Download structural vibrations and more Exams Mechanical Engineering in PDF only on Docsity!
!!!! Structural vibration worked
examples
Natural Frequencies and Mode Shapes
1. Derive the equation of motion and hence find the natural frequencies for the system shown in Figure Q1.
Solution
Free-body diagram
L
k ⋅ K^ ⋅ L θ
Since clockwise rotation has been chosen as the positive direction, when the bar moves down both springs will be put in compression and will exert upward reaction forces on the underside of the bar. For small angles, the change of
length of the left-hand spring will be θ 2
L .
Equation of motion
Taking moments about the pivot in the same direction as the motion coordinate (i.e., clockwise), we get:
− θ − θ =^2 θ 3
KL L mL
L L
k
Rearranging gives:
θ=
θ + + KL
L
mL k
or 0 3 4
θ=
θ + + K
m k
From the coefficients of acceleration and displacement, the natural frequency is:
( ) m
k K m
K
k
n (^) 4
ω =
2. A wheel (radius r , mass m , moment of inertia about its centre I ) can roll without slipping on a horizontal plane. It is restrained by a horizontal spring (stiffness k ) attached at one end to the centre of the wheel and at the other end to a rigid vertical wall, as in Figure Q2. Derive the equation of motion and hence find the natural frequency for the system. What would the natural frequency be if there was no friction between the wheel and the plane?
Solution
Free-body diagram
r
θ
xG
k xG
F (friction)
G
C
Due to the friction at the point of contact with the horizontal plane, there is no slipping and the rotation of the wheel and the translation of its centre of mass are not independent motions, but are directly linked by the expression
X (^) G = r θ (Q2.1)
the difference between to two displacements. The expression for the force in the spring can therefore be written as k 1 (^) ( x 1 − x 2 ). This force will be positive
any time that x 1 > x 2. This would mean that the spring would be in compression and that the forces acting on it would be as shown below.
k 1 (^) ( x 1 − x 2 ) k 1 (^) ( x 1 − x 2 )
This leads to the directions of the reaction forces shown in the free-body diagrams.
An equally correct expression for the force in the spring would be k 1 (^) ( x 2 − x 1 ).
This force will be positive any time that x 2 > x 1 , which would mean that the spring would be in tension and that the forces acting on it would be as shown below.
k 1 (^) ( x 2 − x 1 ) k 1 (^) ( x 2 − x 1 )
That part of the free-body diagram would then be:
m 1 m 2
k 1 (^) ( x 2 − x 1 )
The two alternative force expressions are both correct, provided that the positive directions of the forces are identified correctly.
Equations of motion
Mass m 1 : − k 1 (^) ( x 1 − x 2 ) = m 1 x 1
Mass m 2 : + k (^) 1 ( x 1 − x 2 ) − k 2 ( x 2 − x 3 ) = m 2 x 2
Mass m 3 : + k (^) 2 ( x 2 − x 3 ) = m 3 x 3
Rearranging gives:
3
2
1
2 2
1 1 2 2
1 1
3
2
1
3
2
1
x
x
x
k k
k k k k
k k
x
x
x
m
m
m
Note that the leading diagonal of the mass and stiffness matrices contain only positive terms and that both matrices are symmetric.
4. Find the natural frequencies and mode shapes for the system in Figure Q
when k 1 = 10 kN/m, k 1 = 30 kN/m, m 1 = m 2 = 5 kg and m 3 = 10 kg.
Solution
Put x (^) 1 = X 1 cosω t , etc. into the equations of motion from Question 2 and
substitute the given mass and stiffness values.
− × × − ω
− × − ω − ×
3
2
1
4 4 2
4 4 2 4
4 2 4
x
x
x
Itís convenient to put (^4)
2
10
5 ω β = to simplify the algebra. Hence:
( ) ( ) ( iii )
ii
i
x
x
x
β
β
β LL L
3
2
1
The determinant of the coefficient matrix equated to zero gives the frequency equation, which is:
2 β^2 − 13 β^2 + 12 β = 0
Although this is a cubic, there is no constant term, so that one of the roots will be β = 0 , leaving only a quadratic to solve. The roots are:
β 1 (^) = 0 β (^) 2 = 1. 11 β 2 = 5. 39
Giving: ω 1 = 0 Hz ω 2 = 7. 51 Hz ω 3 = 16. 51 Hz
Mode shapes
From (i), ( 1 − βr ) X 1 − X 2 = 0 (^ )
2
(^1) βr X
X
From (iii), − 3 X (^) 2 +( 3 − 2 βr ) X 2 = 0 (^ )
2
3 βr X
X
Combining the two results gives,
a
k ( a θ + x )
x
k ( a θ+ x )
A
B
m
Solution
Free-body diagrams
Positive rotation of the bar moves end B upwards by aθ (angles are small) and positive displacement of the mass is downwards. The change of length of the spring is therefore the sum of the two displacements, that is, aθ + x.
