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]]>ASCE 7 has a method for calculating wind loads on rooftop Structures and equipment for buildings, and this article will describe that method. Fortunately, the method is relatively straight forward, not too complicated, and is covered in ASCE 7-16 Section 29.4.1

In section 29.4.1 of ASCE 7, it coveres all rooftop structures and equipment **EXCEPT** solar panels and structures identified in section 29.4 (chimneys, tanks, open signs, single-lane open frames, and trussed towers). We have covered the method for calculating the wind load on solar panels in a separate article. The most common rooftop equipment that would fit this criteria would be Heating Ventilation and Air Conditioning (HVAC) units.

The lateral force acting on the equipment is calculated using equation 29.4-2. An corresponding excerpt from ASCE 7-16 Sec 29.4.1 is shown with Equations 29.4-2. This is a very simple equation made up of the terms GCr, qh and Af.

The equation is simple, with the most complicated aspect being that the value of GCr is dependent upon Af, and if 0.1*B*h < Af < B*h then we must determine GCr through linear interpolation.

The vertical uplift force on the equipment is calculated using equation 29.4-3, which is shown in the corresponding excerpt. This equation is very similiar to Equation 29.4-2, which is used for lateral force. The value of GCr is determined based upon the value of Ar, and if 0.1*B*L < Ar < B*L then GCr will be determined using linear interpolation.

Let’s work through an actual example now to see how this all works. We will consider a HVAC unit on top of a flat roof building. Please note that although we are assuming the roof is flat, the standard just requires that we provide the mean roof height, and so this method would apply to any style of roof. We will use ASCE 7-16 for this example and the building parameters are as follows:

**Building Mean Roof Height**: h = 50 ft [15.2 m]

**Building Length**: W = 100 ft [30.48 m]

**Building Width**: W = 50 ft [15.24 m]

**Rooftop Equipment (HVAC) Length**: L = 8 ft [2.43 m]

**Rooftop Equipment (HVAC) Width**: W = 3 ft [0.91 m]

**Rooftop Equipment (HVAC) Height**: W = 4 ft [1.22 m]

*The Rooftop equipment is oriented so that the length of the equipment is aligned with the length of the building. *

**Wind Speed**: V = 150 mph [67.1 m/s]

* (Based upon Category III)*

**Exposure**: C *(Open Terrain)*

**Topography:** Flat, no topographic features

The all references in the following calculations are per ASCE 7-16.

First we need to calculate the velocity pressure at the mean roof height:

Table 26.11-1 for Exposure C:

zg = 900 ft , Alpha = 9.5

Kh = 2.01*(z/zg)^(2/Alpha) {Table 26.10-1}

Kh = 2.01*(50/900)^(2/9.5) = **1.094**

Kzt = **1.0** {Site has no topographic features}

Kd = **0.85** {Table 26.6-1 for Rooftop Equipment}

Ke = **1.0** {Site is near sea level}

qh = 0.00256*Kh*Kzt*Kd*Ke*V^2 {Eqn 26.10-1}

qh = 0.00256*1.094*1.0*0.85*1.0*150^2 = **53.56 psf [2.56 kPa]**

Now we need to determine Af, the vertical projected area normal to the wind. Since our equipment is rectangular, we need to calculate this for both orthogonal directions:

Wind Normal to 3 ft Face: Af = 3 ft * 4 ft = **12 sq ft [1.11 sq m]**

Wind Normal to 8 ft Face: Af = 8 ft * 4 ft = **32 sq ft [2.97 sq m]**

When calculating the horizontal projected area, Ar, the value will be the same regardless of the wind direction:

Ar = 3 ft * 8 ft = **24 sq ft [2.23 sq m]**

For the lateral force calculation we now need to determine the GCr, and to do that we need to determine where Af is in relation to 0.1*Bh and Bh. B is the width of the building normal to the wind direction, and so B will be different for each direction we consider.

Wind Normal to 3 ft Equipment Face: h = 50 ft [15.24 m] and B = 50 ft [15.24 m]

Bh = 50*50 = 2500 sq ft [232.3 sq m]

0.1*Bh = 0.1*2500 = 250 sq ft [23.2 sq m]

Since Af = 12 sq ft [1.11] that is well below 0.1*Bh = 250 sq ft [23.2], and so GCr = **1.9**

Wind Normal to 8 ft Equipment Face: h = 50 ft [15.24 m] and B = 100 ft [30.48 m]

Bh = 100*50 = 5000 sq ft [464.5 sq m]

0.1*Bh = 0.1*5000 = 500 sq ft [46.5 sq m]

Since Af = 32 sq ft [2.97] that is well below 0.1*Bh = 500 sq ft [46.5], and so GCr = **1.9**

For the vertical force calculation, we need to determine GCr, and this is dependent upon Ar. These calculations are also dependent upon B that we used earlier, but it also uses L which is the building dimension in the direction of the wind.

