EN40: Dynamics and Vibrations
Homework 7: Rigid Body Kinematics
School of Engineering
Brown University
1. FORCED VIBRATION OF ENGINE IDLING
The Subaru Legacy has an unusual horizontally-aligned four-cylinder
engine. The engine is connected to
10
lateral motor mounts by 4 springs and 4 dashpots, as shown in the figure. When idling, slightly
differences in the firing of the individual cylinders lead to 9an effective rotor forcing of the engine, with an
effective mass imbalance of eΔm = 0.4kgm at a frequency
corresponding to 600 / π RPM. The total
8
mass of the engine is M=200kg. Brand new, each spring has
stiffness k=12800N/m and each damper has
7
damping coefficient λ=400 N-s/m.
6
X
⎛ Δme ⎞
⎜
⎟
⎝ m ⎠
0.1
5
4
0.15
3
0.2
0.25
2
1
0
0.5
0
1.1 What are the natural frequency and damping coefficient
0.5
1
ω / ωn
1.5
2
2.5
ς for the engine?
From formulas ωn = k / m = 4 × 12800 / 200 = 16rad / s .
The damping coefficient is ζ = c / 2 km = 4 × 400 / 2 4 ×12800 × 200 = 1/ 4
(2 POINTS)
1.2 What is the typical steady-state amplitude of the lateral vibrations of the engine?
The forcing frequency is ω = 2π × 600 / π / 60 = 20rad / s ⇒ ω / ωn = 20 / 16 . Doing the calculation
with the formula,
2
X / (Δme / m) = (ω / ωn ) /
2
(1 − ω 2 / ωn2 )
+ 4ζ 2ω 2 / ωn2 = 1.86 ⇒ X = 1.86Δme / m = 0.0037 m . Or
more quickly reading off the graph X / (Δme / m) is about 2…
(3 POINTS)
1.3 The main problem is not the engine vibration, but the forces caused on the attachment points to the
body of the car. These attachment fixtures fatigue under load. What is the approximate amplitude of
the force exerted on one damper attachment point for the new car?
The force is cdx / dt = cX ω cos(ωt + φ ) ≈ 29.6cos(ωt + φ ) N
(3 POINTS)
3
1.4 As the car ages, the spring stiffness gets smaller. How does this change the vibration amplitude?
(increase, decrease, the same?)
This decreases ω n so the operating point shifts to the right on the curve, and the amplitude
decreases.
(2 POINTS)
B
2. ROBOT ARM KINEMATICS
The figure shows a robot arm. Point C on the arm
is required to move horizontally with constant
speed 1m/s. This is accomplished by rotating
links AB and BC with appropriate angular speeds
ω AB , ωBC and angular accelerations α AB ,α BC .
The goal of this problem is to calculate values for
ω AB , ωBC , α AB ,α BC at the instant shown.
C
1m
A
450
j
VCx = 1m/s
aC = 0
1m
i
2.1 Determine formulas for the velocity vectors v B , vC of points B and C, in terms of ω AB , ωBC .
Applying the rigid body kinematics formula gives
v B = v A + ω AB × rB / A = ωAB k × (i + j) / 2 = ωAB (−i + j) / 2
vC = v B + ωBC k × i = ω AB (−i + j) / 2 + ωBC j
(3 POINTS)
2.2 Determine formulas for the acceleration vectors a B , aC of points B and C in terms of
α AB ,α BC , ω AB , ωBC .
