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
An 11 kW motor running at 1450 rpm drives a pinion with 50 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 11 kWn1 = 1450 rpmN1 = 50N2 = 60η = 98%
T1 = N·m
2.Gears & gear trainsHard
A 1.1 kW motor at 1440 rpm drives gear A (30 teeth), which meshes with gear B (48 teeth). Gear C (36 teeth) on the same shaft as B drives gear D (48 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
P = 1.1 kWnin = 1440 rpmA = 30B = 48C = 36D = 48ηmesh = 97%
Tout = N·m
3.Torsion of shaftsMedium
A stepped solid steel shaft is fixed at A. Segment AB has diameter 30 mm and length 2.5 m; segment BC has diameter 25 mm and length 1.5 m. A torque of 500 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 500 N·md1 = 30 mm, L1 = 2.5 md2 = 25 mm, L2 = 1.5 m
τmax = MPa
4.Torsion of shaftsHard
A hollow shaft transmits 30 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 30 kWN = 750 rpmdo = 60 mmdi = 36 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 110 MPa, σᵧ = −70 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 110 MPaσy = −70 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 140 MPa, σᵧ = 50 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.
σx = 140 MPaσy = 50 MPaτxy = −30 MPaθ = -20°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 75 kNd = 16 mmL = 1.5 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of copper (outer diameter 80 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 250 MPatube: Ø80/Ø50 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
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 = 5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 40 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 40 kNn = 3L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed 2 cycles later by a peak of 5.7 mm. Find the damping ratio ζ.
x0 = 20 mmx2 = 5.7 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 50 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 25 kgk1 = 25 kN/mk2 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 72.44 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 = 11000/151.8 = 72.44 N·m
Answer
T1 = 72.44 N·m
2.Gears & gear trains
Answer: Tout = 14.64 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.
Output speed
nout = 1440·(30/48)·(36/48) = 675 rpm
ωout = 2π·675/60 = 70.69 rad/s
Output power and torque
Pout = 1100·0.97² = 1035 W
Tout = Pout/ωout = 1035/70.69 = 14.64 N·m
Answer
Tout = 14.64 N·m
3.Torsion of shafts
Answer: τmax = 163 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.
Stress in each segment
T = 500 N·m = 500000 N·mm
τAB = 16·500000/(π·30³) = 94.31 MPa
τBC = 16·500000/(π·25³) = 163 MPa
Answer
τmax = 163 MPa
4.Torsion of shafts
Answer: τmax = 10.35 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 = 30000/78.54 = 382 N·m
Polar second moment of area
J = π(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 381972·30/1.107 × 10⁶ = 10.35 MPa
Answer
τmax = 10.35 MPa
5.Mohr’s circle & stress transformation
Answer: θp = −16.85 °
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·−60, 110 − −70) = −33.69°
Angle on the element
θp = −16.85°
Answer
θp = −16.85°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 148.8 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 = 95 + 45·0.766 + −30·−0.6428 = 148.8 MPa
Answer
σθ = 148.8 MPa
7.Axial stress & strain
Answer: δ = 2.798 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·16²/4 = 201.1 mm²
σ = P/A = 75·1000/201.1 = 373 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 373/200000 = 0.001865 = 1865 µε
Elongation
δ = ε·L = 0.001865·1500 mm = 2.798 mm
Answer
δ = 2.798 mm
8.Axial stress & strain
Answer: Pmax = 309.4 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
Ar = π·40²/4 = 1257 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 120·10³·3063 = 6.189 × 10⁸ N
Governing strain
ε = min(0.00125, 0.0005) = 0.0005 (copper)
Largest load
Pmax = 0.0005·6.189 × 10⁸ = 309447 N = 309.4 kN
Answer
Pmax = 309.4 kN
9.Column buckling
Answer: Pcr = 35.88 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/3500² = 35879 N = 35.88 kN
Answer
Pcr = 35.88 kN
10.Column buckling
Answer: d = 61.34 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 = 3·40000 = 120000 N
KL = 1·2000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·2000²/(π³·70000))^(1/4) = 61.34 mm
Answer
d = 61.34 mm
Check: the slenderness KL/r is 130.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.0994
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 = (1/2)·ln(20/5.7) = 0.6276
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6276/√(39.48 + 0.3939) = 0.0994
Answer
ζ = 0.0994
12.Mechanical vibrations
Answer: fn = 4.109 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 = 1/25 + 1/50 → keq = 25·50/(25 + 50) = 16.67 kN/m
Natural frequency
ωn = √(16667/25) = 25.82 rad/s
fn = 25.82/(2π) = 4.109 Hz
Answer
fn = 4.109 Hz
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