Wind Normal to 3 ft Equipment Face: L = 100 ft [30.48 m] and B = 50 ft [15.24 m]

BL = 100*50 = 5000 sq ft [464.5 sq m]

0.1*BL = 0.1*5000 = 500 sq ft [46.4 sq m]

Since Ar = 24 sq ft [2.23] that is well below 0.1*BL = 500 sq ft [46.4], and so GCr = **1.5**

Wind Normal to 8 ft Equipment Face: L = 50 ft [15.24 m] and B = 100 ft [30.48 m]

BL = 50*100 = 5000 sq ft [464.5 sq m]

0.1*BL = 0.1*5000 = 500 sq ft [46.5 sq m]

Since Ar = 24 sq ft [2.23] that is well below 0.1*BL = 500 sq ft [46.5], and so GCr = **1.5**

Now we bring together all of these terms to calculate the lateral force (Fh) and vertical uplift force (Fv) on the rooftop equipment:

Wind Normal to 3 ft Equipment Face:

Fh = qh*GCr*Af

Fh = 53.56*1.9*12 = **1221 lbs [5434 N]**

Fv = qh*GCr*Ar

Fv = 53.56*1.5*24 = **1928 lbs [8580 N]**

Wind Normal to 8 ft Equipment Face:

Fh = qh*GCr*Af

Fh = 53.56*1.9*32 = **3256 lbs [14491 N]**

Fv = qh*GCr*Ar

Fv = 53.56*1.5*24 = **1928 lbs [8580 N]**

That wasn’t too complicated, but it also wasn’t exactly simple either. There is a faster way to do this using MecaWind software. Lets work through this same example using the MecaWind Software. If you would like to follow along using your own copy of MecaWind, here is a link to download the input file; however, since the inputs are simple it’s just as easy to enter your own model using the screen shots provided here.

The first step is to enter our wind parameters and select our structure type as “Other” and then select “Rooftop Equipment”. See inputs in corresponding image.

In the commentary of ASCE 7-16 (Section C29.4.1) there is some discussion about Mechanical Equipment Screens. These are often used to conceal plumbing, electrical, or mechanical equipment from view. They are defined as rooftop structures not covered by a roof and are located away from the edge of the building roof such that they are not considered a parapet. These can be solid or porous panels, the latter of which allows some wind to flow through the panel. The commentary states that little research is available to provide guidance for determining wind loads on these screens as well as the equipment behind the screens. Consequently, ASCE 7-16 recommends that the screens (whether solid or porous) and all of the equipment behind the screens should be designed for the FULL wind load determined in accordance with Section 29.4.1. The only exception to this is when the situation has been appropriately analyzed using the Wind Tunnel Procedure in Chapter 31.

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]]>ASCE 7-16 defines Components and Cladding (C&C) as: “Elements of the building envelope or elements of building appurtances and rooftop structures and equipment that do not qualify as part of the MWFRS (Main Wind Force Resisting System).” In simple terms, C&C would be considered as windows, doors, the siding on a house, roofing material, etc..

We will use ASCE 7-16 for this example and the building parameters are as follows:

**Building Eave Height**: EHt = 40 ft [12.2 m]

**Building Length**: L = 200 ft [60.96 m]

**Building Width**: W = 100 ft [30.48 m]

**Roof Type**: Monoslope with 1:12 Slope

**Enclosure Type: ** Enclosed

**Elevation:** Job site is at sea level

**Wind Speed**: V = 150 mph [67.1 m/s]

* (Based upon Category III)*

**Exposure**: C *(Open Terrain)*

**Topography:** Flat, no topographic features

In order to calculate the wind pressures for each zone, we need to know the effective area of the C&C. To determine the area we need the Width and Length:

Length = Span of the component

Width = The effective width of the component which need not be less than 1/3 of the span length.

As an example, a roof joist that spans 30 ft and are spaced 5 ft apart would have a length of 30 ft and the width would be the greater of 5 ft or 30 ft / 3 = 10 ft. In this case the 1/3 rule would come into play and we would use 10ft for the width.

Eff Area = 30 ft x 10 ft = 300 sq ft

(Note: MecaWind makes this adjustment automatically, you just enter the Width and Length and it will check the 1/3 rule)

When calculating C&C pressure, the SMALLER the effective area the HIGHER the wind pressure.

Most of the figures for C&C start at 10 sq ft [0.9 sq m] and so for the purpose of this example we will consider an effective area of 10 sq ft for all wall and roof wind zones. This will give us the most conservative C&C wind pressure for each zone.

Chapter 30 of ASCE 7-16 provides the calculation methods for C&C, but which of the seven (7) parts in this section do we follow? We just have to follow the criteria for each part to determine which part(s) our example will meet. To do this we first need our mean roof height (h) and roof angle.

