OSE800: Analytical and Computational
Dynamics (Spring, 2013)
Lecture 3: D’Alembert’s Principle
In this lecture we will learn:
• D’Alembert’s Principle
• The concept of virtual displacements
• Why the reactions do disappear in the final equations of motions
• How we can recover reaction forces (and moments, if any) from the Free-Body Diagrams, if needed.
3.1
D’Alembert’s contributions to mechanics
D’Alembert’s own words: The principle of equilibrium, together
with the principle of the force of inertia, and the principle of
compound motion, therefore leads us to the solution of all problems which concern the motion of body in so far as it can be
stopped by an impenetrable and immovable obstacle – that is,
in general, by another body to which it must necessarily impart
motion in order to keep at least a part of its own. From these
(three) principle together can easily be deduced the laws of the
motion of bodies that collide in any manner whatever, or which
affect each other by means of some body placed between them
and to which they are attached.
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Fig. 1. Traité de Dynamique (par Jean D’Alembert), Preleminaire
(page XV))
On the force of inertia: the property of bodies of remaining in
their state of rest or motion.
On time: The nature of time is to run uniformly, and mechanics
supposes this uniformity.
On equilibrium: Lagrange said, ”D’Alembert had reduced dynamics to statics by means of his principle.”
On the center of oscillation: the concept of equilibrium in dynamical systems embodied in D’Alembert’s principle subsumed the
work of Jacques Bernoulli (1654-1705), Jacob Herman (16781733) and Leonhard Euler(1707-1783).
On the laws of impact: the velocity of two bodies after impact
given by
v=
mu + M U
m+M
: The principle of momentum conservation!
(3.1)
On D’Alembert’s paradox: For incompressible and inviscid potentia lflow, the drag force is zero on a body moving with a constant velocity relative to the fluid (due to neglect of the effects of
viscosity!).
Finally, he derived the wave equation while working on the vibration of musical instruments.
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3.2
D’Alembert’s Principle
Principle of Virtual Work for Statics
For statics, we have:
N
X
f i · δri = 0
(3.2)
i=1
where {f i } are impressed forces, and
{δ} designates the virtual character of the instantaneous variations, as opposed to the differential symbol {d} designating the
actual differential of the position vector {ri } taking place during
the time interval {dt}.
Principle of Virtual Work for Dynamics
For dynamics á la d’Alembert:
N
X
(Fi − mi r̈i ) · δri = 0
(3.3)
i=1
where {r̈i } is the acceleration of particle mi .
Generalized Coordinates
Virtual displacements δri may be expressed in terms of the generalized virtual displacements δqk (k = 1, 2, 3, ..., n):
δri =
n
X
∂ri
δqk , i = 1, 2, 3, ..., N
k=1 ∂qk
(3.4)
Let’s consider the following example:
r = xi + yj + L(cos θi + sin θj)
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(3.5)
Fig. 2. A Spring-Mass System
Observe that there are three independent variables, (x, y, θ),
yielding the following virtual displacement:
δr =
∂r
∂r
∂r
δx +
δy + δθ
∂x
∂y
∂θ
(3.6)
= δxi + δyj + L(− sin θi + cos θj) δθ
Note that the dimensional unit of (x, y) is in meter(m) whereas
that of θ is in radian (non-dimensional quantity), suggesting that
the units of generalized coordinates can be different.
3.3
A Spring-Mass-Bar System Revisited:
Let’s revisit the formulation of the equations of motion for the
spring-mass-bar problem studied in the previous lecture. To this
end, first, we must obtain the virtual displacements.