Equations of motion
For the bar: − k ( a θ+ x ). a = I θ
For the mass: − k ( a θ+ x ) = mx
Rearranging gives:
θ
θ
ka k^ x
ka ka m x
I
Note that the leading diagonal of the mass and stiffness matrices contain only positive terms and that both matrices are symmetric.
6. Derive the frequency equation for flexural vibration of a uniform beam that is free at both ends and find an expression for the mode shape function.
Solution
At x = 0 and x = L , 2 0
2 2
2
∂
x
y x
y
. Since y ( x , t ) = Y ( x ) cosω t , 2 0
2 2
2 = = dx
dY dx
d Y .
Since ( x ) = C 1 sin λ x + C 2 cosλ x + C 3 sinhλ x + C 4 coshλ x (Q6.1)
At x = 0: (^2122324)
2
- C. C 0. C. C d x
d Y = −λ + +λ
and (^2212234)
2
. C 0. C. C 0. C dx
d Y =λ + +λ +
At x = L : (^221222324)
2 sin L. C cos L. C sinh x. C cosh x. C dx
d Y =−λ λ −λ λ +λ λ +λ λ
and (^221222324)
2 cos L. C sin L. C cosh L. C sinh L. C dx
d Y =−λ λ +λ λ +λ λ +λ λ
Assembling the equations into matrix format:
C 3 @ C 4
( ) ( ) ( ) ( Q d )
Q c
Q b
Q a
Expanding the determinant of the coefficient matrix and equating to zero gives the frequency equation:
λ^10 (cos λ L .coshλ L − 1 ) = 0
Since a freeñfree beam can have two rigid-body modes, each with a frequency of zero, the frequency equation becomes:
λ^2 ( cosλ L .coshλ L − 1 ) = 0
The mode shapes for the first four modes are shown below.
‐ 3
‐ 2
‐
0
1
2
3
0 0. 1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
Displacement amplitude Y(x)
Axial position x/L
Y 1 Y2 Y3 Y
7. A 25 mm diameter shaft, 1.5 m long, is held by two roller bearings at one end (giving a ëclampedí boundary condition) and by a self-aligning ball bearing at the other end (giving a ëpinnedí boundary condition). Using the roots of the appropriate frequency equation given in Table 6.3 on page 398, find the first three critical speeds of the shaft.
Solution
Recall that the critical speeds of a shaft are numerically identical to the natural frequencies in flexure of an equivalent ëbeamí with a circular cross-section.
The expression for the natural frequencies is A
EI
L
rL r ρ
λ ω =
2 , where the
roots of the frequency equation, λ (^) rL , are given in Table 6.3 in the book for a
clamped-pinned beam.
Taking E = 207 GN/m 2 and ρ = 7800 kg/m 3 , along with the dimensions of the
shaft, the expression for the natural frequencies is ω r (^) = 14. 31 ( λ rL )^2 rad/s
= ( ) 2
- 7 λ (^) rL rev/min.
Results for the first three critical speeds are: 2107, 6981 and 14 245 rev/min.
Response of damped single-degree-of-freedom systems
8. If a heavily damped structure is given an initial displacement Z 0 and then released from rest, find the constants of integration and sketch the graph of z ( t ) against time.
Solution
The general solution for a heavily damped system (equation 6.35) is
z ( ) t = A 1 e λ^1 t^ + A 2 e^ λ^2 t in which λ 1 and λ 2 are both real and negative.
If z ( ) t = Z 0 when t = 0, substitution into the general solution
gives Z (^) 0 = A 1 + A 2.
