Transcript of "Worked example wind loading on portal frame"
1.
dıgest
CI/SfB (J4)
March 1999
Wind loading on
buildings
Digest 436
Part 3
BS 6399-2:1997 worked
examples – loads on a portal
frame building and on an
ofﬁce tower on a podium
This is the last of three Parts giving brief
guidance on the use of BS6399-2.
Part 1 gives advice and guidance on
implementing BS6399-2 and suggests
certain options.
Part 2 demonstrates the recommended
options by example calculations for the
case of a two-storey timber frame house.
This Part gives example calculations for a
further two building types: a steel portal
frame building, and a 15-storey tower
surrounded by a two-storey podium.
This Digest is aimed at architects,
engineers and professionals who need to
know the effect of wind on buildings, and
design options that minimise it.
The full title of BS 6399-2:1997 is Loading for buildings.
Code of practice for wind loads. In this Digest we refer to
it as BS 6399-2 or the Standard. It supersedes
BS 6399-2:1995 which itself was a technical revision of
Loading. Wind loads,CP3:Chapter V:Part 3:1972
(abbreviated here to CP3-V-2 or the Code); references to
Code or Code of practice are to this 1972 document.
The three stages
Part 2 describes how BS 6399-2 is implemented in three distinct stages:
1 the dynamic classification – to ensure that the Standard is suitable for
the structural form,
2 the design dynamic pressure for the site which is independent of the
structural form of the building, and
3 the pressure coefficients and structural loads on the building.
Because Stage 2 is independent of the building, the examples given in
Part 2 of each of the options recommended in Part 1 do not need to be
repeated. Instead, the worked examples of this Part use only the
recommended Option 2(b) – four orthogonal cases using Equation 29 (see
Q16 and Q33 of Part 1) – the ‘hybrid’ method permitted by §3.4.2 that
gives the optimum balance between conservatism and complexity.
Figures and tables in this
Digest are denoted by
letters (eg Figure A, Table
A) to distinguish them from
those in BS 6399-2.
Except where speciﬁcally
noted, any numbered
reference to a Clause (§),
Figure or Table, refers to
that in BS 6399-2:1997.
The notional site location
For the example calculations in this Part, we shall adopt the same site used in Part 2: in the south-west
outskirts of Sheffield at grid reference SK320810. This site was chosen to illustrate the options
recommended in Part 1 because it is typical of sites in inland towns. The relevant values for this
location are given in Table A from which values for the calculations will be taken. The altitude of the
site is 120 m above mean sea level and the site is not in the zone of significant topography, defined in
Figure 7. See Part 2 for more guidance on these site parameters.
Table A Distances-to-sea and in-town for the site at SK320810
Direction (°)
0
30
60
90
120
150
180
210
240
270
300
330
Distance-to-sea (km)
200
128
110
110
200
200
200
200
200
200
112
200
Distance-in-town (km)
7.5
12.5
9.5
3.5
2.5
1.5
1.5
3.5
2.5
2.5
1.5
3.5
2.
2
Portal frame building
The principal dimensions of the portal frame building in this example are given in Figure A.
Stage 1: Dynamic classiﬁcation
This is a test to determine whether it is
appropriate to use BS 6399-2 and to
obtain the value of Cr, and is demonstrated in
Calculation 1.
Note A
The value of Cr given by
the classiﬁcation
indicates that the
building is 3.5%
dynamic.
α = 10˚
6m
H =7m
W = 30 m
L = 60 m
Figure A Principal dimensions of the portal frame building
Calculation 1 Dynamic classiﬁcation of the portal frame building
Clause
Action
Notes
1.3.3.2
Building height above its base
H=7m
Use height to eaves
1.6.1
Read value of Kb from Table 1
Kb = 2
When in doubt, take next larger value
1.6.1
Using H and Kb, read Cr from Figure 3 Cr = 0.035
1.6.2
Check Cr < 0.25
If Cr > 0.1, get better value from Annex C
If Cr > 0.25, BS 6399-2 is not applicable
BS 6399 can be used
Yes
Stage 2: Design wind speed and dynamic pressure
Notional site layout
In accordance with §1.7.2 and the distances in Table A, the site is categorised as in-town terrain. We
shall assume that the building is in an industrial estate, surrounded by buildings of a similar height,
and is aligned with the ridge at 45° east of north. This gives four orthogonal cases – NW, NE, SE and
SW (shown in Calculation 2) – that are the same as for the house in Part 2 (Figure C of Part 2). We shall
assume that the spacing of the other buildings in the estate is unknown but typical. This gives an
obstruction height Ho = 9.7 m and, from Q8 of Part 1, a spacing X = 20 m.
Owing to the symmetry of the house, we need to consider only two orthogonal directions:
q wind normal to the ridge θ = 0°, and
q wind parallel to the ridge θ = 90°.
Accordingly, the orthogonal cases SW for θ = 90° and NW for θ = 0° control the design, and the
dynamic pressures for the other cases are not required.
Stage 3: Pressure coefﬁcients and design loads
Pressure coefficients on walls
The scaling dimensions for the walls of the portal frame building are determined in Calculation 3 from
the dimensions given in Figure A. The pressure coefficients for the walls of the building are shown in
Figure B for θ = 0° (Case NW) and in Figure C for θ = 90° (Case SW).
