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
Two parallel shafts are 80 mm apart. They must be connected by a pair of spur gears with a module of 2 mm and a speed reduction of 4 : 1. How many teeth N2 must the gear have?
a = 80 mmm = 2 mmi = 4
N2 =
2.Gears & gear trainsHard
An open belt connects a 100 mm pulley and a 250 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 100 mmd2 = 250 mmC = 500 mm
θ = °
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 = 500 N·md = 30 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 100 kW at 500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 100 kWN = 500 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = 60 MPa and τₓᵧ = −10 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 = 60 MPaτxy = −10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −70 MPa, σᵧ = −80 MPa and τₓᵧ = 10 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = −70 MPaσy = −80 MPaτxy = 10 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 85 kNd = 10 mmL = 3 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 400 mm, E = 100 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 140 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 140 kNn = 2L = 1.5 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 one cycle later by a peak of 23.3 mm. Find the damping ratio ζ.
x0 = 30 mmx1 = 23.3 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 30 kN/m and k₂ = 5 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 100 kgk1 = 30 kN/mk2 = 5 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N2 = 64
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.
Total number of teeth
N1 + N2 = 2a/m = 2·80/2 = 80
Split by the ratio
N1 = 80/(1 + 4) = 16
N2 = 4·16 = 64
Answer
N2 = 64
2.Gears & gear trains
Answer: θ = 162.7 °
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(0.15) = 8.627°
Angle of wrap
θ = 180 − 2·8.627 = 162.7°
Answer
θ = 162.7°
3.Torsion of shafts
Answer: φ = 6.755 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·1500/(80000·79522) = 0.1179 rad
φ = 0.1179·180/π = 6.755°
Answer
φ = 6.755°
4.Torsion of shafts
Answer: d = 57.94 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·500/60 = 52.36 rad/s
T = P/ω = 100000/52.36 = 1910 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·1.91 × 10⁶/(π·50)) = 57.94 mm
Answer
d = 57.94 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −82.98 °
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·−10, −20 − 60) = −166°
Angle on the element
θp = −82.98°
Answer
θp = −82.98°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 5.939
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = −75 MPa, R = 11.18 MPa
σ1 = −63.82 MPa, σ2 = −86.18 MPa, σ3 = 0
Von Mises
σvM = √((−63.82)² − (−63.82)·(−86.18) + (−86.18)²) = 77.46 MPa
n = 460/77.46 = 5.939
Answer
n = 5.939
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: ε = 10823 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·10²/4 = 78.54 mm²
σ = P/A = 85·1000/78.54 = 1082 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 1082/100000 = 0.01082 = 10823 µε
Answer
ε = 10823 µε
8.Axial stress & strain
Answer: σ2 = 23.69 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 = 200000·706.9/400 = 353429 N/mm
k₂ = 100000·1963/400 = 490874 N/mm
Share of segment 2
F2 = 80000·490874/844303 = 46512 N
Stress
σ2 = 46512/1963 = 23.69 MPa (compression)
Answer
σ2 = 23.69 MPa
9.Column buckling
Answer: Pcr = 48.83 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 = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/3000² = 48835 N = 48.83 kN
Answer
Pcr = 48.83 kN
10.Column buckling
Answer: d = 65.65 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 = 2·140000 = 280000 N
KL = 1·1500 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·280000·1500²/(π³·70000))^(1/4) = 65.65 mm
Answer
d = 65.65 mm
Check: the slenderness KL/r is 91.39, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.04019
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 = (1/1)·ln(30/23.3) = 0.2527
Damping ratio
ζ = δ/√(4π² + δ²) = 0.2527/√(39.48 + 0.06388) = 0.04019
Answer
ζ = 0.04019
12.Mechanical vibrations
Answer: fn = 2.978 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 30 + 5 = 35 kN/m
Natural frequency
ωn = √(35000/100) = 18.71 rad/s
fn = 18.71/(2π) = 2.978 Hz
Answer
fn = 2.978 Hz
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