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 200 mm pulley at 960 rpm drives a 300 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 200 mmd2 = 300 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 12 teeth and each planet gear has 12 teeth. The planet carrier is held fixed and the sun is driven at 1800 rpm. Find the speed nr of the ring gear (magnitude).
Ns = 12Np = 12ns = 1800 rpm
nr = 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 = 100 N·md = 25 mmL = 1.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 75 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa, σᵧ = 10 MPa and τₓᵧ = 40 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 = −40 MPaσy = 10 MPaτxy = 40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 40 MPa, σᵧ = 10 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.
σx = 40 MPaσy = 10 MPaτxy = −30 MPaθ = -20°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 10 kNd = 25 mmL = 1.5 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 55 kNsegment 1: Ø35 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 5 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 5 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
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 15.1 mm. Find the damping ratio ζ.
x0 = 25 mmx2 = 15.1 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 10 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 5 kN/mk2 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 620.8 rpm
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Speed without slip
960·200/300 = 640 rpm
With slip
n2 = 640·0.97 = 620.8 rpm
Answer
n2 = 620.8 rpm
2.Gears & gear trains
Answer: nr = 600 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 = 12 + 2·12 = 36
Willis equation with the carrier fixed
With nc = 0 the sun drives the ring through the planets, which act as idlers.
nr = −ns·Ns/Nr = −1800·12/36 = −600 rpm
|nr| = 600 rpm
Answer
nr = 600 rpm
The ring turns opposite to the sun.
3.Torsion of shafts
Answer: φ = 5.746 °
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) = 100·1000·1500/(39000·38350) = 0.1003 rad
φ = 0.1003·180/π = 5.746°
Answer
φ = 5.746°
4.Torsion of shafts
Answer: d = 45.99 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·500/60 = 52.36 rad/s
T = P/ω = 75000/52.36 = 1432 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·1.432 × 10⁶/(π·75)) = 45.99 mm
Answer
d = 45.99 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = 16 °
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 = 40/47.17 = 0.848, sin 2θs = −(−25)/47.17 = 0.53
2θs = atan2(25, 40) = 32.01°
Angle on the element
θs = 16°
check: θs − θp = 16 − 61 = −45°
Answer
θs = 16°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 55.77 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.
Transformation equation
σθ = 25 + 15·0.766 + −30·−0.6428 = 55.77 MPa
Answer
σθ = 55.77 MPa
7.Axial stress & strain
Answer: ε = 291 µε
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 = 10·1000/490.9 = 20.37 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 20.37/70000 = 0.000291 = 291 µε
Answer
ε = 291 µε
8.Axial stress & strain
Answer: σmax = 273.5 MPa
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·35²/4 = 962.1 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 57.17 MPa
σ2 = 273.5 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 273.5 MPa
Answer
σmax = 273.5 MPa
9.Column buckling
Answer: Pcr = 313.9 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 = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN
Answer
Pcr = 313.9 kN
10.Column buckling
Answer: d = 49.28 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 = 2.5·5000 = 12500 N
KL = 1·4000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·12500·4000²/(π³·70000))^(1/4) = 49.28 mm
Answer
d = 49.28 mm
Check: the slenderness KL/r is 324.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.04009
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(25/15.1) = 0.2521
Damping ratio
ζ = δ/√(4π² + δ²) = 0.2521/√(39.48 + 0.06355) = 0.04009
Answer
ζ = 0.04009
12.Mechanical vibrations
Answer: fn = 1.299 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 = 1/5 + 1/10 → keq = 5·10/(5 + 10) = 3.333 kN/m
Natural frequency
ωn = √(3333/50) = 8.165 rad/s
fn = 8.165/(2π) = 1.299 Hz
Answer
fn = 1.299 Hz
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