2.8 m
+0.6
Windward face
Windward4.3) (D/H = 4.3)
( D/H = face
-0.4
-0.8
-0.1
-1.3
14 m
Side face (isolated)
Side face (isolated)
Leeward face
(Leeward face (D/H = 4.3)
D/H = 4.3)
Figure B Pressure coefﬁcients for walls of the portal frame building θ = 0°
Pressure coefficients on the roof
The scaling dimensions for the walls of the portal
frame building are determined in Calculation 4
from the dimensions given in Figure A.
The pressure coefficients for the roof of the
building are shown in Figure D (on page 4) for
both orthogonal cases.
Calculation 2 Dynamic pressures for the portal
3.9 m
+0.6
Windward face
Windward face (D/H = 4.3)
( D/H = 6.2)
-0.4
Side face
(isolated)
Side face (isolated)
-0.8
19.3 m
Figure C Pressure coefﬁcients for walls of the portal frame building θ = 90°
-1.3
-0.1
Leeward face
Leeward face (D/H = 4.3)
( D/H = 6.2)
3.
3
Note B
Ready-reckoner gives
values of Sb directly,
avoiding the use of Sc,
St, Tc and Tt factors.
frame building by Option 2(b)
Clause
Action
2.2.1
Notes
From Figure 6, basic wind speed Vb = 22.6 m/s
Hourly mean speed
10 m above ﬂat open country
From site plan, altitude of site ∆ = 120 m
1.3.3.1
2.2.2.2.1
2.2.2.2
2.1.1.2
2.2.2.3
Site less than halfway up hill
From Equation 9, altitude factor Sa = 1.120
Orthogonal direction : NE
SE
SW
0.85
1.00
Topography not signiﬁcant
Range of NE includes 0°, 30°, 60° and 90°
NW
0.99
NE value is biggest of
From Table 3, select biggest
direction factor in range Sd = 0.78
2.2.2.1
Take from OS 1:50,000 mapping
Using Figure 7, check for signiﬁcant topography = No
0.78, 0.73, 0.73 & 0.74
Ss and Sp taken as unity
From Equation 8,
site wind speed Vs = 19.74 21.52 25.31 25.06 m/s
1.3.3.4
Surrounded by similar buildings
Select lowest obstruction
height in range Ho = 9.7
9.7
9.7
9.7
m
20
20
20
m
7.6
7.6
7.6
m
Spacing unknown but typical.
Select furthest obstruction
separation in range Xo = 20
E.2.1
displacement height Hd = 7.6
2.2.3.3
110
200
112
km
1.5
1.5
1.5
km
Hr = 9.7
9.7
9.7
9.7
m
Larger of He = Hr – Hd = 2.1
2.1
2.1
2.1
m
or He = 0.4 Hr = 3.9
3.9
3.9
3.9
m
Selected for range from Table A
From Table A, shortest
distance-in-town in range = 3.5
1.7.3.3
Hd = 1.2Ho – 0.2 Xo
Selected for range from Table A
From Table A, closest
distance-to-sea in range = 110
1.7.3.1
See Q8 of Part 1
Deﬁned in Annex E.
From Q10 of Part 1,
Safe assumption for whole building
Reference height at ridge
See Q10 of Part 1
Option 2(b): use Equation 29 or ready-reckoner
3.2.3.2.3
From Table 22, factor Sc = 0.839 0.839 0.839 0.839
Logarithmic interpolation used in
From Table 22, factor S t = 0.197 0.197 0.197 0.197
Tables 22 and 23, but linear interpolation
From Table 23, factor Tc = 0.702 0.722 0.722 0.722
is adequate
From Table 23, factor Tt = 1.687 1.687 1.687 1.687
3.4.2.1
Standard value of factor g t = 3.44
2.2.3.3
From Equation 29, terrain-&-
3.44
3.44
Equivalent to CP3 Class A
3.44
Equation 29 allows for
building factor S b = 1.260 1.296 1.296 1.296
2.2.3.1
Now gust speed, equivalent to CP3 Class A
effective wind speed Ve = 24.9
2.1.2.1
actual distance-in-town
From Equation 12,
27.9
32.8
32.5
m/s
Gust dynamic pressure,
From Equation 1,
dynamic pressure qs = 379.4 476.5 659.5 646.4 Pa
equivalent to CP3 Class A
Calculation 3 Scaling dimensions for walls of the portal frame building
Clause
Wind parallel to ridge, θ = 90°, case SW
Wind normal to ridge, θ = 0°, case NW
1.3.3.2
Height H = 9.65 m
Height of gable
Height H = 7 m
1.3.4.3
Breadth B = 30 m
Figure A, B = W
Breadth B = 60 m
1.3.4.4
Depth D = 60 m
Figure.A, D = L
Depth D = 30 m
2.2.3.2
H < B, one part, Hr = 9.65 m
2.4.1.2
Span ratio D/H = 6.2
2.4.1.3
Scaling length b = 19.3 m
Peak of gable
See Table 5
Smaller of B or 2H
Height of eaves
Figure A, B = L
Figure A, D = W
H < B, one part, Hr = 7 m
Span ratio D/H = 4.3
Scaling length b = 14 m
Eaves
See Table 5
Smaller of B or 2H
Calculation 4 Scaling dimensions for roof of the portal frame building
Clause
Wind parallel to ridge, θ = 90°, case SW
Wind normal to ridge, θ = 0°, case NW
1.3.3.2
Height H = 9.65 m = Hr
2.5.2.2
Scaling length bW = 19.3 m
Height of ridge
Smaller of W or 2H
Height H = 9.65 m = Hr
Scaling length bL = 19.3 m
Height of ridge
Smaller of L or 2H
4.
