Coastal Engineering - Amphibious Housing

## Case Study: Hurricane Jeanne

Credits: NASA
As stated on the previous page regarding hurricane background, each year about one to two major hurricanes occur, with a major hurricane being classified as a category 3 or above. For our case study we chose Hurricane Jeanne; a category 3 storm that hit South East Florida in September of 2004. According to the NOAA, Hurricane Jeanne was the 13th costliest storm to hit the United States with a death toll of 5.

Through the following calculations, we intend to test whether or not a typical amphibious home would withstand major hurricane conditions.
This way we roughly determine whether or not this method of housing would be suitable for areas in the United States that have experienced significant damages due to flooding from hurricanes and tropical storms.

In the event that we had not been given the maximum wind velocity of Hurricane Jeanne, we would have been able to calculate it from one of the following three models:

The Meyers Hurricane Model for a curved isobar is as follows:

Vgr = [(r/Pa)*(dp/dr) + (fr/2)]1/2 - fr/2; given that dp/dr = [(((Pa - P0)*R)/r)*e(-R/r)]

This can be substituted into the first equation. Also, since little “r” is any distance from the center of the eye of the storm, we can chose it to equal R
so that it will cancel out of the equation.

The equation will then become:                      Vgr = [(r/Pa)*(((Pa - P0)*R/r)*e(-R/r)) + (fr/2)]1/2 - fr/2]
Once the calculations are performed:             Vgr = [(r/Pa)*(((Pa- P0)*R/r2)*e(-R/r)) + (fr/2)]1/2 - fr/2]
The equation then becomes:                           Vgr = [(((Pa - Po)/Pa)*e-1)+fr/2]1/2 - fr/2]

Given only the ambient pressure, minimum pressure, and the latitude where the hurricane occurred; the velocity of the hurricane can be calculated.

The Meyers Hurricane Model for a non-curved isobar is as follows:

Vgr = (1/(Pa*fr))*(dp/df)
Since fr = 2*w*sinФ; w = 7.27 x 10-5; and dp/dr = (Pa - P0)*(R/r2)*e(-R/r)
Substituting these values into the equation gives us: Vgr = 1/(Pa*fr)*(Pa-P0)*(R/r2)*e(-R/r)
Since we can chose the value of “R”, can chose the same value as “r” so that the two cancel each other out. The equation after cancellations has been made as follows:

Vgr = 1/(Pa*fr)*(Pa-P0)*(R/r2)*e(-R/r) = Vgr = ((Pa-P0)/(Pa*fr*r)*e(-1)

Given the ambient pressure, the radius of the storm, and the minimum pressure that occurred, the velocity of the hurricane can be calculated.

The Rankine Vortex Model:

Vgr = [1/pair(Pa – Po)]1/2

Given the ambient air pressure, the minimum pressure of the hurricane, and standard air pressure, the velocity of the storm can be calculated.

The Navy had a few sites in Florida where they were taking realtime data during Hurricane Jeanne, and therefore it was not necessary for us to do
the above calculations as we had the actual values. These are listed below:

Average Storm Surge (ft): 2.35 (.716m)
Average Storm Tide (ft): 7.83 (2.386m)
Maximum Wind Speed: 120 mph

Specific Data for Jacksonville Florida includes the following (Data is taken from the NHC Tropical Cyclone Report for Hurricane Jeanne at the
Mayport NAS Naval Station in Jacksonville, Florida)

How would an amphibious home handle in a hurricane?
Assumptions:
House Dimensions: 20 feet high, by 20 feet wide by 30 feet long makes for a two story home of approximately 1200 square feet.

Storm Surge: Assume that the hydrostatic force acting on all of the steel posts will cancel out since thewater is surrounding it.
Since the maximum wind speed reached was 120mph, this will be the wind speed used in our calculations

Wave Height:
C = √(g*d) given that the water depth is .716m, and g = 9.81m/s
C = √(9.81*.716) = 2.65 m/s
C = 2л/T
T = 2 л/C  given that c = 2.65 m/s
T = 2.371 seconds

Assume that the period stays the same…

L0 = 1.56T2
L0 = 1.56*(2.371)2
L0 = 8.77m

Using Iteration:

L1 = L0√tanh((2л/L0)*d)
L1 = 8.77*√tanh((2л/8.77)*.716)
L1 = 6.03m
L2 = L0* tanh((2л/L0)*d)
L2 = 8.77*tanh((2 л/8.77)*.716)
L2 = 4.14m
L = (L1 + L2)/2
L = (6.03+4.14)/2
L = 5.09m

Because the wave length is 5.09m and the house will be rising with the waves, it can be assumed that the shear force from the waves will be negligible.Therefore the determining factor as to whether or not the amphibious home will withstand a hurricane will be the wind speed it experiences.

