6161103 11.7 stability of equilibrium

11.7 Stability of Equilibrium

Once the equilibrium configuration for a body or
system of connected bodies are defined, it is
sometimes important to investigate the type of
equilibrium or the stability of the configuration
Example
Consider a ball resting on each
of the three paths
Each situation represent an
equilibrium state for the ball


When the ball is at A, it is at stable
equilibrium
Given a small displacement up the hill, it will
always return to its original, lowest, position
At A, total potential energy is a minimum
When the ball is at B, it is in neutral
equilibrium
A small displacement to either the left or
right of B will not alter this condition
The balls remains in equilibrium in the
displaced position and therefore, potential
energy is constant


When the ball is at C, it is in unstable
equilibrium
A small displacement will cause the ball’s
potential energy to be decreased, and so it
will roll farther away from its original, highest
position
At C, potential energy of the ball is maximum

Types of Equilibrium
1. Stable equilibrium occurs when a small
displacement of the system causes the system
to return to its original position. Original
potential energy is a minimum
2. Neutral equilibrium occurs when a small
to remain in its displaced state. Potential
energy remains constant
3. Unstable equilibrium occurs when a small
to move farther from its original position.
Original potential energy is a maximum


System having One Degree of Freedom
For equilibrium, dV/dq = 0
For potential function V = V(q), first
derivative (equilibrium position) is
represented as the slope dV/dq which is
zero when the function is maximum,
minimum, or an inflexion point
Determine second derivative and evaluate
at q = qeq for stability of the system


If V = V(q) is a minimum,
dV/dq = 0
d2V/dq2 > 0
stable equilibrium
If V = V(q) is a maximum
dV/dq = 0
d2V/dq2 < 0
unstable equilibrium


If d2V/dq2 = 0, necessary to investigate higher-
order derivatives to determine stability
Stable equilibrium occur if the order of the
lowest remaining non-zero derivative is even and
the is positive when evaluated at q = qeq
Otherwise, it is unstable
For system in neutral equilibrium,
dV/dq = d2V/dq2 = d3V/dq3 = 0


System having Two Degree of Freedom
A criterion for investigating the stability becomes
increasingly complex as the number for degrees of
freedom for the system increases
For a system having 2 degrees of freedom,
equilibrium and stability occur at a point (q1eq, q2eq)
when
δV/δq1 = δV/δq2 = 0
[(δ2V/δq1δq2)2 – (δ2V/δq12)(δ2V/δq22)] < 0
δ2V/δq12 > 0 or δ2V/δq22 >0


System having Two Degree of Freedom
Both equilibrium and stability occur when
δV/δq1 = δV/δq2 = 0
[(δ2V/δq1δq2)2 – (δ2V/δq12)(δ2V/δq22)] < 0
δ2V/δq12 > 0 or δ2V/δq22 >0

Procedure for Analysis
Potential Function
Sketch the system so that it is located at some
arbitrary position specified by the independent
coordinate q
Establish a horizontal datum through a fixed point
and express the gravitational potential energy Vg
in terms of the weight W of each member and its
vertical distance y from the datum, Vg = Wy
Express the elastic energy Ve of the system in
terms of the sketch or compression, s, of any
connecting spring and the spring’s stiffness, Ve =
½ ks2


Potential Function
Formulate the potential function V = Vg + Ve
and express the position coordinates y and s
in terms of the independent coordinate q

Equilibrium Position
The equilibrium position is determined by
taking first derivative of V and setting it to
zero, δV = 0

Stability
Stability at the equilibrium position is determined
by evaluating the second or higher-order
derivatives of V
If the second derivative is greater than zero, the
body is stable
If the second derivative is less than zero, the
body is unstable
If the second derivative is equal to zero, the
body is neutral


Example 11.5
The uniform link has a mass of
10kg. The spring is un-stretched
when θ = 0°. Determine the
angle θ for equilibrium and
investigate the stability at the
equilibrium position.