4
1.93 m
Note C
For the wind angle
θ = 0° and roof pitch of
α = 10° the pressure
coefﬁcients on the roof
are always suctions. In
determining horizontal
components the
asymmetric loads
provisions of §2.1.3.7
will apply to the
beneﬁcial action of the
suctions on the upwind
roof pitch. (See Q24 of
Part 1.)
7.5 m -1.8
-0.4
-0.4
-0.6
-0.35
7.5 m -1.3
-0.8
-0.8
-0.6
-0.6
-0.8
1.4 m
-1.3
-0.45
-0.45
-0.6
-0.35
-1.8
-1.55
-1.55
-1.0
-1.0
-1.55
1.4 m
9.65 m
9.65 m
Figure D Pressure coefﬁcients for roof of the portal frame building θ = 0° and 90°
6m
6 m -1.3
-0.8
3.9 m
Figure E Location
and size of roller
shutter door
Internal pressure coefficients
We shall assume that there is a 6 m-high roller shutter door spanning the full width of the end bay on
one of the side walls. This will be assumed to be closed at the ultimate limit state, corresponding to
Calculation 5. We will treat the door open as a serviceability limit state, corresponding to Calculation
6 (see Q46 of Part 1). Figure E shows the external pressure coefficients over the area of the door at
wind angle θ = 90°. We shall take the typical porosity for curtain walling from Q44 of Part 1 as 3.5 ×
10-4.
We shall assume that all walls are equally permeable and the roof is impermeable. As the internal
pressure, door closed, is set by the average flow of wind in and out of the distributed porosity, it is
reasonable to take the average height of the walls as the reference height for internal pressure. Because
the area of the side walls is twice the area of the gable walls, the average height is close to the eaves
height and we shall use the eaves height as the reference height for the internal pressure. This is a
pragmatic engineering decision.
The reference height for the internal pressure caused by a dominant opening is the reference height
for the wall in which the opening occurs, which is also the eaves height.
Dynamic pressure
Reference heights for the portal frame building are ridge height for the roof and the gable walls and
eaves height for the side walls. The values at ridge height were derived in Calculation 2.
Corresponding values of dynamic pressure at eaves height and for the serviceability limit are
given in Table B.
Table B Values of dynamic pressure for the portal frame building
Reference height
Ultimate Sp = 1
Serviceability Sp = 0.8
Application
θ = 0°
θ = 90°
θ = 0°
θ = 90°
Hr = 7 m
547 Pa
558 Pa
350 Pa
357 Pa
pe side walls, pi (see above)
Hr = 9.7 m
646 Pa
660 Pa
414 Pa
422 Pa
pe roof and gable
Loading of roller shutter door
Clearly wind loads occur on the roller shutter door only when it is closed, and values are derived in
Calculation 7. We shall treat the door as a single structural element and use the average pressure
coefficient over the area of the door. (See Q38 of Part 1.)
Note E
At higher values of roof
pitch α the highestloaded purlin will be in
the ﬁrst (ridge) or
second purlin on the
downwind slope.
Remember that the
loaded zones are
deﬁned in plan so that
the dimensions up the
roof slope depend on
the pitch angle.
Highest-loaded purlin
We shall assume that there are eight lines of purlins on each roof slope, spaced equally from eaves to
ridge, giving a purlin spacing of 2.17 m in the plane of the roof.
Inspection of the pressure
coefficients in Figure D reveals that
6m
6m
the highest-loaded purlin will be either
the first (eaves) or second purlin on the
upwind roof slope of the end bay, as
2.17 m
2.17 m
shown in
1.42 m
Calculations 8a (opposite) and 8b
1.93 m
9.65 m
(on page 6). The relevant tributary
Wind angleangle θ0˚ 0°
Wind angle θθ= 90˚
= =
Wind θ
Wind angle = 90°
areas are shown hatched in Figure F.
Figure F Highest-loaded purlin
5.
5
Note D
With the door open, the
large positive value of
internal pressure for
wind angle θ = 0°
increases the net uplift
on the roof, while the
large negative value
θ = 90° increases the
load on the windward
wall.
