Determination of Pressure Coefficients to Wind Loading Code

Determining the pressure coefficients on the surfaces of a structure using the wind loading code AS1170.2 is not an easy task, as the code has very little coverage of real world buildings, and thus is a relatively deficient code, and little wind loading research is being done to fill in the gaps. So the following example will hopefully identity some of the issues involved.
Site Plan of Example building for determining wind loading pressure coefficients

Site Plan of Example building for determining wind loading pressure coefficients

The drawing is taken from a real project undertaken back in 2005, it has been modified slightly for the current exercise. The original project was for the design of the proposed canopy, resulting in a 39 page report, a large portion of which was concerned with under-roof pressure for the canopy, discussing airflow, and otherwise assessing strengthening requirements for the house. In terms of an exercise in determining pressure coefficients, we have an example which includes enclosed building, a canopy (free roof), and enclosed space with dominant openings. 

The pressure coefficients I am about to determine are unlikely to be an exact match to the original project, for the following reasons:

  1. The original project was to AS1170.2:2002, the current example will be done to AS1170.2:2011
  2. I have around 12 years extra experience
  3. My original concern was only in terms of the new canopy, and the transfer of loads from canopy to the house. The current example I am going to consider how the presence of the canopy changes the wind pressure on the house: therefore the reserve capacity of the house structure will be different than if calculate its wind loading based on original shape. Also going to give some consideration to the difference in the pressure coefficients adopted by AS1684.1 to design a range of buildings, and the coefficients which would be selected if used AS1170.2 directly for a given building.
  4. With original project I was under pressure and time constraint to come up with an answer. Also assuming the house has no reserve structural capacity the objective was to minimise the strengthening requirements.

Getting Started

The first thing to note is that the drawing lacks adequate information. For the current exercise I added the information about the materials used for the house construction as it was on a different drawing, but its still lacking adequate information. The following information is missing:

  1. Roof Pitch for the canopy (22.5 degrees)
  2. Eaves Height for the canopy (2.4 m)

The following additional information would also be useful:

  1. Dimensions of the house
  2. Eaves overhang Dimension at House
  3. Roof pitch to house (22.5 degrees)
  4. Fence height (assume 1.8 m)
  5. Boundary clearances
  6. Location of doors and windows in the house

Since I produced a CAD drawing of the house, I assume it is one of the rare occasions I got the builder to provide some information about the existing building. So for current exercise I can get additional dimensional information from the CAD drawing as and when needed. Whilst the local council reviewing the original project could scale dimensions of the drawing to check the calculations.

Once we have adequate information about the structures (eg. dimensions: alpha, b, d, h) then we can look at the wind loading code to get Cfig, though I tend to continue using the symbols from AS1170.2:1989, and refer to Cpe, Cpi and Cpn.

A Brief Look At the Wind Loading Code AS1170.2

The two main roof shapes covered by AS1170.2 are doubly pitched (eg. gable) and monoslope (eg. skillion), these can be free roofs (no surrounding walls), or part of an enclosed building.

Roofs Shapes covered by AS1170.2
Roofs Shapes covered by AS1170.2

Additional to the basic shapes, is consideration of combined shapes, with provision for multispan pitched roofs, and multispan sawtooth roofs for enclosed buildings. Away from the ends of a building these may be suitable for multispan free roofs.

Multispan Roof Shapes Covered by AS1170.2
Multispan Roof Shapes Covered by AS1170.2

Hipped roofs weren’t explicitly covered in AS1170.2:1989, however AS1170.2:2002 introduced an explicit statement of how pressure coefficients areto be applied to hipped roofs both with a simple rectangular floor plan and L-Shaped floor plan.

Attached canopies: that is a free roof attached to an enclosed building isn’t covered for the two basic roof shapes. The code only provides coverage for a monoslope roof which is considered to be flat (roof pitch less than 10 degrees). From the diagrams the coverage is for an attached canopy attached to the sidewall of a building with a doubly pitched roof and no eaves overhang. That is the wind flow is either transverse across the canopy and doubly pitched roof of the building, or longitudinal and parallel to the ridge of the doubly pitched building.  Thus attaching to the gable end wall not covered, nor attaching to a building with a monoslope roof  or more complex roof shape.

For the attached canopy which is considered there is also consideration of adjacent walls. The canopy is taken to be attached near the end of the building. There can be one wall near the gable end of the house, so that airflow parallel to the house ridge is blocked under the canopy. Alternatively there can be two walls, with an open end at the house gable end: so airflow transverse across the house is blocked and flow parallel to the house ridge is blocked from one direction. There is no allowance for the walls having openings such as an adjacent fence with a height less than the canopy.

