Homework Prob. 3.1 Consider the system below, whose motion is described by the absolute coordinates shown.

a) Write down the potential energy function U for this four-DOF system and use the following results from lecture to develop the stiffness matrix for the system:

Kij =∂2U

∂qi ∂q j q0

b) Use the method of influence coefficients to develop the flexibility matrix

A⎡⎣ ⎤⎦ = K⎡⎣ ⎤⎦

−1⎛⎝

⎞⎠ .

c) Check your results in a) and b) above by verifying that A⎡⎣ ⎤⎦ K⎡⎣ ⎤⎦ = I⎡⎣ ⎤⎦ .

k

G

y1 smooth

no slipR

G

B

A

m

2m3m

2k3k

y2 y3

C

m

y4

k

Homework Prob. 3.2 A system is made up of two identical pendulums, with each pendulum comprised of a particle of mass m and a massless link of length L. The first pendulum is pinned to ground at point O. The second pendulum is pinned to the first at particle A, as shown below. The motion of the system is to be described by the angles θ1 and θ2 , where these angles describe the orientation of the pendulum links measured counterclockwise from the vertical. A constant, horizontal force F acts at particle A.

a) Write down an expression for the kinetic energy T in terms of the generalized coordinates θ1 and θ2 and their time derivatives. From this expression, identify

the elements mij , where:

T = 1

2mij!θi!θ j

j=1

2

∑i=1

2

∑

b) Determine the expression for potential energy U and the generalized forces corresponding to F for the generalized coordinates. Use these results to determine the angles θ1 and θ2 corresponding to static equilibrium. Leave these angles in terms of the ratio F/mg.

c) Determine the mass matrix [M] and the stiffness matrix [K] corresponding small oscillations about the equilibrium state for F/mg = 0.

d) Determine the mass matrix [M] and the stiffness matrix [K] corresponding small oscillations about the equilibrium state for F/mg = 2. Compare these with those found in part c).

g

m

O

θ1

L

m

θ2 L

A

B

F

y

x

Kinematics Using the rigid body kinematics equations (see Help Desk for background and notation):

�

v A = v 0 + ˙ θ 1k( ) × r A /O

= 0 + ˙ θ 1k( ) × Lcosθ1i + Lsinθ1 j( )= −L ˙ θ 1 sinθ1i + L ˙ θ 1 cosθ1 j

�

v B = v A + ˙ θ 2k( ) × r B /A

= −L ˙ θ 1 sinθ1i + L ˙ θ 1 cosθ1 j( ) + ˙ θ 2k( ) × Lcosθ2i + Lsinθ2 j( )= −L ˙ θ 1 sinθ1 + ˙ θ 2 sinθ2( )i + L ˙ θ 1 cosθ1 + ˙ θ 2 cosθ2( )i

Therefore,

�

vA2 = v A •v A = L ˙ θ 1 sinθ1( )2

+ L ˙ θ 1 cosθ1( )2

= L ˙ θ 1( )2sin2θ1 + cos2θ1( ) = L2 ˙ θ 1

2

�

vB2 = v B •v B = L ˙ θ 1 sinθ1 + L ˙ θ 2 sinθ2( )2

+ L ˙ θ 1 cosθ1 + L ˙ θ 2 cosθ2( )2

= L2 ˙ θ 12 sin2θ1 + 2 ˙ θ 1 ˙ θ 2 sinθ1 sinθ2 + ˙ θ 2

2 sin2θ2( ) + L2 ˙ θ 12 cos2θ1 + 2 ˙ θ 1 ˙ θ 2 cosθ1 cosθ2 + ˙ θ 2

2 cos2θ2( )= L2 ˙ θ 1

2 sin2θ1 + cos2θ1( ) + 2L2 ˙ θ 1 ˙ θ 2 sinθ1 sinθ2 + cosθ1 cosθ2( ) + L2 ˙ θ 22 sin2θ2 + cos2θ2( )

= L2 ˙ θ 12 + 2L2 ˙ θ 1 ˙ θ 2 cos θ2 −θ1( ) + L2 ˙ θ 2

2

Kinetic energy Since A and B are to be treated as particles (zero moment of inertia about their centers of mass):

�

T = 12mvA

2 + 12mvB

2

= 12mL2 ˙ θ 1

2 + 12m L2 ˙ θ 1

2 + 2L2 ˙ θ 1 ˙ θ 2 cos θ2 −θ1( ) + L2 ˙ θ 22( )

= 12

2mL2( ) ˙ θ 12 + 1

22mL2 cos θ2 −θ1( )( ) ˙ θ 1 ˙ θ 2 + 1

2mL2( ) ˙ θ 2

2

= 12m11

˙ θ 12 + 1

2m12 + m21( ) ˙ θ 1 ˙ θ 2 + 1

2m22

˙ θ 22

Therefore,

�

m11 = 2mL2

m12 = mL2 cos θ2 −θ1( ) = m21m22 = mL2

Potential energy Using a gravitational datum line through point O:

�

U = −mgLcosθ1 −mg Lcosθ1 + Lcosθ2( )= −2mgLcosθ1 −mgLcosθ2

Differential work Finding the work done by force F:

�

dW = F • dr A= F j( ) • d Lcosθ1i + Lsinθ1 j( )= F j( ) • d −Lsinθ1 dθ1 i + Lcosθ1 dθ1 j( )= FLcosθ1( ) dθ1 = Q1 dθ1 + Q2 dθ2