Calculation 5 Internal pressure coefﬁcient for the case with the door closed
Clause
Action
Notes
Internal pressure coefﬁcient Cpi = –0.3
2.6.1.1
See Q43 of Part 1
From Equation 13, diagonal dimension a
= 10 × 3√(volume of storey) = 10 × 3√(60 × 30 × (7 + 2.65/2)) = 248 m
Includes roof space
From Figure 4, size effect factor Ca = 0.646
2.1.3.4
Site in-town, line C
Calculation 6 Internal pressure coefﬁcient for the case with the door open
Clause
Action
From Q46 of Part 1, probability factor Sp = 0.8
2.2.2.5
2.6.2
Notes
2 year return
From Equation 15, diagonal dimension is greater of diagonal
Opening controls
dimension of opening = √(62 + 62) = 8.5 m,
the value
or 0.2 ×3√(volume of storey) = 5.0 m, a = 8.5 m
From Figure 4, size effect factor Ca = 0.952
2.1.3.4
Site in-town, line C
Area of opening A1 = 6 × 6 m = 36 m2
Total surface area A2 = 3176 m2
Ratio, opening area to sum of remaining openings
= 36 / ( 3125 × 3.5 × 10–4) = 32
Much greater than 3
From Table 17, internal pressure coefﬁcient Cpi = 0.9 × Cpe
Wind angle θ = 0°, Cpi = +0.6 × 0.9 = +0.54
See Figure E
Wind angle θ = 90°, Cpi = (–1.3 × 3.9 – 0.8 × 2.1) / 6 × 0.9 = –1.01
Calculation 7 Loads on the roller shutter door
Clause
Action
Notes
Diagonal dimension of door a = √(62 + 62) = 8.5 m
2.1.3.4
Site in-town, line C
From Figure 4, size effect factor Ca = 0.952
θ = 0°
Wind angle :
2.4.1
θ = 90°
External pressure coefﬁcient
averaged over door Cpe = 0.6
2.1.3.1
From Equation 2, pe = qs × Cpe × Ca = 547 × 0.6
558 × –1.125
× 0.952 = 312 Pa
2.1.3.2
× 0.952 = –598 Pa
From Equation 3, pi = qs × Cpi × Ca = 547 × –0.3
558 × –0.3
× 0.645 = –106 Pa
2.1.3.3
Cpi and Ca from
× 0.645 = –108 Pa
earlier calculation
From Equation 4,
net pressure p = pe – pi = 418 Pa
2.1.3.5
See Figure E
–1.125
–490 Pa
Load on door P = p × A = 15.0 kN
–17.6 kN
Ultimate limit
Calculation 8a Highest-loaded purlin (ultimate limit state)
Clause
Action
Notes
Eaves purlin
Diagonal dimension a = 6.1 m
2.1.3.4
6.4 m
From Figure 4, size effect factor Ca = 0.982
0.978
Site in-town, line C
13.02 m2
Loaded area A = 6.51 m2
2.1.3.5
Second purlin
True area on slope
θ = 0°
θ = 90°
From Table B, External qs =
646 Pa
660 Pa
Internal qs =
547 Pa
558 Pa
Ultimate limit, door closed
2.1.2
Loaded purlin
Eaves purlin
Second purlin Second purlin
2.5.2.4
Averaged Cpe = –1.55
–0.62
–0.99
2.1.3.1
pe = qs × Cpe × Ca = –983 Pa
–392 Pa
–639 Pa
pi = qs × Cpi × Ca = –106 Pa
–106 Pa
–108 Pa
Equation 4, net pressure, p = pe – pi = –877 Pa
–286 Pa
–531 Pa
–3.72 kN
–6.91 kN
2.1.3.2
2.1.3.3
2.1.3.5
(Ca =0.646)
Load on purlin, P = p × A = –5.71 kN
6.
6
Calculation 8b Highest-loaded purlin (serviceability limit state)
Clause
Action
Notes
θ = 0°
θ = 90°
From Table B, External qs =
414 Pa
422 Pa
Internal qs =
350 Pa
357 Pa
Serviceability limit, door open
2.1.2
Loaded purlin
Eaves purlin
Second purlin Second purlin
Averaged Cpe = –1.55
–0.62
–0.99
(From Calculation 6) Averaged Cpi = +0.54
+0.54
–1.01
pe = qs × Cpe × Ca = –630 Pa
–253 Pa
–412 Pa
–343 Pa
2.5.2.4
2.1.3.1
2.1.3.2
(Ca = 0.952)
pi = qs × Cpi × Ca = +180 Pa
+180 Pa
2.1.3.3
Equation 4, net pressure p = pe – pi = –810 Pa
–431 Pa
–66 Pa
–5.61 kN
–0.86 kN
2.1.3.5
Load on purlin, P = p × A = –5.27 kN
Loads on portal frames, wind normal to ridge
We will calculate the individual member loads and the overall
horizontal base shear for wind normal to the ridge (θ = 0°). The wind
loading is transmitted to the portal frames as a series of point loads
through the rail and purlin fixings as shown in Figure G.
It is possible to sum the individual standard rail and purlin loads,
then to apply the appropriate size effect factor for the frame. However,
we shall adopt the quicker and commonly used practice of summing
Figure G Rail and purlin loads on frame
the load over the tributary area of cladding to give an equivalent
uniformly distributed load (UDL). (See Q38 of Part 1.)
Figure H shows the tributary area (hatched) for the horizontal base
Note F
6m
(Figure H)
shear of a frame viewed in elevation and the corresponding diagonal
Because pressure is a
dimension for base shear, as recommended in Q20 and Figure C of
scalar, we may
determine the horizontal
Part 1. We shall assume no load sharing between
component of load on
9.65 m
a
frames. The load on the upwind roof slope is a beneficial component
the roof by applying the
roof pressures to the
of the horizontal base shear, so we will need to apply the provisions of
horizontally resolved
§2.1.3.7.
components of the
surface area, as shown
Inspection of the pressure coefficient zones leads us to expect the
in Figure H. This is
Figure H Tributary area for
highest loading to occur on Frame 2 because the tributary area spans
exactly the same as
base shear of frame in elevation
resolving the pressure
the A–C and E–G zone boundaries, but we shall check also the loading
load normal to the roof
into its horizontal
on the central frames (using Frame 4). First we obtain the unfactored
component.
internal and external pressures in Calculation 9. This and later calculations omit some of the
intermediate steps already demonstrated (eg determining the zone areas, to save space and avoid
unnecessary duplication).
