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 7.5 kW motor running at 960 rpm drives a pinion with 12 teeth, which meshes with a gear with 18 teeth. The gear pair has an efficiency of 96%. Find the output torque T2 on the gear shaft.
P = 7.5 kWn1 = 960 rpmN1 = 12N2 = 18η = 96%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 3000 rpm drives gear B (80 teeth). Gear C (40 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
nin = 3000 rpmA = 12B = 80C = 40D = 72
nout = rpm
3.Torsion of shaftsMedium
A solid aluminium 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 = 200 N·md = 20 mmL = 0.8 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 30 mm is fixed at A. A torque of 1000 N·m acts at B, 2 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the maximum shear stress τmax in the shaft.
d = 30 mmTB = 1000 N·mTC = 250 N·mAB = 2 mBC = 1.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 10 MPa, σᵧ = 100 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 10 MPaσy = 100 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 0 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = 0 MPaσy = 50 MPaτxy = 30 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 200 mm, E = 120 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, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø60 mm, 200 mm, E = 120 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 200 mm, E = 70 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: aluminium, Ø30 mm, 200 mm, E = 70 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 40 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 3 cycles later by a peak of 1.5 mm. Find the damping ratio ζ.
x0 = 10 mmx3 = 1.5 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 40 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 = 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 = 20 kgk1 = 30 kN/mk2 = 25 kN/mk3 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 107.4 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 7500/100.5 = 74.6 N·m
Output torque
T2 = 74.6·18/12·0.96 = 107.4 N·m
Answer
T2 = 107.4 N·m
2.Gears & gear trains
Answer: nout = 250 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 = 3000·12/80 = 450 rpm = nC
Second stage
nout = 450·40/72 = 250 rpm
Answer
nout = 250 rpm
total ratio = 12 : 1
3.Torsion of shafts
Answer: φ = 22.45 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 200·1000·800/(26000·15708) = 0.3918 rad
φ = 0.3918·180/π = 22.45°
Answer
φ = 22.45°
4.Torsion of shafts
Answer: τmax = 235.8 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 250 N·m
TAB = TC + TB = 250 + 1000 = 1250 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 1250 N·m
τmax = 16·1.25 × 10⁶/(π·30³) = 235.8 MPa
Answer
τmax = 235.8 MPa
5.Mohr’s circle & stress transformation
Answer: θp = 73.15 °
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.
Angle on the circle
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·30, 10 − 100) = 146.3°
Angle on the element
θp = 73.15°
Answer
θp = 73.15°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −13.48 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 + −25·0.5 + 30·−0.866 = −13.48 MPa
Answer
σθ = −13.48 MPa
7.Axial stress & strain
Answer: P = 110.6 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
A₁ = 706.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(120000·2827) = 5.895 × 10⁻⁷ mm/N
Load
P = 0.3/2.712 × 10⁻⁶ = 110639 N = 110.6 kN
Answer
P = 110.6 kN
8.Axial stress & strain
Answer: σ2 = 46.6 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 = 70000·706.9/200 = 247400 N/mm
Share of segment 2
F2 = 80000·247400/600830 = 32941 N
Stress
σ2 = 32941/706.9 = 46.6 MPa (compression)
Answer
σ2 = 46.6 MPa
9.Column buckling
Answer: Pcr = 39.48 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 = 80·30³/12 = 180000 mm⁴
Effective length
KL = 0.5·6000 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·180000/3000² = 39478 N = 39.48 kN
Answer
Pcr = 39.48 kN
10.Column buckling
Answer: σcr = 7.896 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 1·5000 = 5000 mm
KL/r = 5000/10 = 500
Critical stress
σcr = π²E/(KL/r)² = π²·200000/500² = 7.896 MPa
Answer
σcr = 7.896 MPa
11.Mechanical vibrations
Answer: ζ = 0.1001
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 = (1/3)·ln(10/1.5) = 0.6324
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6324/√(39.48 + 0.3999) = 0.1001
Answer
ζ = 0.1001
12.Mechanical vibrations
Answer: fn = 5.416 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 + 25 = 55 kN/m
keq = 55·40/(55 + 40) = 23.16 kN/m
Natural frequency
ωn = √(23158/20) = 34.03 rad/s
fn = 34.03/(2π) = 5.416 Hz
Answer
fn = 5.416 Hz
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