Machine Design: free practice, theory and problems

1. Stress and sizing

What is it about?

Real machine parts are rarely loaded by just one type of load at a time. A bolt may have both preload (tension) and bending from a slightly misaligned joint; a shaft may have both a bending moment from its own weight and torsion from the driving torque. Stress and sizing is about combining these contributions into one representative stress at the critical point, comparing it with the material's yield strength using a sensible safety factor, and accounting for the fact that sharp transitions and varying load can cause failure far below what a static tensile test would suggest.

Concepts and formulas

• Combined axial and bending stress at an outer fiber: $\sigma = \dfrac{F}{A} \pm \dfrac{M}{W}$ (same sign on the compression side, opposite on the tension side).
• Safety factor against yielding: $n = R_e/\sigma$. Typically 1.5–3 in machine design, higher for uncertain loads or serious consequences of failure.
• Stress concentration factor $K_t$: at a hole, a fillet or a notch the local stress becomes $\sigma_{local} = K_t\sigma_{nom}$, where $\sigma_{nom}$ is the stress without the concentration.
• Fatigue: under varying load, failure can occur at a stress well below $R_e$, especially at notches. The endurance limit $\sigma_e$ is the stress amplitude the material withstands for "infinitely" many cycles.
• Bolt preload: a torque wrench controls the preload force through $T \approx KFd$, where $K \approx 0.2$ for dry threads and $d$ is the nominal diameter.

How to solve the problems

1. Identify every type of load acting at the critical section (axial, bending, torsion).
2. Compute the stress contribution from each load separately, with the correct sign.
3. Add contributions of the same type at the same point (for example axial and bending at the same fiber). At a notch: multiply the nominal stress by $K_t$.
4. Compare the total stress with $R_e$ and find $n = R_e/\sigma$, or check that $n$ is large enough.
5. For variable load: consider whether fatigue may govern, not just static yielding.

Example

A short column with a rectangular cross-section of 40 × 30 mm carries an axial compressive force of 40 kN acting 15 mm outside the centroidal axis. $W = bh^2/6 = 40\cdot 30^2/6 = 6000$ mm³ and $A = 1200$ mm². Find the largest compressive stress.

1. Axial stress: $\sigma_{ax} = F/A = 40,000/1200 \approx 33.3$ MPa.
2. Bending moment from the eccentricity: $M = Fe = 40,000\cdot 15 = 600,000$ Nmm $= 600$ Nm.
3. Bending stress: $\sigma_{bend} = M/W = 600,000/6000 = 100$ MPa.
4. On the side where both contributions are compressive, they add up: $\sigma_{max} = 33.3+100 = 133.3$ MPa.

Answer: $\sigma_{max} \approx 133.3$ MPa.

Common mistakes

• Forgetting the bending that appears when an axial force does not act exactly at the centroidal axis.
• Adding stresses from different points or different directions as if they were the same type.
• Applying $K_t$ to the wrong stress – it must multiply the nominal stress at the same point, not a different load.
• Believing a static safety factor is enough when the load actually varies cyclically.

Concepts in this part

Practise stress and sizing in the app →

2. Tolerances and fits

What is it about?

No part can be made to an exact dimension – everything has a manufacturing tolerance. Tolerances and fits are about controlling this inexactness on purpose: how much deviation from the nominal size is acceptable, and how two parts (typically a shaft in a hole) should fit together – with clearance, with an interference (press) fit, or with a mix of the two. The right choice decides whether a shaft spins freely in a bearing, whether a hub stays fixed on a shaft without a key, or whether manufacturing costs become unnecessarily high.