The loads in each member, expressed as a UDL, are found in Calculation 10 by applying the
unfactored pressures from Calculation 9 to the relevant tributary areas and size effect factors.
Similarly, Calculation 11 develops the horizontal shear at the base of each frame.
We shall also calculate the member loads at the serviceability limit with the door open on the
windward face. Calculation 12 gives the unfactored pressures and Calculation 13 develops these into
member loads expressed as a uniform distributed load.
Comparing Calculation 13 with the previous Calculation 10 we find, as expected, that the
Note G
When summing loads on
a number of
components (eg gable,
top and bottom storeys)
it is convenient to
determine the standard
loads on each
component by taking
Ca = 1, then to apply the
value of Ca appropriate
to the total loaded area
after summation.
Calculation 9 Unfactored pressures for θ = 0°, case NW, ultimate limit, door closed
Clause
Action
2.1.3.2
2.1.3.4
2.1.3.1
Unfactored* internal pressure, pi = qs Cpi = 547 × –0.3 = –164 Pa
Size effect factor for internal pressure, Ca = 0.646
Averaged unfactored* external pressures (UDL):
Frame 2
Frame 4
on upwind wall, pe front = qs × Cpe front = 328.2 Pa
–54.7 Pa
*Apply Ca later.
See Note G
328.2 Pa
on downwind wall, pe rear = qs × Cpe rear = –54.7 Pa
Notes
on upwind roof slope, pe upw
= qs × (CpeA × AA + CpeB × AB + CpeC × AC) / A = –358.0 Pa
walls,qs = 646 Pa
–324.3 Pa
on downwind roof slope, pe dnw
= qs × (CpeE × AE + CpeF × AF + CpeG × AG) / A = –282.9 Pa
qs = 547 Pa for
for roof,
from Table B
–270.6 Pa
7.
7
Calculation 10 Member loads expressed as UDL for θ = 0°, case NW, ultimate limit, door closed
Clause
Action
Notes
Frame 2
Frame 4
Columns: diagonal dimension a = √(72 + 62) = 9.2 m
0.945
Rafters: a = √ ((15/cos(10°))2 + 62) = 16.4 m
16.4 m
From Figure 4, size effect factor Ca = 0.892
2.1.3.5
9.2 m
From Figure 4, size effect factor Ca = 0.945
2.1.3.4
0.892
Upwind column UDL = (pe front × Ca – pi × Ca) × 6 = 2497 N/m
Downwind column UDL = (pe rear × Ca – pi × Ca) × 6 = 326 N/m
Upwind rafter UDL = (pe upw × Ca – pi × Ca) × 6 = –1280 N/m
Downwind rafter UDL = (pe dnw × Ca – pi × Ca) × 6 = –878 N/m
Note H
The tributary area of
Frame 2 spans the A–C
and E–G zone
boundaries, so the
members of Frame 2
are more highly loaded
than Frame 4. But the
higher A zone values are
beneﬁcial and are
reduced to 60% by
§2.1.3.7, so the base
shear for Frame 4 and
the other central frames
is marginally higher than
for Frame 2.
The internal pressure
has no effect on the
base shear because its
effect cancels out over
the inside surfaces.
between frames
would reduce Ca
2497 N/m
326 N/m
Expressed as force
–1100 N/m
per unit length
–813 N/m
Calculation 11 Horizontal base shear for for θ = 0°, case NW, ultimate limit, door closed
Clause
Action
Notes
Frame 2
Diagonal dimension a = √(9.652 + 62) = 11.4 m
2.1.3.4
From Figure 4, size effect factor Ca = 0.925
2.1.3.5
Frame 4
11.4 m
0.925
Site in-town, line C
*Apply Ca later
Unfactored* loads:
Pfront = pe front × Afront = pe front × 6 × 7 = 13 784 N
13 784 N
Adverse
Prear = pe rear × Arear = pe rear × 6 × 7 = –2297 N
–2297 N
Adverse
Pupw = pe upw × Aupw = pe rear × 6 × 15 × tan(10°) = –5681 N
–5146 N
Beneﬁcial
Pdnw = pe dnw × Adnw = pe rear × 6 × 15 × tan(10°) = –4489 N
–4294 N
Adverse
2.1.3.6
Note I
(Calculation 11)
Load sharing
Base shear P = 0.85 × (ΣPfront – ΣPrear) × Ca × (1 + Cr) =
2.1.3.7
0.85 × (Pfront + Pupw × 0.6 – Pdnw – Prear) × Ca × 1.035 =
13.97 kN
14.07 kN
See Note G
See Note H
Calculation 12 Unfactored pressures for θ = 0°, case NW, serviceability limit, door open
Clause
Action
Notes
Unfactored* internal pressure pi = qs × Cpi = 350 × 0.54 = 189 Pa
2.1.3.2
Size effect factor for internal pressure Ca = 0.955
2.1.3.4
2.1.3.1
Averaged unfactored* external pressures (UDL) :
Frame 2
Frame 4
on upwind wall pe front = qs × Cpe front = 210 Pa
–35 Pa
see Note G
210 Pa
on downwind wall pe rear = qs × Cpe rear = –35 Pa
*Apply Ca later,
on upwind roof slope pe upw
qs = 350 Pa for
walls, qs = 414 Pa
= qs × (Cpe A × AA + Cpe B × AB + Cpe C × AC) / A = –229.4 Pa
–207.9 Pa
on downwind roof slope pe dnw
for roof,
from Table B
= qs × (Cpe E × AE + Cpe F × AF + Cpe G × AG) / A = –181.3 Pa
–173.4 Pa
Calculation 13 Member loads expressed as UDL for θ = 0°, case NW, serviceability limit, door open
Member loads, expressed as UDL
Clause
Notes
Action
Frame 2
Frame 4
Columns: diagonal dimension a = √(72 + 62) = 9.2 m
9.2 m
From Figure 4, size effect factor Ca = 0.945
2.1.3.4
0.945
Rafters: a = √((15/cos(10))2 + 62) = 16.4 m
From Figure 4, size effect factor Ca = 0.892
2.1.3.5
Upwind column UDL = (pe front × Ca – pi × Ca) × 6 = 111 N/m
16.4 m
0.892
111 N/m
Downwind column UDL = (pe rear × Ca – pi × Ca) × 6 = –1278 N/m
–1278 N/m
Upwind rafter UDL = (pe upw × Ca – pi × Ca) × 6 = –2307 N/m
–2192 N/m
Downwind rafter UDL = (pe dnw × Ca – pi × Ca) × 6 = –2050 N/m
–2008 N/m
Load sharing
between frames
would reduce Ca
Expressed as
force per
unit length
dominant opening on the windward face alters the balance of loads between the members. The
windward column is less highly loaded, but the leeward column and rafters are more highly loaded.
