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 15 kW motor running at 960 rpm drives a pinion with 25 teeth, which meshes with a gear with 36 teeth. The gear pair has an efficiency of 95%. Find the torque T1 on the motor shaft.
P = 15 kWn1 = 960 rpmN1 = 25N2 = 36η = 95%
T1 = N·m
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 100 mmn1 = 720 rpmd2 = 250 mmC = 500 mmμ = 0.35T1 = 1500 N
P = kW
3.Torsion of shaftsMedium
A solid aluminium shaft 3 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 1500 N·mL = 3 mG = 26 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 10 kWN = 3000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 80 MPa, σᵧ = −20 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 80 MPaσy = −20 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 100 MPa and τₓᵧ = −10 MPa (σz = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −80 MPaσy = 100 MPaτxy = −10 MPaσY = 275 MPa
n =
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 40 kNsegment 1: Ø30 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)
δ = mm
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 = 4 mE = 70 GPa (aluminium)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 6 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 500 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 6930 N·s/m. Find the damping ratio ζ.
m = 500 kgk = 150 kN/mc = 6930 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 25 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 100 kN/mk2 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 149.2 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 = 15000/100.5 = 149.2 N·m
Answer
T1 = 149.2 N·m
2.Gears & gear trains
Answer: P = 3.562 kW
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.
Angle of wrap
θ = 2.84 rad
Slack-side tension
eμθ = e0.9942 = 2.702, T₂ = 1500/2.702 = 555.1 N
Power
P = (1500 − 555.1)·3.77 = 3.562 kW
Answer
P = 3.562 kW
3.Torsion of shafts
Answer: d = 100.3 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.
Required polar moment
T = 1500 N·m = 1.5 × 10⁶ N·mm
L = 3 m = 3000 mm
G = 26 GPa = 26000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 1.5 × 10⁶·3000/(26000·0.01745) = 9.917 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 100.3 mm
Answer
d = 100.3 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 12.93 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·3000/60 = 314.2 rad/s
T = P/ω = 10000/314.2 = 31.83 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·31831/(π·75)) = 12.93 mm
Answer
d = 12.93 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 53.85 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.
Radius of Mohr’s circle
τmax = R = √(((80 − −20)/2)² + −20²) = 53.85 MPa
Answer
τmax = 53.85 MPa
6.Mohr’s circle & stress transformation
Answer: n = 1.75
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 10 MPa, R = 90.55 MPa
σ1 = 100.6 MPa, σ2 = −80.55 MPa, σ3 = 0
Von Mises
σvM = √(100.6² − 100.6·(−80.55) + (−80.55)²) = 157.2 MPa
n = 275/157.2 = 1.75
Answer
n = 1.75
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: P = 113.1 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N
Load
P = 0.3/2.653 × 10⁻⁶ = 113097 N = 113.1 kN
Answer
P = 113.1 kN
8.Axial stress & strain
Answer: δ = 5.861 mm
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 = 56.59 MPa
σ2 = 353.7 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 56.59·1000/70000 = 0.8084 mm
δ2 = σ2L2/E2 = 353.7·1000/70000 = 5.053 mm
δ = δ1 + δ2 = 5.861 mm
Answer
δ = 5.861 mm
9.Column buckling
Answer: Pcr = 7.931 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 = 40·30³/12 = 90000 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·90000/2800² = 7931 N = 7.931 kN
Answer
Pcr = 7.931 kN
10.Column buckling
Answer: σcr = 25.18 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.7·6000 = 4200 mm
KL/r = 4200/15 = 280
Critical stress
σcr = π²E/(KL/r)² = π²·200000/280² = 25.18 MPa
Answer
σcr = 25.18 MPa
11.Mechanical vibrations
Answer: ζ = 0.4001
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.
Critical damping
cc = 2·√(k·m) = 2·√(150000·500) = 17321 N·s/m
Damping ratio
ζ = c/cc = 6930/17321 = 0.4001
Answer
ζ = 0.4001
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 7.118 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 = 1/100 + 1/25 → keq = 100·25/(100 + 25) = 20 kN/m
Natural frequency
ωn = √(20000/10) = 44.72 rad/s
fn = 44.72/(2π) = 7.118 Hz
Answer
fn = 7.118 Hz
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