For canopies in general there is no allowance for consideration of surrounding walls, either full walls or partial walls. For example agricultural shelters with two adjacent walls or three adjacent walls are not considered, nor parallel walls. Similarly shelters where walls have gaps at the bottom and top are not covered.

Partially Enclosed Buildings
Partially Enclosed Buildings

Whilst the code does consider canopies which are blocked under the blockage consider is a volumetric blockage, which blocks 75% of the cross section. Such blockage alters the airflow under a canopy differently than a simple blockage of the cross-section and also differently than a full sectional blockage at the end of the canopy. So without further information, canopies with attached or adjacent walls are best treated as enclosed buildings with dominant openings.

Blocked Airflow Under a Free Roof
Blocked Airflow Under a Free Roof

However, the problem with treating canopies with walls as enclosed buildings with dominant openings is the applicability of the simple rule: which makes the internal pressure coefficient match the external pressure coefficient of the surface which contains the opening. The applicability of this rule is questionable because a missing wall cannot develop such pressure coefficient. So it is not entirely valid to suggest that if we remove the windward then the internal pressure coefficient is +0.7 matching the wall. The wall isn’t there so it cannot develop such surface pressure, and therefore such pressure cannot be transferred to an opening in the wall. However, we have no other guidelines, so that it is the approach taken.

to be continued …

More Limitations

{So got slightly side tracked and delayed}

One common prescriptive solution is the timber framing code (AS1684). Contrary to popular opinion and common practice, the code is not suitable for design/specification of houses of any shape. Not only is AS1684 limited by the constraints on dimensions contained within its own pages, but also limited by the constraints with AS4055 wind loads for housing and AS1170.2 Wind Actions. In particular the reference to rectangular, is only valid if a section through the building produces a simple doubly pitched or monoslope profile: with the roof sat on two walls. This restriction is partly because the code doesn’t cover more complex structural forms, and partly because the wind loading code cannot provide consistent and reliable pressure coefficients for more complex shapes.

If rectangles with the required profile are placed end to end, or side by side then the structural form and wind pressures are covered. if the rectangles are assembled into L-shaped, T-Shaped and U-Shaped floor plans then the structural form at the intersection of the rectangles is not fully covered. Depending on the relative dimensions of each of the rectangular segments, the wind pressures taken from AS1170.2 may not be reliable. If the floor plan is developed from overlapping rectangles then chances are it is not within the scope of the codes to provide reliable prediction of the distribution of wind load on its surfaces.

The following diagrams illustrate, the situation of overlapping rectangles (squares in this case), and then two possible roof forms. One with gable roof segments and the other with hipped roof segments.  Whilst there are clearly segments where can take sections with the required profile, the intersection of the segments is complex (or at least involved), it is questionable as to how practical it is to install a practical structure to support such roof shape.

Also whilst can apply pressures to the sloping surfaces directly from AS1170.2, the interaction of those roof shapes at the intersection is not covered. The interaction of shapes can cause turbulence which alters the fluid flow, such turbulence can generate suction where otherwise assigned a positive pressure, and likewise where assigned suction the turbulence may produce positive pressure. More importantly turbulence can cause oscillation between positive and negative (suction) pressures: the resulting vibration can then cause fatigue of the materials.

Floor Plan with Overlapping Rectangles
Floor Plan with Overlapping Rectangles


Gable Roof 3D
Gable Roof 3D
Hipped Roof 3D
Hipped Roof 3D


If something is declared to be code compliant then it should be fully within the scope of the code, otherwise it is guided by the code but not compliant with it.

Wind Loading American Barn Style Building
Wind Loading American Barn Style Building

Another common situation is the American Barn Style cold-formed steel shed, commonly used for holiday shacks. Is the pressure on the wall above the skillion negative or positive? At the junction between the two segments there will be turbulence, how much will depend on the relative height of the gable to the skillion. If there is little difference in height, then it is likely the wall will experience pressure which will match that of the skillion roof. Also what dimension do we use for the parameter (d) when selecting pressure coefficients: that for the individual segment or the over all width of the building? If there is little difference in the heights, then the whole profile could be enveloped by a single gable profile: then use the pressures on the enveloping profile to assign pressures to adjacent surfaces on original profile. Such an approach will assign suction to the small vertical wall above the skillion roof.

