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 1450 rpm drives a pinion with 25 teeth, which meshes with a gear with 50 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 1450 rpmN1 = 25N2 = 50η = 100%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (48 teeth) on the input shaft turning at 1450 rpm drives gear B (50 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (32 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 48B = 50C = 16D = 32
nout = rpm
3.Torsion of shaftsMedium
A stepped solid brass shaft is fixed at A. Segment AB has diameter 30 mm and length 0.5 m; segment BC has diameter 25 mm and length 0.5 m. A torque of 250 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.
T = 250 N·md1 = 30 mm, L1 = 0.5 md2 = 25 mm, L2 = 0.5 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 1000 N·m acts at B, 2.5 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 3 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 1000 N·mTC = 100 N·mAB = 2.5 mBC = 3 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 110 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 130 MPa?
σx = 110 MPaσy = 100 MPaσ1 = 130 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial tensile force of 200 kN together with a torque of 2000 N·m. At the surface of the shaft, find the maximum in-plane shear stress τmax.
d = 50 mmF = 200 kNT = 2000 N·m
τmax = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 105 kNd = 20 mmL = 2 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 30 kNsegment 1: Ø35 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, brass (E = 100 GPa)
σmax = MPa
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 = 5 mE = 70 GPa (aluminium)d = 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 = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 5 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 40 N·s/m. Find the damping ratio ζ.
m = 5 kgk = 40 kN/mc = 40 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 40 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 50 kgk1 = 10 kN/mk2 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 39.51 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 = 3000/151.8 = 19.76 N·m
Output torque
T2 = 19.76·50/25·1 = 39.51 N·m
Answer
T2 = 39.51 N·m
2.Gears & gear trains
Answer: nout = 696 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 = 1450·48/50 = 1392 rpm = nC
Second stage
nout = 1392·16/32 = 696 rpm
Answer
nout = 696 rpm
total ratio = 2.083 : 1
3.Torsion of shafts
Answer: φC = 7.098 °
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 moments
J1 = π·30⁴/32 = 79522 mm⁴
J2 = π·25⁴/32 = 38350 mm⁴
Twist of each segment
T = 250 N·m = 250000 N·mm
G = 39 GPa = 39000 MPa
φAB = 250000·500/(39000·79522) = 0.04031 rad
φBC = 250000·500/(39000·38350) = 0.08358 rad
Total twist
φC = 0.04031 + 0.08358 = 0.1239 rad = 7.098°
Answer
φC = 7.098°
4.Torsion of shafts
Answer: τmax = 44.82 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 = 100 N·m
TAB = TC + TB = 100 + 1000 = 1100 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 1100 N·m
τmax = 16·1.1 × 10⁶/(π·50³) = 44.82 MPa
Answer
τmax = 44.82 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 24.49 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 = (110 + 100)/2 = 105 MPa
R = σ1 − C = 130 − 105 = 25 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(25² − 5²) = 24.49 MPa
Answer
τxy = 24.49 MPa
6.Mohr’s circle & stress transformation
Answer: τmax = 96.09 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·50²/4) = 200000/1963 = 101.9 MPa
τ = 16T/(πd³) = 16·2000000/(π·50³) = 81.49 MPa
Mohr’s circle
C = σ/2 = 50.93 MPa
R = √(50.93² + 81.49²) = 96.09 MPa
Answer
τmax = 96.09 MPa
7.Axial stress & strain
Answer: δ = 5.713 mm
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·20²/4 = 314.2 mm²
σ = P/A = 105·1000/314.2 = 334.2 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 334.2/117000 = 0.002857 = 2857 µε
Elongation
δ = ε·L = 0.002857·2000 mm = 5.713 mm
Answer
δ = 5.713 mm
8.Axial stress & strain
Answer: σmax = 95.49 MPa
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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 31.18 MPa
σ2 = 95.49 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 95.49 MPa
Answer
σmax = 95.49 MPa
9.Column buckling
Answer: Pcr = 13.89 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/2500² = 13891 N = 13.89 kN
Answer
Pcr = 13.89 kN
10.Column buckling
Answer: σcr = 44.22 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 1·2500 = 2500 mm
KL/r = 2500/20 = 125
Critical stress
σcr = π²E/(KL/r)² = π²·70000/125² = 44.22 MPa
Answer
σcr = 44.22 MPa
11.Mechanical vibrations
Answer: ζ = 0.04472
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.
Critical damping
cc = 2·√(k·m) = 2·√(40000·5) = 894.4 N·s/m
Damping ratio
ζ = c/cc = 40/894.4 = 0.04472
Answer
ζ = 0.04472
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.013 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/10 + 1/40 → keq = 10·40/(10 + 40) = 8 kN/m
Natural frequency
ωn = √(8000/50) = 12.65 rad/s
fn = 12.65/(2π) = 2.013 Hz
Answer
fn = 2.013 Hz
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