a B = a A + α AB k × (i + j) / 2 − ω AB k × ω AB k × (i + j) / 2
2
= α AB (−i + j) / 2 − ω AB
(i + j) / 2
aC = a B + α BC k × i − ωBC k × ωBC k × i
2
2
= α AB (−i + j) / 2 − ω AB
(i + j) / 2 + α BC j − ωBC
i
(3 POINTS)
2.3 Hence, calculate the required values of ω AB , ωBC , α AB ,α BC
We know that vC = i . Using the i and j components of 12.1 gives two equations for ω AB , ωBC
−ω AB = 2rad / s
We also know that aC = 0 which gives
ω AB / 2 + ωBC = 0 ⇒ ωBC = 1rad / s
2
2
α AB (−i + j) / 2 − ω AB
(i + j) / 2 + α BC j − ω BC
i=0
⇒ α AB (−i + j) / 2 + α BC j − 2(i + j) / 2 − i = 0
(
⇒ α AB = − 2 + 2
α BC =
1
4 2
−
α AB
2
)
=
rad / s 2
2
2
+
1
(2 + 2 )= 1 + 2
2
2 rad / s 2
(4 POINTS)
3. PRIUS POWER SPLIT DEVICE (PSD)
In class we saw a demonstration
of the Prius’ Planetary Gear Set
(http://eahart.com/prius/psd/).
3.1 At the lowest speeds (<42
mph), the ICE does not have to
provide any power. Which
components of the PSD are
spinning, and in which direction?
Gear plate – NOT moving.
Outer Ring CW (-)
Sun Ring CCW (+)
Planets CW (-)
(1 POINT)
3.2 In this configuration, what is the gear ratio between the sun gear (rotational speed ωs) and the outer
ring (rotational speed ωr) in terms of the radius of the sun gear, rs, and the radius of the planetary gear, rp?
Since the plate is not moving, we can say:
v A = ω sr s= ω p r p
v B = ω rr r= ω p r p
Thus, ωsrs= ωrrr
From geometry, rr=rs+2rp
ωs= [(rs+2rp)/ rs]ωr
Gear Ratio: [(rs+2rp)/ rs]
Also accepted: [rs /(rs+2rp)]
(3 POINTS)
3.3 Now the outer ring is not rotating (the car is not moving)
but the ICE engine continues to run! What is the gear ratio
between the sun gear and the gear plate (rotational speed ωpp)
in terms of rs and rp?
vA= ωsrs as before (because O remains fixed)
From our kinematics equations:
yB= yC + |rB/C| sin θ
vB= vC + rp dθ/dt cos θ
however let’s examine the point at which θ=0:
vB= vC + rp dθ/dt
vB = ωpp(rs+rp) + rp dθ/dt
We can also write down how B depends on A:
yB= yA + |rB/A| sin θ
vB= vA + 2rp dθ/dt cos θ
however let’s examine the point at which θ=0:
vB= vA + 2rp dθ/dt
vB = ωsrs + 2rp dθ/dt
If the outer ring remains fixed: vB = ωrrr = 0
-2ωpp(rs+rp) = 2 rp dθ/dt
ωsrs = - 2rp dθ/dt
eliminating dθ/dt:
2ωpp(rs+rp) = ωsrs
ωs = [2(rs+rp) / rs]ωpp
gear ratio: [2(rs+rp)/ rs] (or inverse)
(3 POINTS)
3.4 For the configuration where all components are rotating, derive a relationship between ωr, ωs, and ωpp
in terms of rs and rp.
We can use the same solution method as above except B is not stationary:
Starting with these 2 equations from 3.3:
vB = ωpp(rs+rp) + rp dθ/dt
vB = ωsrs + 2rp dθ/dt
Eliminate dθ/dt and substitute vB=rr ωr = (rs+2rp) ωr
(rs+2rp) ωr = 2(rs+rp)ωpp - rs ωs
(3 POINTS)
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cl-ose af f ;
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ic: IthetaO (i) , O] ;
options:odeset ( 'Event' , Gdetect zero crossing) ;
Ic, w, tevent, wevent, index] :ode45(Geom, [0, 5], ic, options),.
% time difference between two successive events ls t reaf
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plot (thetaO, t
real-
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? fol-lowing 1s the code to detect the range of theta 0 for which
? l_real is within 5% and 1? of t app
for 1:1:100
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if (t_reaf (i) <1.01)
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end
%
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thetaO
thetaO
5
1
? function of the eom in terms of the normafized time "tau"
function dwdt:eom (t, w)
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(w (1 ) ) I ;
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slope
function Ivalue, stop, dir]:detecL zero crossing (t, w)
vafue:w(1);
rc+nn-A.
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end
end
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