Shafts, Keys, Keyseats, Circlips and Seals: Failure Mode 2

Failure Mode 2: Fatigue (The Silent Assassin)

This is what killed the practitioner's shaft. And it kills more shafts than any other mechanism.

For shafts that perform a large number of revolutions during their working life, fatigue is the dominant failure mode. Fatigue failure can occur even though stresses are well below the yield point — and that's what makes it so dangerous.

Fatigue happens when stress reversal occurs. And stress reversal can happen in two ways:

1. Change in direction of the load: Think about a car's transmission shaft. Every time you lift off the accelerator and the engine acts as a brake, the torsional shear stress reverses direction. Every gear change into reverse does the same thing.

2. Rotation of the shaft itself: This is the more common — and more insidious — mechanism. Consider a horizontal shaft with a flywheel or pulley creating a downward bending load. The top half of the shaft is in compression. The bottom half is in tension. Now rotate the shaft 180°. The half that was in compression is now in tension, and vice versa.

The direction of the axial stress due to bending reverses twice for every revolution of the shaft, even though the load direction and the direction of rotation do not change.

At 1450 rev/min (a common motor speed), that's 2,900 stress reversals per minute. 174,000 per hour. Over 4 million per day.

Each reversal creates a microscopic crack that grows imperceptibly. The beach marks the practitioner found on the practitioner's shaft were the forensic evidence — each line representing a period of crack growth, like rings in a tree trunk.

The shaft was confessing its murder for months. Nobody listened.

Two Approaches to Shaft Design

the practitioner explained that there are fundamentally two philosophies for shaft design:

Approach 1: Endurance Limit Method

Calculate peak loads, stress concentrations, and allowable shaft stresses as accurately as possible. Peak loads include inertia loads and shock loads. Stress concentrations include allowances for keyways, shoulders, grooves, holes, and shrink fits.

The allowable stresses include an allowance for shaft size (larger diameters have lower allowable stress). For shafts doing many revolutions, the allowable stress is based on the endurance limit, determined by fatigue tests on polished test specimens of 8–10 mm diameter.

A relatively small factor of safety (typically 1.2) is applied based on the endurance limit.

This approach follows standards such as AS 1403-1985 Design of Rotating Steel Shafts.

Approach 2: ASME Code Method (Recommended)

Calculate the maximum likely load under operating conditions (the design load). Then apply shock factors and fatigue factors along with a relatively generous factor of safety to account for inertia loads and non-uniformity of shaft material properties.

Basic strength of materials formulas are then used to calculate shaft stresses or shaft diameter.

This is the approach used throughout this guide because it provides a more fundamental understanding of the stresses involved and avoids complex formulas whose derivation and assumptions are often opaque.

The Core Formulas Every Engineer Must Know

These formulas are the backbone of shaft design. They apply to solid circular shafts subjected to combined torsion and bending — which covers the vast majority of engineering applications.

Combined Shear Stress (Formula 1)

$$f_s = \frac{16 , T_E}{\pi , d^3}$$

Where:

• f_s = combined shear stress (MPa)
• T_E = equivalent torque (Nm)
• d = shaft diameter (mm)

Use N, mm, MPa units consistently

Combined Axial Stress (Formula 2)

$$f_s = \frac{32 , M_E}{\pi , d^3}$$

Where:

• f_s = combined axial (bending) stress (MPa)
• M_E = equivalent moment (Nm)
• d = shaft diameter (mm)

Equivalent Torque (Formula 3)

$$T_E = \sqrt{T^2 + M^2}$$

Where:

• T = design torque at the critical location (Nm)
• M = design bending moment at the critical location (Nm)

This single formula combines bending and torsion into one equivalent torsional load. It's the workhorse of shaft design.

Equivalent Moment (Formula 4)

$$M_E = 0.5 , (T_E + M)$$

This companion formula gives the equivalent bending moment.

Accounting for Reality: Shock and Fatigue Factors

The formulas above assume perfect, steady, smooth loading. Real life is messy.

To account for shock and fatigue effects, apply shock/fatigue loading factors to the steady torque and moment:

$$T = K_T \times T_S$$

$$M = K_M \times M_S$$

Where:

• T_S and M_S are the steady torque and moment
• K_T is the shock/fatigue factor for torsion
• K_M is the shock/fatigue factor for bending

Why is K_M = 1.5 even for a static load? Because the bending stress reverses due to rotation of the shaft (remember the compression-to-tension flip discussed earlier), even though the load itself is constant in direction and strength. This reversal effect is accounted for in the factor.

