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 15 kW motor running at 960 rpm drives a pinion with 30 teeth and a module of 2 mm. Find the tangential force Ft on the pinion teeth.
P = 15 kWn = 960 rpmN1 = 30m = 2 mm
Ft = N
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 300 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 300 mmC = 600 mm
θ = °
3.Torsion of shaftsMedium
A stepped solid brass shaft is fixed at A. Segment AB has diameter 40 mm and length 1.2 m; segment BC has diameter 30 mm and length 1.2 m. A torque of 750 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.
T = 750 N·md1 = 40 mm, L1 = 1.2 md2 = 30 mm, L2 = 1.2 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 1450 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 5 kWN = 1450 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = 30 MPa and τₓᵧ = −20 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 = 30 MPaτxy = −20 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 80 MPa, σᵧ = 120 MPa and τₓᵧ = 50 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 80 MPaσy = 120 MPaτxy = 50 MPaθ = -30°
τθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: brass, Ø40 mm, 500 mm, E = 100 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 50 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 500 mm, E = 110 GPaP = 60 kN
σ2 = 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 = 6 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 5 kNn = 3L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 5 cycles later by a peak of 2 mm. Find the damping ratio ζ.
x0 = 25 mmx5 = 2 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 30 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: Ft = 4974 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·960/60 = 100.5 rad/s
T = P/ω = 15000/100.5 = 149.2 N·m
Pitch diameter
d = m·N = 2·30 = 60 mm
Tangential force
Ft = 2T/d = 2·149.2/0.06 = 4974 N
Answer
Ft = 4974 N
2.Gears & gear trains
Answer: θ = 165.6 °
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(0.125) = 7.181°
Angle of wrap
θ = 180 − 2·7.181 = 165.6°
Answer
θ = 165.6°
3.Torsion of shafts
Answer: φC = 21.89 °
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 moments
J1 = π·40⁴/32 = 251327 mm⁴
J2 = π·30⁴/32 = 79522 mm⁴
Twist of each segment
T = 750 N·m = 750000 N·mm
G = 39 GPa = 39000 MPa
φAB = 750000·1200/(39000·251327) = 0.09182 rad
φBC = 750000·1200/(39000·79522) = 0.2902 rad
Total twist
φC = 0.09182 + 0.2902 = 0.382 rad = 21.89°
Answer
φC = 21.89°
4.Torsion of shafts
Answer: d = 14.97 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·1450/60 = 151.8 rad/s
T = P/ω = 5000/151.8 = 32.93 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·32929/(π·50)) = 14.97 mm
Answer
d = 14.97 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −70.67 °
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·−20, −20 − 30) = −141.3°
Angle on the element
θp = −70.67°
Answer
θp = −70.67°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = 7.679 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 = −−20·−0.866 + 50·0.5 = 7.679 MPa
Answer
τθ = 7.679 MPa
7.Axial stress & strain
Answer: P = 16.76 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 = 314.2 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(100000·1257) = 3.979 × 10⁻⁶ mm/N
Load
P = 0.2/1.194 × 10⁻⁵ = 16755 N = 16.76 kN
Answer
P = 16.76 kN
8.Axial stress & strain
Answer: σ2 = 15.51 MPa
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.
Stiffnesses
k₁ = 200000·1257/600 = 418879 N/mm
k₂ = 110000·1963/500 = 431969 N/mm
Share of segment 2
F2 = 60000·431969/850848 = 30462 N
Stress
σ2 = 30462/1963 = 15.51 MPa (compression)
Answer
σ2 = 15.51 MPa
9.Column buckling
Answer: Pcr = 24.92 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·6000 = 4200 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/4200² = 24916 N = 24.92 kN
Answer
Pcr = 24.92 kN
10.Column buckling
Answer: d = 39.67 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 = 3·5000 = 15000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·4000²/(π³·200000))^(1/4) = 39.67 mm
Answer
d = 39.67 mm
Check: the slenderness KL/r is 403.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.08014
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(25/2) = 0.5051
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5051/√(39.48 + 0.2552) = 0.08014
Answer
ζ = 0.08014
12.Mechanical vibrations
Answer: keq = 130 kN/m
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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 30 + 100 = 130 kN/m
Answer
keq = 130 kN/m
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