Internal pressures Wind loading and structural response Lecture 16 Dr. J.D. Holmes

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


• 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


• 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


• 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


• ‘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


• ‘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


• 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


• 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


• 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


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]

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Internal pressures Wind loading and structural response Lecture 16 Dr. J.D. Holmes