Important notes:

• These formulas cover bending, torsion, and combined bending-and-torsion — the most common loading scenarios for shafts in engineering.
• In cases where significant axial loading exists (e.g., vertical shafts, propeller shafts), more complex ASME code formulas incorporating axial and column effects must be used.

Allowable Stresses and Factors of Safety

For steel shafts using the fundamental design approach:

Bending (tension or compression): the smaller of:

• 40% of yield strength (f_y), OR
• 24% of ultimate tensile strength (f_ult)

Torsion (shear): the smaller of:

• 30% of yield strength (f_y), OR
• 18% of ultimate tensile strength (f_ult)

The allowable shear stress is based on the assumption that the strength in shear is 75% of the strength in tension.

Worked Example 1: Determining Allowable Stresses

Problem: A shaft steel has an ultimate tensile strength of 700 MPa and a yield strength of 470 MPa. Determine the allowable bending and shear stress.

Solution:

Allowable bending stress = smaller of:

• 0.4 × 470 = 188 MPa
• 0.24 × 700 = 168 MPa

Allowable bending stress = 168 MPa ✓

Allowable shear stress = smaller of:

• 0.3 × 470 = 141 MPa
• 0.18 × 700 = 126 MPa

Allowable shear stress = 126 MPa ✓

Stress Concentration: The Keyway Problem

This is where the practitioner's shaft failed. And it's one of the most important concepts in shaft design.

A keyway is one of the most significant stress concentrations in a shaft. Keyways are located where a gear or pulley is fitted — and these are precisely the locations where the shaft is most highly stressed.

The design rule of thumb:

Allowable stresses at a keyway location = 75% of the allowable stresses without the keyway.

This single multiplier accounts for the stress concentration effect. Miss it, and you've effectively designed a shaft that's 25% weaker than you think at its most critical location.

Worked Example 1a: Stresses at a Keyway

For the shaft steel in Example 1 (allowable bending = 168 MPa, allowable shear = 126 MPa), if there is a keyway at a critical location:

Allowable bending at keyway: 0.75 × 168 = 126 MPa

Allowable shear at keyway: 0.75 × 126 = 94.5 MPa

For other stress concentration effects (shoulders, steps, holes), consult AS 1403 or engineering design handbooks.

Estimating Shaft Loads: The Four Common Sources

Before you can size a shaft, you need to know the forces acting on it. the practitioner walked the practitioner through the four most common sources of shaft loading.

Assumption: The shaft is supported by low-friction bearings, so frictional torque at the bearings is negligible.

Load Source 1: Weight

When a heavy pulley, flywheel, or gear sits on a shaft (non-vertical), its weight creates a transverse bending force:

$$F = m \times g$$

Where:

• F = force (N)
• m = mass (kg)
• g = gravitational acceleration (9.81 m/s²)

Straightforward — but don't forget that multiple components each add their own bending contribution at different locations along the shaft.

Load Source 2: Chain Drive

When a chain drive sprocket sits on a shaft, the chain tension creates a transverse force.

Force at a sprocket (Formula 5):

$$F = T_2 + T_1$$

Where:

• T_2 = tight side tension (N)
• T_1 = slack side tension (N)

Torque at a sprocket (Formula 6):

$$T = (T_2 - T_1) \times \frac{d}{2}$$

Where d = pitch circle diameter (PCD) of the sprocket (in metres).

Critical simplification: When the chain drive transmits power, the slack side tension is usually negligible. Therefore T_1 ≈ 0 and:

$$F = T_2 \quad \text{and} \quad T = T_2 \times \frac{d}{2}$$

Worked Example 2: Chain Drive Force

Problem: A chain drive with sprocket PCD 160 mm transmits 10 kW at 450 rev/min. Calculate the torque and force at the sprocket.

Solution:

From P = T × ω:

$$10,000 = T \times \pi \times \frac{450}{30}$$

$$T = 212.2 \text{ Nm}$$

Using Formula 6 with T_1 = 0:

$$212.2 = T_2 \times \frac{0.16}{2}$$

$$T_2 = 2653 \text{ N} = F \text{ (the force at the sprocket)}$$

Load Source 3: Belt Drive (Wedge or Vee Belt)

Belt drives work similarly to chain drives, except the slack side tension is not zero — because friction between the belt and pulley means the belt must maintain some minimum tension to grip.