Whether this is significant will depend on the action of the partial factors for ultimate and
serviceability loads in combination with the other dead and live loads.
8.
8
Fifteen-storey tower surrounded by a two-storey podium
L = 20 m
The office tower consists of a
15-storey tower surrounded by a
2-storey podium. The storey height is
3 m and the tower has a 2 m-high
parapet. The principal dimensions of
the office tower in this example are
given in Figure I.
W = 15 m
H = 47 m
H= 6 m
W = 40 m
L = 40 m
Figure I Principal dimensions of the ofﬁce tower
Stage 1: Dynamic classiﬁcation
Calculation 14 determines whether it is appropriate to use BS 6399-2 and obtain the value of Cr.
Note J
The value of Cr given by
the classiﬁcation
indicates that the
building is only 5%
dynamic.
Calculation 14 Dynamic classiﬁcation of the ofﬁce building
Clause
Action
Notes
1.3.3.2
Building height above its base
H = 47 m
Use height to top of parapet
1.6.1
Read value of Kb from Table 1
Kb = 1
When in doubt, take next larger value
Using H and Kb, read Cr from Figure 3
Cr = 0.05
If Cr > 0.1, get better value from Annex C.
Check Cr < 0.25
Yes
If Cr > 0.25, BS 6399-2 is not applicable
1.6.2
Note K
The combination of
X = 20 m and maximum
funnelling is not
consistent but, since
the spacing is unknown,
it is a conservative
assumption.
Stage 2: Design wind speed and dynamic pressure
Notional site layout
In accordance with §1.7.2, and the distances in Table A, the site is categorised as in-town terrain. We
shall assume that the building is in the commercial area of town and is aligned with the long axis at 45°
east of north. This gives four orthogonal cases NW, NE, SE and SW that are the same as for the house
in Part 2 (Figure C of Part 2). We shall assume that the surrounding buildings are a mixture of two and
three-storeys (2.5 storeys on average), and that their spacing is unknown but typical. From Q8 of Part
1, this gives an obstruction height Ho = 2.5 × 3 = 7.5 m and a spacing
X = 20 m. We shall further assume, since the spacing is unknown, that funnelling (see Q40 of Part 1) is
at maximum along the sides of the podium.
The design dynamic pressure at the top of the tower, Hr = 47 m, is determined in Calculation 15. As
the tower and podium will be divided into a number of parts, we shall require dynamic pressures at a
number of other reference heights, but their calculation is left as an exercise for
the reader.
Owing to the symmetry of the office tower, we need to consider only two orthogonal directions:
q wind normal to the long face, θ = 0°, and
q wind normal to the short face, θ = 90°.
4m
-1.3
2.4 m
0.8
-0.8
-0.3
-1.6
0.6
-0.9
Windward face face
Windward
Side faceSide face
-0.1
3m
-1.3
2.4 m
0.8
-0.8
-0.3
-0.9
-0.1
-1.6
Windward face face
Windward
Side faceSide face
Accordingly, the orthogonal cases
SW for θ = 90° and NW for θ = 0°
control the design, and the
dynamic pressures for the other
cases are not required.
Leeward face face
Leeward
Wind angle θ = 0°
0.6
BS 6399-2 can be used
Leeward face face
Leeward
Wind angle θ = 90°
Figure J Pressure coefﬁcients for the walls of the ofﬁce tower and podium from Table 5
Stage 3: Pressure coefﬁcients
and design loads
Pressure coefficients on walls
and roofs
The scaling dimensions for the
office tower are determined in
Calculation 16 and for the podium in
Calculation 17 (on page 10).
Figure J shows the pressure
coefficients for the walls of the
office tower and podium.
9.
9
Note L
The ready-reckoner
gives the values of Sb
directly, avoiding the
use of Sc, S t, Tc and Tt
factors.
