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 a 100 mm driving pulley at 2880 rpm. The tension is 1200 N on the tight side and 360 N on the slack side. Find the power P transmitted, in kW.
d1 = 100 mmn1 = 2880 rpmT1 = 1200 NT2 = 360 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 125 mm driving pulley at 720 rpm to a 315 mm pulley; the centres are 800 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 125 mmn1 = 720 rpmd2 = 315 mmC = 800 mmμ = 0.3T1 = 600 N
P = kW
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 100 N·m. The two ends of a gauge length of 300 mm rotate 1.37° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 100 N·md = 20 mmL = 300 mmφ = 1.37°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 300 rpmdo = 75 mmdi = 56 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 120 MPa, σᵧ = −40 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 120 MPaσy = −40 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = 70 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = 50 MPaσy = 70 MPaτxy = 10 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 50 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.
segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: copper, Ø50 mm, 500 mm, E = 120 GPaδ = 0.5 mm
P = kN
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 = 25 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø16 mm, 2 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 45 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 45 kNn = 2L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 5 cycles later by a peak of 4.5 mm. Find the damping ratio ζ.
x0 = 30 mmx5 = 4.5 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 = 25 kN/m and the lower one k₂ = 100 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 20 kgk1 = 25 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 12.67 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 = 15.08 m/s
Power
P = (1200 − 360)·15.08 = 12667 W = 12.67 kW
Answer
P = 12.67 kW
2.Gears & gear trains
Answer: P = 1.644 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 = 2.904 rad
Slack-side tension
eμθ = e0.8711 = 2.389, T₂ = 600/2.389 = 251.1 N
Power
P = (600 − 251.1)·4.712 = 1.644 kW
Answer
P = 1.644 kW
3.Torsion of shafts
Answer: G = 79.87 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 = 1.37·π/180 = 0.02391 rad
Shear modulus
T = 100 N·m = 100000 N·mm
G = TL/(Jφ) = 100000·300/(15708·0.02391) = 79874 MPa
G = 79.87 GPa
Answer
G = 79.87 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 55.76 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·300/60 = 31.42 rad/s
T = P/ω = 100000/31.42 = 3183 N·m
Polar second moment of area
J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 3.183 × 10⁶·37.5/2.141 × 10⁶ = 55.76 MPa
Answer
τmax = 55.76 MPa
5.Mohr’s circle & stress transformation
Answer: θs = −29 °
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.
Angle on the circle
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 50/94.34 = 0.53, sin 2θs = −80/94.34 = −0.848
2θs = atan2(−80, 50) = −57.99°
Angle on the element
θs = −29°
check: θs − θp = −29 − 16 = −45°
Answer
θs = −29°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 63.66 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 = 60 + −10·0.5 + 10·0.866 = 63.66 MPa
Answer
σθ = 63.66 MPa
7.Axial stress & strain
Answer: P = 60.73 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 = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(120000·1963) = 2.122 × 10⁻⁶ mm/N
Load
P = 0.5/8.234 × 10⁻⁶ = 60727 N = 60.73 kN
Answer
P = 60.73 kN
8.Axial stress & strain
Answer: σmax = 124.3 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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 50.93 MPa
σ2 = 124.3 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 124.3 MPa
Answer
σmax = 124.3 MPa
9.Column buckling
Answer: Pcr = 2.221 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 = 50·30³/12 = 112500 mm⁴
Effective length
KL = 2·5000 = 10000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·112500/10000² = 2221 N = 2.221 kN
Answer
Pcr = 2.221 kN
10.Column buckling
Answer: d = 49.43 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 = 2·45000 = 90000 N
KL = 0.5·3000 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·90000·1500²/(π³·70000))^(1/4) = 49.43 mm
Answer
d = 49.43 mm
Check: the slenderness KL/r is 121.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.06028
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(30/4.5) = 0.3794
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3794/√(39.48 + 0.144) = 0.06028
Answer
ζ = 0.06028
12.Mechanical vibrations
Answer: fn = 5.033 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 = 1/25 + 1/100 → keq = 25·100/(25 + 100) = 20 kN/m
Natural frequency
ωn = √(20000/20) = 31.62 rad/s
fn = 31.62/(2π) = 5.033 Hz
Answer
fn = 5.033 Hz
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