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 worm with 3 start(s) turns at 1450 rpm and drives a worm wheel with 45 teeth. Find the speed nw of the wheel.
z = 3Nw = 45n = 1450 rpm
nw = rpm
2.Gears & gear trainsHard
An open belt connects a 150 mm driving pulley at 1500 rpm to a 500 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 150 mmn1 = 1500 rpmd2 = 500 mmC = 800 mmμ = 0.25T1 = 800 N
P = kW
3.Torsion of shaftsMedium
A hollow aluminium shaft (outer diameter 40 mm, inner diameter 28 mm, length 1 m) is fixed at one end and twisted by a torque of 400 N·m at the other. Find the angle of twist φ in degrees.
T = 400 N·mdo = 40 mmdi = 28 mmL = 1 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 75 kW at 3000 rpm. Find the maximum shear stress τmax in the shaft.
P = 75 kWN = 3000 rpmdo = 50 mmdi = 25 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −40 MPa, σᵧ = 120 MPa and τₓᵧ = −30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −40 MPaσy = 120 MPaτxy = −30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = −20 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = 150 MPaσy = −20 MPaτxy = −40 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 45 kNd = 25 mmL = 1 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 60 kNsegment 1: Ø35 mm, 1 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 8 kg mass is supported by a spring with a stiffness of 8 kN/m and a viscous damper with a damping coefficient of 50 N·s/m. Find the damping ratio ζ.
m = 8 kgk = 8 kN/mc = 50 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 10 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: nw = 96.67 rpm
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.
Wheel speed
nw = n·z/Nw = 1450·3/45 = 96.67 rpm
Answer
nw = 96.67 rpm
2.Gears & gear trains
Answer: P = 4.627 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 = 2.701 rad
Slack-side tension
eμθ = e0.6751 = 1.964, T₂ = 800/1.964 = 407.3 N
Power
P = (800 − 407.3)·11.78 = 4.627 kW
Answer
P = 4.627 kW
3.Torsion of shafts
Answer: φ = 4.615 °
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(40⁴ − 28⁴)/32 = 190984 mm⁴
Angle of twist
T = 400 N·m = 400000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = TL/(GJ) = 400000·1000/(26000·190984) = 0.08055 rad
φ = 0.08055·180/π = 4.615°
Answer
φ = 4.615°
4.Torsion of shafts
Answer: τmax = 10.38 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·3000/60 = 314.2 rad/s
T = P/ω = 75000/314.2 = 238.7 N·m
Polar second moment of area
J = π(50⁴ − 25⁴)/32 = 575243 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 238732·25/575243 = 10.38 MPa
Answer
τmax = 10.38 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 85.44 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 = √(((−40 − 120)/2)² + −30²) = 85.44 MPa
Answer
τmax = 85.44 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −12.14 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 = 65 + 85·−0.5 + −40·0.866 = −12.14 MPa
Answer
σθ = −12.14 MPa
7.Axial stress & strain
Answer: ε = 1310 µε
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·25²/4 = 490.9 mm²
σ = P/A = 45·1000/490.9 = 91.67 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 91.67/70000 = 0.00131 = 1310 µε
Answer
ε = 1310 µε
8.Axial stress & strain
Answer: σmax = 191 MPa
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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 62.36 MPa
σ2 = 191 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 191 MPa
Answer
σmax = 191 MPa
9.Column buckling
Answer: Pcr = 9.922 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·125664/5000² = 9922 N = 9.922 kN
Answer
Pcr = 9.922 kN
10.Column buckling
Answer: σcr = 17.27 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 2·1500 = 3000 mm
KL/r = 3000/15 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.09882
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·√(8000·8) = 506 N·s/m
Damping ratio
ζ = c/cc = 50/506 = 0.09882
Answer
ζ = 0.09882
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 8.889 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 = 1/10 + 1/80 → keq = 10·80/(10 + 80) = 8.889 kN/m
Answer
keq = 8.889 kN/m
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