Size shafts, bolted joints and bearings, model parts parametrically, read GD&T, sanity-check FEA, choose manufacturing processes, design for manufacturing, compute Cpk and prepare parts for 3D printing.

Machine Design & Manufacturing

## Why it matters

Almost every machine moves power along a shaft: a motor drives a pump, a gearbox drives a conveyor, a pulley drives a fan. If a shaft is too small it twists, bends, wobbles or cracks; if it is far too large the machine is heavy, expensive and needs bigger bearings. Sizing a shaft is one of the first real design calculations a mechanical engineer does, and it pulls together the statics and strength of materials from Engineering Mechanics.

## Loads on a shaft

A rotating shaft usually carries two things at once:

- **Torque (T)** from the power it transmits. Torque causes torsional shear stress.
- **Bending moment (M)** from belt pulls, gear forces and its own weight between bearings. Bending causes normal stress.

Torque comes from power and speed. Power in watts equals torque in N·m times angular speed in rad/s:

```text
omega = 2 * pi * n / 60        (n in rpm, omega in rad/s)
T     = P / omega
```

For a 5 kW motor at 1450 rpm: omega = 2·pi·1450/60 = 151.8 rad/s, so T = 5000 / 151.8 = **32.9 N·m**. Notice that the same power at a lower speed means more torque: a gearbox output shaft turning slowly is thicker than the motor shaft feeding it.

## Stresses in a solid round shaft

The largest stresses are at the outer surface. For a solid shaft of diameter d:

```text
bending stress    sigma = 32 * M / (pi * d^3)
torsional shear   tau   = 16 * T / (pi * d^3)
```

Both scale with 1/d³, so a small increase in diameter buys a lot of strength: going from 20 mm to 25 mm cuts both stresses roughly in half (factor (25/20)³ = 1.95).

To compare the combined state of stress with a material's yield strength, ductile steels use the **von Mises** equivalent stress:

```text
sigma_vm = sqrt(sigma^2 + 3 * tau^2) = 32 * sqrt(M^2 + 0.75 * T^2) / (pi * d^3)
```

## Worked example: a pulley shaft

Our 5 kW shaft also carries a bending moment of M = 50 N·m at the pulley. It is made of a plain carbon steel with a yield strength of about 250 MPa, and we want a static factor of safety of 2.5, so the allowable stress is 250 / 2.5 = 100 MPa.

Check a 25 mm shaft (d = 0.025 m):

```text
sigma    = 32 * 50 / (pi * 0.025^3)     = 32.6 MPa
tau      = 16 * 32.9 / (pi * 0.025^3)   = 10.7 MPa
sigma_vm = sqrt(32.6^2 + 3 * 10.7^2)    = 37.5 MPa   (well under 100 MPa)
```

Or solve for the smallest diameter that just reaches the allowable stress:

```text
d = ( 32 * sqrt(M^2 + 0.75 * T^2) / (pi * sigma_allow) ) ^ (1/3)
  = ( 32 * sqrt(50^2 + 0.75 * 32.9^2) / (pi * 100e6) ) ^ (1/3)
  = 0.0180 m  =  18.0 mm
```

So why choose 25 mm? Because the static check is only the start.

## What the static check leaves out

- **Fatigue.** A rotating shaft under a steady bending load sees fully reversed stress: every point on the surface goes from tension to compression once per revolution. A motor at 1450 rpm reaches a million cycles in under 12 hours. Fatigue strength is well below yield, so real shaft design uses fatigue criteria (such as the Goodman or ASME-elliptic methods taught in machine design textbooks) with endurance-limit correction factors.
- **Stress concentrations.** Shoulders, grooves for retaining rings and keyseats raise the local stress by a factor Kt, often 1.5 to 3. Generous fillet radii at shoulders help a lot.
- **Deflection and slope.** Gears and bearings need the shaft to stay stiff. Bearing makers and gear standards set limits on slope at bearings and deflection at gear meshes, and these limits often govern the diameter rather than strength.
- **Critical speed.** A long, slender shaft can resonate. Keep the operating speed well away from its first natural frequency.