To apply the second boundary condition, first differentiate the general solution
to give z &^ ( ) t =λ 1 A 1 e λ^1 t^ +λ 2 A 2 e^ λ^2 t.
Since the question specifies that z &( )^ t = 0 when t = 0, we find that
0 = λ 1 A 1 +λ 2 A 2.
Solving for A 1 and A 2 gives:
2 1
0 2 1 λ −λ
λ
Z
A and 2 1
0 1 2 λ −λ
− λ
Z
A
In order to sketch the response, we will take as an example Z 0 (^) = 1 , λ 1 =− 1
and λ 2 =− 2. In this case, we find that A 1 = 2 and A 2 = ñ1. If these values are
put into the general solution, we get the following graph. Note that the overall response (the black curve) is consistent with the given initial conditions that Z (^) 0 = 1 and z &^ ( ) t = 0 when t = 0.
Displacement
Time
V 0
10. The rigid beam shown in Figure Q10 has a moment of inertia of 10 kgm 2 about the pivot at A. End C is displaced downwards by 10 mm from its equilibrium position and then released from rest. Find the maximum upward displacement of C from its equilibrium position and the elapsed time at which this occurs.
Solution
Free-body diagram (AC= L )
θ A B
θ 3
2 L &
c ⋅ k^ ⋅ L θ
C
Equation of motion
For positive (anti-clockwise) motion about the pivot at A,
⋅ −^ (^ θ)^ ⋅ = θ
− θ&^ kL L IA &
L L
c 3
Rearranging, 0 9
θ + θ+ kL θ =
L
I (^) A c &
Or 10 θ& + 200 θ&+ 56250 θ= 0
Hence, the undamped natural frequency, ω n = 75 rad/s and damping ratio,
γ = 0. 133.
The damped natural frequency, Ω n = 74. 3 rad/s
Since γ < 1, the system is lightly-damped and the general solution (equation
6.42) is:
θ ( ) t = e −^ γω n^ t ( B 1 cos Ω nt + B 2 sinΩ nt ) (Q10.1)
Initial conditions: θ =Θ 0 , θ = 0 at t = 0
Hence, B 1 (^) =Θ 0 and 2 Θ 0 Ω
γω
n
B n
Here 6. 67
- 5
Θ = mrad
The resulting response is plotted below.
‐ 8
‐ 6
‐ 4
‐ 2
0
2
4
6
0 0.02 0.04 0.06 0.08 0. 1 0. 12 0. 14 0. 16
Angular displacement (mrad)
Time (s)
The response is a maximum when θ&^ = 0. Differentiating equation (Q10.1) and equating to zero leads to the condition that tan Ω (^) n t = 0. The first non-zero root
is Ω (^) n t =π or t = 0.0423s. Substituting this time into equation (Q10.1) gives
θ =+ 4. 369 mrad and a corresponding tip displacement of 6.55 mm.
rocker by the spring ñ damper model of the block are as shown in the free- body diagram.
Equation of motion
Taking moments of the forces about the pivot (anticlockwise positive):
− ( ka θ+ ca θ&^ ) a +( k ( y − b θ) + c ( y &− b θ&)) b = I 0 θ&&
Rearranging, I (^) 0 θ&& + c ( a 2 + b^2 )θ &+ k ( a 2 + b^2 )θ = cby + kby
Substitutions: y ( ) t = Yet ω t and θ( ) t =Θ• et ω t
Hence,
( ) ( ( ) ) ( )
γ
+ω Θ • = 2 2 0 ka^2 b^22 I t ca b
kb t cb
The amplitude of the response is given by:
( ) ( ) ( ( ) ) ( ( ))
γ
−ω +ω +
ω Θ • = 2 (^222) 0
2 2 2
2 2
ka b I ca b
kb cb
The amplitude of the displacement at A is a Θ•
12. For an undamped system with n degrees of freedom, show that the steady-state response in coordinate j due to a sinusoidal force of amplitude P and frequency ω applied in coordinate k is given by
( ) P t
u u x t r (^) r
jr kr j ω
ω − ω
= (^) ∑
sin 1 2 2
Solution
Starting with the general equation of motion (equation 6.61), we uncouple the equations by making the substitution { } x = [ ]φ{ } q (equation 6.62). Assuming
that the modal vectors are scaled to unit modal mass and using their orthogonality properties, we get:
[ ]{ } I q (^) r { } q =[ ]φ { p ( ) t } ={ f ( ) t }
O
2
In this case, only coordinate k has any excitation, so the vector { p ( ) t }has only
one entry, which is in row k. The modal space excitation vector is therefore,
{ ( )}
ω
ω
ω
u P t
u P t
u P t ft kn
kr
k
sin
sin
1 sin
The modal space equation for a typical mode r is: q (^) r + ω r^2 qr = ukrP sinω t.
