I’m not a civil engineer, part 1, wind force
2020-07-19
One of my next substantial projects is building a fence. Because of the type of person I am (insufferable) I want a fence that’s better than the standard ‘stick a 4×4 in the ground and nail some wood onto it’ type of fence. Not that there is anything wrong with that type of fence, I just want better. As part of the project I though it would be fun to go through and do the engineering due-diligence to make sure that my fence will be able to withstand the elements. The starting point was Googling ‘wind load’ which ultimately let me to ASCE-7 or American Society of Civil Engineers Minimum Design Loads for Building and Other structures. This is something that you can buy from the fine folks at American Society of Civil Engineers for a handsome $240, but I found a copy on https://law.resource.org, direct link to file here. To me it only makes sense that standards which are incorporated into law should be made available to people who are expected to follow these laws for no additional costs beyond those already collected through taxes. Hence I do believe that money would be better spent on supporting organizations such as resource.org. Enough soapboxing.
The project consists of three distinct stages
- Calculating the loads
- Selecting materials and design based on #1
- Getting ‘er done
Next several posts will be about stage 1 – calculating the loads. I am basing all of this on ASCE-7 2002 edition as that’s the full text I was able to find. I will try to keep my blithering to a minimum during this series of posts. These posts are intended more as my notes on my thought process in case I ever have to justify my decisions or laugh at a toppled fence after a light hurricane hits it. The interpretations are mine alone. I did not consult a lawyer or an engineer. I mean this is a privacy fence we’re talking about here. If this is something that interests you there is no replacement for reading the standard it self. Yes it’s a long and dense read, but it is quite interesting in its own right.
How much force does the wind exert on a thing
In order to design a structure to accommodate the force acting upon it we must define the force. The primary force acting on a fence is wind. Hence Standard 6 – Wind Loads. Since the fence fits the definition of an open building, the applicable section is 6.5.13, p34. Equation for force (Eq. 6-25) is defined as:
$$ F = q_z\times G \times C_f \times A_f $$
Where
\(F\) = Wind force in \(lb/ft^2\)
\(q_z\) = Velocity pressure
\(G\) = Gust effect factor
\(C_f\) = Net force coefficients
\(A_f\) = Projected area.
Lets address each variable in turn.
\(q_z\) – velocity pressure
The focus on this post is \(q_z\) – velocity pressure. It’s defined in 6.5.10, p31 as (Eq. 6-15):
$$q_z = 0.00256 \times K_z \times K_{zt} \times K_d \times V^2 \times I $$
Where
\(0.00256\) = Climactic constant
\(K_z\) = Velocity pressure exposure coefficient
\(K_{zt}\) = Topographic factor
\(K_d\) = Wind directionality factor
\(V\) = Wind velocity
\(I\) = Importance factor
Lets address each one of those variables directly.
\(K_z = 0.5747\) – Velocity pressure exposure coefficient
This is defined in 6.5.6.6 p29, referencing table 6-3 p75, 6-2 p74, and exposure categories 6.5.6 p28.
Table 6-3 p75, note 2 defines \(K_z\) as
$$ K_z = 2.01 \times (15/Z_g)^\frac {2}{\alpha}$$
\(\alpha\) and \(Z_g\) are defined in table 6-2 p74. Since we’re in exposure category ‘B’ (urban/suburban etc), the values are as follows:
\(\alpha\) = 7.0
\(Z_g\) = 1200
The complete equation then becomes:
$$ K_z = 2.01 \times (15/1200)^\frac {2}{7.0} = 0.5747 $$
One down – our \(K_z = 0.5747\).
\(K_{zt} = 1\) – Topographic factor
This is defined in 6.5.7.2 p30 (Eq. 6-3) as
$$ K_{zt} = (1 + K_1 \times K_2 \times K_3)^2$$
With \(K_x\) provided in figure 6-4 p47
Since \(K_1\) is defined by \(H/L_h\) and linear interpolation is allowed by note 1 and our \(H\) is zero (no height difference relative to uphill terrain) and 0 divided by anything is zero then linear interpolation dictates that at \(H/L_h = 0 \therefore K_1 = 0\) therefore the product of \( K_1 \times K_2 \times K_3 \) is zero and the whole equation evaluates to \(K_{zt} = 1^2 = 1\)
\(K_d = 1 \) – Wind directionality factor
This is defined in 6.5.4.4 p28 and references table 6-4 p76. We’re going to go with 1 even though it’s not specified in the table because all of the factors in the table are \(<1\) which reduce the loading calculation and I’m choosing to err on the side of safety (and simplicity)
\(I = 1.15\) – Importance factor
This is defined in section 6.2, reference table 1-1 p4 and the factor is defined in table 6-1 p73. We’re choosing category IV because we’re silly and think that our privacy fence is at least as important as … hospital, power generation stations … orphanages.
Putting together \(q_z = 24.336 lb/ft^2 \)
Given that we’ve accounted for most of the variables (listed below) the equation becomes a lot less hairy
\(0.00256\) = Climactic constant
\(K_z = 0.5747 \)
\(K_{zt} = 1\)
\(K_d = 1\)
\(V\) = Wind velocity
\(I = 1.15\)
$$q_z = 0.00256 \times 0.5747 \times \times V^2 \times 1.15 = 0.00256 \times 0.6607 \times V^2 = 0.00169 \times V^2$$
Super simple. I’m choosing to base our calculations on basic wind speed of 120mph for margin of safety even though the map (see at end of post) puts me in the 100mph-110mph zone.
Given all that \(q_z = 0.00169 \times 120^2 = 24.336 lb/ft^2 \).
Our initial equation just got simpler:
\(q_z = 24.336 lb/ft^2\)
\(G\) = Gust effect factor
\(C_f\) = Net force coefficients
\(A_f\) = Projected area.
$$ F = 24.336 \times G \times C_f \times A_f $$
\(G = 0.85\) – Gust effect factor
This is defined in 6.5.8.1 p30 as 0.85 for rigid structures. Good enough for me.
\(C_f = 2.0 \) – Net force coefficients
Since we’re building a fence, figure 6-20 p70 applies. We’re just going with the worst case scenario of \(C_f = 2.0 \).
Putting together \(F = 41.3712 \times A_f\)
We’re down to one variable – \(A_f\) and it’s dependent on the actual area of the fence. The formula for calculating the load that a section of the fence must withstand is then: \(F = 41.3712 \times A_f\). Neato.
