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 1440 rpm and drives a worm wheel with 45 teeth. Find the speed nw of the wheel.
z = 4Nw = 45n = 1440 rpm
nw = rpm
2.Gears & gear trainsHard
An 11 kW motor at 1440 rpm drives gear A (20 teeth), which meshes with gear B (40 teeth). Gear C (18 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 11 kWnin = 1440 rpmA = 20B = 40C = 18D = 80ηmesh = 98%
Tout = N·m
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 400 mm rotate 2.88° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 50 N·md = 15 mmL = 400 mmφ = 2.88°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 75 kW at 1450 rpm. Find the maximum shear stress τmax in the shaft.
P = 75 kWN = 1450 rpmdo = 60 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 110 MPa, σᵧ = 90 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 110 MPaσy = 90 MPaτxy = 10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 160 MPaσy = −60 MPaτxy = 30 MPaθ = 50°
τθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 60 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.
P = 250 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 70 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 500 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: copper, Ø40 mm, 500 mm, E = 120 GPaP = 80 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 = 1.5 mE = 200 GPa (steel)b × h = 25 × 100 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 = 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 50 mm is followed one cycle later by a peak of 26.6 mm. Find the damping ratio ζ.
x0 = 50 mmx1 = 26.6 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 20 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 = 20 kgk1 = 25 kN/mk2 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: nw = 128 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 = 1440·4/45 = 128 rpm
Answer
nw = 128 rpm
2.Gears & gear trains
Answer: Tout = 622.7 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.
Output speed
nout = 1440·(20/40)·(18/80) = 162 rpm
ωout = 2π·162/60 = 16.96 rad/s
Output power and torque
Pout = 11000·0.98² = 10564 W
Tout = Pout/ωout = 10564/16.96 = 622.7 N·m
Answer
Tout = 622.7 N·m
3.Torsion of shafts
Answer: G = 80.06 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·15⁴/32 = 4970 mm⁴
φ = 2.88·π/180 = 0.05027 rad
Shear modulus
T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·400/(4970·0.05027) = 80056 MPa
G = 80.06 GPa
Answer
G = 80.06 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 12.42 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·1450/60 = 151.8 rad/s
T = P/ω = 75000/151.8 = 493.9 N·m
Polar second moment of area
J = π(60⁴ − 30⁴)/32 = 1.193 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 493929·30/1.193 × 10⁶ = 12.42 MPa
Answer
τmax = 12.42 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 14.14 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 = √(((110 − 90)/2)² + 10²) = 14.14 MPa
Answer
τmax = 14.14 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −113.5 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 = −110·0.9848 + 30·−0.1736 = −113.5 MPa
Answer
τθ = −113.5 MPa
7.Axial stress & strain
Answer: δ = 0.5079 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·30²/4 = 706.9 mm²
At = π(80² − 60²)/4 = 2199 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 70·10³·2199 = 2.953 × 10⁸ N
Shortening
δ = 250000·600/2.953 × 10⁸ = 0.5079 mm
Answer
δ = 0.5079 mm
8.Axial stress & strain
Answer: σ2 = 29.31 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 = 200000·706.9/400 = 353429 N/mm
k₂ = 120000·1257/500 = 301593 N/mm
Share of segment 2
F2 = 80000·301593/655022 = 36835 N
Stress
σ2 = 36835/1257 = 29.31 MPa (compression)
Answer
σ2 = 29.31 MPa
9.Column buckling
Answer: Pcr = 114.2 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 1·1500 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·130208/1500² = 114232 N = 114.2 kN
Answer
Pcr = 114.2 kN
10.Column buckling
Answer: σcr = 9.064 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 = 0.7·5000 = 3500 mm
KL/r = 3500/7.5 = 466.7
Critical stress
σcr = π²E/(KL/r)² = π²·200000/466.7² = 9.064 MPa
Answer
σcr = 9.064 MPa
11.Mechanical vibrations
Answer: ζ = 0.09994
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 = (1/1)·ln(50/26.6) = 0.6311
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6311/√(39.48 + 0.3983) = 0.09994
Answer
ζ = 0.09994
12.Mechanical vibrations
Answer: fn = 7.549 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 = 25 + 20 = 45 kN/m
Natural frequency
ωn = √(45000/20) = 47.43 rad/s
fn = 47.43/(2π) = 7.549 Hz
Answer
fn = 7.549 Hz
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