According to Fundamentals of Structural Analysis [insert reference], the equation for static wind pressure for SI units is as follows:

qz = 0.613V2IKzKztKd  where V = 81.4km/h or 22.6m/s

I = 1; for a residential home
Kz = Varies depending on height (For areas exposed to winds moving across open waters of about 5000ft or more)
Kzt = 1 (assume flat and level terrain since we are taking all water to be on the same height)
Kd = .85 (Main wind force-resisting system)

The design wind pressure will vary for different heights, therefore…

For 0 to 4.6m (0 to 15feet):

qz = 0.00256 V2IKzKztKd  where V = 81.4km/h or 74.3ft/s or 50.6 mph
I = 1; for a residential home
Kz = 1.03 (For areas exposed to winds moving across open waters of about 5000ft or more)
Kzt = 1 (assume flat and level terrain since we are taking all water to be on the same height)
Kd = .85 (Main wind force-resisting system)
qz = 0.00256(120mph)2*1*1.03*1*.85
qz = 32.27lb/ft

For 6.1m (20feet):
qz = 0.00256 V2IKzKztKd  where V = 81.4km/h or 74.3ft/s or 50.6 mph
I = 1; for a residential home
Kz = 1.03 (For areas exposed to winds moving across open waters of about 5000ft or more)
Kzt = 1 (assume flat and level terrain since we are taking all water to be on the same height)
Kd = .85 (Main wind force-resisting system)
qz = 0.00256(120mph)2*1*1.08*1*.85
qz = 33.84lb/ft

Next; we can find the design wind pressure acting on each side of a building.

For 0 to 15feet:

Windward Side: Cp = .8
p = qzGCp where G = .85 (a rigid structure)
p = 21.94 lb/ft2

For 15 to 20 feet:
Windward Side: Cp = .8
p = qzGCp where G = .85 (a rigid structure)
p = 23.01 lb/ft2

Leeward Side: Cp = -.5 (since L/B = .67)
p = qzGCp where G = .85 (a rigid structure)
p = -14.38 lb/ft2

Moment Calculations:

Windward side of the building:

0 to 15 feet:
15ft (21.94)lb/ft = 329.1 lb acting at 7.5 feet above base of home

15 to 20 feet:
5ft (23.01)lb/ft = 115.05lb acting at 17.5 feet above base of home

Leeward side of the building:
20ft (14.38)lb/ft = 287.6 lb acting at 10 feet above base

Assuming an idealized situation where the wind forces will only act on their respective faces of the building; assume that the force from the windward
side does not impact the force on the leeward side of the building, and the vice versa.

Therefore the total force on the windward side of the building is:
329.1lb(7.5ft) + 115.05lb(17.5ft) = 4481.6lb*ft or 373.49lb*in

And the total force on the leeward side of the building is:
287.6lb(10ft) = 2876lb*ft or 239.6 lb*in

Assuming that two by fours made of spruce pine are being used to construct the home…

τ = P/A where τ is the shear force, P = force on the pine, and A = area over which the force is acting.

If the shear force on the windward side of the building does not exceed the maximum shear force for spruce pine, then it can be assumed that the shear force for the pine will not be exceeded on the leeward side of the building since the force per unit length is less than that of the windward side.

Because the maximum shear force for spruce pine is 1490 psi; the wind forces on the house in this hurricane would not be enough to cause the wood to
fail.

Other Considerations:

Another consideration that would have to be accounted for are the joints at where the house meets the floating base. In the event of a hurricane, these joints could be potential weak points in the structure that would account for failure.

After extensive research we were unable to determine the exact design of how the amphibious homes were attached to the floating bases. Therefore we were unable to perform the necessary calculations.

The prediction would be that these joints would be experience a strong shear force due to the high winds; this would could cause the joint to fail in shear with this added force.