## Keys

A **key** locks a pulley or gear to the shaft so torque passes through it. A common parallel key for a 25 mm shaft is 8 mm wide and 7 mm tall (from the standard key tables for that shaft size). The torque becomes a force at the shaft surface, F = 2T/d = 2·32.9/0.025 = 2.63 kN. With a 28 mm key length:

```text
key shear   tau = F / (w * L)       = 2630 / (8 * 28)   = 11.8 MPa
key bearing     = F / ((h/2) * L)   = 2630 / (3.5 * 28) = 26.9 MPa
```

Keys are often made slightly weaker than the shaft on purpose, so an overload shears a cheap key instead of breaking the shaft.

## Common mistakes

- **Forgetting rpm to rad/s.** Using rpm directly in T = P/omega gives a torque about 10 times too small.
- **Mixing N·mm and N·m.** Keep metres and pascals together, or millimetres and MPa (N/mm²).
- **Stopping at the static check.** For anything that rotates under load, fatigue and deflection decide the final size.
- **Ignoring the keyseat.** The stress concentration at the end of a keyseat is a classic fatigue crack starter.

## A note on real design

This lesson teaches the mechanics so you can understand and sanity-check a shaft. Real shafts are designed to recognised methods and standards, use verified material data, and are reviewed by qualified engineers. Where public safety depends on the design, a licensed professional engineer takes responsibility for it.

## Recap

- T = P / omega, with omega = 2·pi·n/60.
- sigma = 32M/(pi·d³), tau = 16T/(pi·d³), and von Mises combines them.
- Diameter matters as d³, so small changes help a lot.
- Fatigue, stress concentrations, deflection and critical speed usually set the final size.

## Try it

On the Do tab, finish a Python function that solves for the minimum shaft diameter, and check it against the worked example.


Turn power and speed into torque, combine bending and torsion with von Mises, size a solid shaft and check its key.

Shafts and keys

Understanding Torsion

Shaft Design for INFINITE LIFE and Fatigue Failure in Just Over 10 Minutes!

## Why it matters

Machines are held together by bolts, screws, pins, rivets, welds and adhesives. Bolted joints are everywhere because they can be taken apart for maintenance. They are also a common cause of failures: joints that loosen, bolts that snap in fatigue, flanges that leak. Most of those failures trace back to one idea people skip: **preload**.

## How a bolted joint really works

When you tighten a nut, the bolt stretches like a stiff spring and the clamped parts (the **members**) squeeze together. The tension in the bolt after tightening is the **preload**, Fi. A good joint keeps the members clamped under every load it sees.

When an external load P tries to pull the joint apart, it does not all go into the bolt. The bolt and the members share it according to their stiffness. The **joint constant** C is the bolt's share:

```text
C = kb / (kb + km)        kb = bolt stiffness, km = member stiffness
bolt tension    Fb = Fi + C * P
member clamp    Fm = Fi - (1 - C) * P
```

Members are usually much stiffer than the bolt, so C is often around 0.2 to 0.3. That is the key insight: in a well-preloaded joint, the bolt only feels a small part of the external load. The members simply unload. If P grows until the clamp force reaches zero, the joint **separates**, and from then on the bolt carries the whole load and every load cycle hits it fully. That is how bolts fail in fatigue.

## Choosing a preload

A common textbook recommendation for reusable joints is a preload of about **75 % of the proof load**:

```text
Fi = 0.75 * At * Sp
At = tensile stress area of the thread
Sp = proof strength of the bolt's property class
```

For an **M10 x 1.5** bolt, At = 58.0 mm². For property class 8.8 at this size, ISO 898-1 gives a proof stress of about 580 MPa (check the current standard for your size and class):

```text
Fi = 0.75 * 58.0 * 580 = 25,230 N  =  25.2 kN
```

## From preload to tightening torque

In the shop you rarely measure preload; you set a torque. A simple estimate is:

```text
T = K * Fi * d          d = nominal diameter, K = nut factor (about 0.2 for dry steel)
T = 0.2 * 25,230 * 0.010 = 50.5 N·m
```

Be careful: K depends on friction, so it scatters a lot. Lubricant, plating, thread condition and washers can change the preload from the same torque by roughly 25 % or more either way. Critical joints use the fastener maker's torque data, torque-angle methods, direct tension indicators or ultrasonic bolt measurement.