Roof Angle = arctan(1/12) = 4.76 Deg

Sec 2.62 defines the mean roof height as the average of the roof eave height and the height to the highest point on the roof surface, except that, for roof angles less than or equal to 10 deg, the mean roof height is permitted to be taken as the roof eave height.

Since our Roof Angle (4.76 Deg) <= 10 Deg, then we can take h as the eave height (EHt).

h = EHt = 40 ft [12.2 m]

The other determination we need to make is whether this is a low rise building. To be considered a low rise, the building must be enclosed (this is true), the h <= 60 ft [18] (this is true) and the h<= least horizontal width. Our least horizontal dimension is the width of 100 ft [30.48] and our h is less than this value, so this criteria is met as well. Therefore this building **is a low rise building.**

Using all of this criteria, we can then determine that the only two methods of Chapter 30 where we meet all criteria are Part 1 and 4 (see chart).

We now follow the steps outlined in Table 30.3-1 to perform the C&C Calculations per Chapter 30 Part 1:

**Step 1: **We already determined the risk category is III

**Step 2**: V = 150 mph

**Step 3**: Determine Wind Load Parameters

Kd = 0.85 (Per Table 26.6-1 for C&C)

Kzt = 1 (There are no topographic features)

Ke = 1 (Job site is at sea level)

GCpi = +/-0.18 (Tabel 26.13-1 for enclosed building)

**Step 4**: Determine Velocity pressure exposure coefficient

zg = 900 ft [274.32] (Table 26.11-1 for Exposure C)

Alpha = 9.5 (Table 26.11-1 for Exposure C)

Kh = 2.01*(40 ft / 900 ft)^(2/9.5) = 1.044

**Step 5**: Determine velocity pressure

qz = 0.00256*Kh*Kzt*Kd*Ke*V^2

= 0.00256*(1.044)*(1)*(0.85)*(1.0)*(150^2) = 51.1 psf

**Step 6**: Determine External Pressure Coefficient (GCp).

We are looking at pressures for all zones on the wall and roof. For the wall we follow Figure 30.3-1:

For 10 sq ft, we get the following values for GCp. Note 5 of Figut 30.3-1 indicates that for roof slopes <= 10 Deg that we reduce these values by 10%, and since our roof slope meets this criteria we multiply the figure values by 0.9

Zone 4: GCp = +1.0*0.9 = +0.9 / -1.1*0.9 = -0.99

Zone 5: GCp = +1.0*0.9 = +0.9 / -1.4*0.9 = -1.26

A Monoslope roof with a slope between 3 deg and 10 deg follows Fig 30.3-5A. For each zone, we get the following values:

Zone 1: GCp = +0.3 / -1.1

Zone 2: GCp = +0.3 / -1.3

Zone 2′: GCp = +0.3 / -1.6

Zone 3: +0.3 / -1.8

Zone 3′: +0.3 / -2.6

**Step 7**: Calculate wind pressure

We can then use all of these values to calculate the pressures for the C&C. Since we have GCp values that are postive and negative, and our GCpi value is also positive and negative, we take the combinations that produce the largest positive value and negative value for pressure:

p1 = qh*(GCp – GCpi)

= 51.1 * (0.3 – (-0.18)) = 24.53 psf (Zone 1)

p2 = 51.1*(-1.1 – (+0.18)) = -65.41 (Zone 1)

The calculations for Zone 1 are shown here, and all remaining zones are summarized in the adjacent tables.

These pressures follow the normal ASCE 7 convention, Positive pressures are acting TOWARD the surface, and Negative Pressures are acting AWAY from the surface.

Chapter 30 Part 4 was the other method we could use. This is considered a “Simplified” method and is supposed to be easier to calculate by looking up values from tables. Using the same information as before we will now calculate the C&C pressures using this method.

**Step 1**: The category is III

**Step 2**: V = 150 mph

**Step 3**: Wind load parameters are the same as earlier

**Step 4**: For walls and roof we are referred to Table 30.6-2.

Table 30.6-2 (above) refers us to Fig 30.4-1, which is shown below.

Referring back to Table 30.6-2, it indicates in note 5 that when Fig 30.4-1 applies then we must use the adjustment factor Lambda for building height and exposure.

Referring to this table for a h = 40 ft and Exposure C, we get a Lambda value of 1.49. This value is then multiplied by the value obtained from Fig 30.4-1.

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]]>The old wind standards (ASCE 7-88, ANSI A58.1 and UBC 97) defined wind speed as the “Fastest Mile Wind” velocity. The Fastest Mile, is not some record set by Olympic Track Star Usain Bolt (shown in the post image above), but rather a definition for wind speed that was used for many years in wind standards. It is defined as follows:

“** The wind speed corresponding to the whole mile of wind movement that has passed by the 1 mile contact anemometer in the least amount of time.” **per US-EPA AP-42

I’ve read this definition many times over the years, and I’m still not sure I understand it. Fortunately, I don’t need to understand the definition as long as I know what to do with the fastest-mile wind velocity.

