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PACIFIC CUP
1998 - Preparation Seminar No. 1
Emergency
Rudder Design and
Construction
Berkeley
Yacht Club February 28
1998
DESIGN
REQUIREMENTS:
- Cheap and easy to
build
- Light weight
SOLUTION:
Build blade like surfboard.
Thick blade for strength and light weight. Moderately rough
surface okay.
Keep gudgeons well separated to keep upper gudgeon lightly
loaded.
For swim step transom, use stern pulpit to support top
gudgeon.
MATERIALS AND
SUPPLIERS:
Foam blank - "Lastafoam"
medium density urethane foam boards available from
Svendsen's in 1.5" x 4' x 8' sizes or cut fractions at $8.59
per square ft.
Epoxy: TAP Plastics 314 marine epoxy resin ($50.25/gallon)
and 143 slow hardener ($33.35/half gallon). Or West System
epoxy (West Marine or Svendsen's).
Glass: "Knytex" from Tap Plastics, or similar. This is a
mat-cloth combination totaling 25.3 oz. per sq. yard. $12
per 36" of 50" wide material. Selvege tape lapped around
leading and trailing edges. (Tech. contact at TAP: Russ
Miller, manager at San Leandro,
510-357-3755.)
Rules for fiberglass/resin/foam work:
1) Always make a test patch
2) Cut glass carefully to size before mixing
resin
3) Use a very good particle mask
DEPLOYMENT:
Allow full rotational
degrees of freedom at lower gudgeon during deployment. Only
one bolt in rudder and one bolt in transom, fitted loosely.
Additional bolts added after top gudgeon is in place to
establish alignment.
DESIGN
METHODOLOGY:
1) ESTIMATE DESIGN
SPEED
- This determines the
maximum force on the rudder blade. Suggest 10 knots for
45 ft. boat, 6 knots for 30 ft. boat.
2) DETERMINE LENGTH OF THE
BLADE
- Try to go to at least
half the depth of the original rudder, and up to the
middle or upper stern rail. (Measure depth from the
transom bottom, not from the static
waterline.)
3) CALCULATE FORCE ON THE
BLADE:
- Use the
formula:
F = A * Cl * 1/2 * RHO * V^2
F = force (lb)
A = area below transom (ft^2)
Cl = Coeff. of lift (use 3.0 to allow for pumping
transients)
RHO = density of water (1.9905
slugs/ft^3)
V = design speed (ft/sec)
(1 knot = 1.6878 ft/sec)
F = 8.5 * A * V^2
F = force (lb)
A = area below transom (ft^2)
V = design speed (knots)
[example: 1 ft. x 4 ft. blade, 7 knots: F = 1,666
lb.]
4) DETERMINE BENDING MOMENT
AT THE LOWER GUDGEON:
- Assume the force is
centered between the lower gudgeon and the blade tip. if
this distance is L, then:
M = 1/2 * L * F
M = bending moment (ft-lb)
L = distance from lower gudgeon to tip
(ft)
F = maximum blade force at design speed
[example: L = 4 ft, so M = 3,332
ft-lb)]
5) DETERMINE THE REQUIRED
SECTION MODULUS:
- Use 10,000 psi as design
stress in low-tech laminate.
Required "section modulus" = M*12/10,000 (the 12 is to
change moment from ft-lb to in.-lb)
[example: SM required = 4.0 in^3
6) DETERMINE THE REQUIRED
THICKNESS OF FIBERGLASS LAMINATE:
- SM = W * (T^3 -
t^3) / (6 * T)
(section inertia divided by half of max
thickness)
SM = section modulus (in.^3)
W = width of blade (in.)
T = overall thickness of blade (in.)
t = thickness of core material (in.)
[example: blade is 12" wide (but use 10" to account
for shaping), core is 1.5" thick: By trial and error, use
T = 2.02". SM = 4.02 in^3. So required thickness of
fiberglass = 1/2 (2.02 - 1.50) = 0.26
in.]
7) CALCULATE LOAD ON UPPER
GUDGEON:
- Upper gudgeon force: FU
= M/D
FU = force on upper gudgeon (lb)
M = Bending moment at lower gudgeon
(ft-lb)
D = distance between gudgeons (ft)
[example: For D = 6.0, FU = 3,332/6 = 555
lb]
8) CALCULATE LOAD ON LOWER
GUDGEON:
- Lower gudgeon force: FL
= FU + F
F = force on blade (lb)
FU = force on upper gudgeon (lb)
[example: FL = 555 + 1666 = 2221 lb.]
9) SIZE
PINTLES:
For pins in double shear (as
in turnbuckle clevis pins) use safety factor of 5 and look
in rigging catalog for appropriate turnbuckle size. Or use
allowable shear stress of 6,000 psi for same
result.
A = 1/2 * FP/sigma (for double shear)
sigma = allowable shear stress (use 6,000 psi for 316
stainless)
FP = force on pintle (upper or lower, lb)
A = required area of pintle pin (in.^2)
Solve for required pin diameter = sqrt(4 * A /
PI)
[example: A = 1/2 * 2,221/6,000 = 0.1851 in.^2; pin
diameter = 0.486 in., use 1/2 in. diameter pin for bottom
pintle. For top, 1/4 in. diameter is sufficient, but use 3/8
in. for easier alignment.]
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