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 1.5 kW motor running at 960 rpm drives a pinion with 16 teeth, which meshes with a gear with 20 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 1.5 kWn1 = 960 rpmN1 = 16N2 = 20η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turns at 1440 rpm and drives gear B (80 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 67.5 rpm. How many teeth ND must gear D have?
nin = 1440 rpmA = 15B = 80C = 18nout = 67.5 rpm
ND =
3.Torsion of shaftsMedium
A solid steel 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 = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 75 kWN = 500 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 0 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 = 0 MPaσy = 100 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −30 MPa, σᵧ = 30 MPa and τₓᵧ = 40 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 = 30 MPaτxy = 40 MPaθ = -45°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 35 kNd = 10 mmL = 1.5 mE = 200 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 = 85 kNsegment 1: Ø35 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 2 m, aluminium (E = 70 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)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 3 cycles later by a peak of 3.1 mm. Find the damping ratio ζ.
x0 = 30 mmx3 = 3.1 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 30 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 15 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 = 50 kN/mk2 = 15 kN/mk3 = 30 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 14.92 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 = 1500/100.5 = 14.92 N·m
Answer
T1 = 14.92 N·m
2.Gears & gear trains
Answer: ND = 72
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.
Intermediate shaft speed
nB = nC = 1440·15/80 = 270 rpm
Teeth on gear D
ND = C·nC/nout = 18·270/67.5 = 72
Answer
ND = 72
3.Torsion of shafts
Answer: φ = 0.5699 °
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) = 400·1000·500/(80000·251327) = 0.009947 rad
φ = 0.009947·180/π = 0.5699°
Answer
φ = 0.5699°
4.Torsion of shafts
Answer: d = 56.71 mm
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·500/60 = 52.36 rad/s
T = P/ω = 75000/52.36 = 1432 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·1.432 × 10⁶/(π·40)) = 56.71 mm
Answer
d = 56.71 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = 74.52 °
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, 0 − 100) = 149°
Angle on the element
θp = 74.52°
Answer
θp = 74.52°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = −30 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 = −−30·−1 + 40·0 = −30 MPa
Answer
τθ = −30 MPa
7.Axial stress & strain
Answer: δ = 3.342 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·10²/4 = 78.54 mm²
σ = P/A = 35·1000/78.54 = 445.6 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 445.6/200000 = 0.002228 = 2228 µε
Elongation
δ = ε·L = 0.002228·1500 mm = 3.342 mm
Answer
δ = 3.342 mm
8.Axial stress & strain
Answer: σmax = 270.6 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 = 88.35 MPa
σ2 = 270.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 270.6 MPa
Answer
σmax = 270.6 MPa
9.Column buckling
Answer: Pcr = 4.422 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 = 60·20³/12 = 40000 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·40000/2500² = 4422 N = 4.422 kN
Answer
Pcr = 4.422 kN
10.Column buckling
Answer: σcr = 35.25 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 = 0.7·1500 = 1050 mm
KL/r = 1050/7.5 = 140
Critical stress
σcr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa
Answer
σcr = 35.25 MPa
11.Mechanical vibrations
Answer: ζ = 0.1196
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(30/3.1) = 0.7566
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7566/√(39.48 + 0.5724) = 0.1196
Answer
ζ = 0.1196
12.Mechanical vibrations
Answer: fn = 5.099 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 = 50 + 15 = 65 kN/m
keq = 65·30/(65 + 30) = 20.53 kN/m
Natural frequency
ωn = √(20526/20) = 32.04 rad/s
fn = 32.04/(2π) = 5.099 Hz
Answer
fn = 5.099 Hz
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