Today’s standards do not use fastest-mile wind speed. For example, ASCE 7-16 uses a 3 second gust at an Mean Recurrence Interval (MRI), which depends upon the structural category. This is basically a period of time over which you look at data to determine a wind speed. I understand this as a wind speed that occurs once every 300 yrs (ASCE 7-16 Category I) would be lower than a wind that occurs once every 3,000 years (ASCE 7-16 Category IV). Looking at the corresponding data in the table, you can see that as you increase the category, then the MRI gets larger and produces more conservative (higher) wind speeds.

ASCE 7-05 was a little different because in that standard the MRI was 50 yrs for all wind speeds, but a factor called the “Importance Factor” was used to adjust the pressures and increase them with the higher categories.

**ASCE 7-16:**

Category I – MRI = 300 yrs

Category II – MRI = 700 yrs

Category III – MRI = 1,700 years

Category IV – MRI = 3,000 years

**ASCE 7-10:**

Category I – MRI = 300 yrs

Category II – MRI = 700 yrs

Category III and IV – MRI = 1,700 yrs

**ASCE 7-05:**

All Categories – MRI = 50 yrs (Importance factors used for different categories)

To convert from fastest-mile wind speeds to something useful in today’s standards, we make use of a figure contained in the commentary section of ASCE 7-05. In ASCE 7-95, this was Figure C6-1. Some version of this same graph exists in every version of ASCE 7. Please refer to the corresponding calculations to show an example of how one would convert from 90 mph using fastest-mile wind speed to 3 second Gust and 50 year MRI. Using this graph, we determine that the equivalent wind velocity for a 50 yr MRI and 3 sec gust would be 107 mph. This would be a wind speed that we would use if we were following ASCE 7-05.

3) Open Meca Wind Conversion tool and enter this value into the gust, and 50 yrs in the recurrence period. (See below).

The result from the tool is 106 mph and we got 107 mph in the our previous calculation above. Since we are reading values from a graph, there can be some slight variation in the values obtained due to inaccuracies.

At Meca, we work with wind codes from around the world and it’s often necessary to try to convert wind speeds from one code to another. In order to do this, we have summarized the following parameters for our use and we offer them here for your convenience. You are responsible for verifying the accuracy of these values before you use them, but they might prove useful to you in certain circumstances.

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]]>MecaStack allows the user to enter ladders and platforms on their stack. These are considered an appurtenance, and not adding any structural stiffness to the stack. MecaStack does consider the wind area and weight added by these ladders and platforms, which can be significant on a stack.

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]]>ASCE 7-16 Figure 29.4-1 indicates that it covers chimneys, tanks and similiar structures. Since most chimneys and tanks are cylindrical, this would imply that it only applies to round structures; however, if you look at Fig 29.4-1 then you see that it covers more than just round structures, there are also shape factors for square, hexagonal and octagonal structures. There are many times when a structure may not fit into any other classification within ASCE 7, and this “Chimney” section may be the best option. At Meca, we have applied this criteria to wind load calculations for Air Cooled Heat Exchangers, heaters, and various other structures.

Since Sect 29.4 is written for chimneys, lets work an example of a cylindrical chimney. Assume that we have a 100 ft [30.48 m] overall height chimney that is 3 ft [0.914 m] OD at the top, and 5 ft [1.524 m] at the bottom. There is a conical transition from elevation 55 ft [16.764 m] to 50 ft [15.24 m]. The material is A-36 and the thickness of the stack is 0.25 in [6.35 mm], both of which would be needed to determine the natural frequency.

h = Stack Height in ft [m]

E = Modulus of Elasticity in psi [Pa]

I = Moment of Inertia in in^4 [m^4]

ma = Average mass per unit length of stack in lbm/in [kg/m]

*Note the 386.4 is only needed for Imperial units in order to resolve the units correctly.*

In this case running MecaStack indicates that the natural frequency is 1.84 Hz for this stack. Using this formula and the base diameter of the stack to determine I and ma, we get an estimated frequency of 1.64 Hz. This is about 11% lower than the actual value of 1.84 Hz, but it’s not a horrible estimate. Please note that even though the units below are Imperial, we do not show the 386.4 factor indicated above for Imperial units, because MathCad will automatically adjust the units accordingly.

Now that we have the flexible gust factor, the rest of the wind force calculations are somewhat easy. We will use Equation 29.4-1 for this calculation.

