6161103 11.7 stability of equilibrium
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Transcript of 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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
� 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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
� 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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
System having One Degree of Freedom
� 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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
System having One Degree of Freedom
� 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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
2 >0
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
2 >0
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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.
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
θcos22
1 2
+−=
+=
lsWks
VVV ge
( )
( ) ( )θθ
θθ
cos22
cos12
1
,coscos
22
22 −−−=
−=+=
WlklV
llsorlslSince
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Equilibrium Position
� For first derivative of V,
or
� Equation is satisfied provided
( )
( )
o
o
8.53)6.0)(200(2
)81.9(101cos
21cos
0,0sin
0sin2
cos1
0sin2
sincos1
11
2
=
−=
−=
==
=
−−
=−−=
−−
kl
W
Wkll
Wlkl
d
dV
θ
θθ
θθ
θθθθ
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Stability
� For second derivative of V,
( ) Wlklkl
Vdcossinsincoscos1 22
2−+−= θθθθθ
� Substituting values for constants
( )
( )
( )
( )mequilibriustable
d
Vd
mequilibriuunstable
d
Vd
Wlkl
Wlklkl
d
Vd
09.46
8.53cos2
)6.0)(81.9(108.53cos8.53cos)6.0(200
04.29
0cos2
)6.0)(81.9(100cos0cos)6.0(200
cos2
2coscos
cos2
sinsincoscos1
28.532
2
202
2
2
222
>=
−−=
<−=
−−=
−−=
−+−=
=
=
ooo
ooo
o
o
θ
θ
θ
θ
θθθ
θθθθθθ
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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.
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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)
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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
θ
θθθ
sin60.369.392.1)81.9(2
sin5.181.9
sin60.369.392.1'
sin60.369.3sin2.1cos5.1
92.12.15.1'
22
22
−−−
=
−−=−=∆
−=−+=
=+=
mV
BCCBy
BC
CB
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Equilibrium Position
� For first derivative of V,mdV cos60.3)81.9(
=−=θ
θ
� For mass,
kgm
md
dV
m
d
dV
53.658.1014.69
058.1014.69
0sin60.369.3
cos60.32
)81.9(cos6.73
20
==
=−=
=
−
−=
= oθθ
θθ
θθ
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Stability
� For second derivative of V,
mVd )cos60.3(1)81.9( 22 − − θ
� For equilibrium position,
( )
mequilibriuunstabled
Vd
m
m
d
Vd
06.47
sin60.369.3
sin60.3
2
)81.9(
sin60.369.3
)cos60.3(
2
1
2
)81.9(sin6.73
2
2
2/3
2
2
2
<−=
−
−
−
−
−
−
−−=
θ
θθ
θθ
θθ
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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.
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
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]
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Equilibrium Position
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
11.7 Stability of Equilibrium11.7 Stability of Equilibrium
Solution
Stability
� For second derivative of V,
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