Admissions in india 2015

Admissions in India 2015
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Wind loading and structural response - Lecture 11
Dr. J.D. Holmes
Basic structural dynamics II

• Topics :
• multi-degree-of freedom structures - free vibration
• multi-degree-of freedom structures - forced vibration
• response of a tower to vortex shedding forces

Basic structural dynamics I
• Multi-degree of freedom structures - :
• Consider a structure consisting of many masses connected
together by elements of known stiffnesses
xn
x3
x2
x1
mn
m3
m2
m1
The masses can move independently with displacements x1, x2 etc.

• Multi-degree of freedom structures – free vibration :
• Each mass has an equation of motion
For free vibration:
m1x1 + k11x1 + k12x 2 + k13x3 +.......k1n x n = 0
m2 x 2 + k 21x1 + k 22x 2 + k 23x3 +.......k 2n x n = 0
…………………….
mn x n + k n1x1 + k n2x 2 + k n3x3 +.......k nn x n = 0
mass m1:
mass m2:
mass mn:
Note coupling terms (e.g. terms in x2, x3 etc. in first equation)
stiffness terms k12, k13 admission.edhole.co metc. are not necessarily equal to zero

In matrix form :
[m]{x} + [k]{x} = {0}
Assuming harmonic motion : {x }= {X}sin(wt+f)
ω2 [m]{X} = [k]{X}
[k]-1[m]{X} = (1/ω2 ){X}
This is an eigenvalue problem for the matrix [k]-1[m]

There are n eigenvalues, lj and n sets of eigenvectors {fj}
for j=1, 2, 3 ……n
Then, for each j :
[ k ] -1 [ m ]{ f } j = λ { j f } j = (1/ω 2
{ } j
) f
j
wj is the circular frequency (2pnj); {fj} is the mode shape for mode j.
They satisfy the equation :
[ ]{ } [ ]{ } j j
2j
ω m f = k f
The mode shape can be scaled arbitrarily - multiplying both sides of the
equation by a constant will not affect the equality

• Mode shapes - :
m3
m2
m1
mn
m3
m2
m1
Mode 1 Mode 2
Mode 3
mn
mn
m1
m2
m3
Number of modes, frequencies = number of masses = degrees of freedom

• Multi-degree of freedom structures – forced vibration
• For forced vibration, external forces pi(t) are applied
to each mass i:
m3
m2
m1
mn
xn
x3
x2
x1
Pn
P3
P2
P1

to each mass i:
m1x1 + k11x1 + k12x 2 + k13x3 +.......k1n x n = p1 (t)
m2 x 2 + k 21x1 + k 22x 2 + k 23x3 +.......k 2n x n = p2 (t)
…………………….
mn x n + k n1x1 + k n2x 2 + k n3x3 +.......k nn x n = pn (t)
• These are coupled differential equations of motion

to each mass i:
m1x1 + k11x1 + k12x 2 + k13x3 +.......k1n x n = p1 (t)
m2 x 2 + k 21x1 + k 22x 2 + k 23x3 +.......k 2n x n = p2 (t)
…………………….
mn x n + k n1x1 + k n2x 2 + k n3x3 +.......k nn x n = pn (t)
• These are coupled differential equations

• In matrix form :
[m]{x} + [k]{x} = {p(t)}
Mass matrix [m] is diagonal
Stiffness matrix [k] is symmetric
{p(t)} is a vector of external forces –
each element is a function of time

• Modal analysis is a convenient method of solution of
the forced vibration problem when the elements of
the stiffness matrix are constant – i.e.the structure is
linear
The coupled equations of motion are transformed
into a set of uncoupled equations
Each uncoupled equation is analogous to the equation
of motion for a single d-o-f system, and can be solved
in the same way

Assume that the response of each mass can be written as:
for i = 1, 2, 3…….n
xi (t) f ij.a j (t)
fij is the mode shape coordinate representing the position of the
ith mass in the jth mode. It depends on position, not time
aj(t) is the generalized coordinate representing the variation of the
response in mode j with time. It depends on time, not position
mi
xi(t)
å=
=
n
j 1
fi1
= a1(t) ´
+ a2(t) ´ fi2
Mode 1
+ a3(t) ´
Mode 2
fi3
Mode 3

