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 4 kW motor running at 1440 rpm drives a pinion with 18 teeth and a module of 2.5 mm. Find the tangential force Ft on the pinion teeth.
P = 4 kWn = 1440 rpmN1 = 18m = 2.5 mm
Ft = N
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 720 rpm to a 250 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 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 = 100 mmn1 = 720 rpmd2 = 250 mmC = 600 mmμ = 0.35T1 = 600 N
P = kW
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 = 50 mmL = 1.2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 3000 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 10 kWN = 3000 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 150 MPa, σᵧ = −70 MPa and τₓᵧ = −40 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 = 150 MPaσy = −70 MPaτxy = −40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 20 MPa, σᵧ = 100 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.
σx = 20 MPaσy = 100 MPaτxy = −30 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 105 kNd = 25 mmL = 2 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 total elongation δ.
P = 95 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 0.5 m, brass (E = 100 GPa)
δ = mm
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 = 5 mE = 70 GPa (aluminium)b × h = 20 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 45 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 45 kNn = 3L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
An 80 kg mass is supported by a spring with a stiffness of 20 kN/m and a viscous damper with a damping coefficient of 130 N·s/m. Find the damping ratio ζ.
m = 80 kgk = 20 kN/mc = 130 N·s/m
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 25 kN/m and k₂ = 15 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 = 25 kgk1 = 25 kN/mk2 = 15 kN/mk3 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Ft = 1179 N
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Torque on the pinion
ω = 2π·1440/60 = 150.8 rad/s
T = P/ω = 4000/150.8 = 26.53 N·m
Pitch diameter
d = m·N = 2.5·18 = 45 mm
Tangential force
Ft = 2T/d = 2·26.53/0.045 = 1179 N
Answer
Ft = 1179 N
2.Gears & gear trains
Answer: P = 1.44 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 = 2.891 rad
Slack-side tension
eμθ = e1.012 = 2.751, T₂ = 600/2.751 = 218.1 N
Power
P = (600 − 218.1)·3.77 = 1.44 kW
Answer
P = 1.44 kW
3.Torsion of shafts
Answer: φ = 1.05 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1200/(80000·613592) = 0.01833 rad
φ = 0.01833·180/π = 1.05°
Answer
φ = 1.05°
4.Torsion of shafts
Answer: d = 15.94 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 = 10000/314.2 = 31.83 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·31831/(π·40)) = 15.94 mm
Answer
d = 15.94 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −54.99 °
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 = −40/117 = −0.3417, sin 2θs = −110/117 = −0.9398
2θs = atan2(−110, −40) = −110°
Angle on the element
θs = −54.99°
check: θs − θp = −54.99 − (−9.992) = −45°
Answer
θs = −54.99°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 37.4 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 + −40·−0.1736 + −30·0.9848 = 37.4 MPa
Answer
σθ = 37.4 MPa
7.Axial stress & strain
Answer: ε = 1070 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 = 105·1000/490.9 = 213.9 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 213.9/200000 = 0.00107 = 1070 µε
Answer
ε = 1070 µε
8.Axial stress & strain
Answer: δ = 8.071 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·25²/4 = 490.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 193.5 MPa
σ2 = 840 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 193.5·2000/100000 = 3.871 mm
δ2 = σ2L2/E2 = 840·500/100000 = 4.2 mm
δ = δ1 + δ2 = 8.071 mm
Answer
δ = 8.071 mm
9.Column buckling
Answer: Pcr = 3.76 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 = 100·20³/12 = 66667 mm⁴
Effective length
KL = 0.7·5000 = 3500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·66667/3500² = 3760 N = 3.76 kN
Answer
Pcr = 3.76 kN
10.Column buckling
Answer: d = 84.16 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 = 3·45000 = 135000 N
KL = 1·6000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·135000·6000²/(π³·200000))^(1/4) = 84.16 mm
Answer
d = 84.16 mm
Check: the slenderness KL/r is 285.2, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05139
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·√(20000·80) = 2530 N·s/m
Damping ratio
ζ = c/cc = 130/2530 = 0.05139
Answer
ζ = 0.05139
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 5.38 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 = 25 + 15 = 40 kN/m
keq = 40·100/(40 + 100) = 28.57 kN/m
Natural frequency
ωn = √(28571/25) = 33.81 rad/s
fn = 33.81/(2π) = 5.38 Hz
Answer
fn = 5.38 Hz
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