WIND INDUCED STATIC AND DYNAMIC RESPONSE CALCULATIONS - Theory of Bridge Aerodynamics - page 123

Let us allot the following values to the remaining constants that are necessary for a numerical
calculation of

( )

xL 2

σ = :

r r

y

ρ
(kg/m 3 )

B
(m)

D (m)

m
(kg/m)

C

ω ζ

y y

D

(rad/s)

1.25 0.7 20 4 10000 0.4 0.005

m is constant along the span, then the modally equivalent and evenly distributed mass

Since

mm = . Finally, let us adopt quasi-static values to the aerodynamic derivatives, in which case

y y

0 κ = and the aerodynamic damping ζ is given by (see Eqs. 5.26 and 6.24)

ae y ae y

2 2

⎛ ⎞

ρ

B

ρ

B D V

ρ

DC V

* 4

P 2 C

D

4.375 10

−

V ζ = = ⎜ − ⎟ = − ≈ − ⋅ ⋅

ae 1

⎜

D

⎟

y

4 m 4 m B B

ω

2 m

ω

⎝ ⎠

y y y y y

The non-dimensional joint acceptance function ˆ J may readily be obtained by numerical
calculations. However, as shown by Davenport [14], in many cases closed form solutions may be
obtained. The situation that

ˆ

() ( ) ( )

φ

y x

=

sin

π

xL

and

Co

ω Δ

, x

is a simple exponential

uu

function is such a case. Substituting

x

=

x

,

x

=+Δ

xx

,

x

= π

xL

, Δ=Δ

x xL

and

2

ωω

ˆ

=

ux CLV

, then

2

⎛ ⎞

ˆ ˆ

() () ( )

2 2

J

∫∫

φ

x

φ

x Co x ,

ω

dx dx

∫

φ

dx

= ⋅ ⋅ Δ ⎜ ⎟

y y 1 y 2 uu 1 2 y

⎜ ⎟

⎝ ⎠ L L

exp

2

LL x

⎡

−Δ

⎤

exp exp

⎛

L

⎞

ˆ

( ) ( ) ( )

2

∫ ∫

⎢ ⎥

∫

=

2

φ

xx

+ Δ ⋅ φ

x x o xdx

Δ

,

ω Δ ⎜ φ

x

⎟

y y uu y

⎜ ⎟

⎢ ⎥

⎝ ⎠

0 0 0

⎣ ⎦

2

LL x

⎡

−Δ

⎤ exp exp L

⎛ ⎞

π π π

( ) ( )

2

∫ ∫

⎢ ⎥

∫

=

2 sin x

+ Δ

x

⋅

sin xdx

⋅

exp

−

C

ω Δ

x V d x

Δ ⎜

sin xdx

⎟

ux

⎜ ⎟

⎢

L L

⎥

L

⎝ ⎠ 0 0 0

⎣ ⎦

( )

Using that sin 2 α =

2sin

α

cos

α and that

sin

α += β

sin

α ⋅

cos

β +

cos

α ⋅

sin

β , then this

may be expanded into

LL x

⎡

−Δ

⎤

exp exp

8 1 2 Cx

⎛ π π π π ⎞ ⎛ ω Δ ⎞

ˆ

2

⎢

2

⎥

ux

J

∫ ∫

cos x sin x sin x sin x dx exp d x = Δ ⋅ + Δ ⋅ ⋅ − Δ

⎜ ⎟ ⎜ ⎟ y

2

L L 2 L L V

L

⎢ ⎥

⎝ ⎠ ⎝ ⎠

0 0

⎣ ⎦

( ) ( )

⎛

1 x ˆ 1 x ˆ

⎞

π −Δ π −Δ

1

8

π

1

π

⎜

2

⎟

( )

∫

cos x ˆ

∫

sin xdx ˆ ˆ sin x ˆ

∫

sin 2 xdx ˆ ˆ exp ˆ x d x ˆ ˆ

= Δ + Δ ⋅ − ω Δ Δ

π ⎜

L 2 L

⎟

0

⎝

0 0

⎠

1

8 1

⎧ π

ˆ

2

( ) ( ) ( )

J

=

∫

1

− Δ

x ˆ cos

π Δ

x ˆ

− ⎡

cos

π Δ

x ˆ

⋅

sin 2

π

1

− Δ

x ˆ

−

sin

π Δ

x ˆ

⋅

cos 2

π

1

− Δ

x ˆ

⎤

⎨

⎣ ⎦

y

2 4

π

⎩

0

1 ⎫

( )

ˆ ˆ ˆ ˆ +

sin

π Δ

x exp

− ω Δ

x d x

Δ

⎬ ⎭

4