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 0.75 kW motor running at 1440 rpm drives a pinion with 20 teeth, which meshes with a gear with 45 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 0.75 kWn1 = 1440 rpmN1 = 20N2 = 45η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1440 rpm drives gear B (30 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (30 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 12B = 30C = 15D = 30
nout = 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 = 750 N·md = 50 mmL = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 10 kWN = 1000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −30 MPa, σᵧ = 30 MPa and τₓᵧ = −50 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −30 MPaσy = 30 MPaτxy = −50 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 80 MPa, σᵧ = 50 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = 80 MPaσy = 50 MPaτxy = −10 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 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 0.5 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: brass, Ø60 mm, 200 mm, E = 100 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaP = 50 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 2 cycles later by a peak of 5.5 mm. Find the damping ratio ζ.
x0 = 15 mmx2 = 5.5 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 20 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 80 kN/mk2 = 20 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 10.85 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 = 750/150.8 = 4.974 N·m
Output torque
T2 = 4.974·45/20·0.97 = 10.85 N·m
Answer
T2 = 10.85 N·m
2.Gears & gear trains
Answer: nout = 288 rpm
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.
First stage
nB = 1440·12/30 = 576 rpm = nC
Second stage
nout = 576·15/30 = 288 rpm
Answer
nout = 288 rpm
total ratio = 5 : 1
3.Torsion of shafts
Answer: φ = 4.489 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·2500/(39000·613592) = 0.07835 rad
φ = 0.07835·180/π = 4.489°
Answer
φ = 4.489°
4.Torsion of shafts
Answer: d = 18.65 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·1000/60 = 104.7 rad/s
T = P/ω = 10000/104.7 = 95.49 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·95493/(π·75)) = 18.65 mm
Answer
d = 18.65 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 58.31 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 = √(((−30 − 30)/2)² + −50²) = 58.31 MPa
Answer
τmax = 58.31 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 48.84 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 = 65 + 15·−0.5 + −10·0.866 = 48.84 MPa
Answer
σθ = 48.84 MPa
7.Axial stress & strain
Answer: P = 132.9 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·2827) = 7.074 × 10⁻⁷ mm/N
Load
P = 0.5/3.763 × 10⁻⁶ = 132868 N = 132.9 kN
Answer
P = 132.9 kN
8.Axial stress & strain
Answer: σ2 = 23.7 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 = 200000·490.9/400 = 245437 N/mm
k₂ = 70000·706.9/400 = 123700 N/mm
Share of segment 2
F2 = 50000·123700/369137 = 16755 N
Stress
σ2 = 16755/706.9 = 23.7 MPa (compression)
Answer
σ2 = 23.7 MPa
9.Column buckling
Answer: Pcr = 19.62 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 = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/2000² = 19621 N = 19.62 kN
Answer
Pcr = 19.62 kN
10.Column buckling
Answer: σcr = 100.7 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.7·1500 = 1050 mm
KL/r = 1050/7.5 = 140
Critical stress
σcr = π²E/(KL/r)² = π²·200000/140² = 100.7 MPa
Answer
σcr = 100.7 MPa
11.Mechanical vibrations
Answer: ζ = 0.07959
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(15/5.5) = 0.5017
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5017/√(39.48 + 0.2517) = 0.07959
Answer
ζ = 0.07959
12.Mechanical vibrations
Answer: keq = 16 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/80 + 1/20 → keq = 80·20/(80 + 20) = 16 kN/m
Answer
keq = 16 kN/m
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