we can express the total differential of the function U as
dU= ∂U
∂u1
du1+ ∂U
∂u2
du2+ ∂U
∂u3
du3. (2.85)
Equating equations (2.84) and (2.85), we see that f1 = ∂U
∂u1
f2 = ∂U
∂u2
f3 = ∂U
∂u3
. (2.86)
Equation (2.86) is an important relationship between the energy function and the forces that act on a system. The expressions state that the force in a particular coor- dinate direction is equal to the partial derivative of the energy function with respect to the displacement in that direction. This result provides another interpretation of a force that is dependent on the motion of the particle because it allows us to deriveall of the forcesfrom a single scalar energy function.
The relationship between energy and force can be visualized if we restrict ourselves to a single dimension. Let us assume for the moment that the motion of the particle consists of only a single dimension denoted by the displacementu. Figure 2.10a is a plot of a generic force-to-displacement relationship for a single dimension. The differential unit of energy is defined as the rectangle shown in the figure, which represents the quantity f(u). Integrating this function from u1 to u2 yields the area under the curve as shown in Figure 2.10b.
The relationships described by equation (2.86) can be visualized by drawing a generic energy function U as shown in Figure 2.11. Since we are assuming that the system is a function of only a single dimension, equation (2.86) is reduced to
f = dU
du. (2.87)
u f(u)
du dU = f(u) du
u f (u)
(a) (b)
u2 u1
total work
Figure 2.10 (a) Differential unit of energy in a generic force-to-displacement relationship; (b) total work performed as the area under the force–displacement curve.
The force is visualized as the slope of the curve at a particular point:
f(u1)= dU du
u1
. (2.88)
This relationship is also shown in Figure 2.11. It is interesting to note that in this example it is clear that the slope, and hence the force, varies as a function ofu.
We will find that there are a number of energy functions that will be useful to us in our study of active materials. Another such function is thepotential energyof a system, defined as the negative of the energy function U and denoted
−V=U. (2.89)
The potential energy is defined in this manner due simply to the convention that the potential energy function of a systemdecreaseswhen work is performed on the
u U
u1 f(u1)
Figure 2.11 Relationship between energy and force.
system. The differential of the potential energy is defined as
dV= −f(u)·du. (2.90) Combining the relationships in equation (2.86) with the definition of potential energy allows us to write a succinct relationship between force and potential energy:
f= −∇V. (2.91)
Equation (2.91) states that the force vector is the negative gradient of the potential function V.
Example 2.6 The potential energy function for a nonlinear spring is V=1
2u2+1
4u4. (2.92)
Plot the potential energy function over the range−2 to 2 and determine the force at u= 1 andu = −1.5.
Solution Equation (2.87) defines the relationship between the internal force and the energy function U. Combining this expression with equation (2.89) yields
−dV
d x = f(u)= −u−u3. (2.93)
Substitutingu =1 andu = −1.5 into this expression, we have
f(1)= −1−(1)3= −2 (2.94)
f(−1.5)=1.5−(−1.5)3 =4.875. (2.95) Plots of the potential energy and force are shown in Figure 2.12. As shown, the force can be interpreted as the negative of the slope of the potential energy function at the elongation specified. As the potential energy function illustrates, the slope of V changes at u=0; therefore, the sign of the force changes as the nonlinear spring changes from extension to compression.
The discussion so far has centered on the work done to a particle by an applied force. If we consider the situation where we have stress applied at the body of an elastic body (as shown in Figure 2.1), we define the workper unit volumeperformed by the applied stress as
W˜12 = S2
S1
T·dS. (2.96)
–2 –1 0 1 2 0
1 2 3 4 5 6
(a) (b)
V
u
–10 –8 –6 –4 –2 0 2 4 6 8 10
f
–2 –1 0 1 2
u
Figure 2.12 Plots of the (a) potential energy and (b) force associated withV= 12u2+14u4.
The integral is written in indicial notation as W˜12=
S2
S1
TidSi (2.97)
and the units of this work term are joules per unit volume, J/m3.
The energy expression for an elastic body is derived by incorporating the relation- ship between stress and strain into the work expression. For a general elastic body the stress is written as a function of the strain,T(S), and the integration in equation (2.96) can be expressed as
U˜ = S2
S1
T(S)·dS. (2.98)
For a linear elastic body the stress–strain relationship is written as equation (2.31), and the energy function is
U˜ = S2
S1
cS·dS. (2.99)
The energy function for a linear elastic body can be integrated to yield U˜ = 1
2ScS S2
S1
, (2.100)
which demonstrates that the energy function for a linear elastic body is a quadratic function of the strain.