In matrix form :
{x} = [f ]{a(t)}
[f] is a matrix in which the mode shapes are written as
columns
([f]T is a matrix in which the mode shapes are written as
rows)
Differentiating with respect to time twice :
{x} = [f ]{a(t)}

By substitution, the original equations of motion reduce
to:
[G]{a} + [K]{a} = [f ]T {p(t)}
The matrix [G] is diagonal, with the jth term equal
to :
Gj is the generalized mass in the jth
mode
i j m G f å=
2
ij
n
i 1
=
The matrix [K] is also diagonal, with the jth term equal to :
j
j G ω m ω K = = å=
2j
2
ij
n
i 1
i
2j
f

[G]{a} + [K]{a} = [f ]T {p(t)}
The right hand side is a single column, with the jth
term equal to :
n
P (t) { } {p(t)} .pi (t)
T
j j å=
= f = f
i 1
ij
Pj(t) is the generalized force in the jth mode

[G]{a} + [K]{a} = [f ]T {p(t)}
We now have a set of independent uncoupled
equations. Each one has the form :
G ja j + K ja j = Pj (t)
Gen. mass
This is the same in form as the equation of motion of a
single d.o.f. system, and the same solutions for aj(t)
can be used

[G]{a} + [K]{a} = [f ]T {p(t)}
Gen. stiffness
admiscsaionn b.eed uhsoelde.com

[G]{a} + [K]{a} = [f ]T {p(t)}
Gen.
force
can be used

[G]{a} + [K]{a} = [f ]T {p(t)}
Gen.
coordinate
can be used

• Cross-wind response of slender towers
f(t)
Cross-wind force is
approximately
sinusoidal in low
turbulence conditions

Sinusoidal excitation model :
Assumptions :
• sinusoidal cross-wind force variation with time
• full correlation of forces over the height
• constant amplitude of fluctuating force coefficient
‘Deterministic’ model - not random
Sinusoidal excitation leads to sinusoidal response (deflection)

Equation of motion (jth mode):
Gj is the ‘generalized’ or effective mass = òh
fj(z) is mode shape
Pj(t) is the ‘generalized’ or effective force =
0
2
m(z) f j (z) dz
òh
f(z, t) f j (z) dz
0

Applied force is assumed to be sinusoidal with a frequency
equal to the vortex shedding frequency, ns
Maximum amplitude occurs at resonance when ns=nj
Force per unit length of structure
=
ö çè
æ p 
C = cross-wind (lift) force coefficient
1
ρ C U (z) b sin(2 n t ψ)
2
j
2
a + ÷ø
b = width of tower

Then generalized force in jth mode is :
P (t) f(z, t) f (z) dz 1 f 
ö çè= = æ
ò ò + ÷ø
= Pj,maxsin(2pn j t + ψ)
Pj,max is the amplitude of the sinusoidal generalized force
h
0
j
2
a j
h
0
j j ρ C b sin(2π n t ψ) U (z) (z) dz
2
1 f 
ö çè
ò ÷ø
= æ
h
0
j
2
ρaC b U (z) (z) dz
2

Then, maximum amplitude
P
a = =
max 8π n G ζ
j j
2
j
2
j,max
j,max
j j
P
2K ζ
Note analogy with single d.o.f system result
(Lecture 10)
Substituting for Pj,max :
z2
z1
1
æ
=
ö çè
ò ÷ø
f 
ρ C b U (z) (z) dz
2
max 8π n G ζ
2
j
2
2
a
a
Then, maximum deflection on structure at height, z,
j j
j
x max (z) =f j (z).a max
(Slide 14 - considering only 1st mode contribution)

Maximum deflection at top of structure
(Section 11.5.1 in ‘Wind Loading of Structures’)
h
0
ò ò = = L
C (z) dz
ρ C b (z) dz
x (h)
where zj is the critical damping ratio for the jth mode, equal to
j
G K
j j
C
2
n b
U(z )
St n b
= s =
U(z )
e
j
e
Strouhal Number for vortex shedding
ze = effective height (» 2h/3)
(Scruton Number or mass-damping parameter)
m = average mass/unit height
2
4 m
a
j
ρ b
Sc
p z
=
ò
0
2
j
2
j
2
j j
2
h
0
j
2
a
max
4π Sc St (z) dz
16π G ζ St
b
f
f f  

End of Lecture
John Holmes
225-405-3789 JHolmes@lsu.edu

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