American Barn Style Building With Enveloping Gable
American Barn Style Building With Enveloping Gable

The illustration shows the enveloping profile developed from the central gable segment of roof, an alternative is to envelope the profile by extending the skillion roofs and raising above the gable.

Irrespective of whether the skillion roofs are canopies or enclosed buildings, the shape of the building is not strictly within the scope of the wind loading code. So any pressure coefficients we assign to the surfaces are a matter of judgement and derivation, backed by qualitative understanding of fluid mechanics. So for example it does not seem reasonable that a small segment of wall above the skillion roof would experience positive pressure when the larger adjacent surfaces of the roofs  experience suction, and such segment is in the middle of such roofs. Therefore need to assess a value for the suction: multispan roof profiles give some guidance, though the profiles considered in AS1170.2 don’t match the American barn style building the information does tend to suggest adjacent surfaces experience similar pressures.

Back to The Example

In the original project I didn’t check the pressures on the house. First because only really interested in the design of the canopy, and houses are typically designed using simplistic approaches such as AS1684. The approach used by AS1684, being part documented in AS1684.1: Design Criteria. Due to the structural form of timber framed houses, most of the structural members can be taken in isolation from other members. That is to say that rafters, wall studs and lintels can be designed as beams independent of one another. Due to the need to produce span tables and cover the suitability for a large range of houses, the pressure coefficients which may be experienced by a structural element in a given surface of a house of a given shape is not considered. Instead values for the pressure coefficients which are likely to cover most common applications are selected. The values are not the highest values which could be chosen from AS1170.2 but they are in most cases likely higher than would get if used AS1170.2 directly: in a few situations they may be lower.

In particular in non-cyclonic regions low internal pressure coefficients are adopted without any real justification, whilst in cyclonic regions high internal pressure coefficients are adopted with questionable justification. Questionable because if the house envelope is as fragile as implied then are walls are equally likely to be damaged, and only openings on the windward wall produce internal pressure, all other walls produce suction. Airflow into the building has to be balanced by airflow out, and the areas of outflow likely exceed the inflow. Furthermore pressure coefficients for canopies are lower than for enclosed building with dominant windward opening. So if don’t design for the high pressure then failure of windward door/window/wall will result in the wind rushing in and then bursting the windows on its way out, venting any further airflow and pressure. It is not conservative to arbitrarily adopt +0.7 as an internal pressure coefficient when assessing a building, especially if considering the whole building: it may be suitable for member in isolation but not for the whole building. {eg. a windward wall would have near zero pressure: fine if no wall at all, but not fine if have some wall with small opening}

So for example AS4055:1992 has Cpe=(-0.9, +0.4) and Cpi=(0.2,-0.3) which would give Cpn=(-1.1,+0.7) compared to AS1684.1:1999 which has Cpn=(-1.1,+0.56) for rafters with Cpn=(+0.56,-1.6) for overhangs. So first thing is, I am most likely to use Cpi=(+0.4,-0.3) for which I have a derivation and justification other than housing therefore use AS4055 Appendix B (informative): since its informative not strictly part of the code (eg. cannot be mandated). However whilst my Cpi may be higher the value of Cpe for the actual dimension and shape of the house is likely lower magnitude than -0.9. A value of Cpe=-0.9 is typically only for relatively flat  roofs (pitch less than 10 degrees) or it is the longitudinal wind loading near the windward edge, away from the windward edge the load decreases. {NB: There is more than one way to get a net pressure coefficient Cpn=-1.1: the question is can we justify something of lower magnitude or do we need to adopt something higher?}

Therefore if we determine the wind loading for the house without the attached canopy it is likely less than the house was actually designed for using AS1684. The house structure will therefore have reserve structural capacity that we would otherwise assume it doesn’t have.

Irrespective of such assessment, connections specified to AS1684.2 tend to have reserve for wind class N2, but not for wind class N1 or N3. Wind class N1 typically uses nominal connections and may have already allowed a 10% overstress: since already overloaded there is no reserve to accommodate an attached canopy. Whilst with wind class N3, starting to get to the limits of the connection types specified in AS1684.2, so little to no reserve. Whilst for wind class N2, the connections available in AS1684.2 typically have adequate reserve. There is only any need to assess the house, if attempting to get rid off strengthening requirements which are impractical to install, when attaching a canopy.

