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 150 mm driving pulley at 960 rpm. The tension is 600 N on the tight side and 120 N on the slack side. Find the power P transmitted, in kW.
d1 = 150 mmn1 = 960 rpmT1 = 600 NT2 = 120 N
P = kW
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 30 teeth and each planet gear has 20 teeth. The ring gear is held fixed and the sun is driven at 3000 rpm. Find the speed nc of the planet carrier.
Ns = 30Np = 20ns = 3000 rpm
nc = rpm
3.Torsion of shaftsMedium
A solid steel shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 750 N·md = 25 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft transmits 50 kW at 1500 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 0.5° per m of length. Find the smallest diameter d that satisfies both limits.
P = 50 kWN = 1500 rpmτallow = 75 MPaθallow = 0.5°/mG = 39 GPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −60 MPa, σᵧ = 90 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −60 MPaσy = 90 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 0 MPa, σᵧ = 120 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = 0 MPaσy = 120 MPaτxy = −60 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 15 kNd = 12 mmL = 0.5 mE = 200 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 = 30 kNsegment 1: Ø40 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 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 = 1.5 mE = 200 GPa (steel)d = 20 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 75 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 75 kNn = 3L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 10 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 2 kgk = 10 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 20 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 10 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 = 20 kgk1 = 30 kN/mk2 = 10 kN/mk3 = 20 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 3.619 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 = 7.54 m/s
Power
P = (600 − 120)·7.54 = 3619 W = 3.619 kW
Answer
P = 3.619 kW
2.Gears & gear trains
Answer: nc = 900 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Ring gear teeth
Nr = Ns + 2·Np = 30 + 2·20 = 70
Willis equation with the ring fixed
Set nr = 0 and solve for the carrier speed.
(3000 − nc)/(0 − nc) = −70/30
nc = ns·Ns/(Ns + Nr) = 3000·30/100 = 900 rpm
Answer
nc = 900 rpm
The carrier turns in the same direction as the sun.
3.Torsion of shafts
Answer: φ = 21.01 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1500/(80000·38350) = 0.3667 rad
φ = 0.3667·180/π = 21.01°
Answer
φ = 21.01°
4.Torsion of shafts
Answer: d = 55.56 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 = 50000/157.1 = 318.3 N·m
Diameter for strength
T = 318.3 N·m = 318310 N·mm
dτ = ∛(16·318310/(π·75)) = 27.86 mm
Diameter for stiffness
θ = 0.5·π/180/1000 = 8.727 × 10⁻⁶ rad/mm
G = 39 GPa = 39000 MPa
dφ = (32·318310/(π·39000·8.727 × 10⁻⁶))1/4 = 55.56 mm
Governing diameter
Both limits must hold, so the larger diameter governs.
d = max(27.86, 55.56) = 55.56 mm
Answer
d = 55.56 mm
Here stiffness governs.
5.Mohr’s circle & stress transformation
Answer: θp = −75.96 °
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·−40, −60 − 90) = −151.9°
Angle on the element
θp = −75.96°
Answer
θp = −75.96°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −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
σθ = 60 + −60·0.5 + −60·0.866 = −21.96 MPa
Answer
σθ = −21.96 MPa
7.Axial stress & strain
Answer: ε = 663.1 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·12²/4 = 113.1 mm²
σ = P/A = 15·1000/113.1 = 132.6 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 132.6/200000 = 0.0006631 = 663.1 µε
Answer
ε = 663.1 µε
8.Axial stress & strain
Answer: σmax = 265.3 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 23.87 MPa
σ2 = 265.3 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 265.3 MPa
Answer
σmax = 265.3 MPa
9.Column buckling
Answer: Pcr = 6.89 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 1·1500 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·7854/1500² = 6890 N = 6.89 kN
Answer
Pcr = 6.89 kN
10.Column buckling
Answer: d = 46.19 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 = 3·75000 = 225000 N
KL = 0.7·2000 = 1400 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·1400²/(π³·200000))^(1/4) = 46.19 mm
Answer
d = 46.19 mm
Check: the slenderness KL/r is 121.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 9.746 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.
Undamped natural frequency
ωn = √(10000/2) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 11.25·√(1 − 0.5²) = 9.746 Hz
Answer
fd = 9.746 Hz
12.Mechanical vibrations
Answer: fn = 4.109 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 = 30 + 10 = 40 kN/m
keq = 40·20/(40 + 20) = 13.33 kN/m
Natural frequency
ωn = √(13333/20) = 25.82 rad/s
fn = 25.82/(2π) = 4.109 Hz
Answer
fn = 4.109 Hz
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