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 worm with 4 start(s) turns at 960 rpm and drives a worm wheel with 45 teeth. Find the speed nw of the wheel.
z = 4Nw = 45n = 960 rpm
nw = rpm
2.Gears & gear trainsHard
A 4 kW motor at 960 rpm drives gear A (18 teeth), which meshes with gear B (20 teeth). Gear C (50 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.
P = 4 kWnin = 960 rpmA = 18B = 20C = 50D = 60ηmesh = 96%
Tout = N·m
3.Torsion of shaftsMedium
A solid steel shaft 1 m long carries a torque of 250 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 250 N·mL = 1 mG = 80 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 5 kWN = 750 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −50 MPa, σᵧ = 10 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −50 MPaσy = 10 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −40 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = −40 MPaσy = 90 MPaτxy = −10 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø20 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 70 GPaL = 500 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 = 105 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 0.5 m, copper (E = 117 GPa)
σmax = 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 = 2 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 3 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 2 cycles later by a peak of 4.7 mm. Find the damping ratio ζ.
x0 = 10 mmx2 = 4.7 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 25 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 25 kN/mk2 = 25 kN/mk3 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: nw = 85.33 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.
Wheel speed
nw = n·z/Nw = 960·4/45 = 85.33 rpm
Answer
nw = 85.33 rpm
2.Gears & gear trains
Answer: Tout = 48.89 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.
Output speed
nout = 960·(18/20)·(50/60) = 720 rpm
ωout = 2π·720/60 = 75.4 rad/s
Output power and torque
Pout = 4000·0.96² = 3686 W
Tout = Pout/ωout = 3686/75.4 = 48.89 N·m
Answer
Tout = 48.89 N·m
3.Torsion of shafts
Answer: d = 36.75 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.
Required polar moment
T = 250 N·m = 250000 N·mm
L = 1 m = 1000 mm
G = 80 GPa = 80000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 250000·1000/(80000·0.01745) = 179049 mm⁴
Diameter
d = (32J/π)1/4 = 36.75 mm
Answer
d = 36.75 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 16.29 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·750/60 = 78.54 rad/s
T = P/ω = 5000/78.54 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·75)) = 16.29 mm
Answer
d = 16.29 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 67.08 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 = √(((−50 − 10)/2)² + 60²) = 67.08 MPa
Answer
τmax = 67.08 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 26.44 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 = 25 + −65·−0.1736 + −10·0.9848 = 26.44 MPa
Answer
σθ = 26.44 MPa
7.Axial stress & strain
Answer: δ = 0.08545 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·20²/4 = 314.2 mm²
At = π(120² − 70²)/4 = 7461 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 70·10³·7461 = 5.851 × 10⁸ N
Shortening
δ = 100000·500/5.851 × 10⁸ = 0.08545 mm
Answer
δ = 0.08545 mm
8.Axial stress & strain
Answer: σmax = 334.2 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·30²/4 = 706.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 148.5 MPa
σ2 = 334.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 334.2 MPa
Answer
σmax = 334.2 MPa
9.Column buckling
Answer: Pcr = 5.426 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·7854/1000² = 5426 N = 5.426 kN
Answer
Pcr = 5.426 kN
10.Column buckling
Answer: σcr = 34.27 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·3000 = 1500 mm
KL/r = 1500/6.25 = 240
Critical stress
σcr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa
Answer
σcr = 34.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.05997
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(10/4.7) = 0.3775
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3775/√(39.48 + 0.1425) = 0.05997
Answer
ζ = 0.05997
12.Mechanical vibrations
Answer: fn = 2.418 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
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 25 + 25 = 50 kN/m
keq = 50·15/(50 + 15) = 11.54 kN/m
Natural frequency
ωn = √(11538/50) = 15.19 rad/s
fn = 15.19/(2π) = 2.418 Hz
Answer
fn = 2.418 Hz
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