EQUIVALENT SYSTEMS METHOD

2. 4 OTHER SOURCES OF POTENTIAL ENERGY

2.12 EQUIVALENT SYSTEMS METHOD

and

(2.82) Equation (2.79) becomes

(2.83) Equation (2.80) has two solutions: (the static case) and x. This satisfies

(2.84) Equation (2.84) is the differential equation for any linear, single degree-of-freedom system. It only requires identification of , and . That is, any linear SDOF system is modeled by a mass-spring and viscous-damper system with equivalent coeffi- cients, as in Figure 2.44. The equivalent mass is identified from the quadratic form of kinetic energy in . The equivalent stiffness is identified from the quadratic form of potential energy in . The equivalent viscous-damping coefficient is identified from the energy dissipation in The work done by external forces, shown as is used to calculate .

If an angular coordinate is chosen as the generalized coordinate, the appropriate form of Equation (2.84) is

(2.85) The appropriate equivalent systems model is a thin disk of moment of inertia I_eq attached to a shaft of torsional stiffness k_t,eqin parallel with a torsional viscous-damper coefficient c_t,eqas shown in Figure 2.45.

I_equ

$

+ c_t,equ

#

+ k_t,equ = M_eq(t)

F_eq(t) 1t^t₁²F_eqx#

dt,

U₁:2 = -1x^x₁²c_eqx# dt.

V = ¹

2k_eqx² T = ¹

2m_eqx#

2

F_eq(t) m_eq, c_eq, k_eq

m_eqx$ + c_eqx#

+ k_eqx = F_eq(t) x#

= 0 F_eqx#

- c_eqx#

= m_eqx$ x#

+ k_eqxx# d

dtaL

x

x₁

c_eqx#

dxb = d dtaL

t

t₁

c_eq#x

2dtb = c_eqx#

2 x

c_eq k_eq

m_eq F_eq (t)

FIGURE 2.44

Equivalent mass-spring and viscous-damper system when a linear displacement xis chosen as the generalized coordinate.

k_t,eq c_t,eq

I_eq M_eq (t)

θ

FIGURE 2.45

Equivalent torsional system used when an angular coordinateuis chosen as the generalized coordinate.

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Use the equivalent systems method to derive the differential equation governing the motion of the bar of Figure 2.43(a) and Example 2.24 using as the clockwise angular dis- placement of the bar from the system’s equilibrium position and as the chosen generalized coordinate. Assume small .

S O L U T I O N

The kinetic energy of the bar at an arbitrary instant is

(a) Thus, . The potential energy of the system at an arbitrary instant is

(b) The equivalent torsional stiffness is . The work done by the viscous damper between an initial position and an arbitrary position is

(c)

Hence, the equivalent torsional stiffness is . The differential equation governing is (d) Equation (d) reduces to Equation (b) of Example 2.24.

1 9 mL²u

$ +c L²

36 u

# + 5

9kL²u=0

c_t,eq = c₃₆^L2 u W₁:2= -

L

u

u₁ ac L 6 u

#b d aL 6 ub = -

L

u

u₁ ac L² 36 u

#bdu k_t,eq = ⁵

9kL² V =

1 2kaL

3 ub²+ 1

2ka2L 3 ub²=

1 2a5

9kL²bu² I_eq=¹

9 mL²

= 1 2 a1

9 mL²bu# T = 2

1 2mv² +

1 2Iv² =

1 2maL

6 u

#b² + 1

2a 1 12mL²bu

#2

u

u

E X A M P L E 2 . 2 6

Use the equivalent system method to derive the differential equation governing the free vibrations of the system of Figure 2.46. Use x, the displacement of the mass center of the disk from the system’s equilibrium position, as the generalized coordinate. The disk rolls without slipping, no slip occurs at the pulley, and the pulley is frictionless. Include an approximation for the inertia effects of the springs. Each spring has a mass m_s.

S O L U T I O N

Let be the clockwise angular rotation of the pulley from the system’s equilibrium position and x_Bbe the downward displacement of the block, also measured from equilibrium. Then (a) Eliminating between these equations leads tox_B2x. Since the disk rolls without slip, its angular velocity is The inertia effect of each spring is approximated by plac- ing a particle of mass m_s/3 at the location where the spring is attached to the system.

v_D = x# /r_D. u

x = ru x_B = 2ru u

E X A M P L E 2 . 2 5

To this end it is imagined that a particle of mass m_s/3 is attached to the center of the disk and a particle of mass m_s/3 is attached to the block. The total kinetic energy of the system, including the kinetic energies of the imagined attached particles is

(b) The equivalent mass is

(c) The potential energy of the system at an arbitrary instant is

(d) Comparison to the quadratic form of potential energy leads to k_eq5k.

The work done by the viscous dampers between two arbitrary instants is

Comparison with the general form of work done by a viscous damper leads to c_eq5c.