Force at a pulley (Formula 5):

$$F = T_2 + T_1$$

Torque at a pulley (Formula 6):

$$T = (T_2 - T_1) \times \frac{d}{2}$$

Where d = PCD of the pulley (in metres).

For parallel belts, this is the arithmetic sum. For non-parallel belts, use the vector sum (but the arithmetic sum errs on the safe side).

Since both T_2 and T_1 are unknown, you need additional information. Three approaches exist:

Method (a): Assume a Slack Side Tension

Use published values for slack side tension under average loading:

These values are based on mid-load power for each belt section at 1000 rev/min, assuming a belt tension ratio of 12:1 and pulley PCD equal to the smallest recommended for the belt size.

Method (b): Assume a Belt Tension Ratio

The ratio of belt tensions doesn't vary greatly with belt section or number of belts — it depends primarily on the drive ratio. Suggested values:

These are based on 90% of the tension ratio at the point of slipping at the smaller pulley, with a wedge angle of 38°, coefficient of friction 0.3, and centre distance equal to the sum of pulley PCDs.

Method (c): Assume a Drive Application Factor

If the slack side tension were zero, then from Formulas 5 and 6: F = T_2 and T = T_2 × d/2, giving:

$$F = \frac{2T}{d}$$

But since T_1 is not zero, the actual force is greater. This is accounted for with a drive application factor f:

$$F = \frac{2 , f , T}{d} \quad \text{(Formula 7)}$$

For vee or wedge belt drives, f is typically taken as 1.5.

Notes:

• Each of the three methods gives a slightly different answer due to different assumptions.
• Using f = 1.5 usually errs on the safe side and slightly overstates the shaft load.
• Formula 7 can also be used for chain drives — in that case, T_1 = 0, so f = 1.
• If f = 1.25 is used instead of 1.5, much closer correlation with the other methods is obtained.

Worked Example 3: Belt Drive — Three Methods Compared

Problem: An SPA section wedge belt drive with pulley PCD 125 mm transmits 10 kW at 950 rev/min to another pulley with PCD 400 mm. Calculate the force on the shaft using all three approaches.

Solution:

First, calculate torque:

$$10,000 = T \times \pi \times \frac{950}{30}$$

$$T = 100.5 \text{ Nm}$$

(a) Using assumed slack side tension:

From the table: T_1 = 150 N for SPA section.

Using Formula 6: T = (T_2 - T_1) × d/2

$$100.5 = (T_2 - 150) \times \frac{0.125}{2}$$

$$T_2 = 1758 \text{ N}$$

Using Formula 5: F = T_2 + T_1 = 1758 + 150 = 1908 N

(b) Using belt tension ratio:

Drive ratio = 400/125 = 3.2

From the table (by interpolation), belt tension ratio = 10.2

So T_2 = 10.2 × T_1

Substituting into Formula 6:

$$100.5 = (10.2,T_1 - T_1) \times \frac{0.125}{2}$$

$$9.2,T_1 = 1608$$

$$T_1 = 174.8 \text{ N} \quad \text{and} \quad T_2 = 10.2 \times 174.8 = 1783 \text{ N}$$

F = T_2 + T_1 = 1783 + 174.8 = 1958 N

(c) Using drive application factor (f = 1.5):

$$F = \frac{2 \times 1.5 \times 100.5}{0.125} = \textbf{2412 N}$$

Comparison: Method (c) with f = 1.5 gives ~20% higher results than the other methods. With f = 1.25, much closer correlation is obtained.

Load Source 4: Gear Drive

When power is transmitted through gears, the force on each shaft acts at the gear tooth contact point. Understanding this force requires resolving it into components.

Gear Tooth Forces — The Three Components

At the point of contact between meshing gear teeth:

• F_t = tangential force — acts along the pitch circle, creating the torque
• F_s = separating (radial) force — acts along the line connecting gear centres, pushing the gears apart
• F = resultant transverse force — the vector sum of F_t and F_s
• θ = pressure angle (the angle between F and F_t) — assume 20° unless stated otherwise

Spur Gear Formulas

Tangential force (Formula 8):

$$F_t = \frac{2T}{d}$$

Where T = torque (Nm) and d = PCD of the gear (m).

Separating force (Formula 9):

$$F_s = F_t \times \tan\theta$$

Resultant transverse force (Formula 10):

$$F = \sqrt{F_t^2 + F_s^2}$$

Worked Example 4: Spur Gear Loading

Problem: A spur gear PCD 100 mm transmits 800 Nm of torque. Determine the load on the shaft.

