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 7.5 kW motor running at 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 7.5 kWn1 = 1440 rpmN1 = 12N2 = 36η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turns at 1440 rpm and drives gear B (15 teeth). Gear C (36 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 864 rpm. How many teeth ND must gear D have?
nin = 1440 rpmA = 12B = 15C = 36nout = 864 rpm
ND =
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 400 mm rotate 7.04° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 50 N·md = 12 mmL = 400 mmφ = 7.04°
G = GPa
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 1000 N·m acts at B, 1.2 m from A, and a torque of 1500 N·m in the opposite direction acts at the free end C, a further 0.5 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 1000 N·mTC = 1500 N·mAB = 1.2 mBC = 0.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 10 MPa, σᵧ = 120 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 10 MPaσy = 120 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −40 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = −40 MPaσy = 100 MPaτxy = −10 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 300 mm, E = 70 GPaP = 100 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 50 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 500 kg mass on a spring with a stiffness of 200 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
m = 500 kgk = 200 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 15 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 100 kgk1 = 15 kN/mk2 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 144.7 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 7500/150.8 = 49.74 N·m
Output torque
T2 = 49.74·36/12·0.97 = 144.7 N·m
Answer
T2 = 144.7 N·m
2.Gears & gear trains
Answer: ND = 48
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.
Intermediate shaft speed
nB = nC = 1440·12/15 = 1152 rpm
Teeth on gear D
ND = C·nC/nout = 36·1152/864 = 48
Answer
ND = 48
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 = 7.04·π/180 = 0.1229 rad
Shear modulus
T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·400/(2036·0.1229) = 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: τmax = 35.37 MPa
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 1500 N·m
TAB = TC − TB = 1500 − 1000 = 500 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 1500 N·m
τmax = 16·1.5 × 10⁶/(π·60³) = 35.37 MPa
Answer
τmax = 35.37 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 58.52 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 = √(((10 − 120)/2)² + −20²) = 58.52 MPa
Answer
τmax = 58.52 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −13.66 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 = 30 + −70·0.5 + −10·0.866 = −13.66 MPa
Answer
σθ = −13.66 MPa
7.Axial stress & strain
Answer: P = 52.95 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 = 490.9 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·490.9) = 4.074 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·1257) = 1.592 × 10⁻⁶ mm/N
Load
P = 0.3/5.666 × 10⁻⁶ = 52948 N = 52.95 kN
Answer
P = 52.95 kN
8.Axial stress & strain
Answer: σ2 = 58.64 MPa
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.
Stiffnesses
k₁ = 200000·314.2/600 = 104720 N/mm
k₂ = 70000·1257/300 = 293215 N/mm
Share of segment 2
F2 = 100000·293215/397935 = 73684 N
Stress
σ2 = 73684/1257 = 58.64 MPa (compression)
Answer
σ2 = 58.64 MPa
9.Column buckling
Answer: Pcr = 179.9 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 100·50³/12 = 1.042 × 10⁶ mm⁴
Effective length
KL = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·1.042 × 10⁶/2000² = 179915 N = 179.9 kN
Answer
Pcr = 179.9 kN
10.Column buckling
Answer: σcr = 69.09 MPa
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
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.5·3000 = 1500 mm
KL/r = 1500/15 = 100
Critical stress
σcr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa
Answer
σcr = 69.09 MPa
11.Mechanical vibrations
Answer: fd = 2.273 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 = √(200000/500) = 20 rad/s
fn = 20/(2π) = 3.183 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 3.183·√(1 − 0.7²) = 2.273 Hz
Answer
fd = 2.273 Hz
12.Mechanical vibrations
Answer: fn = 2.757 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 15 + 15 = 30 kN/m
Natural frequency
ωn = √(30000/100) = 17.32 rad/s
fn = 17.32/(2π) = 2.757 Hz
Answer
fn = 2.757 Hz
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