Springs, gears, and bearings

Why it matters

Shafts and fasteners are rarely designed alone. A gearbox is shafts, gears and bearings working together; a valve is a spring, a seat and a stem. Most of these parts are bought, not designed from scratch. A mechanical engineer's job is often to work out the loads and then select a standard part from a catalogue with the right margin. This lesson covers the three most common: springs, gears and rolling bearings, with bearing selection as the worked skill.

Springs

A helical compression spring stores energy by twisting its wire. Its rate (stiffness) is:

k = G * d^4 / (8 * D^3 * Na)
G  = shear modulus (about 79 GPa for spring steel)
d  = wire diameter, D = mean coil diameter, Na = number of active coils

Example: d = 2 mm, D = 16 mm, Na = 10, G = 79,300 MPa:

k = 79,300 * 2^4 / (8 * 16^3 * 10) = 1,268,800 / 327,680 = 3.87 N/mm

The ratio C = D/d is the spring index (here 8). Values of roughly 4 to 12 are easy to make; outside that range springs are hard to coil or buckle easily. The wire stress is tau = Kw · 8FD/(pi·d³), where Kw (the Wahl factor) corrects for curvature. Springs that cycle millions of times are designed for fatigue, and manufacturers publish stock springs with their rate, maximum deflection and solid height.

Gears

Gears transmit torque between shafts at a fixed speed ratio. For spur gears in metric sizes, the module m sets the tooth size, and the pitch diameter is d = m · z (z = number of teeth). Two gears only mesh if they share the same module and pressure angle (usually 20°).

ratio i = z_driven / z_driver
speed out  = speed in / i
torque out ~= torque in * i * efficiency     (a spur stage is typically around 97-99 %)
tooth force Wt = 2 * T / d

Example: a 20-tooth pinion (m = 2 mm, d = 40 mm) drives a 60-tooth gear. The ratio is 3, so a 1450 rpm motor gives about 483 rpm, and 32.9 N·m becomes roughly 96 N·m after losses. The tooth force on the pinion is Wt = 2·32.9/0.040 = 1.65 kN. That force is also a load on the shaft and its bearings, which is how the parts connect. Gear tooth strength and surface durability are rated with AGMA or ISO 6336 methods.

Rolling bearings

Ball and roller bearings support shafts with low friction. Their life is limited by rolling contact fatigue, and catalogues rate it statistically. The basic rating life L10 is the life that 90 % of a large group of identical bearings will reach or exceed:

L10 = (C / P) ^ p          millions of revolutions
p = 3 for ball bearings, 10/3 for roller bearings
C = basic dynamic load rating (catalogue), P = equivalent dynamic load
L10h = L10 * 1,000,000 / (60 * n)    hours at n rpm

When there is both radial (Fr) and axial (Fa) load, the catalogue gives factors so P = X·Fr + Y·Fa.

Worked example: select a deep-groove ball bearing

A shaft bearing carries a steady radial load of P = 2.5 kN at n = 1500 rpm. The machine should run 20,000 hours (roughly 10 years of single-shift use) before 10 % of bearings would be expected to fail.

needed life  L10 = 20,000 * 60 * 1500 / 1,000,000 = 1,800 million revolutions
needed rating  C = P * L10^(1/3) = 2.5 * 1800^(1/3) = 2.5 * 12.16 = 30.4 kN

Now go to the catalogue. Here is a short, illustrative list of deep-groove ball bearings (real ratings differ slightly between manufacturers):

The 6306 (C = 32.5 kN) is the smallest that meets 30.4 kN. Its life is (32.5/2.5)³ = 2,197 million revolutions, or about 24,400 hours. If the shaft must be 30 mm at the bearing, you could also compare the 6206 (too weak) with the 6306 (heavier series, same bore): this is why bearing series exist.

What else selection checks

• Speed limits in the catalogue (lubrication and heat).
• Static rating C0 for shock loads or slow rotation.
• Lubrication and contamination: modern catalogues adjust L10 with life-modification factors; dirt can cut life dramatically.
• Fits on the shaft and in the housing, and how the bearing is located axially (one fixed, one floating bearing is common).

Common mistakes

• Forgetting the exponent changes. Doubling the load cuts ball bearing life by 2³ = 8 times, not 2.
• Using the static rating C0 in the life equation instead of the dynamic rating C.
• Ignoring axial load. Even a modest axial load can raise P a lot.
• Treating L10 as a guarantee. One bearing in ten is expected to fail before L10.

A note on real design

Use the manufacturer's current catalogue and selection tools, and their application engineers for unusual duty. The method here is for learning and checking; machinery that affects safety is reviewed by qualified engineers.

Recap

• Spring rate k = G·d⁴/(8·D³·Na); keep the spring index in a practical range.
• Gear ratio = z_driven / z_driver; tooth force Wt = 2T/d loads the shafts and bearings.
• L10 = (C/P)^p million revolutions; solve for the C you need, then pick from the catalogue.

Try it

On the Do tab, finish the required-rating function and let the code pick a bearing from the catalogue.