Dave Gerr - Boat Mechanical Systems Handbook-How to Design, Install, and Recognize Proper Systems in Boats

Chapter 12:Rudder-Stock Angle, Control, and Installation Considerations

(706 kgm) of steering torque, 60 percent of
which is 37,181 in.-lb. (424 kgm).
The tiller arm length that the drag link is
fastened to on each rudder is 10. 3 in. (262 mm).
Thus the design compression load on the
drag link from the rudder torque is 37,181
in.-lb. ÷ 10. 3 in.= 3,610 lb., or 424 kgm ÷ 0. 262
m= 1,618 kg.
The moment of inertia of the two-pin-end
column (our drag link) required to accept the
compression load without buckling is found
with Formula 12-3.

Formula 12-3. Two-Pin-End-Column
Moment of Inertia

Where

I= moment of inertia, in.^4 or cm^4

P= compression load, lb. or N

E= modulus of elasticity, psi or MPa

(see Table 12-2)

L= drag-link length, pin center to pin

center, in. or mm

SF= safety factor, use 4

Example: In our case, the length of the
drag link from pin center to pin center is 116
inches (2,946 mm). You can use solid rod,
but hollow pipe or tube is lighter, and the
larger diameter required isn’t a drawback. In
fact, it’s an advantage as it makes the drag
link more resistant to being bent out of
shape by someone leaning on it. You can use
any material, but 6061 aluminum is usually
best in this application and is highly corro-
sion resistant (unless there are bronze clevis
end fittings, in which case stainless or

bronze tube or pipe would be superior).

Using aluminum, we find

Referring to a standard pipe table, we find

that a nominal 2^1 / 2 - inch schedule 80 pipe

has an I of 1. 92 in.^4 and is 2. 875 inches OD,

0. 276 inches wall. This section, or any pipe

section with a greater moment of inertia,

would serve. Don’t use thinner wall than

schedule 40 (standard weight).

Or

1,618 kg× 9. 8066 = 15,867 N

Note that the entered units were all mm.

To convert to cm^4 , we divided by 10,000 or by

(10 mm/cm)^4.

Referring to a standard DIN 17172 pipe

table, we find that a DIN 17172 76. 1 × 6. 3

aluminum pipe (76.1 mm OD, 6.3 mm wall)

has a moment of inertia of 84.8 cm^4 (see

Formula 12-4). Don’t use thin-walled pipe or

tube.

The moment of inertia of a hollow, round

section can be found using Formula 12-4.

Formula 12-4. Moment of Inertia for a

Hollow, Round Section

For the metric DIN pipe, we get:

76 .1 mm OD – (2 × 6 .3 mm wall)=

63 .5 mm ID

Twin Rudders and

No Drag Link?

There is a drawback to a drag link connecting

twin rudders—it takes up space. Drag links

Z

64

= OD^44 ID

⎛

⎝⎜

⎞

⎠⎟

×−

π

()

I

4 15,867 N (2,946 mm)

10,000 71,700 N

2

= 2

× ×

××π //mm

2 =^77 .84 cm^4

I

4 3,610 lb. (116in.)

10,400,000 psi

2

== 2

××

×π

11. 89 in.^4

I

SF PL

E

2

= 2

×

π

Z

64

= 7 .61 cm OD^44 (6.35 cm ID)

⎛

⎝⎜

⎞

⎠⎟×−( )=

π

() 884 .8cm^4

TABLE 12-2.MODULUS, E

Modulus, E

psi MPa*

Aluminum 10,400,000 71,700
Silicon Bronze 16,000,000 110,300
Steel and Stainless 28,000,000 193,000
Steel

*MPa = N/mm^2 (megapascals = newtons per
millimeter squared)

Formula 12-3.

Formula 12-4.