We already calculated the gust factor, G, above, and so now we only need to determine qz, Cf, and Af. The determination of these values, and then the final calculation of F is as follows. Please note that we have calculated these at several different values of z along the height of the stack. Since the Kz factor changes at different elevations, we calculate the wind pressures at more elevations in order to give a more accurate wind loading on the overall stack.

An excel spreadsheet with a summary of the wind pressure calculations is available for free download at this link. There is also a free pdf of the calculations shown in this article at this link.

Yes, MecaStack handles this calculation easily. The flexible gust factor can be involved and for most realistic stack/chimney arrangements the rule of thumb presented is probably not going to be accurate. The easier way to do this stack would be with MecaStack, which would calculate the natural frequency and wind loads. Here are the wind calculations produced by MecaStack for this same example, and here is a copy of the input file which can be opened if you have a license to MecaStack.

MecaWind can be used, but it’s got some issues with this particular example. Within MecaWind we only have the approximate method to estimate the frequency. For example, if we use this tool, we get 1.64 Hz, which is exactly what we got when we performed our manual calculation earlier.

The problem is that the software is written to follow ASCE 7, and in ASCE 7 it states that if the frequency is greater than 1 Hz then the structure is rigid, and the rigid gust factor is used. In order to use the flexible gust factor we would have to enter it in the “Override the Gust Factor”. This isn’t ideal because in order to override you would first have to calculate it by hand and we want the software to do that. We will consider adding this capability in the future, but at the time of this article being published, the capability doesn’t exist.

For the purposes of this example, since we already know the natural frequency is 1.84 hz, that is the value that has been used in the MecaWind analysis. We also forced “G” to equal 0.991 to match our previous calculations. Here is the output for MecaWind, and also the MecaWind input file which you can open if you have a licensed copy of MecaWind.

This procedure can be applied to other “Similar Structures”. The same basic procedure followed here would apply, except that if the structure is not flexible then the much simpler rigid gust factor (G = 0.85) can be used. In ASCE 7, it also mentions that structures in Petrochemical and Other Industrial facilities that are not otherwise addressed in ASCE 7 can be found in the “Wind Loads for Petrochemical and Other Industrial Facilities” which is published by ASCE. If you design structures found in these facilities, this can be a useful reference. It is available at www.asce.org, here is the link. The guide discusses open frame structures, partially cladded structures, pressure vessels, cooling towers, and air cooled heat exchangers.

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]]>There is a book titled “Tubular Steel Structures Theory and Design” by M.S. Troitsky. Mr. Troitsky was a Professor of Engineering at Concordia University in Montreal, and this book was originally published in 1982 and the 2nd edition was released in 1990. This book was originally published by “The James F. Lincoln Arc Welding Foundation”, but at the time of this article a visit to their website indicates that it is not currently being offered by them. Performing a Google search there do appear to be some sellers offering used copies of the book. Since it’s not in publication any longer, we have scanned the breach section and included it here for your download.

On the “Breach Reinforcement” screen, you can select the “Troitsky” as the breach design analysis for that breach by selecting the option shown in the image.

When a hole is cut into the shell, we reduce the buckling capacity of that shell. Let’s assume that a perfect cylinder (one without any cutouts) has a capacity of Pcl. Pcl is the compressive load that the perfect cylinder can take before it starts to buckle. Now, lets assume we cut a rectangular hole in the shell. It’s safe to assume that Pcl won’t be it’s limit any longer and that the compressive capacity will be some value less than Pcl. Let’s call this reduced capacity of our cylinder with a cutout, P. Below is Figure 5.23, taken from Troitsky, that shows the relationship of P/Pcl versus rbar. The term rbar is a non-dimensional parameter that takes into account the cylinder geometry and cutout dimensions as follows:

rbar = r / (R*t)^0.5

r = (a+b)/4

a= Width of Opening

b = Height of Opening

R = Radius of the Cylinder

t = Thickness of the cylinder

All units must be the same (i.e. inches, mm, etc..)

Using Figure 5.23 we can find the reduced capacity of the shell section due to the cutout. In the Fig 5.23 there are a couple of curves shown, but in MecaStack we use the lower bound curve to be conservative.

In MecaStack we calculate the rbar and then the P/Pcl ratio. Then, we use whatever code has been specified by the user (i.e. ASME STS-1, CICIND, Euro, etc..) to calculate the permissible stress for a perfect cylinder and reduce it by multiplying it by the ratio P/Pcl to get the new capacity of the section with the breach cutout.

Traditionally Round Openings are referred to as Nozzles, and typically these are reinforced using reinforcement pads (Repads). When the Breach becomes large, then it may be more practical to use Vertical and Circumferential stiffeners for reinforcent. For this situation we have added a option on the Breach Screen to indicate that the opening is round rather than rectangular. This will show the graphics correctly. Also, when the round option is selected, then we follow this curve from Troitsky for the buckling capacity. The remainder of the analysis is the same regardless of whether then opening is round or rectangular.

