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 2.2 kW motor running at 1500 rpm drives a pinion with 16 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 1500 rpmN1 = 16N2 = 36η = 97%
T2 = N·m
2.Gears & gear trainsHard
A 1.1 kW motor at 1440 rpm drives gear A (15 teeth), which meshes with gear B (80 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (80 teeth) on the output shaft. Each gear mesh has an efficiency of 97%. Find the output torque Tout.
P = 1.1 kWnin = 1440 rpmA = 15B = 80C = 16D = 80ηmesh = 97%
Tout = N·m
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 = 250 N·md = 25 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 1000 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 5 kWN = 1000 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −20 MPa and σᵧ = 50 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 65 MPa?
σx = −20 MPaσy = 50 MPaσ1 = 65 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 30 mm carries a bending moment of 100 N·m together with a torque of 100 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 30 mmM = 100 N·mT = 100 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 200 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: copper, Ø50 mm, 200 mm, E = 120 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 50 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø120/Ø50 mm, E = 110 GPa, σallow = 100 MPa
Pmax = kN
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 = 4 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 6 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 3 cycles later by a peak of 3.9 mm. Find the damping ratio ζ.
x0 = 10 mmx3 = 3.9 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 15 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 25 kgk1 = 20 kN/mk2 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 30.57 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 2200/157.1 = 14.01 N·m
Output torque
T2 = 14.01·36/16·0.97 = 30.57 N·m
Answer
T2 = 30.57 N·m
2.Gears & gear trains
Answer: Tout = 183 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·(15/80)·(16/80) = 54 rpm
ωout = 2π·54/60 = 5.655 rad/s
Output power and torque
Pout = 1100·0.97² = 1035 W
Tout = Pout/ωout = 1035/5.655 = 183 N·m
Answer
Tout = 183 N·m
3.Torsion of shafts
Answer: φ = 2.334 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·500/(80000·38350) = 0.04074 rad
φ = 0.04074·180/π = 2.334°
Answer
φ = 2.334°
4.Torsion of shafts
Answer: d = 18.25 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·1000/60 = 104.7 rad/s
T = P/ω = 5000/104.7 = 47.75 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·47746/(π·40)) = 18.25 mm
Answer
d = 18.25 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τxy = 35.71 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 = (−20 + 50)/2 = 15 MPa
R = σ1 − C = 65 − 15 = 50 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 35²) = 35.71 MPa
Answer
τxy = 35.71 MPa
6.Mohr’s circle & stress transformation
Answer: σ1 = 45.54 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 = 32M/(πd³) = 32·100000/(π·30³) = 37.73 MPa
τ = 16T/(πd³) = 16·100000/(π·30³) = 18.86 MPa
Mohr’s circle
C = σ/2 = 18.86 MPa
R = √(18.86² + 18.86²) = 26.68 MPa
σ1 = C + R = 45.54 MPa
σ2 = C − R = −7.813 MPa
Answer
σ1 = 45.54 MPa
7.Axial stress & strain
Answer: P = 88.08 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
A₁ = 1257 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(120000·1963) = 8.488 × 10⁻⁷ mm/N
Load
P = 0.25/2.838 × 10⁻⁶ = 88082 N = 88.08 kN
Answer
P = 88.08 kN
8.Axial stress & strain
Answer: Pmax = 675.8 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·25²/4 = 490.9 mm²
At = π(120² − 50²)/4 = 9346 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·9346 = 1.126 × 10⁹ N
Governing strain
ε = min(0.0006, 0.0009091) = 0.0006 (steel)
Largest load
Pmax = 0.0006·1.126 × 10⁹ = 675757 N = 675.8 kN
Answer
Pmax = 675.8 kN
9.Column buckling
Answer: Pcr = 313.9 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 = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN
Answer
Pcr = 313.9 kN
10.Column buckling
Answer: σcr = 34.27 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.5·6000 = 3000 mm
KL/r = 3000/12.5 = 240
Critical stress
σcr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa
Answer
σcr = 34.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.04989
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(10/3.9) = 0.3139
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3139/√(39.48 + 0.09851) = 0.04989
Answer
ζ = 0.04989
12.Mechanical vibrations
Answer: fn = 5.955 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 20 + 15 = 35 kN/m
Natural frequency
ωn = √(35000/25) = 37.42 rad/s
fn = 37.42/(2π) = 5.955 Hz
Answer
fn = 5.955 Hz
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