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 125 mm pulley at 2880 rpm drives a 500 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.
n1 = 2880 rpmd1 = 125 mmd2 = 500 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 400 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 400 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 75 N·m. The two ends of a gauge length of 400 mm rotate 4.21° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 75 N·md = 20 mmL = 400 mmφ = 4.21°
G = GPa
4.Torsion of shaftsHard
A solid shaft must transmit 15 kW at 3000 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 15 kWN = 3000 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 120 MPa, σᵧ = 20 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 120 MPaσy = 20 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −40 MPa and τₓᵧ = 60 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −10 MPaσy = −40 MPaτxy = 60 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 30 kNd = 16 mmL = 3 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 total elongation δ.
P = 115 kNsegment 1: Ø40 mm, 1 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 4 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 3 cycles later by a peak of 5.1 mm. Find the damping ratio ζ.
x0 = 50 mmx3 = 5.1 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 5 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 12 kN/mk2 = 5 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 698.4 rpm
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.
Speed without slip
2880·125/500 = 720 rpm
With slip
n2 = 720·0.97 = 698.4 rpm
Answer
n2 = 698.4 rpm
2.Gears & gear trains
Answer: θ = 168 °
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(0.1042) = 5.979°
Angle of wrap
θ = 180 − 2·5.979 = 168°
Answer
θ = 168°
3.Torsion of shafts
Answer: G = 25.99 GPa
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 moment and angle
J = π·20⁴/32 = 15708 mm⁴
φ = 4.21·π/180 = 0.07348 rad
Shear modulus
T = 75 N·m = 75000 N·mm
G = TL/(Jφ) = 75000·400/(15708·0.07348) = 25992 MPa
G = 25.99 GPa
Answer
G = 25.99 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: d = 16.94 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 = 15000/314.2 = 47.75 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·47746/(π·50)) = 16.94 mm
Answer
d = 16.94 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 58.31 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 = √(((120 − 20)/2)² + 30²) = 58.31 MPa
Answer
τmax = 58.31 MPa
6.Mohr’s circle & stress transformation
Answer: n = 4.182
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 = −25 MPa, R = 61.85 MPa
σ1 = 36.85 MPa, σ2 = −86.85 MPa, σ3 = 0
Von Mises
σvM = √(36.85² − 36.85·(−86.85) + (−86.85)²) = 110 MPa
n = 460/110 = 4.182
Answer
n = 4.182
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 2132 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·16²/4 = 201.1 mm²
σ = P/A = 30·1000/201.1 = 149.2 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 149.2/70000 = 0.002132 = 2132 µε
Answer
ε = 2132 µε
8.Axial stress & strain
Answer: δ = 5.608 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·40²/4 = 1257 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 91.51 MPa
σ2 = 366.1 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 91.51·1000/100000 = 0.9151 mm
δ2 = σ2L2/E2 = 366.1·1500/117000 = 4.693 mm
δ = δ1 + δ2 = 5.608 mm
Answer
δ = 5.608 mm
9.Column buckling
Answer: Pcr = 107.9 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 60·50³/12 = 625000 mm⁴
Effective length
KL = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·625000/2000² = 107949 N = 107.9 kN
Answer
Pcr = 107.9 kN
10.Column buckling
Answer: σcr = 26.99 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 = 0.5·4000 = 2000 mm
KL/r = 2000/12.5 = 160
Critical stress
σcr = π²E/(KL/r)² = π²·70000/160² = 26.99 MPa
Answer
σcr = 26.99 MPa
11.Mechanical vibrations
Answer: ζ = 0.1202
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(50/5.1) = 0.7609
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7609/√(39.48 + 0.579) = 0.1202
Answer
ζ = 0.1202
12.Mechanical vibrations
Answer: fn = 5.669 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
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 12 + 5 = 17 kN/m
keq = 17·50/(17 + 50) = 12.69 kN/m
Natural frequency
ωn = √(12687/10) = 35.62 rad/s
fn = 35.62/(2π) = 5.669 Hz
Answer
fn = 5.669 Hz
ExamProblems · examproblems.com/courses/machine-design