If the reduced shell buckling capacity is GREATER THAN the stress which is placed on the section, then the section is adequate and technically requires no reinforcement. However, it’s always desireable to add reinforcement even if this check passes because it’s only going to improve the buckling capacity if you decide to use reinforcement anyway.

If the reduced shell buckling capacity is LESS THAN the stress being placed on the section, then you definitely have to provide reinforcement of the section. The remaining setps in the flowchart (3 through 6) will now be checked.

When the cutout occurs, a certain amount of material is removed. In this check, we make sure that the area removed (the removed area is highlighted in yellow in the image) is completely replaced by the stiffener area.

Similar to the area, we also check to make sure that the moment of inertia, which was lost due to the cutout, is completely replaced by the vertical stiffeners.

The axial load and moment is calculated on each vertical stiffener, using the equations in Para 5.7 of Troitsky. The vertical stiffener is treated as a pinned-pinned column for the analysis. Then using the criteria in AISC 360-16 the vertical stiffener is checked for combined axial compression and flexure using the appropriate section of AISC 360-16. The stiffener shape used and configuration determine which sections of AISC are used for the calculations.

These AISC checks are quite involved, and require a lot of calculations. In the case of structural shapes used from AISC, the properties are obtained using the values in the AISC tables. For structural shapes which are included in MecaStack that are taken from other country’s standards (i.e. British, Australian, Brazilian, etc..) the shapes are approximated by using a collection of rectangles. This method should give reasonably accurate results, but it does not take into account fillet radius on corners as may be found in the actual shape, and so it is an approximate result.

The Troitsky Breach Analysis didn’t discuss designing the rings to keep the stack from ovaling, but that seems to be a logical requirement of the ring; therefore, in MecaStack we have added a check that verifies that the moment of inertia of the ring (in bending) is completely replacing the moment of inertia of the cylinder opening that is lost due to the breach cutout.

The Troitsky Breach Analysis assumes that circumferential stiffeners are full rings. There are some examples in industry of the circumferential stiffeners only being above the opening, and not wrapping 360 degrees around the stack. This is discussed briefly in the commentaries of CICIND “Model Code for Steel Chimneys”. They state that when the width of the opning is less than 40% of the stack diameter locally, that it is not necessary to provide a horizontal stiffener extending around the full circumference and a more local arrangement may be used. Vertical reinforcement should be continued above and below the opening to a point where the added stress is insignificant. The code recommends continuing the reinforcement at least 0.5 times the width of the opening.

MecaStack does not have an option to indicate this type of reinforcement, but we offer this information as information only. Typically, Meca doesn’t use this type of partial ring reinforcement, we recommend full rings.

In addition to the Troitsky Breach Analysis, MecaStack is also going to modify the section properties to consider the removal of the cutout and the addition of the vertical stiffeners. The circumferential stiffeners for the breach are not considered in the section properties calculation; however, they are considered as “stiffening” rings for the stack and are considered in all stiffening ring checks that are performed.

There are endless combinations of openings that can occur, and it’s impossible to give a solution that addresses all possibilities. The one thing to remember is that even though MecaStack may graphically show the multiple openings (see corresponding image), it doesn’t mean that it’s properly analyzing multiple openings. The section properties will be modified to include all of the openings, but as shown in this discussion of the Troitsky method, there are a lot more considerations. Each Troitsky Breach Analysis that MecaStack performs, assumes that the breach is the ONLY breach that exists. It does not consider any other breach. The Troitsky method does assume the full area and moment of inertia is replaced with the vertical stiffeners; therefore, basically each breach designed using Troitsky assumes that the reinforcement makes the section equal to or greater than the original perfect cylinder. If you have multiple breach openings and each is designed using Troitsky, then it would seem that each breach should be sufficiently designed to be independent of the other breach; however, it’s not always that simple. The situations can become very complicated and when in doubt refer to experts or perform finite element analysis to validate results.

This all makes more sense when you apply it to an actual example. To illustrate this, we took “Numerical Example No 1” from Troitsky, and we have simulated the exact same problem using MecaStack. Please refer to the example No 1 in the text, and the PDF of this file can be downloaded here. Please note that in RED we have added comments in a couple of places where more explanation is required. Here you can download a PDF of the comparable MecaStack Output for the same problem, and the results can be compared. The results are very close, and where they differ we have explained why in the comments of the Troitsky PDF.