Solution
Potential Function
Datum established at the top of the
link when the spring is un-stretched
When the link is located at arbitrary
position θ, the spring increases its
potential energy by stretching and
the weight decreases its potential
energy


Solution
V = Ve + Vg
1 2  l 
= ks − W  s + cos θ 
2  2 
Since l = s + l cos θ or s = l (l − cos θ ),
1 2 Wl
V = kl (1 − cos θ ) − (2 − cosθ )
2

2 2


Solution
For first derivative of V,
dV Wl
= kl 2 (1 − cos θ )sin θ − sin θ = 0
dθ 2
or
 W
l kl (1 − cos θ ) −  sin θ = 0
 2
Equation is satisfied provided
sin θ = 0, θ = 0o
 W   10(9.81) 
θ = cos −1 1 −  = cos −1 1 −  = 53.8o
 2(200)(0.6) 
 2kl   

Solution
Stability
For second derivative of V,
d 2V Wl
= kl 2 (1 − cosθ )cosθ + kl 2 sin θ sin θ − cosθ
dθ 2 2
Wl
= kl 2 (cosθ − cos 2θ ) − cosθ
2
Substituting values for constants
d 2V
dθ 2 θ = 0o
= ( )
200(0.6) 2 cos 0o − cos 0o −
10(9.81)(0.6)
2
cos 0o

= −29.4 < 0 unstable equilibrium
d 2V
dθ 2 θ =53.8o
= (
200(0.6) 2 cos 53.8o − cos 53.8o − ) 10(9.81)(0.6)
2
cos 53.8o

= 46.9 > 0 stable equilibriu m


Example 11.6
Determine the mass m of the
block required for equilibrium
of the uniform 10kg rod when θ
= 20°. Investigate the stability
at the equilibrium position.


Solution
Datum established through
point A
When θ = 0°, assume block
to be suspended (yw)1 below
the datum
Hence in position θ,
V = Ve + Vg
= 9.81(1.5sinθ/2) –
m(9.81)(∆y)


Solution
Distance ∆y = (yw)2 - (yw)1 may be related to
the independent coordinate θ by measuring
the difference in cord lengths B’C and BC
B' C = (1.5)2 + (1.2)2 = 1.92
BC = (1.5 cos θ )2 + (1.2 − sin θ )2 = 3.69 − 3.60 sin θ
∆y = B ' C − BC = 1.92 − 3.69 − 3.60 sin θ
 1.5 sin θ 
V = 9.81 (
 − m(9.81) 1.92 − 3.69 − 3.60 sin θ )
 2 


Solution
For first derivative of V,
3.60 cosθ
= 73.6 cosθ − 
dV m(9.81)  =0
dθ  2  3.69 − 3.60 sin θ 
 


dV
= 69.14 − 10.58m = 0
dθ θ = 20 o

For mass,
69.14
m= = 6.53kg
10.58

Solution
Stability
d 2V  m(9.81)  − 1  − (3.60 cos θ ) 
2

dθ 2
= −73.6 sin θ −   2  (3.69 − 3.60 sin θ )3 / 2 
  
 2   
 m(9.81)  − 3.60 sin θ 
−  
 2  3.69 − 3.60 sin θ 

For equilibrium position,
d 2V
= −47.6 < 0 unstable equilibrium
dθ 2


Example 11.7
The homogenous block having a mass m
rest on the top surface of the cylinder. Show
that this is a condition of unstable equilibrium
if h > 2R.

Solution
Datum established at the base of the cylinder
If the block is displaced by an amount θ from the
equilibrium position, for potential function,
V = Ve + Vg
= 0 + mgy
y = (R + h/2)cosθ + Rθsinθ
Thus,
V = mg[(R + h/2)cosθ + Rθsinθ]


Solution
dV/dθ = mg[-(R + h/2)sinθ + Rsinθ
+Rθcosθ]=0
=mg[-(h/2)sinθ + Rθsinθ] = 0
Obviously, θ = 0° is the equilibrium position
that satisfies this equation

Solution
Stability
d2V/dθ2 = mg[-(h/2)cosθ + Rcosθ - Rθsinθ]
At θ = 0°,
d2V/dθ2 = - mg[(h/2) – R]
Since all the constants are positive, the block is in
unstable equilibrium
If h > 2R, then d2V/dθ2 < 0

6161103 11.7 stability of equilibrium

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