Note that if optimise the structural design of the house in the first instance then it will lack reserve capacity to later attach a canopy. Far better to consider designing the house for attached canopies in the first place. The house design can be issued with guidance on maximum size canopy which can be attached. Irrespective of what AS1684.2 permits, it is recommended that bottom plates are bolted to the slab, and that studs are connected to plates using straps or framing brackets, and similarly rafters are installed using framing brackets or straps. The relative dimensions of house and block of land give some indication of the size of canopy which may be installed in the future: and currently the main issue for councils is tie-down connections the possibility of overloading the wall studs and buckling them is seldom considered. But these are structural design issues for another post. {NB: Wall studs are laterally restrained by noggins and plasterboard lining. A builder can use AS1684 span tables to estimate the size of installed wall stud, then identify the maximum roof load width (RLW) for which the stud is suitable. For example F7 70×35 wind class N1/N2 @600 c/c stud height 2.4m, with rafters @1200 c/c supporting sheet metal roof then maximum RLW=6500 mm. (Table:7). Therefore if house 6m wide with full span roof trusses, then used RLW=6/2=3m (approx.), therefore remaining RLW = 6.5-3=3.5m. Therefore first estimate of maximum width of canopy which can be attached is less than 3.5m. RLW should be along slope, so if increase roof pitch this number will decrease.}

Back to determining the pressure coefficients.

Given that pressure coefficients are documented relative to roof profiles, some profiles of the building would be more useful than a plan drawing.


Elevation Views on House
Elevation Views on House


House Elevations
House Elevations

So the basic shape is doubly pitched with hipped ends. However being U-Shaped floor plan, we don’t really know what happens as wind flows directly into the opening: it is a small opening and could be turbulent in the region. Similarly what happens when wind flows pass the opening. Then there is airflow across the roof and opening, is it appropriate to consider the two legs of the U-shape as independent or part of a multispan roof with a hole: when the canopy is installed it will be a multispan roof. To start with just go with independent segments then adjust as judged necessary.

So the relevant segments are as shown in the following figure.

Segments of House and Dimensions Needed for WInd Load Assessment
Segments of House and Dimensions Needed for WInd Load Assessment

In all case the eaves overhang has been ignored and the walls located at the edge of the roof. There is a common roof pitch of 22.5 degrees and common eaves height (he) of 2.4m. To determine pressure coefficients for each segment, we required the downwind distance for the transverse (theta=0) direction and the average height (h or h[avg]). To get ‘h’ need to calculate the ridge height or height to the top (ht) and take the average: or we can just calculate the mid height of the roof based on half the rise. The old code however used ‘he’ and ‘ht’ and which was used for ‘h’ depended on direction of wind loading: so I am in the habit of calculating the two so that’s how I will get h[avg].

For each segment need to calculate h/d ratio to get roof coefficients, whilst need d/b ratio for the walls.

Pressure Coefficients: Walls

For the d/b ratio, will just take the overall dimensions of the building at the walls. The two dimensions are 18.264 m and 17.870 m. The d/b ratio depends on direction therefore d/b = 18.264/17.870=1.0220  and d/b=17.870/18.264=0.9784.

So the building isn’t elevated and not going to vary Cpe with height therefore Cpe[W]=+0.7. For the leeward wall also need roof pitch alpha (=22.5 degrees).  So have a problem with Table 5.2(B), common roof pitch of 22.5 degrees not in the table. The table has 20 degrees for all values of d/b, then jumps to angles greater than 25 degrees. For the current exercise d/b is greater than 0.3, so adopt Cpe[L]=-0.5 for any direction.

As for the consideration of segments. If the wind is considered to flow from within the small alcove of the building outwards across the building, then it may be sensible to reduce the value of the parameter ‘b’ to the length of the segment, whilst also reducing ‘d’ to the width of the segment. However it seems unreasonable to considered that wind can flow out off that segment over the two legs. Something is unknown is going to happen to the airflow as the fluid flows over one leg of the U-Shape and then encounters the open alcove below. The pressures on those walls therefore cannot be grabbed directly from the code.

The wind blowing into the alcove and across the wide roof may pose a difference. But is that difference large enough to reduce the d/b so as to get a different Cpe from the code. So taking b=3m , and d=10.564 m from above, d/b =10.564/3 =3.52, definitely a major difference, but still bigger than 0.3, therefore still get Cpe[L]=-0.5. As for the direction at 90 degrees to this leg, then d/b is as for the building as a whole.

