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 15 kW motor running at 1450 rpm drives a pinion with 30 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
P = 15 kWn1 = 1450 rpmN1 = 30N2 = 40η = 96%
T2 = N·m
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 30 teeth and each planet gear has 20 teeth. The ring gear is held fixed and the sun is driven at 1500 rpm. Find the speed nc of the planet carrier.
Ns = 30Np = 20ns = 1500 rpm
nc = rpm
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 75 N·m. The two ends of a gauge length of 250 mm rotate 6.6° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 75 N·md = 12 mmL = 250 mmφ = 6.6°
G = GPa
4.Torsion of shaftsHard
A solid shaft must transmit 15 kW at 750 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 15 kWN = 750 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 80 MPa, σᵧ = −30 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 80 MPaσy = −30 MPaτxy = −50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −30 MPa, σᵧ = −50 MPa and τₓᵧ = 40 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −30 MPaσy = −50 MPaτxy = 40 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa). Both are 400 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
P = 150 kNrod: Ø25 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 70 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 250 MPatube: Ø100/Ø50 mm, E = 110 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 45 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 45 kNn = 2L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 15 kg mass is supported by a spring with a stiffness of 200 kN/m and a viscous damper with a damping coefficient of 350 N·s/m. Find the damping ratio ζ.
m = 15 kgk = 200 kN/mc = 350 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 100 kN/mk2 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 126.4 N·m
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.
Motor torque
ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 15000/151.8 = 98.79 N·m
Output torque
T2 = 98.79·40/30·0.96 = 126.4 N·m
Answer
T2 = 126.4 N·m
2.Gears & gear trains
Answer: nc = 450 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Ring gear teeth
Nr = Ns + 2·Np = 30 + 2·20 = 70
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(1500 − nc)/(0 − nc) = −70/30
nc = ns·Ns/(Ns + Nr) = 1500·30/100 = 450 rpm
Answer
nc = 450 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: G = 79.96 GPa
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 moment and angle
J = π·12⁴/32 = 2036 mm⁴
φ = 6.6·π/180 = 0.1152 rad
Shear modulus
T = 75 N·m = 75000 N·mm
G = TL/(Jφ) = 75000·250/(2036·0.1152) = 79957 MPa
G = 79.96 GPa
Answer
G = 79.96 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: d = 25.31 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·750/60 = 78.54 rad/s
T = P/ω = 15000/78.54 = 191 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·60)) = 25.31 mm
Answer
d = 25.31 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −66.14 °
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.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −50/74.33 = −0.6727, sin 2θs = −55/74.33 = −0.7399
2θs = atan2(−55, −50) = −132.3°
Angle on the element
θs = −66.14°
check: θs − θp = −66.14 − (−21.14) = −45°
Answer
θs = −66.14°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 5.578
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = −40 MPa, R = 41.23 MPa
σ1 = 1.231 MPa, σ2 = −81.23 MPa, σ3 = 0
Tresca
σmax − σmin = 82.46 MPa
n = 460/82.46 = 5.578
Answer
n = 5.578
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.238 mm
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·25²/4 = 490.9 mm²
At = π(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 70·10³·2199 = 2.521 × 10⁸ N
Shortening
δ = 150000·400/2.521 × 10⁸ = 0.238 mm
Answer
δ = 0.238 mm
8.Axial stress & strain
Answer: Pmax = 407 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·25²/4 = 490.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N
Governing strain
ε = min(0.00125, 0.0005455) = 0.0005455 (bronze)
Largest load
Pmax = 0.0005455·7.461 × 10⁸ = 406979 N = 407 kN
Answer
Pmax = 407 kN
9.Column buckling
Answer: Pcr = 109.9 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 round section has the same I about every axis.
I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/2000² = 109878 N = 109.9 kN
Answer
Pcr = 109.9 kN
10.Column buckling
Answer: d = 90.25 mm
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.
Required critical load
Pcr = n·P = 2·45000 = 90000 N
KL = 2·2500 = 5000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·5000²/(π³·70000))^(1/4) = 90.25 mm
Answer
d = 90.25 mm
Check: the slenderness KL/r is 221.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.101
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.
Critical damping
cc = 2·√(k·m) = 2·√(200000·15) = 3464 N·s/m
Damping ratio
ζ = c/cc = 350/3464 = 0.101
Answer
ζ = 0.101
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 4.594 Hz
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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/100 + 1/50 → keq = 100·50/(100 + 50) = 33.33 kN/m
Natural frequency
ωn = √(33333/40) = 28.87 rad/s
fn = 28.87/(2π) = 4.594 Hz
Answer
fn = 4.594 Hz
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