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 150 mm pulley at 960 rpm drives a 250 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 150 mmd2 = 250 mms = 2.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 250 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 250 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
A solid aluminium shaft 1 m long carries a torque of 500 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.
T = 500 N·mL = 1 mG = 26 GPaφallow = 0.5°
d = mm
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 1500 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 10 kWN = 1500 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = 80 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −20 MPaσy = 80 MPaτxy = −30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −50 MPa, σᵧ = 70 MPa and τₓᵧ = −60 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −50 MPaσy = 70 MPaτxy = −60 MPaθ = 30°
τθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 40 mm, length 400 mm, E = 70 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, Ø40 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 400 mm, E = 70 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 = 45 kNsegment 1: Ø40 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
σmax = 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 = 4 mE = 70 GPa (aluminium)d = 30 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 = 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 10 mm is followed 4 cycles later by a peak of 0.8 mm. Find the damping ratio ζ.
x0 = 10 mmx4 = 0.8 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 60 kN/m and k₂ = 50 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 = 60 kN/mk2 = 50 kN/mk3 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 561.6 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
960·150/250 = 576 rpm
With slip
n2 = 576·0.975 = 561.6 rpm
Answer
n2 = 561.6 rpm
2.Gears & gear trains
Answer: θ = 175.2 °
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(0.04167) = 2.388°
Angle of wrap
θ = 180 − 2·2.388 = 175.2°
Answer
θ = 175.2°
3.Torsion of shafts
Answer: d = 68.83 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.
Required polar moment
T = 500 N·m = 500000 N·mm
L = 1 m = 1000 mm
G = 26 GPa = 26000 MPa
φ = 0.5·π/180 = 0.008727 rad
J = TL/(Gφ) = 500000·1000/(26000·0.008727) = 2.204 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 68.83 mm
Answer
d = 68.83 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 17.55 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·1500/60 = 157.1 rad/s
T = P/ω = 10000/157.1 = 63.66 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·63662/(π·60)) = 17.55 mm
Answer
d = 17.55 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −74.52 °
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
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·−30, −20 − 80) = −149°
Angle on the element
θp = −74.52°
Answer
θp = −74.52°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = 21.96 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
τθ = −−60·0.866 + −60·0.5 = 21.96 MPa
Answer
τθ = 21.96 MPa
7.Axial stress & strain
Answer: P = 76.49 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 = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(70000·1257) = 4.547 × 10⁻⁶ mm/N
Load
P = 0.5/6.537 × 10⁻⁶ = 76491 N = 76.49 kN
Answer
P = 76.49 kN
8.Axial stress & strain
Answer: σmax = 223.8 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 35.81 MPa
σ2 = 223.8 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 223.8 MPa
Answer
σmax = 223.8 MPa
9.Column buckling
Answer: Pcr = 3.504 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·39761/2800² = 3504 N = 3.504 kN
Answer
Pcr = 3.504 kN
10.Column buckling
Answer: σcr = 1.687 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 = 2·4000 = 8000 mm
KL/r = 8000/12.5 = 640
Critical stress
σcr = π²E/(KL/r)² = π²·70000/640² = 1.687 MPa
Answer
σcr = 1.687 MPa
11.Mechanical vibrations
Answer: ζ = 0.09999
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 = (1/4)·ln(10/0.8) = 0.6314
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6314/√(39.48 + 0.3987) = 0.09999
Answer
ζ = 0.09999
12.Mechanical vibrations
Answer: fn = 5.782 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 = 60 + 50 = 110 kN/m
keq = 110·15/(110 + 15) = 13.2 kN/m
Natural frequency
ωn = √(13200/10) = 36.33 rad/s
fn = 36.33/(2π) = 5.782 Hz
Answer
fn = 5.782 Hz
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