The side wall pressures vary depending on the distance from the windward edge, they vary from Cpe[S]=-0.65 to Cpe[S]=-0.2. The distance from the windward edge (dwe) is measured in multiples of ‘h’, which haven’t calculated yet. But estimating just using ‘he’, then 3.he = 7.2m, the building is longer than this therefore a large portion of the building will have Cpe[s]=-0.2. The legs of the U-shape  (segments 3 & 4) will otherwise experience higher pressures. Though around the location of the alcove, no matter which direction along this axis of flow, the pressure coefficient at the alcove will be around -0.2. Which may be indicative that the pressure coefficient on all walls within the alcove are also -0.2: assuming the alcove is enclosed space with dominant opening (ignoring the roof opening at this point in time).

… to be continued.

Pressure Coefficients: Roof

To get pressure coefficients for roof need to calculate the aforementioned h/d ratio’s, which requires calculation of ‘h’. So use a bit of trig’ to get the roof rise and then calculate h/d. Of course I could just measure heights in CAD, however most of the time I don’t have a CAD drawing just a builders freehand sketch and they don’t want nor need formal drawings: so better to have a spreadsheet (or other eWorkbook), which takes common known dimensions and calculates needed dimensions (can otherwise check trig’ calcs by drawing in CAD and vice versa). So the h/d ratios are:

Example Wind Calc's Step 1
Example Wind Calc’s Step 1

So with these dimensions can look at the standard and get the external pressure coefficients (Cpe) for the surfaces U, D and R. so looking at Table 5.3(A), our roof pitch alpha=22.5 degrees and for direction theta=0 the table is not relevant has our pitch is greater than 10 degrees. For direction theta=90, the table is relevant to get Cpe[R] which varies with distance from windward edge. With the exception of segment  4, all the h/d ratio’s are less than 0.5, therefore maximum Cpe[R]=-0.9. Segment 4 has a h/d=0.51 this is greater than 0.5 but less than 1, there is thus a gap in the code. However the note(1) in the code indicates linear interpolation can be used for intermediate values of roof slope and h/d . However our current situation is relatively close to 0.5, especially if rounded to 1 decimal place, therefore will adopt  column with Cpe[R]=-0.9.

Next table is Table 5.3(B), direction theta=0, and alpha greater than 10 degrees. Alpha=22.5 not in the tables, but linear interpolation is permitted. The h/d ratio’s not directly given and linear interpolation also permitted for these values also. But interpolation requires more effort, so won’t do that here: given I have written a function library to do that for me.

So scanning the tables has h/d increases Cpe increases, therefore adopt h/d=0.5 for looking up values. Further scanning indicates that has increase roof pitch the magnitude of negative pressures decreases, whilst that of positive pressures increases.  So will adopt 20 degrees to get negative value and 25 degrees to get positive value. If I interpolated the values then the results would also show similar bias. So I adopt these values for all segments. Interpolation however may be preferable to get lower Cpe for segments 1 and 2. So Cpe[U]= (-0.4,+0.3), given a further check indicates positive values of Cpe decrease as h/d increases, therefore use h/d=0.25 to get positive values.

Then move onto Table 5.3(C) which will provide Cpe[D] and Cpe[R, hipped]. The table is for alpha greater than or equal to 10 degrees. ok! Previous versions of the code, table was clearly for direction theta=0, and no reference to hips. I take it that crosswind is normal to the direction of airflow. mmh!

I take transverse (theta=0) to be across the profile of the roof, and longitudinal (theta=90) to be parallel to the ridge line. Fluid flowing parallel to the ridge line is relatively stable, fluid flowing up {My 2002 version of code is printed, I put theta back on the diagrams, my 2011 version is pdf file, and I keep forgetting I have. Whilst I do have vba functions, I still reference the code as there are gaps in AS1170.2’s coverage, and I seem to get projects which identify them.}

ok! My interpretation. We have common roof planes and hip roof planes. The hip roof planes are the full triangular segments of roof only. In the current example, whilst the whole roof is “hipped” only the legs of the U-shape plan have hip planes. These hip planes only have Cpe[R] from table 5.3(C) when wind is direction theta=90 relative to these planes, flow parallel to these planes, to which earlier codes would have applied values from the equivalent of Table 5.3(A). This new approach only applies to hip planes with angle alpha greater than 10 degrees, otherwise its Table 5.3(A) as in previous codes. So code is indicating, with small hip planes, the pressure is influenced more by the downwind slope than the upwind slope.

So in current example wind normal (theta=90) to segment 1, then Cpe[R] comes from Table 5.3(A). For the hip planes need some additional sectional profiles.  Figure 5.2 in AS1170.2 indicates the parameters ‘b’ and ‘d’ are taken for the whole of the L-Shaped building. Whilst this may seem appropriate for determining pressure coefficients for walls, as shown above, it seems inappropriate for determining the pressure coefficients for roof planes. Though does pose the problem of what is ‘d’ for the hip plane? Here will consider similar to monoslope roof, and therefore ‘d’ is the projected length of the hip plane: horizontal distance from eaves to apex (ridge).

