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 5.5 kW motor running at 960 rpm drives a pinion with 32 teeth, which meshes with a gear with 48 teeth. Ignore losses. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 960 rpmN1 = 32N2 = 48η = 100%
T2 = N·m
2.Gears & gear trainsHard
An open belt connects a 200 mm pulley and a 500 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 200 mmd2 = 500 mmC = 1000 mm
θ = °
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 2000 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.
T = 2000 N·mdi/do = 0.7τallow = 40 MPa
do = mm
4.Torsion of shaftsHard
A hollow shaft transmits 50 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 50 kWN = 750 rpmdo = 50 mmdi = 35 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −70 MPa, σᵧ = 100 MPa and τₓᵧ = 50 MPa (tension positive). The element is rotated 20° counter-clockwise. Find the normal stress σy′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.
σx = −70 MPaσy = 100 MPaτxy = 50 MPaθ = 20°
σy′ = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 60 mm carries a bending moment of 750 N·m together with a torque of 500 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 60 mmM = 750 N·mT = 500 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 500 mm, E = 70 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø50 mm, E = 100 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 100 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 100 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 4 mm. Find the damping ratio ζ.
x0 = 50 mmx5 = 4 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 60 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 30 kN/mk2 = 60 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 82.06 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 = 5500/100.5 = 54.71 N·m
Output torque
T2 = 54.71·48/32·1 = 82.06 N·m
Answer
T2 = 82.06 N·m
2.Gears & gear trains
Answer: θ = 162.7 °
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt angle
asin(0.15) = 8.627°
Angle of wrap
θ = 180 − 2·8.627 = 162.7°
Answer
θ = 162.7°
3.Torsion of shafts
Answer: do = 69.46 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.
Outer diameter from the allowable stress
T = 2000 N·m = 2 × 10⁶ N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·40·0.7599)) = 69.46 mm
di = 0.7·69.46 = 48.62 mm
Answer
do = 69.46 mm
A solid shaft would need d = 63.38 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 34.13 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 = 50000/78.54 = 636.6 N·m
Polar second moment of area
J = π(50⁴ − 35⁴)/32 = 466269 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 636620·25/466269 = 34.13 MPa
Answer
τmax = 34.13 MPa
5.Mohr’s circle & stress transformation
Answer: σy′ = 47.97 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′ = 15 − (−85)·0.766 − 50·0.6428 = 47.97 MPa
Check
The sum of the normal stresses does not change with rotation.
σx′ = −17.97 MPa, σx′ + σy′ = 30 = σx + σy
Answer
σy′ = 47.97 MPa
6.Mohr’s circle & stress transformation
Answer: σ1 = 38.94 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) = 32·750000/(π·60³) = 35.37 MPa
τ = 16T/(πd³) = 16·500000/(π·60³) = 11.79 MPa
Mohr’s circle
C = σ/2 = 17.68 MPa
R = √(17.68² + 11.79²) = 21.25 MPa
σ1 = C + R = 38.94 MPa
σ2 = C − R = −3.569 MPa
Answer
σ1 = 38.94 MPa
7.Axial stress & strain
Answer: P = 35.83 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 = 490.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·2827) = 2.526 × 10⁻⁶ mm/N
Load
P = 0.2/5.582 × 10⁻⁶ = 35829 N = 35.83 kN
Answer
P = 35.83 kN
8.Axial stress & strain
Answer: Pmax = 521.5 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·20²/4 = 314.2 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 100·10³·5890 = 6.519 × 10⁸ N
Governing strain
ε = min(0.0008, 0.001) = 0.0008 (steel)
Largest load
Pmax = 0.0008·6.519 × 10⁸ = 521504 N = 521.5 kN
Answer
Pmax = 521.5 kN
9.Column buckling
Answer: Pcr = 17.27 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 = 60·50³/12 = 625000 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·625000/5000² = 17272 N = 17.27 kN
Answer
Pcr = 17.27 kN
10.Column buckling
Answer: d = 56.68 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 = 2.5·100000 = 250000 N
KL = 1·2000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·250000·2000²/(π³·200000))^(1/4) = 56.68 mm
Answer
d = 56.68 mm
Check: the slenderness KL/r is 141.1, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.08014
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 = (1/5)·ln(50/4) = 0.5051
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5051/√(39.48 + 0.2552) = 0.08014
Answer
ζ = 0.08014
12.Mechanical vibrations
Answer: keq = 20 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/30 + 1/60 → keq = 30·60/(30 + 60) = 20 kN/m
Answer
keq = 20 kN/m
ExamProblems · examproblems.com/courses/machine-design