6161103 11.7 stability of equilibrium

11.7 Stability of Equilibrium11.7 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 configurationequilibrium 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� 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 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 Equilibrium1. Stable equilibrium occurs when a small

displacement of the system causes the system to return to its original position. Original potential energy is a minimumpotential energy is a minimum

2. Neutral equilibrium occurs when a small displacement of the system causes the system to remain in its displaced state. Potential energy remains constant

3. Unstable equilibrium occurs when a small displacement of the system causes the system 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 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 > 0d2V/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 � 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/δq1

2)(δ2V/δq22)] < 0

δ2V/δq12 > 0 or δ2V/δq2

System having Two Degree of Freedom

� Both equilibrium and stability occur when

δV/δq1 = δV/δq2 = 0

[(δ2V/δq1δq2)2 – (δ2V/δq1

2)(δ2V/δq22)] < 0 [(δ2V/δq1δq2)

2 – (δ2V/δq12)(δ2V/δq2

2)] < 0

δ2V/δq12 > 0 or δ2V/δq2

Procedure for AnalysisPotential Function� Sketch the system so that it is located at some

arbitrary position specified by the independent coordinate qcoordinate q

� Establish a horizontal datum through a fixed point and express the gravitational potential energy Vgin 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

Procedure for Analysis

Potential Function

� Formulate the potential function V = Vg + Ve

and express the position coordinates y and s 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

Procedure for AnalysisStability� Stability at the equilibrium position is determined

by evaluating the second or higher-order derivatives of Vderivatives 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 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-stretchedlink 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

θcos22

VVV ge

( ) ( )θθ

,coscos

22 −−−=

−=+=

llsorlslSince

Solution

� For first derivative of V,

� Equation is satisfied provided

8.53)6.0)(200(2

)81.9(101cos

0,0sin

sincos1

−−

=−−=

−−

θθθθ

Solution

Stability

� For second derivative of V,

( ) Wlklkl

Vdcossinsincoscos1 22

2−+−= θθθθθ

� Substituting values for constants

( )mequilibriustable

mequilibriuunstable

Wlklkl

8.53cos2

)6.0)(81.9(108.53cos8.53cos)6.0(200

)6.0)(81.9(100cos0cos)6.0(200

2coscos

sinsincoscos1

28.532

−−=

−+−=

θθθ

θθθθθθ

Example 11.6

Determine the mass m of the

block required for equilibrium 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 � 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 BCthe difference in cord lengths B’C and BC

( ) ( )

( )θθ

θθθ

sin60.369.392.1)81.9(2

sin5.181.9

sin60.369.392.1'

sin60.369.3sin2.1cos5.1

92.12.15.1'

−−−

−−=−=∆

−=−+=

Solution

� For first derivative of V,mdV cos60.3)81.9(

=−=θ

� For mass,

53.658.1014.69

058.1014.69

0sin60.369.3

cos60.32

)81.9(cos6.73

= oθθ

Solution

Stability

mVd )cos60.3(1)81.9( 22 − − θ

� For equilibrium position,

mequilibriuunstabled

sin60.369.3

sin60.3

)81.9(

sin60.369.3

)cos60.3(

)81.9(sin6.73

−−=

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 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 = V + VV = 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+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] d V/dθ = 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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