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 belt runs on an 80 mm driving pulley at 1500 rpm. The tension is 800 N on the tight side and 240 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 1500 rpmT1 = 800 NT2 = 240 N
P = kW
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turns at 3000 rpm and drives gear B (24 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 1600 rpm. How many teeth ND must gear D have?
nin = 3000 rpmA = 16B = 24C = 12nout = 1600 rpm
ND =
3.Torsion of shaftsMedium
A solid brass shaft 1.2 m long carries a torque of 100 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.
T = 100 N·mL = 1.2 mG = 39 GPaφallow = 2°
d = mm
4.Torsion of shaftsHard
A solid steel shaft transmits 5 kW at 300 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.
P = 5 kWN = 300 rpmτallow = 75 MPaθallow = 1°/mG = 80 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −20 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 30 MPa?
σx = −20 MPaσy = −20 MPaσ1 = 30 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 5 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.
D = 1000 mmt = 5 mmp = 2.5 MPaα = 50°
|τn| = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 100 kNd = 30 mmL = 3 mE = 200 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 300 mm, E = 110 GPaP = 20 kN
σ2 = MPa
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 = 1.5 mE = 70 GPa (aluminium)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 150 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 620 N·s/m. Find the damping ratio ζ.
m = 150 kgk = 100 kN/mc = 620 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 5 kN/m and k₂ = 100 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 = 50 kgk1 = 5 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 3.519 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.
Belt speed
v = 6.283 m/s
Power
P = (800 − 240)·6.283 = 3519 W = 3.519 kW
Answer
P = 3.519 kW
2.Gears & gear trains
Answer: ND = 15
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.
Intermediate shaft speed
nB = nC = 3000·16/24 = 2000 rpm
Teeth on gear D
ND = C·nC/nout = 12·2000/1600 = 15
Answer
ND = 15
3.Torsion of shafts
Answer: d = 30.78 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 = 100 N·m = 100000 N·mm
L = 1.2 m = 1200 mm
G = 39 GPa = 39000 MPa
φ = 2·π/180 = 0.03491 rad
J = TL/(Gφ) = 100000·1200/(39000·0.03491) = 88147 mm⁴
Diameter
d = (32J/π)1/4 = 30.78 mm
Answer
d = 30.78 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 32.83 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·300/60 = 31.42 rad/s
T = P/ω = 5000/31.42 = 159.2 N·m
Diameter for strength
T = 159.2 N·m = 159155 N·mm
dτ = ∛(16·159155/(π·75)) = 22.11 mm
Diameter for stiffness
θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm
G = 80 GPa = 80000 MPa
dφ = (32·159155/(π·80000·1.745 × 10⁻⁵))1/4 = 32.83 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(22.11, 32.83) = 32.83 mm
Answer
d = 32.83 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: τxy = 50 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 + (−20))/2 = −20 MPa
R = σ1 − C = 30 − (−20) = 50 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 0²) = 50 MPa
Answer
τxy = 50 MPa
6.Mohr’s circle & stress transformation
Answer: |τn| = 61.55 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 1000/2 = 500 mm
σh = pr/t = 2.5·500/5 = 250 MPa
σa = pr/(2t) = 125 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
|τn| = (250 − 125)·sin 50°·cos 50° = 61.55 MPa
Answer
|τn| = 61.55 MPa
7.Axial stress & strain
Answer: δ = 2.122 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 = 100·1000/706.9 = 141.5 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 141.5/200000 = 0.0007074 = 707.4 µε
Elongation
δ = ε·L = 0.0007074·3000 mm = 2.122 mm
Answer
δ = 2.122 mm
8.Axial stress & strain
Answer: σ2 = 7.89 MPa
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.
Stiffnesses
k₁ = 200000·314.2/300 = 209440 N/mm
k₂ = 110000·1963/300 = 719948 N/mm
Share of segment 2
F2 = 20000·719948/929388 = 15493 N
Stress
σ2 = 15493/1963 = 7.89 MPa (compression)
Answer
σ2 = 7.89 MPa
9.Column buckling
Answer: Pcr = 81.59 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 0.7·1500 = 1050 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·130208/1050² = 81594 N = 81.59 kN
Answer
Pcr = 81.59 kN
10.Column buckling
Answer: σcr = 4.441 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 1·5000 = 5000 mm
KL/r = 5000/7.5 = 666.7
Critical stress
σcr = π²E/(KL/r)² = π²·200000/666.7² = 4.441 MPa
Answer
σcr = 4.441 MPa
11.Mechanical vibrations
Answer: ζ = 0.08004
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·√(100000·150) = 7746 N·s/m
Damping ratio
ζ = c/cc = 620/7746 = 0.08004
Answer
ζ = 0.08004
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 7.293 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 = 5 + 100 = 105 kN/m
Natural frequency
ωn = √(105000/50) = 45.83 rad/s
fn = 45.83/(2π) = 7.293 Hz
Answer
fn = 7.293 Hz
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