Calculation 15 Dynamic pressures for ofﬁce building by Option 2(b)
Clause
Action
Notes
2.2.1
From Figure 6, basic wind speed Vb = 22.6 m/s
Hourly mean speed
10 m above ﬂat open country
1.3.3.1
2.2.2.2.1
2.2.2.2
2.1.1.2
2.2.2.3
From site plan, altitude of site ∆ = 120 m
Site less than halfway up hill
From Equation 9, altitude factor Sa = 1.120
Orthogonal direction : NE
Topography not signiﬁcant
Range of NE includes 0°, 30°, 60° and 90°
SE
SW
NW
0.85
1.00
0.99
NE value is biggest of
From Table 3, select biggest
direction factor in range Sd = 0.78
2.2.2.1
Take from OS 1:50,000 mapping
Using Figure 7, check for signiﬁcant topography = No
0.78, 0.73, 0.73 and 0.74
Ss and Sp taken as unity
From Equation 8,
site wind speed Vs = 19.74 21.52 25.31 25.06 m/s
1.3.3.4
Surrounded by similar buildings
Select lowest obstruction
height in range Ho = 7.5
7.5
7.5
7.5
m
20
20
20
m
5.0
5.0
5.0
m
Spacing unknown, but typical.
Select furthest obstruction
separation in range X o = 20
E.2.1
displacment height Hd = 5.0
2.2.3.3
110
200
112
km
1.5
1.5
1.5
km
47
47
47
m
42
42
42
m
Selected for range from Table A
From Table A, shortest
distance-in-town in range = 3.5
1.7.3.1
1.7.3.3
Other reference heights will be needed
Reference height at parapet
Hr = 47
See Q10 of Part 1
From Q10 of Part 1,
effective height He = Hr – Hd = 42
Hd = 1.2Ho – 0.2 Xo
Selected for range from Table A
From Table A, closest
distance-to-sea in range = 110
See Q8 of Part 1
Deﬁned in Annex E.
From Q10 of Part 1,
Option 2(b): use Equation 29 or ready-reckoner
3.2.3.2.3
From Table 22, factor Sc = 1.263 1.263 1.263 1.263
Logarithmic interpolation used in
From Table 22, factor S t = 0.151 0.151 0.151 0.151
Tables 22 and 23, but linear interpolation
From Table 23, factor Tc = 0.911 0.937 0.937 0.937
is adequate
From Table 23, factor Tt = 1.286 1.197 1.197 1.197
3.4.2.1
Standard value of factor g t = 3.44
2.2.3.3
From Equation 29, terrain-&-
3.44
3.44
Equivalent to CP3 Class A
3.44
Equation 29 allows for
building factor S b = 1.918 1.918 1.918 1.918
2.2.3.1
effective wind speed Ve = 37.9
2.1.2.1
actual distance-in-town
Now gust speed, equivalent to CP3 Class A
From Equation 12,
41.3
48.5
48.1
m/s
Gust dynamic pressure,
From Equation 1,
dynamic pressure qs = 879
1044 1444 1416 Pa
Calculation 16 Scaling dimensions for the ofﬁce tower
Clause
Wind angle, θ = 0°
equivalent to CP3 Class A
90°
Notes
From roof of podium
1.3.3.2
Height of tower H = 41 m
41 m
1.3.4.3
Breadth B = 20 m
15 m
Depth D = 15 m
20 m
1.3.4.4
2.2.3.2
Aspect ratio H/B = 2.1
2.7
Both H/B > 2. See Figure T
2.4.1.2
Span ratio D/H = 0.37
0.49
D/H < 1. See Table 5
2.4.1.3
Scaling length b = 20 m
15 m
Smaller of B or 2H
parapet height h = 2 m
2m
Assume 2 m high parapet.
0.13
See Table 8
2.5.1.4
parapet height ratio h/b = 0.1
Figure K (on page 10) shows the zones of pressure coefficient on the roof of the tower and the
podium for each wind direction. Note that the parapet of the tower reduces values in the A and B zones
by a different amount in each wind direction because the ratio h/b differs.
10.
10
Calculation 17 Scaling dimensions of the podium
Clause
Wind angle, θ
= 0°
= 90°
Notes
From roof of podium
1.3.3.2
Height of podium H
=6m
=6m
1.3.4.3
Breadth B
= 40 m
= 40 m
1.3.4.4
Depth D
= 40 m
= 40 m
2.2.3.2
Aspect ratio H/B
= 0.15
= 0.15
= 10
= 10
D/H > 4. See Table 5
= 12 m
= 12 m
Smaller of B or 2H
2.4.1.2
Span ratio D/H
2.4.1.3
Scaling length b
Clause 2.5.1.7 requires that we consider the
effect of the presence of the tower on the
pressure coefficients for the podium roof. The
zones of additional pressure coefficients from
§2.5.1.7 are given in Figure L and must be
used with the dynamic pressure for the tower
walls. Because the height of the tower gives a
large dynamic pressure, these additional zones
may control the design pressures on the
podium roof.
