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 125 mm pulley at 720 rpm drives a 400 mm pulley by a belt with 2 % slip. Find the speed n2 of the driven pulley.
n1 = 720 rpmd1 = 125 mmd2 = 400 mms = 2 %
n2 = rpm
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 15 teeth and each planet gear has 12 teeth. The ring gear is held fixed and the sun is driven at 3000 rpm. Find the speed nc of the planet carrier.
Ns = 15Np = 12ns = 3000 rpm
nc = rpm
3.Torsion of shaftsMedium
A solid steel 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 = 200 N·md = 20 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 100 kW at 300 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 100 kWN = 300 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 0 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 0 MPaσy = 110 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 100 MPa, σᵧ = 70 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = 100 MPaσy = 70 MPaτxy = −60 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 25 kNd = 10 mmL = 2.5 mE = 100 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 = 20 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 0.5 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 200 GPa (steel)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 30 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 30 kNn = 3L = 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 30 mm is followed 4 cycles later by a peak of 18.1 mm. Find the damping ratio ζ.
x0 = 30 mmx4 = 18.1 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 5 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 40 kN/mk2 = 5 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: n2 = 220.5 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
720·125/400 = 225 rpm
With slip
n2 = 225·0.98 = 220.5 rpm
Answer
n2 = 220.5 rpm
2.Gears & gear trains
Answer: nc = 833.3 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 = 15 + 2·12 = 39
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(3000 − nc)/(0 − nc) = −39/15
nc = ns·Ns/(Ns + Nr) = 3000·15/54 = 833.3 rpm
Answer
nc = 833.3 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 4.559 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·500/(80000·15708) = 0.07958 rad
φ = 0.07958·180/π = 4.559°
Answer
φ = 4.559°
4.Torsion of shafts
Answer: d = 68.7 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·300/60 = 31.42 rad/s
T = P/ω = 100000/31.42 = 3183 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·3.183 × 10⁶/(π·50)) = 68.7 mm
Answer
d = 68.7 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = 75.69 °
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
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·30, 0 − 110) = 151.4°
Angle on the element
θp = 75.69°
Answer
θp = 75.69°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 40.54 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 = 85 + 15·0.5 + −60·0.866 = 40.54 MPa
Answer
σθ = 40.54 MPa
7.Axial stress & strain
Answer: ε = 3183 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·10²/4 = 78.54 mm²
σ = P/A = 25·1000/78.54 = 318.3 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 318.3/100000 = 0.003183 = 3183 µε
Answer
ε = 3183 µε
8.Axial stress & strain
Answer: σmax = 99.47 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 15.92 MPa
σ2 = 99.47 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 99.47 MPa
Answer
σmax = 99.47 MPa
9.Column buckling
Answer: Pcr = 49.35 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 = 60·50³/12 = 625000 mm⁴
Effective length
KL = 2·2500 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·625000/5000² = 49348 N = 49.35 kN
Answer
Pcr = 49.35 kN
10.Column buckling
Answer: d = 67.54 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 = 3·30000 = 90000 N
KL = 0.7·4000 = 2800 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·2800²/(π³·70000))^(1/4) = 67.54 mm
Answer
d = 67.54 mm
Check: the slenderness KL/r is 165.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.0201
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 = (1/4)·ln(30/18.1) = 0.1263
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1263/√(39.48 + 0.01596) = 0.0201
Answer
ζ = 0.0201
12.Mechanical vibrations
Answer: keq = 45 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 = 40 + 5 = 45 kN/m
Answer
keq = 45 kN/m
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