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 3 kW motor running at 2900 rpm drives a pinion with 16 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 3 kWn1 = 2900 rpmN1 = 16N2 = 50η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 1450 rpm drives gear B (30 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (36 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 15B = 30C = 12D = 36
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 = 500 N·md = 40 mmL = 1.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft transmits 5 kW at 300 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 0.25° per m of length. Find the smallest diameter d that satisfies both limits.
P = 5 kWN = 300 rpmτallow = 60 MPaθallow = 0.25°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 50 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 130 MPa?
σx = 50 MPaσy = 100 MPaσ1 = 130 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 80 MPa, σᵧ = −20 MPa and τₓᵧ = 10 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 80 MPaσy = −20 MPaτxy = 10 MPaθ = 45°
τθ = MPa
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 50 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
P = 150 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 300 mm
δ = mm
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 50 mm, E = 110 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø50 mm, E = 110 GPa, σallow = 70 MPa
Pmax = kN
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 = 5 mE = 70 GPa (aluminium)d = 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 40 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 40 kNn = 2.5L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
An 8 kg mass is supported by a spring with a stiffness of 5 kN/m and a viscous damper with a damping coefficient of 160 N·s/m. Find the damping ratio ζ.
m = 8 kgk = 5 kN/mc = 160 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 5 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 = 30 kN/mk2 = 5 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 9.879 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 3000/303.7 = 9.879 N·m
Answer
T1 = 9.879 N·m
2.Gears & gear trains
Answer: nout = 241.7 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 = 1450·15/30 = 725 rpm = nC
Second stage
nout = 725·12/36 = 241.7 rpm
Answer
nout = 241.7 rpm
total ratio = 6 : 1
3.Torsion of shafts
Answer: φ = 4.384 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·1500/(39000·251327) = 0.07652 rad
φ = 0.07652·180/π = 4.384°
Answer
φ = 4.384°
4.Torsion of shafts
Answer: d = 55.56 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 = 5000/31.42 = 159.2 N·m
Diameter for strength
T = 159.2 N·m = 159155 N·mm
dτ = ∛(16·159155/(π·60)) = 23.82 mm
Diameter for stiffness
θ = 0.25·π/180/1000 = 4.363 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·159155/(π·39000·4.363 × 10⁻⁶))1/4 = 55.56 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(23.82, 55.56) = 55.56 mm
Answer
d = 55.56 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τxy = 48.99 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.
Required radius
C = (50 + 100)/2 = 75 MPa
R = σ1 − C = 130 − 75 = 55 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(55² − 25²) = 48.99 MPa
Answer
τxy = 48.99 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −50 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
τθ = −50·1 + 10·0 = −50 MPa
Answer
τθ = −50 MPa
7.Axial stress & strain
Answer: δ = 0.0807 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 − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 100·10³·3063 = 5.576 × 10⁸ N
Shortening
δ = 150000·300/5.576 × 10⁸ = 0.0807 mm
Answer
δ = 0.0807 mm
8.Axial stress & strain
Answer: Pmax = 474.8 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 − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N
Governing strain
ε = min(0.0008, 0.0006364) = 0.0006364 (bronze)
Largest load
Pmax = 0.0006364·7.461 × 10⁸ = 474809 N = 474.8 kN
Answer
Pmax = 474.8 kN
9.Column buckling
Answer: Pcr = 33.91 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·306796/2500² = 33913 N = 33.91 kN
Answer
Pcr = 33.91 kN
10.Column buckling
Answer: d = 82.88 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 = 2.5·40000 = 100000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·100000·4000²/(π³·70000))^(1/4) = 82.88 mm
Answer
d = 82.88 mm
Check: the slenderness KL/r is 193.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.4
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.
Critical damping
cc = 2·√(k·m) = 2·√(5000·8) = 400 N·s/m
Damping ratio
ζ = c/cc = 160/400 = 0.4
Answer
ζ = 0.4
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 3.23 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 = 30 + 5 = 35 kN/m
keq = 35·50/(35 + 50) = 20.59 kN/m
Natural frequency
ωn = √(20588/50) = 20.29 rad/s
fn = 20.29/(2π) = 3.23 Hz
Answer
fn = 3.23 Hz
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