## Worked example: a tension joint

The M10 bolt above holds a cover that sees an external separating load of P = 8 kN per bolt, and C = 0.25.

```text
bolt tension    Fb = 25.23 + 0.25 * 8      = 27.23 kN
bolt stress        = 27,230 / 58.0          = 469 MPa   (below the 580 MPa proof stress)
yield factor    nL = (Sp*At - Fi) / (C*P)   = (33,640 - 25,230) / 2,000 = 4.2
separation      n0 = Fi / (P * (1 - C))     = 25,230 / 6,000             = 4.2
```

Both factors are comfortable here. Now imagine the bolt was only snugged to 5 kN preload: n0 = 5,000 / 6,000 = 0.83, so the joint separates, and the bolt sees the full 8 kN swing on every cycle. Same bolt, very different life.

## Shear joints

Many joints carry load sideways, such as a bracket bolted to a frame. A preloaded joint resists slip by **friction**: the slip resistance is roughly mu x Fi x (number of bolts) x (number of friction surfaces), with mu often around 0.2 to 0.5 depending on the surface condition. If it slips, the bolt shanks bear on the hole edges, which is a different (and usually worse) way to carry load. Dowel pins are often added to locate parts precisely so the bolts only clamp.

## Practical details that matter

- **Thread engagement.** In a tapped steel hole, engage roughly one diameter; softer materials such as aluminium need more (often 1.5 to 2 diameters) or threaded inserts.
- **Washers** spread the load under the head and protect soft surfaces.
- **Loosening** comes mainly from transverse slip under vibration. Adequate preload is the first defence; thread-locking adhesive and prevailing-torque nuts help. Spring (split) washers are not reliable for this.
- **Torque-to-yield bolts** (common in engines) are designed to be stretched past yield and must not be reused.

## Common mistakes

- **Using the nominal area.** Bolt stress uses the tensile stress area At, not pi·d²/4.
- **Trusting torque as if it were preload.** Torque is an indirect, scattered measure.
- **Mixing grades.** Class 8.8 and 10.9 (or SAE Grade 5 and 8) look similar; always check the head marking.
- **Under-tightening "to be safe".** Low preload is what lets a joint separate and fail in fatigue.

## A note on real design

Real joints follow fastener standards (ISO, ASME, SAE), the manufacturer's data and the design rules of the industry involved (pressure vessels, structures and aerospace each have their own). Use these calculations to understand joints and sanity-check them; safety-critical joints need review by a qualified engineer, and where public safety is involved, a licensed professional engineer.

## Recap

- Preload clamps the joint; the bolt sees only C·P of an external load.
- Target about 75 % of proof load for reusable joints; T = K·Fi·d estimates torque.
- Separation factor n0 = Fi / (P(1 − C)) must stay above 1 with margin.
- Low preload is the usual root cause of loosening and bolt fatigue.

## Try it

On the Do tab, finish the separation-factor function and test how preload and joint stiffness change it.


Choose a bolt preload, estimate tightening torque, and check a tension joint for bolt load and separation.

Fasteners and joints

The Incredible Strength of Bolted Joints

BOLT TENSION and Tension at Non-Permanent Joints in Just Over 10 MINUTES!

## 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:

```text
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:

```text
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°).

```text
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:

```text
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.

```text
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):

| bearing | bore (mm) | C (kN) |
|---|---|---|
| 6205 | 25 | 14.8 |
| 6206 | 30 | 20.3 |
| 6207 | 35 | 27.0 |
| 6306 | 30 | 32.5 |
| 6307 | 35 | 35.1 |

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.


Compute a spring rate and a gear ratio, then select a ball bearing from a catalogue using the L10 life equation.

Springs, gears, and bearings

BEARINGS BASICS and Bearing Life for Mechanical Design in 10 Minutes!

Mechanical Springs - Stress, Deflection, and Spring Constant in Just Over 10 MINUTES!

Size the parts that carry power and load: shafts and keys for torque and bending, preloaded bolted joints, and springs, gears and rolling bearings selected from a catalogue.