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
Two parallel shafts are 225 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 4 : 1. How many teeth N2 must the gear have?
a = 225 mmm = 5 mmi = 4
N2 =
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 450 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 450 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
A solid aluminium 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 = 500 N·md = 30 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 1450 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 5 kWN = 1450 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −70 MPa, σᵧ = 30 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −70 MPaσy = 30 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 60 mm carries a bending moment of 500 N·m together with a torque of 1000 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 60 mmM = 500 N·mT = 1000 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 500 mm, E = 70 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø60 mm, E = 110 GPa, σallow = 50 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 = 1 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 65 kNn = 2.5L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 200 kg mass on a spring with a stiffness of 50 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 = 200 kgk = 50 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 10 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 25 kgk1 = 40 kN/mk2 = 10 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: N2 = 72
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.
Total number of teeth
N1 + N2 = 2a/m = 2·225/5 = 90
Split by the ratio
N1 = 90/(1 + 4) = 18
N2 = 4·18 = 72
Answer
N2 = 72
2.Gears & gear trains
Answer: θ = 145.1 °
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(0.3) = 17.46°
Angle of wrap
θ = 180 − 2·17.46 = 145.1°
Answer
θ = 145.1°
3.Torsion of shafts
Answer: φ = 11.08 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·800/(26000·79522) = 0.1935 rad
φ = 0.1935·180/π = 11.08°
Answer
φ = 11.08°
4.Torsion of shafts
Answer: d = 13.08 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·1450/60 = 151.8 rad/s
T = P/ω = 5000/151.8 = 32.93 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·32929/(π·75)) = 13.08 mm
Answer
d = 13.08 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 53.85 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 = √(((−70 − 30)/2)² + −20²) = 53.85 MPa
Answer
τmax = 53.85 MPa
6.Mohr’s circle & stress transformation
Answer: σ1 = 38.15 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) = 32·500000/(π·60³) = 23.58 MPa
τ = 16T/(πd³) = 16·1000000/(π·60³) = 23.58 MPa
Mohr’s circle
C = σ/2 = 11.79 MPa
R = √(11.79² + 23.58²) = 26.36 MPa
σ1 = C + R = 38.15 MPa
σ2 = C − R = −14.57 MPa
Answer
σ1 = 38.15 MPa
7.Axial stress & strain
Answer: P = 30.93 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
A₁ = 706.9 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·706.9) = 1.011 × 10⁻⁵ mm/N
Load
P = 0.4/1.293 × 10⁻⁵ = 30925 N = 30.93 kN
Answer
P = 30.93 kN
8.Axial stress & strain
Answer: Pmax = 296 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 − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5027 = 6.511 × 10⁸ N
Governing strain
ε = min(0.0008, 0.0004545) = 0.0004545 (bronze)
Largest load
Pmax = 0.0004545·6.511 × 10⁸ = 295952 N = 296 kN
Answer
Pmax = 296 kN
9.Column buckling
Answer: Pcr = 27.47 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 1·1000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·39761/1000² = 27470 N = 27.47 kN
Answer
Pcr = 27.47 kN
10.Column buckling
Answer: d = 31.17 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 = 2.5·65000 = 162500 N
KL = 0.5·1500 = 750 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·162500·750²/(π³·200000))^(1/4) = 31.17 mm
Answer
d = 31.17 mm
Check: the slenderness KL/r is 96.26, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 2.401 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 = √(50000/200) = 15.81 rad/s
fn = 15.81/(2π) = 2.516 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.516·√(1 − 0.3²) = 2.401 Hz
Answer
fd = 2.401 Hz
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 = 40 + 10 = 50 kN/m
Answer
keq = 50 kN/m
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