So need to back-track to the drawings to get more information to determine Cpe[R], however should be able to continue and get Cpe[D].  Once again h/d ratio and roof pitch alpha are not directly available in the code. Scanning table indicates has h/d increases then negative Cpe magnitude increases, and only negative values given. For low h/d as alpha increases magnitude of Cpe increases: no consistent trend here. For alpha greater than 25 degrees also need to calculate b/d ratio’s. Interpolation is also permitted so if interpolated between 20 and 25, would need b/d. Will just adopt Cpe[D]=-0.6 has h/d<=0.5, and alpha >20.

So the pressure coefficients determined so far are:

Example Wind Calculations Step 2
Example Wind Calculations Step 2

So if use AS4055 Cpi=+0.2, then maximum net got so far is Cpn=-0.6-(+0.2)=-0.8, or using my preference of Cpi=+0.4 then Cpn=-0.6-(+0.4)=-1.0, which is closer to the Cpn=-1.1 adopted by AS1684.1 (Now AS1720.?).

Well thus far certainly takes a lot longer to explain than do! Only got pressure coefficients for part of the house and still got to find coefficients for the canopy, and then combine the house and canopy together.

So iterative process need more sections. So here is a 3D line drawing of the roof, something which can be drawn in Acad LT, just to check 2D development of roof layout. The lines do not represent rafters or purlins, they are just lines I scribed on the surfaces of the planes to assist find the line of intersection of the planes. Lines at the same height, lie in the same plane and the intersection of such lines identifies a point of intersection for the planes.

House Roof Shape View1
House Roof Shape View1

So the whole process is improved if have a complete picture of the building, including appropriate sections. If I produce drawings I typically draw all four elevations (N,E,S,W), and then visualise sections millimetre by millimetre, if it is different then I draw the section. Four elevations and a typical section is seldom adequate description of a building, especially a house. For the original project all I had was enough information to draw an outline of the building and then develop the roof layout: as the job at hand was to design the attached canopy. So simplistically only needed enough information about the house roof at the point of attachment.

For the current exercise may have been more productive to do some preparation and get all drawings as needed: but using pencil and paper don’t really have that problem, I produce sketches on the fly as needed. On the other hand for the original project I did produce the following drawing to aid with hand written calcs.

Wind Loading Sketch 1
Wind Loading Sketch 1

The sketch covers the two axes of wind action and the two directions along these two axes: to provide a total of 4 directions to consider (Theta=0,90,180,270). The primary purpose of the diagrams to assist determine under roof venting.

To further assist with determining wind loading I have a few standard calc sheets which can be printed out from CAD, more than anything they were drawn to assist with distributed loads, for roof pitch less than 10 degrees and longitudinal wind loading.

Wind Loading Sketch 2
Wind Loading Sketch 2

Another example worksheet for wind loading is the following, which is for assessing internal pressure coefficients due to one or more openings.

Wind Loading Sketch 3
Wind Loading Sketch 3

I rarely use these worksheets, I drew them up whilst learning the wind loading code and developing my spreadsheets and VBA functions {which probably around 20 years ago}. i could produce freehand sketches on an as needs basis, however these worksheets save time and provide consistency.

To write this post however, and produce digital work, I am messing around with ProgeCAD 2016, pdffactory, Bluebeam, paint, MS Picture Manager, and MS Excel. Either producing sketches directly in bluebeam, or plotting from ProgeCAD to pdffactory then using bluebeam to get a snapshot saved in paint and cropped and resized with MS Picture Manager. Which is a lot of messing around compared to scribbling some numbers on a piece of paper.

Using spreadsheets I can just input building dimensions and it determines pressure coefficients by interpolation in a few seconds, though need to repeat for each building segment. Whilst using pen and paper can write conservative pressure coefficients down from memory (eg. W=+0.7, U,D,R=-0.9, S=-0.65, L=-0.5 or Cpn=-1.2, -1.6).

So will go away do some more sketches for the example, and then return and continue with the canopy, and house canopy combination.

… to be continued.


Will write article on pressure coefficients on hipped roofs before completing.



  1. c/c = centre to centre


  1. [21/05/2018]: First draft (brief look at code)
  2. [23/05/2018]: More limitations and wall coefficients
  3. [24/05/2018]: