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

Chapter 11: Rudder-Stock Size, Construction, and Bearing Specification

the stock and the bearing must resist the total
side pressure of the water.
Example:If we assume that our 5.28 sq.ft.
(0.49 m^2 ) rudder were on a single-screw 28-
knot boat, the force on the one rudder would
be the same as we’ve already found: 8,512 lb.
(3,839 kg, 37,647 N). Each bearing takes half
this load, or 4,256 lb. (18,823 N).
You would find the required bearing
size (for allowable bearing stress) as in
Formula 11-7 above; however, it makes sense
to turn down the lower end of the rudder
stock to fit into a smaller, more streamlined
bearing/skeg below, as that portion of the
stock is taking only the shear resulting from
the side load; it’s not also transmitting the
torque from the steering gear.
You are then forming a pintlefor the
lower rudder bearing. Generally, this should
be approximately 44 percent of overall shaft
diameter, but you need to check that it has
adequate shear area to take the side load. Use
a large safety factor of 5 to allow for the
stress concentrations of abrupt change in
shape between the main rudder stock and the
lower pintle.
Example:Now for this rudder—with
bearings top and bottom—the stock diameter
will be considerably smaller, with no bending
load. Working through the previous formulas
(and using 316L stainless, instead of Aqualoy
22 HS), we found the stock diameter required
was 2^3 / 8 - or 2^1 / 2 - inch 316L SS. We’ll use 2^1 / 2
inches as being more readily available and
having a larger safety factor. For the metric
shaft, we found 60 mm.

2. 5 in. dia.× 0. 44 = 1. 1 in.; try 1 in. dia.

or

60 mm dia.× 0. 44 = 26 .4 mm; try 25 mm

Check shear stress (single shear):

Area =π(1. 0 in. ÷2)^2 = 0 .78 sq. in.

4,256 lb. ÷ 0 .78 sq. in. =5,456 psi

51,000 psi ÷5,456 psi =SF 9. 3 ; this is well

over 5 SF and so acceptable

or

Area = π(25 mm ÷2)^2 =491 mm^2

18,823 N ÷491 mm^2 = 38 .3 N/mm^2

352 N/mm^2 ÷ 38 .3 N/mm^2 =SF 9. 2 ; this is

well over 5 SF and so acceptable

Gudgeon-and-Pintle

Proportions

If this were a traditional rudder hung on several

gudgeons and pintles instead of using a

stock, the load is divided by the number of

gudgeons and pintles. The pintle diameter

can be as small as practical, as long as the

safety factor in shear is greater than 5. Divide

the side load by the number of pintles and

gudgeons to get the load on each.

Pintle length in the gudgeon or bearing

should be at least 1.2 times pintle diameter.

The pintle may be tapered at a ratio of

approximately 1 to 6 from maximum diameter

at the top, to minimum diameter at the bottom.

The height of the gudgeon housing the pintle

must be at least 1.2 times pintle diameter.

The gudgeon wall thickness is to be 0. 5

times the pintle diameter or more.

Fastening of the gudgeons into the hull or

transom must take the side load from the pintle

with a safety factor of 5. Check:

• fasteners in shear


• fasteners in tension


• fasteners in bearing in the hull, transom,
  or skeg as applicable

Axial Bearing Loads

and Shaft Collars

Rudders have weight that pulls them down

out of the boat. This axial load must be resisted

with a shaft collar inside the boat at one of

the bearings or by an axial bearing at the

lower rudder bearing (Figure 11-5).

Where the lower rudder bearing is

necked down to form a pintle, as described

TABLE 11-4. ULTIMATE SHEAR
STRENGTH, USS

Material psi MPa*

Aluminum 5000 Series 20,400 140
Aluminum 6082 27,600 190
Silicon Bronze 36,000 248
Stainless Steel (316L) 51,000 352
Carbon Composite Not recommended
in this application
Aqualoy 22 HS 78,000 538

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