This is a question that has puzzled stack designers since the beginning of time, is it Breach or Breech? Lets look at the definition for each according to Dictionary.com:

**Breech** – *The lower, rear part of the trunk of the body. In an ordnance, the rear part of the bore of a gun, especially the opening and associated mechanism that permits insertion of a projectile. *

**Breach** – *The act or result of breaking; break or rupture. A gap made in a wall or fortification. To make a breach or opening in.*

According to the definitions, it seems that BREACH is the most accurate for the context of an opening in a cylidndrical shell; however, I have been using breech to describe this situation for my entire career. The book “Tubular Steel Structures” by Troitsky uses the term breech, and ASME STS-1 uses breech. Consulting with colleages they have stated that they believe that the industry has used breech and breach interchangeably, so that either is acceptable.

At the end of the day, we at Meca have decided that breach is more consistent with the actual situation in which the word is used and so moving forward we will use breach; however, don’t be surprised if you see breech in various locations in our website and software, it’s going to take a while to locate and update all references.

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]]>MecaWind has many features and options within the software, which can be both a blessing and a curse. It gives the user greater control over the results they obtain, but for the novice user it can be overwhelming to understand what is being requested with each option. This is the single biggest challenge facing new users, where do I begin. To help with this confusion, in v2342 of MecaWind we added a new “Wizard” option. This option is intended to help the user get the basic parameters of their analysis setup quickly, and then they can spend more time entering only the parameters needed for their analysis. Let’s go over the basics of how this works.

The first step is now to click the big button that says “Wizard”. This is all it takes to get started with the wizard.

The first screen of the wizard will ask you which standard you wish to use. Please note that we have made the most common choices Green. This doesn’t mean they will always be correct, but at the time this article is written ASCE 7-10 is the most common code used throughout the US, and so it is Green. Also, beneath each option there is additional information to help you select the correct option.

On the next screen you can see that we make heavy use of visual images, which are easier to grasp. You select the Exposure and Category for the structure, and there are images that represent each option as well as a brief description of each. Also, the wind speed is entered on this screen and there is some help to give you more info.

On this screen of the wizard you reach a fork in the road, meaning that depending upon which option you select the subsequent screens will change. We have either a “Building” or “Other [Structure] (Not a Building)” options. Also on this screen you indicate if you want a 0.6 factor applied to your wind pressures and it gives you some explanation as to why you would or wouldn’t want to do that.

First, I will show the screens assuming that you select a “Building” and then I will show the screen if you select “Other [Structure]”. For the type of building, you have some options which are an indication of the porosity of the building envelope (Open, Enclosed, etc..). Please note that the Partially Open option is only applicable when ASCE 7-16 was introduced, so it won’t be available if you select ASCE 7-10 or ASCE 7-05. One of the most frequent questions we get is how do I know if my building is flexible or rigid and we give some simple guidance on that option.

On this tab you will select whether you are interested in analyzing the Main Wind Force System (MWFRS), Components and Cladding (C&C), or both. You must select at least one of these options, otherwise you won’t have any analysis results.

As promised, let’s go back to the Type of Structure screen where we selected the Building Option and lets look at what the screen would look like if “Other” was selected. Please note that there are some options on this screen that are only applicable for ASCE 7-16 because these options weren’t added until this version of the standard (Tanks, Silos, Canopy and Solar Panels).

At the end, you click the “Finish” and return to the main MecaWind program. The selections you made in the wizard are all transferred into the main MecaWind program. Here you will still need to go through each tab and enter the necessary data. The main menu should be nearly complete. You just need to scan and check to see if there are any other details that are needed. The building or other structure screens are where you will enter the details about the structure. Finally, you will need to specify any Components and Cladding (C&C) since none of that info was entered in the wizard.

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]]>The typical dimensions for a residential solar panel is 65″ long and 39″ wide.

The design wind speed for zip code 74011 with exposure C and risk category III is 115.0 mph, and the elevation above sea level is 755 ft. We won’t go in depth in this article on the procedure for calculating the velocity pressure, but for h=30ft it comes out to be qh = 32.28 psf.

In ASCE 7-16 there are two sections that cover rooftop solar panels: 29.4.3 and 29.4.4. Section 29.4.3 addresses buildings of all heights with a flat roof or gable/hip roofs with a slight slope (less than 7 degrees). Section 29.4.4 covers the other roof slopes. Each have their own restrictions on the parameters that can be used due to wind tunnel testing done to develop the procedures and normal use of solar panel arrays. In this example we will focus on section 29.4.3, which utilizes figure 29.4-7.

We will use the typical residential solar panel dimensions and we will do 3 rows of 8 panels in the array. The bottom edge of the panel will have a 1 ft gap between the roof surface and the top edge will have a 3ft gap, giving a tilt angle of 20.3 degs. The minimum horizontal clear distance between the panels and the edge of the roof is the larger of 2(h2-hpt) and 4ft. In our case, there is no parapet (hpt) and the clear distance comes out to be 6ft. The minimum gap between panels is 0.25 inches, but we will use a spacing of 1.5 inches in this example. For the spacing between each row (array) of solar panels, we want to optimize this so the rows aren’t in the shadow of the adjacent row. We will need to know the solar elevation and azimuth for our location, and this can be found using a Solar Chart Program from the University of Oregon, as seen below. We are going to use a time range of 10am to about 2:45pm in the Winter season.