Solution:

$$F_t = \frac{2 \times 800}{0.1} = 16 \text{ kN}$$

$$F_s = 16 \times \tan 20° = 5.82 \text{ kN}$$

$$F = \sqrt{16^2 + 5.82^2} = \textbf{17 kN}$$

Helical Gear Formulas

Helical gears have teeth cut at an angle (the helix angle, α). This makes them stronger and quieter than spur gears, but introduces an additional axial force along the shaft.

Separating force on a helical gear (Formula 11):

$$F_s = \frac{F_t \times \tan\theta}{\cos\alpha}$$

Axial force on a helical gear (Formula 12):

$$F_a = F_t \times \tan\alpha$$

The resultant transverse force is still calculated using Formula 10 (with the modified F_s).

Worked Example 5: Helical Gear Loading

Problem: A helical gear PCD 100 mm, helix angle 20°, transmits 800 Nm of torque. Determine transverse and axial loads.

Solution:

$$F_t = \frac{2 \times 800}{0.1} = 16 \text{ kN}$$

$$F_s = \frac{16 \times \tan 20°}{\cos 20°} = 6.2 \text{ kN}$$

$$F = \sqrt{16^2 + 6.2^2} = \textbf{17.2 kN}$$

(Not very different from the spur gear — the helix angle doesn't dramatically change transverse loading.)

But the axial load is significant:

$$F_a = 16 \times \tan 20° = \textbf{5.82 kN}$$

This axial force must be accommodated by the bearings and considered in the shaft design if significant.

The Design Procedure: Putting It All Together

After estimating the loads on the shaft, the design procedure involves:

1. Draw a torque diagram showing the torque at each location along the shaft
2. Draw a shear force diagram
3. Draw a bending moment diagram
4. Find the position of maximum stress (usually by inspection — typically where torque and bending moment are both at maximum, which is where gears, sprockets, or pulleys are attached)
5. Apply the shaft sizing formulas (Formulas 1–4) to determine the required diameter

Single-Plane vs. Multi-Plane Bending

Two situations arise in practice:

Single-plane problem: All resultant transverse forces act in the same plane (e.g., all horizontal, or all vertical). Only one bending moment diagram is needed.

Multi-plane problem: Transverse forces act in different planes — some horizontal, some vertical. You must draw bending moment diagrams in both planes, then combine them:

Resultant bending moment (Formula 13):

$$M = \sqrt{M_v^2 + M_h^2}$$

Where:

• M_v = bending moment in the vertical plane
• M_h = bending moment in the horizontal plane

The procedure then continues as for a single-plane problem, using this resultant bending moment.

Important Design Notes

• The treatment in this guide excludes situations with significant direct axial stress (this applies to most shafts encountered in engineering, where direct axial stress is small relative to torsion and bending).
• The design examples apply to single-diameter shafts. For stepped shafts, the same principles apply — calculate the required diameter at each step from the maximum stress at that location. Remember to apply a stress-concentration factor at each step (dependent on the ratio of diameters and the fillet radius).

The Complete Design Example: A Shaft Under Real-World Loading

This is where everything comes together. Follow this example step by step and you'll understand how every concept in this chapter connects.

Example 6: Shaft with Single-Plane Bending

Given information:

• Electric motor: 4 pole, 3 phase, hard-start, rated power 10 kW @ 1450 rev/min
• Chain drive: 15.875 mm pitch, Sprocket C: PCD 131.7 mm, power take-off 6 kW, vertically up
• Sprocket E: PCD 101.5 mm, power take-off 4 kW, vertically down
• Both sprockets keyed to shaft
• Shaft material: heat treated SAE 1035, UTS 600 MPa, yield stress 360 MPa
• Shaft layout: Coupling(A) — 100mm — Bearing(B) — 200mm — Sprocket(C) — 100mm — Bearing(D) — 100mm — Sprocket(E)

Step 1: Calculate input torque

$$T = \frac{P}{\omega} = \frac{10,000}{\pi \times 1450/30} = 65.86 \text{ Nm}$$

Step 2: Calculate torque at each sprocket

Sprocket C (6 kW): T_C = (6/10) × 65.86 = 39.51 Nm

Sprocket E (4 kW): T_E = (4/10) × 65.86 = 26.35 Nm

Step 3: Calculate forces at sprockets

Using F = T/(d/2) with T_1 = 0 (chain drive):

At sprocket C: F = 39.51 / (0.1317/2) = 600 N (vertically up)

At sprocket E: F = 26.35 / (0.1015/2) = 519 N (vertically down)

Step 4: Calculate bearing reactions

Taking moments about D (clockwise positive):

R_B × 0.3 + 600 × 0.1 + 519 × 0.1 = 0

R_B = -373 N (acts downward — our assumed direction was wrong)

Sum of vertical forces: R_D - 373 + 600 - 519 = 0

R_D = 292 N (acts upward)

Step 5: Draw torque, shear force, and bending moment diagrams

Step 6: Identify critical location

By inspection: Sprocket C — both torque and bending moment are at maximum.

