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 0.75 kW motor running at 960 rpm drives a pinion with 24 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 0.75 kWn1 = 960 rpmN1 = 24N2 = 32η = 95%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (24 teeth) on the input shaft turning at 1440 rpm drives gear B (72 teeth). Gear C (36 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 = 1440 rpmA = 24B = 72C = 36D = 72
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 = 750 N·md = 50 mmL = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 20 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 1.5 m from A and 2.5 m from B. Find the maximum shear stress τmax in the shaft.
d = 20 mmT0 = 400 N·ma = 1.5 mb = 2.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = −30 MPa and τₓᵧ = 10 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −10 MPaσy = −30 MPaτxy = 10 MPaθ = -30°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 100 MPa, σᵧ = 20 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° clockwise from the x-axis.
σx = 100 MPaσy = 20 MPaτxy = −10 MPaθ = -45°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 35 kNd = 30 mmL = 0.5 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 = 85 kNsegment 1: Ø30 mm, 0.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1 m, copper (E = 117 GPa)
σmax = 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 = 2 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 70 GPa (aluminium)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 500 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 500 kgk = 50 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 100 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 7.46 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 = 750/100.5 = 7.46 N·m
Answer
T1 = 7.46 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 = 1440·24/72 = 480 rpm = nC
Second stage
nout = 480·36/72 = 240 rpm
Answer
nout = 240 rpm
total ratio = 6 : 1
3.Torsion of shafts
Answer: φ = 4.489 °
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) = 750·1000·2500/(39000·613592) = 0.07835 rad
φ = 0.07835·180/π = 4.489°
Answer
φ = 4.489°
4.Torsion of shafts
Answer: τmax = 159.2 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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 400
TA·a = TB·b
TA = 400·2.5/4 = 250 N·m
TB = 400·1.5/4 = 150 N·m
Maximum shear stress
The larger of the two torques governs (same diameter).
τmax = 16·250000/(π·20³) = 159.2 MPa
Answer
τmax = 159.2 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = −16.34 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′ = −20 − 10·0.5 − 10·(−0.866) = −16.34 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −23.66 MPa, σx′ + σy′ = −40 = σx + σy
Answer
σy′ = −16.34 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 70 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 = 60 + 40·0 + −10·−1 = 70 MPa
Answer
σθ = 70 MPa
7.Axial stress & strain
Answer: δ = 0.2116 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·30²/4 = 706.9 mm²
σ = P/A = 35·1000/706.9 = 49.51 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 49.51/117000 = 0.0004232 = 423.2 µε
Elongation
δ = ε·L = 0.0004232·500 mm = 0.2116 mm
Answer
δ = 0.2116 mm
8.Axial stress & strain
Answer: σmax = 751.6 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·30²/4 = 706.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 120.3 MPa
σ2 = 751.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 751.6 MPa
Answer
σmax = 751.6 MPa
9.Column buckling
Answer: Pcr = 37.85 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/1000² = 37849 N = 37.85 kN
Answer
Pcr = 37.85 kN
10.Column buckling
Answer: σcr = 4.318 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.5·5000 = 2500 mm
KL/r = 2500/6.25 = 400
Critical stress
σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa
Answer
σcr = 4.318 MPa
11.Mechanical vibrations
Answer: fd = 1.518 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 = √(50000/500) = 10 rad/s
fn = 10/(2π) = 1.592 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 1.592·√(1 − 0.3²) = 1.518 Hz
Answer
fd = 1.518 Hz
12.Mechanical vibrations
Answer: keq = 200 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 = 100 + 100 = 200 kN/m
Answer
keq = 200 kN/m
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