Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A worm with 3 start(s) turns at 2900 rpm and drives a worm wheel with 48 teeth. Find the speed nw of the wheel.
z = 3Nw = 48n = 2900 rpm
nw = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (32 teeth) on the input shaft turning at 1450 rpm drives gear B (45 teeth). Gear C (32 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 32B = 45C = 32D = 40
nout = rpm
3.Torsion of shaftsMedium
A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 250 N·md = 25 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft transmits 10 kW at 1000 rpm. The shear stress must not exceed 50 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.
P = 10 kWN = 1000 rpmτallow = 50 MPaθallow = 0.75°/mG = 26 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −80 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −80 MPaσy = 50 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries a bending moment of 1500 N·m together with a torque of 400 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 50 mmM = 1500 N·mT = 400 N·m
σ1 = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 65 kNd = 25 mmL = 0.5 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 200 MPatube: Ø120/Ø50 mm, E = 100 GPa, σallow = 50 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)b × h = 50 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 2 cycles later by a peak of 5.5 mm. Find the damping ratio ζ.
x0 = 25 mmx2 = 5.5 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 10 kN/m and k₂ = 40 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 10 kN/mk2 = 40 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: nw = 181.3 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Wheel speed
nw = n·z/Nw = 2900·3/48 = 181.3 rpm
Answer
nw = 181.3 rpm
2.Gears & gear trains
Answer: nout = 824.9 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
First stage
nB = 1450·32/45 = 1031 rpm = nC
Second stage
nout = 1031·32/40 = 824.9 rpm
Answer
nout = 824.9 rpm
total ratio = 1.758 : 1
3.Torsion of shafts
Answer: φ = 11.49 °
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Polar second moment of area
J = π·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·1200/(39000·38350) = 0.2006 rad
φ = 0.2006·180/π = 11.49°
Answer
φ = 11.49°
4.Torsion of shafts
Answer: d = 41.12 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Torque
ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s
T = P/ω = 10000/104.7 = 95.49 N·m
Diameter for strength
T = 95.49 N·m = 95493 N·mm
dτ = ∛(16·95493/(π·50)) = 21.35 mm
Diameter for stiffness
θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm
G = 26 GPa = 26000 MPa
dφ = (32·95493/(π·26000·1.309 × 10⁻⁵))1/4 = 41.12 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(21.35, 41.12) = 41.12 mm
Answer
d = 41.12 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τmax = 71.59 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Radius of Mohr’s circle
τmax = R = √(((−80 − 50)/2)² + 30²) = 71.59 MPa
Answer
τmax = 71.59 MPa
6.Mohr’s circle & stress transformation
Answer: σ1 = 124.4 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Stresses at the surface
σ = 32M/(πd³) = 32·1500000/(π·50³) = 122.2 MPa
τ = 16T/(πd³) = 16·400000/(π·50³) = 16.3 MPa
Mohr’s circle
C = σ/2 = 61.12 MPa
R = √(61.12² + 16.3²) = 63.25 MPa
σ1 = C + R = 124.4 MPa
σ2 = C − R = −2.136 MPa
Answer
σ1 = 124.4 MPa
7.Axial stress & strain
Answer: δ = 0.9458 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 65·1000/490.9 = 132.4 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 132.4/70000 = 0.001892 = 1892 µε
Elongation
δ = ε·L = 0.001892·500 mm = 0.9458 mm
Answer
δ = 0.9458 mm
8.Axial stress & strain
Answer: Pmax = 498.7 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
Ar = π·20²/4 = 314.2 mm²
At = π(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 100·10³·9346 = 9.975 × 10⁸ N
Governing strain
ε = min(0.001, 0.0005) = 0.0005 (brass)
Largest load
Pmax = 0.0005·9.975 × 10⁸ = 498728 N = 498.7 kN
Answer
Pmax = 498.7 kN
9.Column buckling
Answer: Pcr = 235 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 100·50³/12 = 1.042 × 10⁶ mm⁴
Effective length
KL = 0.7·2500 = 1750 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·1.042 × 10⁶/1750² = 234991 N = 235 kN
Answer
Pcr = 235 kN
10.Column buckling
Answer: σcr = 49.35 MPa
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 1·4000 = 4000 mm
KL/r = 4000/20 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa
Answer
σcr = 49.35 MPa
11.Mechanical vibrations
Answer: ζ = 0.1196
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Logarithmic decrement
δ = (1/2)·ln(25/5.5) = 0.7571
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7571/√(39.48 + 0.5731) = 0.1196
Answer
ζ = 0.1196
12.Mechanical vibrations
Answer: keq = 50 kN/m
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 10 + 40 = 50 kN/m
Answer
keq = 50 kN/m
ExamProblems · examproblems.com/courses/machine-design