The differential equation governing free vibration of the system is a19

2 m + I_P r² +

5 3msbx$

+ 5cx#

+ 5kx = 0 U₁:2 = -

L

x₂ x₁

cx# dx -

L

x₂ x₁

c(2x#

)d(2x) = - L

x₂ x₁

5cx# dx V = ¹

2kx² + ¹

2k(2x)² = ¹

2(5k)x² m_eq =

19 2m + 1P

r² + 5 3m_s

= 1 2a19

2m + I_P r² +

5 3m_sbx#

2

= 1 2mx#

2 + 1 2a1

2 mr²_Dba x#

r_Db² + 1 2I_Pax#

rb² + 1 2 (2m)(2x#

)² + 1 2

m_s 3x#

2 + 1 2

m_s 3(2x#

)² T = 1

2mx#

2 + 1

2I_Dv²_D + 1 2I_Pu

#2 + 1 2(2m)x#

2 B + T_s

1

+ T_s

2

k

k

x_B = 2x c

c x

r_D

2r

2m I_p r

FIGURE 2.46

The system of Example 2.26 is modeled by the equivalent system of Figure 2.44.

E X A M P L E 2 . 2 7

The slender rod of Figure 2.47 will be subject only to small displacements from equilib- rium. Use the equivalent systems method to derive the differential equation governing the motion of the rod using , the counterclockwise angular displacement of the rod from its equilibrium position, as the generalized coordinate.

u

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S O L U T I O N

The kinetic energy of the bar at an arbitrary instant is

(a) Comparison with the quadratic form of kinetic energy leads to I_eq mL²/9.

The potential energy in the system is due to gravity. Choosing the plane of the pin sup- port as the datum, the potential energy of the system at an arbitrary instant is

(b) For small the Taylor series expansion for truncated after the second term leads to an approximation for the potential energy as

(c) Comparison with the quadratic form of potential energy leads to k_t,eq mgL/6. Since the datum was chosen as the plane of the pin support, the system has a potential energy of V₀–mgL/6 when it is in equilibrium.

Equation (2.84) is used to write the differential equation governing the motion of the system as

1 (d) 9mL²u

$ + 1

6mgLu = 0 V = -mg L

6a1 - 1

2u²b = 1 2mg L

6u² - mg L 6 cosu u,

V = -mg L 6 cos u T = = 1

2maL 6u#

b² + 1 2a 1

12mL²bu#

2 = 1 2a1

9mL²bu#

2

E X A M P L E 2 . 2 8

A simplified model of a rack-and-pinion steering system is shown in Figure 2.48. A gear of radius r and polar moment of inertia J is attached to a shaft of torsional stiffness k_t. The gear rolls without slip on the rack of mass m. The rack is attached to a spring of stiff- ness k. Derive the differential equation governing the motion of the system using x, the horizontal displacement of the rack from the system’s equilibrium position, as the general- ized coordinate.

S O L U T I O N

Since there is no slip between the rack and the gear, x/r, where is the angular displace- ment of the gear from equilibrium. The kinetic energy of the system at an arbitrary instant is (a) from which the equivalent mass is determined as m_eq mJ/r². The potential energy of the system at an arbitrary instant is

(b) from which the equivalent stiffness is determined as k_eqkk_t/r². The differential equation is

am + (c) J r²bx$

+ ak_tk_t r²bx = 0 V = = 1

2kx² + 1 2k_tax

r²b = 1 2ak +

k_t r²bx² T = = 1

2mx#

2 + 1 2Jax#

rb² = 1 2am +

J r²bx#

2

u u

2L 3

θ L 3

FIGURE 2.47

The compound pendulum is modeled by the equivalent tor- sional system of Figure 2.45.

Gear of radius r, polar moment of inertia J

Rack of mass m

k k_t

x

FIGURE 2.48

Model of the rack-and-pinion system of Example 2.28.

E X A M P L E 2 . 2 9

A simplified transmission system is shown in Figure 2.49. A motor supplies a torque, which turns a shaft. The shaft has a gear on it, which meshes to a second gear designed such that the speed of the second shaft is greater than the first. The shafts are mounted on identical bearings each with a torsional damping coefficient c_t. Let be the angular velocity of the shaft directly connected to the motor. Derive a differential equation governing , which is angular displacement of the shaft directly connected to the motor.

S O L U T I O N

The meshing gears imply a relationship between the angular velocities of the shafts. The gear equation gives

(a) The total kinetic energy of the shafts is

(b) Thus, the equivalent moment of inertia is The work done by the tor- sional viscous dampers is

(c) The equivalent viscous damping coefficient is

The work done by the external moment supplied by the motor is

(d) The equivalent moment is M_eq(t) M(t).

Thus the differential equation governing the angular displacement of the shaft is cJ₁ + an₁ (e)

n₂b²J₂du

$

1 + c_tc1 + an₁ n₂b²du

#

1 + M(t) W₁:2 =

L

t t₁

M(t)u

#

1dt

c_t,eq = c_tc1 + an₁ n₂b²d. W₁:2 = -

L

u

u₁

c_tu

#

1du₁- L

u

u₁

c_tan₁ n₂ u

#

1b dan₁

n₂ u₁b = - L

u

u₁

c_tc1 + an₁ n₂b²du

#

1du₁ I_eq = J₁ + aⁿ_n¹

2b²J₂. T = = 1

2J₁v²₁ + 1

2J₂v²₂ = 1 2J₁u

#

2 1 + 1

2J₂an₁ n₂ u

#

1b² = 1

2cJ₁ + an₁ n₂b²J₂du

#

2 1

n₁v₁ = n₂v₂

u₁ u

#

1

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