1.2 m
6m
3m
±0.2
-2.0
-1.65
-1.14
±0.2
10 m
-0.7
2m
-0.7
5m
1.2 m
±0.2
5m
-1.75
6m
±0.2
-0.7
-1.4
-2.0
-1.2
5m
-0.7
7.5 m
1.5 m
-2.0
-2.0
Both H/B < 1, one part
3m
Dynamic pressure
Following §2.2.3.2, the value aspect ratio H/B
Wind angle θ = 90°
Wind angle θ = 0°
for each direction allows the tower to be split
Figure K Pressure coefﬁcients for the roofs of the ofﬁce tower and podium,
into the parts shown in Figure M. These parts
are used for determining overall lateral loads
from Table 8
only and not cladding (see Q34 of Part 1). The
Note M
middle part, if it exists, may be split into as many slices as desired. We shall align the slices to the 3 m
(Dynamic pressure)
storey heights. The effective height for each part is taken as the height of the top of the part. Relevant
Clause 2.2.3.2 does not
values of design dynamic pressure for each effective height are given in Table C.
apply to cladding and
3m
3m
-1.4
components. See Q34
of Part 1.
-0.3
10 m
-0.8
-0.8
4m
-1.3
7.5 m
0.8
-1.3
-0.8
7.5 m
7.5 m
-0.3
3m
-1.3
0.8
10 m
-1.3
Wind angle θ = 0°
-0.8
7.5 m
Wind angle θ = 90°
Figure L Additional pressure coefﬁcients for the podium roof from §2.5.1.7
Hr = 47 m
Hr = 47 m
H r = 32 m
H r = 26 m
Hr =
30 m
H r = 27 m
27 m
24 m
H r = 21 m
Hr = 6 m
Hr = 6 m
Wind angle θ = 0°
Figure M Reference heights for parts of the ofﬁce tower
Wind angle θ = 90°
11.
11
Note N
The reader may conﬁrm
the values in Table C by
substituting each
reference height into
Calculation 15 and
recalculating the
results.
Table C Values of dynamic pressure for the ofﬁce building
Reference
θ = 0°
θ = 90°
Application
height
Case NW
Hr = 6 m
Case SW
503 Pa
513 Pa
Podium, structure and cladding
Hr = 21 m
1161 Pa
Lowest part of tower, θ = 90°
Hr = 24 m
1214 Pa
1st storey of middle part of tower, θ = 90°
Hr = 26 m
Lowest part of tower, θ = 0°
1216 Pa
Hr = 27 m
1230 Pa
Middle part of tower, θ = 0°.
1255 Pa
2nd storey of middle part of tower, θ = 90°
Hr = 30 m
1292 Pa
3rd storey of middle part of tower, θ = 90°
Hr = 32 m
1315 Pa
Top of middle part of tower, θ = 90°
1444 Pa
Upper part of tower, θ = 0° & 90° and cladding of tower
Hr = 47 m
1416 Pa
Horizontal shear forces on tower for each storey
In determining the horizontal shear forces, §2.2.3.2 permits the tower to be divided into the parts
defined above. Figure 5(c) of the Standard shows that the diagonal dimension for the horizontal shear
at any level is the diagonal of the loaded area above that level in horizontal projection.
Table D gives the calculation of horizontal shear PH at the base of each storey (including the height
of the parapet) above the level of the podium roof as follows:
q qs is the dynamic pressure for each storey from Table C;
q Pstorey is the unfactored horizontal load on each storey (see Note G);
q a is the diagonal dimension of the loaded area of the loaded area above the base of the storey;
q Ca is the size effect factor of the loaded area, using line B of Figure 4 above 10 m and line C below
10 m (see the key to Figure 4);
q PH is the overall horizontal shear at the base of each storey from Equation 7 of the Standard
(§2.1.3.6), using the dynamic pressure qs for the appropriate part and Cr = 0.05 from the
dynamic classification.
Note O
Loads on the parapets
should be determined
as if they were
freestanding walls
(§2.5.1.4.2) using the
provisions of §2.8.1.
The dynamic
augmentation factor Cr
does not apply to the
parapets.
Table D Horizontal shear at the base of each storey of the ofﬁce tower
Storey
Wind direction 0°
Wind direction 90°
qs
Pstorey
a
(Pa)
(kN)
(m)
15
1416
155.8
20.6
14
1416
93.5
21.5
13
1416
93.5
12
1416
93.5
11
1416
10
9
Ca
PH
qs
Pstorey
a
(kN)
(Pa)
(kN)
(m)
Ca
PH
0.893
124
1444
116.8
15.8
0.913
95
0.890
198
1444
70.1
17.0
0.908
151
22.8
0.885
271
1444
70.1
18.6
0.901
207
24.4
0.880
343
1444
70.1
20.5
0.893
261
93.5
26.2
0.875
414
1444
70.1
22.7
0.886
314
1230
81.2
28.3
0.869
474
1292
60.9
25.0
0.878
359
1266
83.6
30.5
0.864
535
1255
62.7
27.5
0.871
405
8
1216
80.3
32.8
0.858
593
1214
60.2
30.0
0.865
449
7
1216
80.3
35.2
0.853
651
1161
60.2
32.6
0.858
491
6
1216
80.3
37.7
0.847
707
1161
60.2
35.3
0.852
533
5
1216
80.3
40.3
0.842
763
1161
60.2
38.1
0.847
576
4
1216
80.3
42.9
0.805
788
1161
60.2
40.9
0.810
594
3
1216
80.3
45.6
0.800
840
1161
60.2
43.7
0.804
633
(kN)
Loads on cladding panels of tower
We shall assume that the tower is clad with panels that span the 3 m storey height and are either
3 m or 6 m wide. We shall also assume that the porosity of each face is equal and that each storey is
open plan, giving Cpi = –0.3. The maximum loads will occur:
q inwards, at any position on the windward face, and
q outwards, at the upwind edge of the side face.