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 2900 rpm drives a pinion with 12 teeth, which meshes with a gear with 15 teeth. The gear pair has an efficiency of 97%. Find the torque T1 on the motor shaft.
P = 0.75 kWn1 = 2900 rpmN1 = 12N2 = 15η = 97%
T1 = N·m
2.Gears & gear trainsHard
A 15 kW motor at 960 rpm drives gear A (15 teeth), which meshes with gear B (25 teeth). Gear C (32 teeth) on the same shaft as B drives gear D (48 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 15 kWnin = 960 rpmA = 15B = 25C = 32D = 48ηmesh = 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 300 mm rotate 2.16° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 50 N·md = 15 mmL = 300 mmφ = 2.16°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 5 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 5 kWN = 750 rpmdo = 40 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 160 MPa?
σx = −40 MPaσy = 100 MPaσ1 = 160 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 25 mm carries an axial compressive force of 200 kN together with a torque of 100 N·m. At the surface of the shaft, find the maximum in-plane shear stress τmax.
d = 25 mmF = −200 kNT = 100 N·m
τmax = 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 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: brass, Ø60 mm, 500 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 80 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 250 MPatube: Ø80/Ø50 mm, E = 70 GPa, σallow = 60 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 = 2.5 mE = 70 GPa (aluminium)b × h = 25 × 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 4 cycles later by a peak of 0.6 mm. Find the damping ratio ζ.
x0 = 12 mmx4 = 0.6 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 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 = 20 kN/mk2 = 5 kN/mk3 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 2.47 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 = 750/303.7 = 2.47 N·m
Answer
T1 = 2.47 N·m
2.Gears & gear trains
Answer: Tout = 358.2 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·(15/25)·(32/48) = 384 rpm
ωout = 2π·384/60 = 40.21 rad/s
Output power and torque
Pout = 15000·0.98² = 14406 W
Tout = Pout/ωout = 14406/40.21 = 358.2 N·m
Answer
Tout = 358.2 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 = 2.16·π/180 = 0.0377 rad
Shear modulus
T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·300/(4970·0.0377) = 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 = 7.411 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·750/60 = 78.54 rad/s
T = P/ω = 5000/78.54 = 63.66 N·m
Polar second moment of area
J = π(40⁴ − 30⁴)/32 = 171806 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 63662·20/171806 = 7.411 MPa
Answer
τmax = 7.411 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 109.5 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.
Required radius
C = (−40 + 100)/2 = 30 MPa
R = σ1 − C = 160 − 30 = 130 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(130² − 70²) = 109.5 MPa
Answer
τxy = 109.5 MPa
6.Mohr’s circle & stress transformation
Answer: τmax = 206.3 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.
Stresses at the surface
σ = F/A = −200000/(π·25²/4) = −200000/490.9 = −407.4 MPa
τ = 16T/(πd³) = 16·100000/(π·25³) = 32.59 MPa
Mohr’s circle
C = σ/2 = −203.7 MPa
R = √((−203.7)² + 32.59²) = 206.3 MPa
Answer
τmax = 206.3 MPa
7.Axial stress & strain
Answer: P = 62.19 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) = 500/(100000·2827) = 1.768 × 10⁻⁶ mm/N
Load
P = 0.3/4.824 × 10⁻⁶ = 62187 N = 62.19 kN
Answer
P = 62.19 kN
8.Axial stress & strain
Answer: Pmax = 237.6 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·20²/4 = 314.2 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 70·10³·3063 = 2.772 × 10⁸ N
Governing strain
ε = min(0.00125, 0.0008571) = 0.0008571 (aluminium)
Largest load
Pmax = 0.0008571·2.772 × 10⁸ = 237639 N = 237.6 kN
Answer
Pmax = 237.6 kN
9.Column buckling
Answer: Pcr = 23.03 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 = 40·25³/12 = 52083 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·52083/1250² = 23029 N = 23.03 kN
Answer
Pcr = 23.03 kN
10.Column buckling
Answer: σcr = 17.27 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 1·2500 = 2500 mm
KL/r = 2500/12.5 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.1184
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 = (1/4)·ln(12/0.6) = 0.7489
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7489/√(39.48 + 0.5609) = 0.1184
Answer
ζ = 0.1184
12.Mechanical vibrations
Answer: fn = 3.106 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 = 20 + 5 = 25 kN/m
keq = 25·80/(25 + 80) = 19.05 kN/m
Natural frequency
ωn = √(19048/50) = 19.52 rad/s
fn = 19.52/(2π) = 3.106 Hz
Answer
fn = 3.106 Hz
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