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 45 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 = 45N2 = 72η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (45 teeth) on the input shaft turns at 1450 rpm and drives gear B (50 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 652.5 rpm. How many teeth ND must gear D have?
nin = 1450 rpmA = 45B = 50C = 16nout = 652.5 rpm
ND =
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 = 50 mmL = 0.8 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 75 kWN = 750 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 80 MPa, σᵧ = 20 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 80 MPaσy = 20 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 130 MPa, σᵧ = 40 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = 130 MPaσy = 40 MPaτxy = 30 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).
P = 25 kNd = 16 mmL = 2 mE = 117 GPa
ε = µε
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 = 70 kNsegment 1: Ø35 mm, 1 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, brass (E = 100 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 145 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 145 kNn = 3L = 2.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 10 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
m = 10 kgk = 100 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 80 kN/m and the lower one k₂ = 12 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 80 kN/mk2 = 12 kN/m
fn = Hz
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: ND = 32
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 = 1450·45/50 = 1305 rpm
Teeth on gear D
ND = C·nC/nout = 16·1305/652.5 = 32
Answer
ND = 32
3.Torsion of shafts
Answer: φ = 0.7662 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·800/(39000·613592) = 0.01337 rad
φ = 0.01337·180/π = 0.7662°
Answer
φ = 0.7662°
4.Torsion of shafts
Answer: d = 40.18 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·750/60 = 78.54 rad/s
T = P/ω = 75000/78.54 = 954.9 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·954930/(π·75)) = 40.18 mm
Answer
d = 40.18 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −76.72 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −60/67.08 = −0.8944, sin 2θs = −30/67.08 = −0.4472
2θs = atan2(−30, −60) = −153.4°
Angle on the element
θs = −76.72°
check: θs − θp = −76.72 − (−31.72) = −45°
Answer
θs = −76.72°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 88.48 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 = 85 + 45·−0.5 + 30·0.866 = 88.48 MPa
Answer
σθ = 88.48 MPa
7.Axial stress & strain
Answer: ε = 1063 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·16²/4 = 201.1 mm²
σ = P/A = 25·1000/201.1 = 124.3 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 124.3/117000 = 0.001063 = 1063 µε
Answer
ε = 1063 µε
8.Axial stress & strain
Answer: σmax = 348.2 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·16²/4 = 201.1 mm²
σ1 = 72.76 MPa
σ2 = 348.2 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 348.2 MPa
Answer
σmax = 348.2 MPa
9.Column buckling
Answer: Pcr = 2.768 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 0.7·2000 = 1400 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·7854/1400² = 2768 N = 2.768 kN
Answer
Pcr = 2.768 kN
10.Column buckling
Answer: d = 94.62 mm
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.
Required critical load
Pcr = n·P = 3·145000 = 435000 N
KL = 1·2500 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·435000·2500²/(π³·70000))^(1/4) = 94.62 mm
Answer
d = 94.62 mm
Check: the slenderness KL/r is 105.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 14.21 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.
Undamped natural frequency
ωn = √(100000/10) = 100 rad/s
fn = 100/(2π) = 15.92 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 15.92·√(1 − 0.45²) = 14.21 Hz
Answer
fd = 14.21 Hz
12.Mechanical vibrations
Answer: fn = 2.571 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 = 1/80 + 1/12 → keq = 80·12/(80 + 12) = 10.43 kN/m
Natural frequency
ωn = √(10435/40) = 16.15 rad/s
fn = 16.15/(2π) = 2.571 Hz
Answer
fn = 2.571 Hz
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