Steady torque: T_S = 65.86 Nm

Steady bending moment: M_S = 74.6 Nm

Step 7: Apply shock/fatigue factors

The motor is hard-start → load applied suddenly with minor shock.

Using mid-point values: K_T = 1.25 and K_M = 1.75

Design torque: T = 1.25 × 65.86 = 82.3 Nm

Design moment: M = 1.75 × 74.6 = 130.6 Nm

Step 8: Calculate equivalent torque and moment

$$T_E = \sqrt{82.3^2 + 130.6^2} = 154.3 \text{ Nm}$$

$$M_E = 0.5 \times (154.3 + 130.6) = 142.5 \text{ Nm}$$

Step 9: Determine allowable stresses

For SAE 1035, UTS = 600 MPa, f_y = 360 MPa:

Bending: smaller of 0.4 × 360 = 144 MPa or 0.24 × 600 = 144 MPa → 144 MPa

Shear: smaller of 0.3 × 360 = 108 MPa or 0.18 × 600 = 108 MPa → 108 MPa

With keyway at the critical location:

Bending: 0.75 × 144 = 108 MPa

Shear: 0.75 × 108 = 81 MPa

Step 10: Calculate required shaft diameter

From torsional shear stress (Formula 1):

$$81 = \frac{16 \times 154.3 \times 10^3}{\pi \times d^3}$$

$$d = 21.3 \text{ mm}$$

From bending stress (Formula 2):

$$108 = \frac{32 \times 142.5 \times 10^3}{\pi \times d^3}$$

$$d = 23.8 \text{ mm}$$

Bending is more critical (because the bearings are located away from the sprockets). The required shaft diameter is 23.8 mm.

Rounding up to the next standard shaft size: d = 25 mm ✓

Example 7: Shaft with Multi-Plane Bending

Modified problem: Same as Example 6, but replace chain drives with spur gears:

• Gear C: PCD 125 mm, meshes with another gear vertically above it
• Gear E: PCD 100 mm, meshes with another gear vertically below it

What changes: Instead of chain forces acting in one direction, the gear tooth forces have both tangential (horizontal) and separating (vertical) components — creating loading in two planes.

Step 1: Calculate tangential forces (Formula 8)

Gear C: F_t = (2 × 39.51) / 0.125 = 632 N

Gear E: F_t = (2 × 26.35) / 0.1 = 527 N

Step 2: Calculate separating forces (Formula 9, θ = 20°)

Gear C: F_s = 632 × tan 20° = 230 N

Gear E: F_s = 527 × tan 20° = 192 N

Step 3: Draw bending moment diagrams in both planes

Step 4: Calculate resultant bending moment (Formula 13)

$$M = \sqrt{28.2^2 + 77.2^2} = 82.2 \text{ Nm}$$

Step 5: Continue as before

Using shock factors K_T = 1.25, K_M = 1.75:

T = 82.3 Nm (same as before)

M = 1.75 × 82.2 = 143.8 Nm

$$T_E = \sqrt{82.3^2 + 143.8^2} = 165.7 \text{ Nm}$$

$$M_E = 0.5 \times (165.7 + 143.8) = 155 \text{ Nm}$$

From torsional shear stress:

$$81 = \frac{16 \times 165.7 \times 10^3}{\pi \times d^3} \quad \Rightarrow \quad d = 21.8 \text{ mm}$$

From bending stress:

$$108 = \frac{32 \times 155 \times 10^3}{\pi \times d^3} \quad \Rightarrow \quad d = 24.5 \text{ mm}$$

Required standard shaft size: d = 25 mm ✓

Key observation: Although gears increased the shaft loading compared to chain drives, the required standard shaft size is the same. This won't always be the case — in many designs, gear forces can significantly increase the required diameter.

Keys: The Tiny Components That Carry Enormous Loads

What Is a Key?

A key is a small piece of metal that sits in matching slots (keyways) in the shaft and the hub of the attached component (gear, sprocket, pulley). Its purpose: transmit torque from the shaft to the component, or vice versa.