For this undamped system, the steady-state response can be obtained by
making the substitution q (^) r ( ) t = Qr sin ω t to give, (^2 ) ω −ω
r
kr r
u P Q
The response in coordinate is given by equation 6.65 as,
( ) ( ) P t
u u x t u q t
n
r (^) r
n jr kr
r
j jr r ω ω −ω
= (^) ∑ =∑ = =
sin 1 2 2 1
Approximate methods
13. Use Dunkerleyís Method and Rayleighís Method to estimate the lowest natural frequency of the torsional system shown in Figure Q13.
Solution
Dunkerley’s Method
There are three sub-systems in this case.
Sub-system 1 with the left-hand inertia only:
I
2 2^ k ω 1 = 2 k I
Since the right-hand end of the system is free, the first mode shape will be such that θ 3 > θ 2 > θ 1 and all inertias rotate in-phase with each other.
Try θ 3 : θ 2 : θ 1 = 1: 2: 3
This gives I
k ω n = 0. 471
The equivalent of a ìstatic deflectionî can be obtained by applying static torques to each disc proportional to its inertia. This leads to θ 3 : θ 2 : θ 1 = 2: 5: 6
and to I
k ω n = 0. 447
For reference, the exact value is I
k ω n = 0. 445
14. A shaft with universal joints at each end has a length of 6 m, a second moment of area of 0.00025 m 4 and a mass/unit length of 75 kg/m. It carries three discs, which can be regarded as point masses of 100, 150, and 200 kg located 1.2, 3.0 and 4.8 m from the left-hand end. Estimate the lowest critical speed using Dunkerleyís and Rayleighís Methods. Take E =
207 GN/m 2 , ρ = 7800 kg/m 2.
Solution
Dunkerley’s Method
This is a numerical example of the method presented in the book starting on page 438. The intermediate results for the sub-systems are as follows.
With only the 100 kg mass: ω^21 = 2. 8076 × 105 s−^2
With only the 150 kg mass: ω^22 = 7. 667 × 104 s−^2
With only the 200 kg mass: ω^23 = 1. 4038 × 105 s−^2
With only the shaft mass: ω^20 = 5. 186 × 104 s−^2
Combining these in Dunkerleyís formula gives ω n = 24. 27 Hz min
rev = 1456
Rayleigh’s Method
This is a numerical example of the method presented in the book, starting on page 442.
A good choice for the mode shape function is ( ) L
x Y x
π = sin since this is the
exact mode shape of the shaft before the discs are added.
6 3
(^24)
0 2
2
- 834 10 (^24)
= ×
π =
= (^) ∫ L
El dx dx
dY U El
L MAX
[ ( )] [ ( )]
2 2 2
0
3
1
2 2 2
= ω + ω = ω
= ω ∫ ρ + ω ∑
L p
TMAX AY x dx mpY xp
Equating gives min
rev ω n = 24. 85 Hz= 1491
15. Find a single-degree-of-freedom approximate model to analyse the motion of the 1 kg mass in the system in Figure Q15 when it is vibrating near its lower natural frequency.
Solution
The exact solution to this problem is presented as a worked example in the book starting, on page 427.
1 kN/m
x 1
p 1 (^) ( ) t = P 1 sinω t
1 kN/m 1 kN/m
x 2
1 kg 2 kg
k
x 1
p 1 (^) ( ) t = P 1 sinω t
m
Original system Approximate model
Model parameters
Equating the maximum kinetic energy in the real and approximate systems,
[ ]
[ ] 2 1
2 2
2 1
2 2
2 1
2 2 1
2
X
X X
m
T (^) MAX mX X X
∴ =
= ω = ω +
Equating the maximum strain energy in the real and approximate systems,
