What is the maximun windspeed a fence that is constructed this way can
withstand before the posts will break off?  The top height of the
fence is 8 feet from the ground, the fence has sheet steel on it that
is 7 feet tall and starts at the eight foot height and comes down to 1
foot off the ground.  The posts that hold the fence up are 4"x6" Green
Treated SPF Pine.  The posts are orentatied with the 6 inch side
parrallel to the fence's face and are 8 foot on center.  The fence has
2x4 girts on it that are nailed to the posts with the 2 inch side
touching the post and the time is nailed to the girts. The fence is
fully exposed to the wind.Please show your math with this answer or
provide a link to the math.

Hello pamsauto, your question is a fairly simple problem in statics.
You did a very good job of providing all necessary info.

To convert wind speed to force per square foot:
Force (lbs/square foot) = wind speed (miles/hour)^2 x .0027

This force (actually pressure) acts on an area (tributary area) of the
amount of fence supported by each post:
A = 7 ft (material height) x 8 ft (post spacing) = 56 ft^2 (square feet)

The distance from the ground (where the maximum bending stress in the
post occurs) to the vertical center of the sheet steel is:
D = (7 ft / 2) + 1 ft = 3.5 + 1 = 4.5 ft

What we have here is actually a cantilever beam. The post being a beam
fastened at one end (the ground) and free at the other. The formula
for the bending moment in this case is:
M = F x L (or in our case D)

The bending moment (overturning torque) is found by:
M = F (force of wind) x D = wind speed^2 x .0027 x 56 ft^2 x 4.5 ft
Multiplying by 12 (inches per foot) to change the answer into inches^3 we have:
                          = wind speed^2 x 8.16

The strength of a 4x6 post in bending (section modulus) is calculated by:
S (section modulus) = b (base) x d (other dimension)^2 / 6

Since a 4x6 is actually 3.5 x 5.5:
S = 5.5 x 3.5^2 / 6 = 5.5 x 12.25 / 6 = 11.2 inches^3

NOTE: The way you have chosen to orient the 4x6 is the weak direction
for wind loading. If oriented in the other way, S would be:
S = 3.5 x 5.5^2 / 6 = 17.6 inches^3

The formula for the bending stress (Fb) is:
Fb = M / S or bending moment / section modulus

Writing this another way we have:
M = Fb x S

Substituting the values we have calculated above we get:
wind speed^2 x 8.16 = Fb x 11.2 inches^3

Dividing both sides of the equation by 8.16 we have:
wind speed^2 = Fb x 1.37

Taking the square root of both sides we have:
wind speed (miles per hour) = sqrt (Fb x 1.37)

Here is where we must do some estimating. The values published for the
bending strength (Fb) for SPF lumber vary quite a bit depending on
where the wood is harvested and exactly what species of pine is
involved. I think a conservative number might be 800 psi. This is the
value given in my "Machinery's Handbook" for White Pine in an outside
location.

Using this number we have:
wind speed = 33 MPH

If you were to orient the 4x6 in the other direction, you would
increase the wind speed to 42 MPH.

If the actual strength of the 4x6 posts is greater that what I used,
you will also get an increase. For instance the value given for
Southern Yellow Pine is 1100 psi. Using this higher number and
orienting the 4x6 in the stronger direction you would get 48 MPH.

I suspect that you are surprised at the low MPH numbers we calculated.
I know you would like to see a 100 MPH figure instead. One thing in
your favor is that these calculations assume that the wind is acting
exactly perpendicular to the fence. In actual conditions this is hard
to achieve. Also, wind is not a constantly applied force and you can
apply factors which will increase the rating (published numbers range
from 1.3 to 1.6). Applying the 1.3 number would give us an increase to
1.3 x 42 or 55 MPH. Another factor is where you live related to
maximum wind speeds seen in your area. You didn't say how you are
planting the 4x6 posts into the ground. There is the possibility that
the fence could blow over before the posts break.

I hope that this answers your question and helps you in your design
choices. I will assume no liability for these calculations and they
are offered strictly for academic interest with absolutely no
applications to the real world (sounds like the disclaimer at the
beginning of "South Park" doesn't it).

Redhoss

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Request for Answer Clarification by pamsauto-ga on 16 May 2004 12:53 PDT

Exactly the answer I was looking for.  I had thought of the "Ground
Breakout" problem also.  For a 25 dollar tip would you be willing to
figure this out also?   The posts are in the gound 30 inches.  They
are in a 12 inch diameter hole.  The bottom 6 inches of the post are
in packed soil.  The top 24 inches are poured 3000psi concrete.  It
has not been vibrated if this matters.  The soil condidtions are 8
inches of black soil on top and the rest is a very loose and course
gravel.  Some holes when drilled would collapse below the black dirt.

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Clarification of Answer by redhoss-ga on 16 May 2004 15:03 PDT

Thank you for your kind words. It makes a poor old burned out engineer
feel good to be appreciated. I have done many footing calculations and
I don't like doing them (very tedious by hand). They are very inexact,
tend to be very conservative (over designed), and without soil tests
pretty useless. I have built a few fences and many carports. Your
footing design looks adequate to me and I think it would be a toss up
as to whether the footing or 4x6 would fail first. If you are really
interested, you might perform a test. I did this accidently one time.
A city building department questioned my footing design. I met the
city plan checker at the job site. I arrived there first and noticed
that a forklift had accidently driven over one of my columns. The
column was bent over at the base and the footing hadn't budged. I
showed this to the plan checker and he was satisfied.

Very detailed and prcise answer.  Researcher did not miss ANY part of
the orginal question I asked.   Thanks.