Internal pressures
Wind loading and structural response
Lecture 16 Dr. J.D. Holmes
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
• 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
Single Dominant Opening
Internal pressures
• Single opening on windward wall
• Dimensional analysis :
1 = A3/2/Vo - where A is the area of the opening, and Vo is the internal volume
)π,π,π,π,F(πUρ
21
pp(t)C 54321
2a
0pi
i
2a
02
Uρ21
pπ - where po is atmospheric (static) pressure
(related to Mach Number)
3 = aUA1/2/ - where is the dynamic viscosity of air (Reynolds Number)
Uπ u
4
(turbulence intensity)
5 = u/A - where u is the length scale of turbulence
Internal pressures
• Single opening on windward wall
• Helmholtz resonator model :
Air ‘slug’ moves in and out of building in response to external pressures
Air ‘slug’
Mixing of moving air is ignored
e
Internal pressures
• Single opening on windward wall
• Helmholtz resonator model :
)(tΔpAxV
Aγpxx
2k
AρxAρ e
o
2o
2a
ea
inertial term (mass times acceleration) for air slug
damping - energy losses through opening
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
Since i a , pi po
γii constant.ρ ap
ii
ii
1-γiγ
i
ii Δρ
ρ
γpργρ
ρ
pp
1-γi
i
i ρconstant.γ adρ
dp
o
a
i
i
V
Axρ
ρ
γp
o
o
V
Axγp
Resisting force = pi.Ao
2o
V
xAγp
= ratio of specific heats(1.4 for air)
Internal pressures
• Single opening on windward wall
• ‘Damping’ term :
From steady flow through a sharp-edged orifice :
k = discharge coefficient
xx2k
ρUUρ
2
1
k
1p
2a
ooa2i
Theoretically k = 2π
π
• Inertial term : xAρA).Δp(t ea
Theoretically e =
πA/4 (circular opening)
Internal pressures
• Single opening on windward wall
• Converting to pressure coefficients :
Second-order differential equation for Cpi(t)
pepipipi
2
0
oapi
o
oea CCCCAp 2k
UVρC
Aγp
Vρ
Undamped natural frequency (Helmholtz frequency) :oea
oH Vρ
γAp
2π
1n
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 :
KA = bulk modulus of air = pressure change for unit change in volume
= (a p)/, equal to po
KB = bulk modulus for the building
)]/K(K[1Vρ
γAp
2π
1n
BAoea
oH
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 Internal Volume
(m3)
Opening Area (m2)
Stiffness ratio
KA/KB
Helmholtz Frequency
(Hertz) Texas Tech field
building 470 0.73 1.5 1.6
House 600 4 0.2 2.9 Warehouse 5000 10 0.2 1.3 concert hall 15000 15 0.2 0.8
arena (flexible roof) 50000 20 4 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) :
Small opening area - high damping Large opening area - low damping - overshoot and oscillations
Vo = 600 m3. Aw = 1m2. U = 30 m/s.
2.0
1.5
1.0
0.5
0
0 0.5 1.0Time (secs)
Cpi
Vo = 600 m3. Aw = 9m2. U = 30 m/s.
2.0
1.5
1.0
0.5
0
0 0.5 1.0
Time (secs)
Cpi
Internal pressures
• Multiple openings on windward and leeward walls :
Neglecting compressibility in this case (a = 0) :
Can be used for mean internal pressures or peak pressures using quasi-steady assumption. Need iterative solution when N is large.
0Qρ ja
N
1
wherea
ie
ρ
pp2kAQ
(modulus allows for flow from interior to exterior)
0,1 ijej
NppA
N is number of openings
Internal pressures
• Multiple openings on windward and leeward walls :
Consider building with 5 openings :
Q1 Q2
Q3
Q4
Q5
pe,1 pe,2
pe,3
pe,4
pe,5
pi
inflowsoutflows
ieieieieie ppAppAppAppAppA 5,54,43,32,21,1
Internal pressures
• Single windward opening and single leeward opening :
i.e. 2 openings :
in terms of pressure coefficients,
Equation 6.16 in book
LiLiWW ppAppA
pLpiLpipWW CCACCA
2
L
W
pL
2
W
L
pWpi
AA
1
C
AA
1
CC
re-arranging,
Internal pressures
• Single windward opening and single leeward opening :
i.e. comparison with experimental data :
Used in codes and standards to predict peak pressures (quasi-steady principle)
-0.4
0
0.4
0.8
0 2 4 6 8 10AW
/AL
Cpi
MeasurementsEquation (6.16)
Internal pressures
• Multiple windward and leeward openings :
Neglect inertial terms, characteristic response time :
Characteristic frequency, nc = 1/(2c)
pLpW3/22L
2Wo
BALWoac CC
)A(Aγkp
)]/K(K[1AAUVρτ
Aw = combined opening area on windward wall
AL = combined opening area on leeward wall
fluctuating internal pressures :
numerical solutions required if inertial terms are included
Internal pressures
• Multiple windward and leeward openings :
Effective standard deviation of velocity fluctuations filtered by building :
ncu/U
u/u
0
0.5
1
0.001 0.01 0.1 1 10
High characteristic frequency - most turbulence fluctuations appear as internal pressures
Low characteristic frequency - most turbulence fluctuations do not appear as internal pressures
dnnn1(n)Sσ
2
cu
2u
Internal pressures
• Porous buildings :
Treated in same way as multiple windward and leeward openings :
AL = average wall porosity total areas of leeward and side walls
Aw = average wall porosity total windward wall area
End of Lecture 16
John Holmes225-405-3789 [email protected]