Concepts and formulas

• Tolerance (tolerance width) $=$ upper limit deviation $-$ lower limit deviation. A low IT grade (e.g. IT6) means a tighter, more expensive tolerance; a high IT grade (e.g. IT11) means a looser, cheaper tolerance.
• Hole-basis system (most common): the hole is given the letter H, with a lower deviation of $0$ – the smallest hole size is always the nominal size. The desired fit is chosen by changing the shaft's tolerance class.
• Shaft-basis system: the opposite – the shaft's upper deviation is $0$ (letter h), and the hole's tolerance class is chosen to give the desired fit. Used when several parts must fit on the same (purchased) shaft.
• Clearance $=$ hole size $-$ shaft size. Positive clearance means the shaft is smaller than the hole.
• Clearance fit: the clearance is always $\ge 0$ (smallest hole $\ge$ largest shaft). Interference fit: the clearance is always $\le 0$ (smallest shaft $>$ largest hole, i.e. interference). Transition fit: the result can be either a little clearance or a little interference, depending on the actual sizes.
• Surface roughness $R_a$: the arithmetic mean deviation of the surface profile, in µm. A lower $R_a$ means a finer (smoother) surface.
• Geometric dimensioning and tolerancing (GD&T) controls form, orientation and position in addition to just size.

How to solve the problems

1. Read off the upper and lower limit deviations for the hole and the shaft (in µm, relative to the same nominal size).
2. Compute the smallest and largest allowable size for each part: nominal size $+$ deviation.
3. Smallest clearance $=$ smallest hole size $-$ largest shaft size. Largest clearance $=$ largest hole size $-$ smallest shaft size.
4. If both are $\ge 0$: clearance fit. If both are $\le 0$: interference fit. If one is positive and the other negative: transition fit.
5. The average clearance (useful as an expected value) is $(\text{smallest} + \text{largest clearance})/2$.

Example

A hole is marked Ø30 H7 (upper deviation $+21$ µm, lower deviation $0$), and the shaft is marked Ø30 js6 (deviation $\pm 6.5$ µm). Classify the fit and find the average clearance.

1. Hole: $30.000$–$30.021$ mm. Shaft: $29.9935$–$30.0065$ mm.
2. Largest clearance: $21-(-6.5) = 27.5$ µm.
3. Smallest clearance: $0-6.5 = -6.5$ µm (negative, i.e. a bit of interference).
4. Since one extreme is clearance and the other is interference, this is a transition fit.
5. Average clearance: $(27.5+(-6.5))/2 = 10.5$ µm.

Answer: transition fit, with an average clearance of $\approx 10.5$ µm.

Common mistakes

• Mixing up the upper and lower deviations when computing the smallest/largest clearance.
• Believing an H-hole always gives a clearance fit. The type of fit depends on the shaft's tolerance class, not on the hole alone.
• Believing a low $R_a$ means a tight size tolerance. Surface roughness and size tolerance are two different things.
• Forgetting that negative clearance means interference (oversize), not a mistake in the calculation.

Concepts in this part

Practise tolerances and fits in the app →

3. Machine elements

What is it about?

Machine elements are the reusable building blocks a machine designer assembles: gears and belts that transmit and change speed and torque, bolts and keys that hold parts together and transmit force, bearings that let shafts rotate with low friction and a long life, and springs that store energy or absorb motion. This unit collects the formulas that tie them together, so you can work through an entire power-transmission chain from motor to output shaft.

Concepts and formulas

• Gear ratio: $i = n_1/n_2 = z_2/z_1$ (gears) $= d_2/d_1$ (pulleys). For an ideal (lossless) transmission: $T_2 = iT_1$.
• Power and torque: $P = T\omega = \dfrac{2\pi nT}{60}$, with $n$ in rpm and $T$ in Nm giving $P$ in watts.
• Efficiency $\eta = P_{out}/P_{in}$. For several stages in series: $\eta_{tot} = \eta_1\eta_2\cdots$.
• Bearing life (ball bearing): $L_{10} = (C/P)^3$ million revolutions, where $C$ is the dynamic load rating and $P$ is the applied load. Roller bearings use the exponent $10/3$.
• Spring stiffness $k = F/x$. In parallel: $k_{tot} = k_1+k_2$. In series: $1/k_{tot} = 1/k_1+1/k_2$.
• Bolt preload: a torque wrench gives $T \approx KFd$, where $K \approx 0.2$ for dry threads. Strength class X.Y gives roughly $R_m \approx 100X$ MPa and $R_e \approx 0.1XY\cdot R_m$.
• Self-locking threads: a threaded joint is self-locking when the friction angle in the threads is larger than the lead angle, just like a block that stays put on an incline.