Virtual Displacements:
Virtual displacement for mass m:
δrA0 = δx i
4
(3.7)
Virtual displacement for bar M :
δrC 0 = δrA0 + δrA0 C 0
L
δrA0 C 0 = (cos θi + sin θj) δθ
2
(3.8)
Virtual rotation for bar M :
Here, one utilizes the angular velocity ω = θ̇k to obtain the
virtual rotation
δθ = δθk
(3.9)
δθ = δθx i + δθy j + δθz k
(3.10)
In general one has
Acceleration Vectors:
Acceleration vector for the sliding mass m:
aA0 = v̇A0 = ẍi
Acceleration vectors for the pendulum bar M :
aC 0 = aA0 + aA0 C 0
aA0 C 0 =
Lθ̈
(cos θi
2
+ sin θj) −
Lθ̇2
(sin θi
2
− cos θj)
Equilibrium equation for the sliding mass m:
f 0A = −kxi − mẍi − mgj + NA j + XAC i + YAC j = 0
Equilibrium equation for the pendulum bar M :
f 0C = F i − M aC 0 − M gj − XAC i − YAC j = 0
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Fig. 3. Free-body diagram for mass
Fig. 4. Free-body diagram for the bar
2
L θ̈
MC 0 = − M12
k + rC 0 A0 × (−XAC i − YAC j) + rC 0 B 0 × F i = 0
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Application of D’Alembert’s Principle for the springmass-bar problem
3.4
(Fi − mi r̈i ) · δri =f 0A · δr0A + f 0C · δr0C + M0C · δθC0 = 0
X
⇓
{−kxi − mẍi − mgj + NA j
+ XAC i + YAC j} · δr0A
+ {F i − M aC 0 − M gj
− XAC i − YAC j} · δr0C
M L2 θ̈
{−
k + rC 0 B 0 × F i
12
+ rC 0 A0 × (−XAC i − YAC j)} · δθ
(3.11)
Note that f 0A · δr0A 6= 0 and so other two terms!
The first term in (3.11) becomes
f 0A · δr0A = {−kxi − mẍi − mgj + NA j + XAC i + YAC j} · (δxi)
⇓
0
0
f A · δrA = {−kx − mẍ + XAC } δx
(3.12)
Remark: Observe that the reactions forces YAC and NA have
played no role in the resulting virtual work!
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The second term in (3.11) becomes
f 0C · δr0C = {F i − M aC 0 − M gj − XAC i − YAC j} · δr0C
Lθ̈
Lθ̇2
(cos θi + sin θj) −
(sin θi − cos θj)]
2
2
L
− M gj − XAC i − YAC j} · {δxi + (cos θi + sin θj) δθ}
2
⇓
= {F i − M [ẍi +
Lθ̈
Lθ̇2
cos θ −
sin θ] − XAC } δx
2
2
Lθ̈
Lθ̇2
L
+ {F − M [ẍ +
cos θ −
sin θ] − XAC } cos θ δθ
2
2
2
2
Lθ̈
Lθ̇
L
+ {−M [
sin θ −
cos θ] − M g − YAC } sin θδθ
2
2
2
⇓
f 0C · δr0C = {F − M [ẍ +
f 0C · δr0C = {F − M [ẍ +
Lθ̇2
Lθ̈
cos θ −
sin θ] − XAC } δx
2
2
L cos θ
M Lg
ML
M L2
θ̈ −
F−
cos θ ẍ −
sin θ
2
2
4
2
L
L
− cos θ XAC − sin θYAC } δθ
2
2
+{
(3.13)
In evaluating the third term of (3.11), first, carry out:
L
(sin θi − cos θj) × F i
2
FL
cos θk
=
2
rC 0 B 0 × F i =
rC 0 A0 × (−XAC i − YAC j) = {
L
L
cos θXAC + sin θYAC }k
2
2
(3.14)
so that we obtain
M
C0
· δθ
C0
M L2 θ̈ F L
L
L
= {−
+
cos θ + [ cos θXAC + sin θYAC ]}δθ
12
2
2
2
(3.15)
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Substituting (3.12), (3.13) and (3.15) into (3.11), we obtain
N
X
ML
ML
cos θ θ̈ −
sin θ θ̇2 − F } δx
2
2
2
ML
ML
M gL
+{
θ̈ẍ +
cos θ ẍ +
sin θ − L cos θF }δθ = 0
3
2
2
(3.16)
(Fi − mi r̈i ) · δri = {(M + m)ẍ + kx +
i=1
3.5
Coupled equations for x and θ via d’Alembert’s
principle
Since δx and δθ are arbitrary, we obtain:
(M + m)ẍ + kx +
ML
ML
cos θ θ̈ −
sin θ θ̇2 = F
2
2
(3.17)
M L2
ML
M gL
θ̈ +
cos θ ẍ +
sin θ = L cos θF
3
2
2
Observations
1. When using Newton’s second law, (3.17) are obtained by
eliminating the reaction forces XAC and YAC .
2. On the other hand, one does not need to consider reaction
forces in applying d’Alembert’s principle! Only apparent forces
including inertia forces need to be considered. This is a major
advantage of applying energy principles over Newton’s method.
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