Therefore,

�

Q1 = FLcosθ1Q2 = 0

Equilibrium equations

�

∂U∂θ1

= Q1 ⇒ 2mgLsinθ1 = FLcosθ1 ⇒ θ1( )st = tan−1 F2mg

⎛

⎝ ⎜

⎞

⎠ ⎟

�

∂U∂θ2

= Q2 ⇒ mgLsinθ2 = 0 ⇒ θ2( )st = 0

Mass and stiffness matrices For small motion,

�

M[ ] = m[ ]θ st=

2mL2 mL2 cos θ2,st −θ1,st( )

mL2 cos θ2,st −θ1,st( ) mL2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=2mL2 mL2 cos θ1,st( )

mL2 cos θ1,st( ) mL2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

�

K[ ] = ∂2U∂θi∂θ j

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ θ st

=2mgLcosθ1,st 0

0 mgLcosθ2,st

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥ =

2mgLcosθ1,st 0

0 mgL

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

For F/mg = 0:

�

θ1,st = θ2,st = 0 and

�

M[ ] =2mL2 mL2

mL2 mL2

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

�

K[ ] =2mgL 0

0 mgL

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

For F/mg = 2:

�

θ1,st = tan−1 1( ) = π /4θ2,st = 0

and

�

M[ ] =2mL2 mL2 cos π /4( )

mL2 cos π /4( ) mL2

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

=2mL2 mL2 / 2

mL2 / 2 mL2

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

�

K[ ] =2mgLcos π /4( ) 0

0 mgL

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥ =

2mgL 0

0 mgL

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

As we see above, only the off-diagonal terms in [M] (mass coupling} is influenced by the static equilibrium shape. The stiffness matrix [K] is also influenced in the (1,1) position.

Homework Prob. 3.3 Reconsider a problem from Homework Set No. 2. As stated there, the system is made up of a homogeneous wheel (mass m and radius R), block A (mass of 3m) and block B (mass of 2m). The wheel can roll without slipping on A. Springs are attached between A and ground, between B and ground and between B and the centroid G of the wheel, with stiffnesses of k, 2k and 3k, respectively. The absolute coordinate

x1 describes the

position of A, the relative coordinate x2 describes the position of B relative to A and θ

is the rotation of the wheel measured from a vertical line. All springs are unstretched when

x1 = x2 = θ = 0 . A horizontal force F acts at the centroid G of the wheel.

a) Use the Newton-Euler formulation to develop the EOM’s for this three-DOF system using the generalized coordinates of:

x1 , x2 and θ . Write these EOM’s in

the matrix-vector form of: M⎡⎣ ⎤⎦

!""x + K⎡⎣ ⎤⎦!x =!f

where !x = x1, x2 , θ{ }T

.

b) Compare the form of the mass matrix, stiffness matrix and forcing vector with those found using the Lagrangian formulation in Homework Set No. 2. In particular, are they the same? If not, are the matrices symmetric as they were with Lagrange? Which terms in the forcing vector here are zero as compared to those in the forcing vector from Lagrange?

k

G

x1

smooth no slip R

G B

A

m 2m

3m

2k 3k

x2

θ

F

Free body diagrams Kinematics (See discussion in solution for Homework No. 2 on kinematics.)

�

aA = ˙ x 1 i

�

aB = ˙ x 1+ ˙ x 2( ) i

�

aG = ˙ x 1+ R˙ θ ( ) i Newton-Euler equations (See discussion in solution for Homework No. 2 on spring stretch/compression.) Wheel

�

Fx∑ = −3k Rθ − x2( ) + f + F = maG = m ˙ x 1+ R˙ θ ( )

�

MG = − fR = IG ˙ θ = 12mR2˙ θ ∑ ⇒ f = −1

2mR˙ θ

Combining the above gives:

G G F

f

N

3k(Rθ − x2)

mg

kx1 A

N

f

3mg

N1 N2

N4

2k(x1 + x2)

B 3k(Rθ − x2)

2mg

N3

x

y

�

−3k Rθ − x2( ) − 12

mR˙ θ + F = m ˙ x 1+ R˙ θ ( ) ⇒ m x 1+ 32

mR˙ θ − 3kx2 + 3kRθ = F

Block B

�

Fx∑ = 3k Rθ − x2( ) −2k x1+ x2( ) = 2maB = 2m ˙ x 1+ ˙ x 2( ) ⇒

2m x 1+ 2m x 2 + 2kx1+ 5kx2− 3kRθ = 0

Block A

�

Fx∑ = −kx1− f = 3maA = 3m x 1 ⇒

− kx1− −12

mR˙ θ ⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 3m x 1 ⇒

3m x 1−12

mR˙ θ + kx1 = 0

The above three EOM’s can be written in matrix-vector form as:

�

m 0 32

mR

2m 2m 0

3m 0 −12

mR

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

˙ x 1

˙ x 2

˙ θ

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

+

0 −3k 3kR

2k 5k −3kR

k 0 0

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥

x1

x2

θ

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

=

F

0

0

⎧

⎨ ⎪ ⎪

⎩ ⎪ ⎪

⎫

⎬ ⎪ ⎪

⎭ ⎪ ⎪

⇒

M[ ]˙ x + K[ ]x = f

Discussion The above three EOM’s resulted from three force balance equations and one moment balance equation. The final form of these three EOM’s depends on how one goes about eliminating the friction force of constraint, f. Your EOM’s might take on a different form from mine above, depending on how you eliminated f. In final conclusion, the mass matrix [M], stiffness matrix [K] and forcing vector f will likely be different than what was found with Lagrange (as seen by comparing mine above or yours with the result from Homework No. 2). In particular, the mass and stiffness matrices above are NOT symmetric. Recall that using the Lagrangian formulation will ALWAYS produce a symmetric [M] and [K].