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 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 18 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 1440 rpmN1 = 12N2 = 18η = 100%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (25 teeth) on the input shaft turns at 960 rpm and drives gear B (60 teeth). Gear C (24 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 192 rpm. How many teeth ND must gear D have?
nin = 960 rpmA = 25B = 60C = 24nout = 192 rpm
ND =
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 = 100 N·md = 50 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 75 kW at 1800 rpm. Find the maximum shear stress τmax in the shaft.
P = 75 kWN = 1800 rpmdo = 50 mmdi = 35 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 140 MPaσy = 50 MPaτxy = 30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 110 MPa, σᵧ = 80 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 110 MPaσy = 80 MPaτxy = 20 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.
P = 200 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 120 MPatube: Ø100/Ø60 mm, E = 100 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)d = 20 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 55 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 55 kNn = 2.5L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 5 cycles later by a peak of 8 mm. Find the damping ratio ζ.
x0 = 15 mmx5 = 8 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 10 kN/m and k₂ = 60 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 10 kN/mk2 = 60 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 29.84 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 = 3000/150.8 = 19.89 N·m
Output torque
T2 = 19.89·18/12·1 = 29.84 N·m
Answer
T2 = 29.84 N·m
2.Gears & gear trains
Answer: ND = 50
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 = 960·25/60 = 400 rpm
Teeth on gear D
ND = C·nC/nout = 24·400/192 = 50
Answer
ND = 50
3.Torsion of shafts
Answer: φ = 0.4789 °
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) = 100·1000·2000/(39000·613592) = 0.008358 rad
φ = 0.008358·180/π = 0.4789°
Answer
φ = 0.4789°
4.Torsion of shafts
Answer: τmax = 21.33 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.
Torque
ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s
T = P/ω = 75000/188.5 = 397.9 N·m
Polar second moment of area
J = π(50⁴ − 35⁴)/32 = 466269 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 397887·25/466269 = 21.33 MPa
Answer
τmax = 21.33 MPa
5.Mohr’s circle & stress transformation
Answer: τabs = 74.54 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = 95 MPa, R = √(45² + 30²) = 54.08 MPa
σ1 = 149.1 MPa, σ2 = 40.92 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(108.2, 149.1, 40.92)/2 = 74.54 MPa
Answer
τabs = 74.54 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
6.Mohr’s circle & stress transformation
Answer: σθ = 112.1 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 = 95 + 15·−0.1736 + 20·0.9848 = 112.1 MPa
Answer
σθ = 112.1 MPa
7.Axial stress & strain
Answer: δ = 0.1126 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(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 110·10³·5890 = 7.108 × 10⁸ N
Shortening
δ = 200000·400/7.108 × 10⁸ = 0.1126 mm
Answer
δ = 0.1126 mm
8.Axial stress & strain
Answer: Pmax = 452.4 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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 100·10³·5027 = 7.54 × 10⁸ N
Governing strain
ε = min(0.0006, 0.001) = 0.0006 (steel)
Largest load
Pmax = 0.0006·7.54 × 10⁸ = 452389 N = 452.4 kN
Answer
Pmax = 452.4 kN
9.Column buckling
Answer: Pcr = 7.91 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 0.7·2000 = 1400 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·7854/1400² = 7910 N = 7.91 kN
Answer
Pcr = 7.91 kN
10.Column buckling
Answer: d = 77.72 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 = 2.5·55000 = 137500 N
KL = 1·3000 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·137500·3000²/(π³·70000))^(1/4) = 77.72 mm
Answer
d = 77.72 mm
Check: the slenderness KL/r is 154.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.02001
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 = (1/5)·ln(15/8) = 0.1257
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1257/√(39.48 + 0.01581) = 0.02001
Answer
ζ = 0.02001
12.Mechanical vibrations
Answer: keq = 70 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 = 10 + 60 = 70 kN/m
Answer
keq = 70 kN/m
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