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 1.1 kW motor running at 1450 rpm drives a pinion with 30 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 1.1 kWn1 = 1450 rpmN1 = 30N2 = 72η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (30 teeth) on the input shaft turning at 960 rpm drives gear B (50 teeth). Gear C (15 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 = 960 rpmA = 30B = 50C = 15D = 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 = 400 N·md = 40 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 750 rpmdo = 75 mmdi = 56 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −80 MPa, σᵧ = −60 MPa and τₓᵧ = −50 MPa (tension positive). The element is rotated 45° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −80 MPaσy = −60 MPaτxy = −50 MPaθ = -45°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −30 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −30 MPaσy = 50 MPaτxy = 30 MPaθ = -45°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 35 kNd = 25 mmL = 3 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 70 kNsegment 1: Ø25 mm, 2 m, steel (E = 200 GPa)segment 2: Ø20 mm, 1 m, aluminium (E = 70 GPa)
δ = mm
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 = 30 × 100 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 = 5 mE = 70 GPa (aluminium)d = 40 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 5 cycles later by a peak of 2.1 mm. Find the damping ratio ζ.
x0 = 50 mmx5 = 2.1 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 60 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 7.244 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 1100/151.8 = 7.244 N·m
Answer
T1 = 7.244 N·m
2.Gears & gear trains
Answer: nout = 240 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 = 960·30/50 = 576 rpm = nC
Second stage
nout = 576·15/36 = 240 rpm
Answer
nout = 240 rpm
total ratio = 4 : 1
3.Torsion of shafts
Answer: φ = 2.806 °
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) = 400·1000·1200/(39000·251327) = 0.04897 rad
φ = 0.04897·180/π = 2.806°
Answer
φ = 2.806°
4.Torsion of shafts
Answer: τmax = 22.3 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 = 100000/78.54 = 1273 N·m
Polar second moment of area
J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 1.273 × 10⁶·37.5/2.141 × 10⁶ = 22.3 MPa
Answer
τmax = 22.3 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = −120 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 for the y′ face
σy′ = −70 − (−10)·0 − (−50)·(−1) = −120 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −20 MPa, σx′ + σy′ = −140 = σx + σy
Answer
σy′ = −120 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −40 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 = −−40·−1 + 30·0 = −40 MPa
Answer
τθ = −40 MPa
7.Axial stress & strain
Answer: ε = 1019 µε
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 35·1000/490.9 = 71.3 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 71.3/70000 = 0.001019 = 1019 µε
Answer
ε = 1019 µε
8.Axial stress & strain
Answer: δ = 4.609 mm
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
Same force in both segments
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
A1 = π·25²/4 = 490.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 142.6 MPa
σ2 = 222.8 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 142.6·2000/200000 = 1.426 mm
δ2 = σ2L2/E2 = 222.8·1000/70000 = 3.183 mm
δ = δ1 + δ2 = 4.609 mm
Answer
δ = 4.609 mm
9.Column buckling
Answer: Pcr = 99.49 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 = 100·30³/12 = 225000 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·225000/1250² = 99486 N = 99.49 kN
Answer
Pcr = 99.49 kN
10.Column buckling
Answer: σcr = 2.763 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 = π²·70000/500² = 2.763 MPa
Answer
σcr = 2.763 MPa
11.Mechanical vibrations
Answer: ζ = 0.1004
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(50/2.1) = 0.634
Damping ratio
ζ = δ/√(4π² + δ²) = 0.634/√(39.48 + 0.402) = 0.1004
Answer
ζ = 0.1004
12.Mechanical vibrations
Answer: keq = 140 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 = 60 + 80 = 140 kN/m
Answer
keq = 140 kN/m
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