Without a key, your gear would just spin freely on the shaft — or the shaft would spin inside the gear. Neither is useful.

Why Keys Matter More Than You Think

A key is the torque transmission bottleneck. It's typically the smallest, weakest component in the entire power transmission path. A 25 mm shaft made of heat-treated steel can carry enormous loads — but the torque only reaches the gear through a tiny rectangular piece of mild steel sitting in a groove.

Key Design Considerations

• Key size is standardized based on shaft diameter — you don't choose key dimensions arbitrarily
• Key length is the design variable — it must be long enough to handle the torque without shearing or crushing
• Two failure modes: shearing (the key is cut in half along the shaft-hub interface) and crushing (the key is compressed against the side of the keyway)
• The keyway weakens the shaft — always apply the 75% stress concentration reduction at keyway locations

Key Selection Quick Guide

For any given shaft diameter, the standard specifies the key width and height. You then calculate the required key length from:

Shear failure criterion:

$$T = F \times \frac{d}{2} = f_s \times b \times L \times \frac{d}{2}$$

Crushing failure criterion:

$$T = F \times \frac{d}{2} = f_c \times \frac{h}{2} \times L \times \frac{d}{2}$$

Where:

• T = torque to be transmitted (Nm)
• d = shaft diameter (m)
• b = key width (m)
• h = key height (m)
• L = key length (m) — this is what you're solving for
• f_s = allowable shear stress of the key material (MPa)
• f_c = allowable crushing (compressive bearing) stress (MPa)

The required key length is the larger of the two values calculated from shear and crushing.

Circlips: The Retention Rings You'll Forget Until Something Falls Off

What Are Circlips?

Circlips (also called retaining rings or snap rings) are spring-steel rings that snap into grooves machined in shafts or bores. Their job: prevent axial movement of components like bearings, gears, and spacers.

Two Types

• External circlips fit into grooves on the outside of a shaft — they prevent components from sliding off the shaft
• Internal circlips fit into grooves inside a bore (housing) — they prevent components from sliding out of a housing

Why They Matter

A missing or incorrectly installed circlip allows a bearing to walk axially along the shaft. Once the bearing moves out of position, alignment is lost, loads are redistributed to locations that weren't designed for them, and catastrophic failure follows.

Circlips are cheap. The failures they prevent are not.

Circlip Selection

Circlips are selected by:

• Shaft diameter (for external) or bore diameter (for internal)
• Axial load capacity — the maximum force the circlip can resist before the groove or ring fails
• Material — standard carbon spring steel, stainless steel for corrosive environments

Always ensure the groove dimensions match the circlip specification. An undersized groove means the circlip doesn't seat properly. An oversized groove means it doesn't grip.

Seals: The Last Line of Defense

Why Seals Exist

A seal has two jobs:

1. Keep lubricant in — bearings and gears need oil or grease to survive. Without a seal, lubricant leaks out and components run dry.
2. Keep contaminants out — dust, moisture, and process debris destroy bearings. A seal blocks them.

Types of Shaft Seals

Lip Seals (Oil Seals / Radial Shaft Seals)

The most common type for rotating shafts. A flexible lip rides on the shaft surface, creating a dynamic seal.

Key specifications:

• Shaft diameter — the bore of the seal matches the shaft
• Bore diameter — the outer diameter fits the housing
• Width — determines the seal's axial footprint

Common styles include:

• CRW1 — standard rubber lip, spring-loaded
• HMS4 — standard double-lip design
• HMSA7 — advanced design for higher speeds/pressures
• SSLEEVE — includes a wear sleeve for shaft protection
• VR1/VR2 — V-ring designs for specific applications

Material options:

• Nitrile rubber (standard, good for mineral oils, -40°C to +100°C)
• Viton/FKM (for high temperatures, synthetic oils, chemicals)
• PTFE (for extreme temperatures and chemical resistance)
• Stainless steel spring (for corrosive environments)

Selection Parameters

When selecting a shaft seal:

Seal Installation Best Practices

• Never install a lip seal dry — pre-lubricate the lip with the fluid being sealed
• Press the seal squarely into the housing — cocked seals leak from day one
• Protect the seal lip from sharp edges during installation (keyways, splines, threads) — use an installation sleeve if necessary
• Check shaft surface condition — scoring, rust, or wear grooves will cause any seal to leak
• Inspect the seal before installation — damaged lips during shipping/handling are common