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 22 kW motor running at 2900 rpm drives a pinion with 24 teeth and a module of 4 mm. Find the tangential force Ft on the pinion teeth.
P = 22 kWn = 2900 rpmN1 = 24m = 4 mm
Ft = N
2.Gears & gear trainsHard
An open belt connects a 160 mm pulley and a 315 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 160 mmd2 = 315 mmC = 600 mm
θ = °
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 = 250 N·md = 20 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 75 kW at 300 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.
P = 75 kWN = 300 rpmdi/do = 0.75τallow = 50 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 30 MPa, σᵧ = −50 MPa and τₓᵧ = 20 MPa (tension positive). The element is rotated 20° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = 30 MPaσy = −50 MPaτxy = 20 MPaθ = 20°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = 20 MPa and τₓᵧ = 20 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = 60 MPaσy = 20 MPaτxy = 20 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 60 mm, E = 100 GPa). Both are 400 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.
P = 80 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 100 GPaL = 400 mm
δ = mm
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 = 10 kNsegment 1: Ø35 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 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 15 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 15 kNn = 3L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 40 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 40 kgk = 80 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 50 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 = 10 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 100 kN/mk2 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 1509 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·2900/60 = 303.7 rad/s
T = P/ω = 22000/303.7 = 72.44 N·m
Pitch diameter
d = m·N = 4·24 = 96 mm
Tangential force
Ft = 2T/d = 2·72.44/0.096 = 1509 N
Answer
Ft = 1509 N
2.Gears & gear trains
Answer: θ = 165.2 °
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.
Belt angle
asin(0.1292) = 7.421°
Angle of wrap
θ = 180 − 2·7.421 = 165.2°
Answer
θ = 165.2°
3.Torsion of shafts
Answer: φ = 22.8 °
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) = 250·1000·2000/(80000·15708) = 0.3979 rad
φ = 0.3979·180/π = 22.8°
Answer
φ = 22.8°
4.Torsion of shafts
Answer: do = 70.86 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 = 75000/31.42 = 2387 N·m
Outer diameter from the allowable stress
T = 2387 N·m = 2.387 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·2.387 × 10⁶/(π·50·0.6836)) = 70.86 mm
di = 0.75·70.86 = 53.14 mm
Answer
do = 70.86 mm
A solid shaft would need d = 62.42 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: σy′ = −53.5 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 for the y′ face
σy′ = −10 − 40·0.766 − 20·0.6428 = −53.5 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = 33.5 MPa, σx′ + σy′ = −20 = σx + σy
Answer
σy′ = −53.5 MPa
6.Mohr’s circle & stress transformation
Answer: n = 5.534
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 = 28.28 MPa
σ1 = 68.28 MPa, σ2 = 11.72 MPa, σ3 = 0
Von Mises
σvM = √(68.28² − 68.28·11.72 + 11.72²) = 63.25 MPa
n = 350/63.25 = 5.534
Answer
n = 5.534
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.06791 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·40²/4 = 1257 mm²
At = π(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 100·10³·2199 = 4.712 × 10⁸ N
Shortening
δ = 80000·400/4.712 × 10⁸ = 0.06791 mm
Answer
δ = 0.06791 mm
8.Axial stress & strain
Answer: σmax = 88.42 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·12²/4 = 113.1 mm²
σ1 = 10.39 MPa
σ2 = 88.42 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 88.42 MPa
Answer
σmax = 88.42 MPa
9.Column buckling
Answer: Pcr = 102.5 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 = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/3500² = 102511 N = 102.5 kN
Answer
Pcr = 102.5 kN
10.Column buckling
Answer: d = 53.66 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 = 3·15000 = 45000 N
KL = 1·2500 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·45000·2500²/(π³·70000))^(1/4) = 53.66 mm
Answer
d = 53.66 mm
Check: the slenderness KL/r is 186.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 6.79 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.
Undamped natural frequency
ωn = √(80000/40) = 44.72 rad/s
fn = 44.72/(2π) = 7.118 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 7.118·√(1 − 0.3²) = 6.79 Hz
Answer
fd = 6.79 Hz
12.Mechanical vibrations
Answer: fn = 2.146 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 + 1/10 → keq = 100·10/(100 + 10) = 9.091 kN/m
Natural frequency
ωn = √(9091/50) = 13.48 rad/s
fn = 13.48/(2π) = 2.146 Hz
Answer
fn = 2.146 Hz
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