Transcript Internal Pressures - LSU Hurricane Engineering

Wind loading and structural response
Lecture 16 Dr. J.D. Holmes
Internal pressures
Internal pressures
• Wind pressure on a wall cladding or roof is always :
external wind pressure - internal pressure
• wind will affect internal pressure magnitude, except for fully sealed
buildings
• Fully-sealed buildings : assume internal pressure is atmospheric pressure
(po)
• Wind-induced internal pressures significant for dominant openings - e.g.
produced by flying debris
Internal pressures
• Single opening on windward wall
Single Dominant Opening
• air flow into building  increase in density of air within the volume
• external pressure changes produced by wind - typically 1% of absolute air pressure
• internal pressure responds quickly to external flow and pressure changes
Internal pressures
• Single opening on windward wall
• Dimensional analysis :
Cpi (t) 
pi  p 0
 F(π1 , π 2 , π3 , π 4 , π5 )
1
ρa U 2
2
1 = A3/2/Vo - where A is the area of the opening, and Vo is the internal volume
π2 
p0
1
ρa U 2
2
- where po is atmospheric (static) pressure
(related to Mach Number)
3 = aUA1/2/ - where  is the dynamic viscosity of air (Reynolds Number)
π4 
u
U
(turbulence intensity)
5 = lu/A - where lu is the length scale of turbulence
Internal pressures
• Single opening on windward wall
• Helmholtz resonator model :
le
Air ‘slug’
Air ‘slug’ moves in and out of building in response to external pressures
Mixing of moving air is ignored
Internal pressures
• Single opening on windward wall
• Helmholtz resonator model :
damping - energy losses through
opening
ρa A
γpo A 2
ρa Al e x  2 x x 
x  A Δpe (t)
2k
Vo
inertial term (mass
times acceleration) for
air slug
stiffness - resistance of internal
pressure to movement of slug
A = area of opening, Vo = internal volume
a = (external) air density, po = (external) air pressure
Internal pressures
• Single opening on windward wall
• ‘Stiffness’ term :
Assume adiabatic law for internal pressure and density
pi  a constant.ρi
γ
 = ratio of specific heats(1.4 for air)
dp i
γ -1
 a constant.γρi
dρi
pi 
pi
γpi
γ -1
γρ

ρ

Δρi
i
i
γ
ρi
ρi
γpo A x
Resisting force = pi.A 
Vo
2

γpi ρ a Ax
ρi Vo

γpo Ax
Vo
Since i  a , pi  po
Internal pressures
• Single opening on windward wall
• ‘Damping’ term :
From steady flow through a sharp-edged orifice :
ρa
1 1
p i  2 ρ a U o U o  2 x x
k 2
2k
k = discharge coefficient
Theoretically k =
• Inertial term :
π
π  2 
Δp(t).A  ρa Al e x
Theoretically le =
πA/4
(circular opening)
Internal pressures
• Single opening on windward wall
• Converting to pressure coefficients :
2
 ρ a Vo U   
ρ a l e Vo 
 C pi C pi  C pi  C pe
C pi  
γpo A
 2k Ap0 
Second-order differential equation for Cpi(t)
Undamped natural frequency (Helmholtz frequency) : n H 
γAp o
1
2π ρ a l e Vo
Increase internal volume Vo : decrease resonant frequency, increase damping
Increase opening area A : increase resonant frequency, decrease damping
Internal pressures
• Single opening on windward wall
• Helmholtz resonant frequency :
Effect of building flexibility :
nH 
γApo
1
2π ρ a l e Vo [1 (KA /K B )]
KA = bulk modulus of air = pressure change for unit change in volume
= (a p)/, equal to  po
KB = bulk modulus for the building
For low-rise buildings, KA/ KB = 0.2 to 5
(for Texas Tech field building, KA/ KB= 1.5)
Internal pressures
• Single opening on windward wall
• Helmholtz resonant frequency :
Type
Texas Tech field
building
House
Warehouse
concert hall
arena (flexible roof)
Internal
Volume
(m3)
470
Opening Area
(m2)
0.73
Stiffness
ratio
KA/KB
1.5
Helmholtz
Frequency
(Hertz)
1.6
600
5000
15000
50000
4
10
15
20
0.2
0.2
0.2
4
2.9
1.3
0.8
0.23
(measured values for Texas Tech building)
Resonant response is not high because of high damping
Internal pressures
• Single opening on windward wall
Sudden windward opening (e.g. window failure) :
2.0
2.0
1.5
Cpi
1.5
Cpi
1.0
1.0
0.5
0.5
0
0
0
0.5
Vo = 600 m3.
1.0
Time (secs)
Aw = 1m2. U = 30 m/s.
Small opening area - high damping
0
0.5
1.0
Time (secs)
Vo = 600 m3.
Aw = 9m2. U = 30 m/s.
Large opening area - low damping
- overshoot and oscillations
Internal pressures
• Multiple openings on windward and leeward walls :

N
1
where
Q  kA
ρa Q j  0
2 p e  pi
N is number of openings
(modulus allows for flow from interior to exterior)
ρa
Neglecting compressibility in this case (a = 0) :

N
1
Aj
pe, j  pi  0
Can be used for mean internal pressures or peak pressures using quasi-steady
assumption. Need iterative solution when N is large.
Internal pressures
• Multiple openings on windward and leeward walls :
Consider building with 5 openings :
pe,4
Q4
pe,5
pi
pe,3
A1
pe,1  pi  A2
Q3
pe, 2  pi  A3
inflows
Q5
Q1
Q2
pe,1
pe,2
pe,3  pi  A4
pe, 4  pi  A5
pe,5  pi
outflows
Internal pressures
• Single windward opening and single leeward opening :
i.e. 2 openings :
AW
pW  pi   AL pi  pL 
in terms of pressure coefficients,
A W C pW  C pi  A L C pi  C pL
re-arranging,
C pi 
C pW
A
1   L
 AW



2

C pL
A
1   W
 AL



2
Equation 6.16 in book
Internal pressures
• Single windward opening and single leeward opening :
i.e. comparison with experimental data :
Measurements
Equation (6.16)
0.8
Cpi
0.4
0
0
2
4
AW /AL
6
8
10
-0.4
Used in codes and standards to predict peak pressures (quasi-steady principle)
Internal pressures
• Multiple windward and leeward openings :
fluctuating internal pressures :
numerical solutions required if inertial terms are included
Neglect inertial terms, characteristic response time :
τc 
ρa Vo UA W A L [1 (KA /K B )]
CpW  CpL
2
2 3/2
γkpo (AW  A L )
Aw = combined opening area on windward wall
AL = combined opening area on leeward wall
Characteristic frequency, nc = 1/(2c)
Internal pressures
• Multiple windward and leeward openings :
Effective standard deviation of velocity fluctuations filtered by building :
2


σ¢u   Su (n) 1   n  dn
  n c 
2
High characteristic frequency
- most turbulence fluctuations
appear as internal pressures
1
u¢/u
0.5
Low characteristic
frequency - most
turbulence
fluctuations do not
appear as internal
pressures
0
0.001
0.01
0.1
nclu/U
1
10
Internal pressures
• Porous buildings :
Treated in same way as multiple windward and leeward openings :
Aw = average wall porosity  total windward wall area
AL = average wall porosity  total areas of leeward and side walls
End of Lecture 16
John Holmes
225-405-3789 [email protected]