The chart shows a solar elevation of 22 degrees and an azimuth correction of 35 degrees. Now to find the spacing between the arrays we will divide the height difference of the panel edges by the tangent of the solar elevation, and then correct for the azimuth by multiplying this spacing by the cosine of the azimuth correction and rounding our answer up to the nearest whole inch. This process yields a spacing of 49 inches, and is illustrated below.

For the sake of this example, I am going to place the solar panels in the center of the building. Taking into account the panel edge to roof edge (d1=6ft), the spacing between rows (d2=4.083ft), and the spacing between panels (d3=0.125ft), the building width parallel to the solar array is 38.875ft (WL=38.875ft) and the building width perpendicular to the solar array is 36.416ft (WS=36.416ft).

The equation we need to solve for the design wind pressure for rooftop solar panels is:

- yp: minimum of (1.2, 0.9+hpt/h)
- yc: maximum of (0.6+0.06*Lp, 0.8)
- yE: 1.5 for uplift loads on panels that are exposed and within a distance of 1.5*Lp from the end of a row at an exposed edge of an array
- yE: 1.0 elsewhere for uplift loads and for all downward loads, as illustrated in Fig. 29.4-7

A panel is defined as exposed if d1 to the roof edge > 0.5*h and at least one of the following applies:

- d1 to the adjacent array or roof edge > max (4*h2, 4ft)
- d2 to the next adjacent panel > max (4*h2, 4ft)

In our example, the d1 value is set as the minimum horizontal clear edge distance of 6ft, and the height of the building is 30ft. Evaluating the expression for an exposed panel shows that d1 (6ft) is not greater than 0.5*h (0.5*30=15ft), so all the panels will use yE = 1.0.

Now solving for our other gamma values:

yp = min(1.2, 0.9+hpt/h) = min(1.2, 0.9+0/30)

yp = 0.9

yc = max(0.6+0.06*Lp, 0.8) = max(0.6+0.06*5.4167ft, 0.8)

yc = 0.925

We will need to use Fig. 29.4-7 to find our GCrn,nom value to use in our solar panel design wind speed equation. For each solar panel, we will need to establish which zone it is in using the “Building Roof Plan” in the figure. In this example, our boundary for the edges is actually greater than either of the building widths (2*h=2*30=60 >WS and WL), thus every panel is in zone 3. The next thing we need is the “Normalized Wind Area” which is defined as:

- An = (1000/[max(Lb,15)^2])*A

– where A is the effective wind area of the solar panel (Lp*Wp) and Lb is the minimum of (0.4*(h*WL)^0.5, h, Ws) - Lb = (0.4*(30*38.875)^0.5, 30, 36.416)

Lb = 13.66 ft - An = (1000/[max(13.66,15)^2])*(5.4167*3.25)

An = 78.24 ft^2

Finally, to know which “Nominal Net Pressure Coefficient” graph to use, we need to know the tilt angle of our solar panels. This can be found by taking the inverse tangent of the height difference of the panel edges to the roof surface divided by the length of the panel (w=arctan[(h2-h1)/Lp]). Solving for the tilt angle we get 20.27 degrees (arctan[(3-1)/5.4167]), which means we will use the right GCrn,nom graph since it is between 15 and 35 degrees.

From inspection of this graph we see that the y-intercept is 3.5 and that the slope is -1 from An = [<1, 500]. This gives us an equation of the line as GCrn,nom = -log(An) + 3.5. Plugging in 78.24 ft^2 for An, GCrn,nom equates to 1.607.

Now we can equate GCrn and thus the solar panel design wind pressure, p:

GCrn = yp*yc*yE*GCrn,nom

GCrn = 0.9*0.925*1.0*1.607

GCrn = 1.338

p = qh*GCrn

p = 32.284*1.338

**p = 43.191 psf**

So with the parameters and location used in the example, each solar panel would see a design wind pressure of an uplift and downward load of +/- 43.191 psf. Every panel seeing the same wind pressure isn’t usually the case. If the panel is in a different zone than another, then the GCrn value will be different. If the panel is considered exposed, then the yE value will be different giving a different GCrn. So in those cases you would need to go back and redo the calculations for those panels.

You may or may not be wondering, is there a faster way to do this? Luckily, our MecaWind program has these calculations built in and will calculate the design wind pressure on each solar panel in one analysis run (plus you can see everything graphically) instead of having to evalute each solar panel or groupings of the solar panels one at a time. As a comparison, below is the parameters used to calculate the wind pressure on the solar panels for this example (just happened the wind pressure is the same for all the panels) and the output MecaWind gives for the same example.

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