How to solve the problems

1. Map out the power-transmission chain: which elements are in series, and what is the gear ratio at each stage?
2. Compute speed stage by stage with $i = n_{in}/n_{out}$, and subtract losses by multiplying the power by the efficiency of each stage.
3. Find the torque from $T = P/\omega$ wherever you actually need it – do not assume the torque is the same throughout the chain.
4. For individual elements (bearing, spring, bolt): plug the numbers directly into that element's own formula.
5. Check the order of magnitude: a low speed must always come with a high torque (and vice versa) for the same power.

Example

A gearbox has two stages with ratios $i_1 = 3$ and $i_2 = 2.5$, each with an efficiency of $0.97$. The input rotates at 1450 rpm and receives 15 kW. Find the output torque.

1. Output speed: $n_{out} = 1450/(3\cdot 2.5) \approx 193.3$ rpm, so $\omega_{out} = 2\pi\cdot 193.3/60 \approx 20.2$ rad/s.
2. Output power: $P_{out} = 15,000\cdot 0.97\cdot 0.97 \approx 14,114$ W.
3. Output torque: $T_{out} = P_{out}/\omega_{out} \approx 14,114/20.2 \approx 697$ Nm.

Answer: $T_{out} \approx 697$ Nm.

Common mistakes

• Assuming the torque is the same throughout the chain. It is the power (minus losses) that survives almost unchanged, not the torque.
• Forgetting the efficiency in multi-stage systems – the losses multiply, they do not add.
• Adding spring stiffnesses in series (should be in parallel), or the reverse.
• Using the ball-bearing $L_{10}$ exponent (3) for a roller bearing, which uses $10/3$.

Concepts in this part

Practise machine elements in the app →

4. Bolted and welded joints

What is it about?

Almost all machines and steel structures are joined with bolts or welds. You must be able to check that a bolt can take the tension, that the bolts in a joint can take the shear, and that a weld is thick and long enough. The calculations are simple: force divided by area, compared with what the material can take.

Concepts and formulas

• The tensile stress area $A_s$ is the effective cross-section in the threads. Examples: M8: 36.6 mm², M10: 58.0 mm², M12: 84.3 mm², M16: 157 mm², M20: 245 mm², M24: 353 mm².
• Tensile stress in a bolt:

• Property class $a.b$: the tensile strength is $R_m = a\cdot 100$ MPa and the yield strength $R_e = 0.b\cdot R_m$. Class 8.8 gives $R_m = 800$ MPa and $R_e = 640$ MPa. Class 10.9 gives 1000 and 900 MPa.
• Shear in a bolted joint with $n$ bolts and one shear plane per bolt:

• Fillet weld: the throat thickness is about $a = 0.7\,z$, where $z$ is the leg length. The stress in the weld is the force divided by the throat area:

• Safety factor: $n = R_e/\sigma$. It must exceed the requirement, often 1.5–3.
• Preload: the bolt is tightened so the joint is clamped together. Then it doesn't slip, and the bolt sees less alternating load (fatigue).

How to solve the problems

1. Find the area: $A_s$ from the table for tension, the shank area $\pi d^2/4$ for shear, $a\cdot L$ for a weld.
2. Compute the stress: force in N divided by area in mm² gives MPa.
3. Compare with the yield strength and compute the safety factor.

Example

An M12 bolt of class 8.8 is loaded with 20 kN of tension.

1. $\sigma = 20\,000/84.3 \approx 237$ MPa.
2. The yield strength is 640 MPa.
3. The safety factor is $640/237 \approx 2.7$. That is comfortably enough.

Common mistakes

• Using the full diameter instead of the stress area in the threads.
• Forgetting to divide by the number of bolts or shear planes.
• Using the leg $z$ instead of the throat $a$ in the weld stress.
• Mixing kN and N. Remember that N/mm² = MPa.

Concepts in this part

Practise bolted and welded joints in the app →