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 125 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 1.5 : 1. How many teeth N2 must the gear have?
a = 125 mmm = 5 mmi = 1.5
N2 =
2.Gears & gear trainsHard
In a compound gear train, gear A (18 teeth) on the input shaft turning at 2900 rpm drives gear B (25 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (36 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 18B = 25C = 16D = 36
nout = rpm
3.Torsion of shaftsMedium
A hollow brass shaft (outer diameter 50 mm, inner diameter 30 mm, length 1 m) is fixed at one end and twisted by a torque of 400 N·m at the other. Find the angle of twist φ in degrees.
T = 400 N·mdo = 50 mmdi = 30 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 3000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 10 kWN = 3000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −30 MPa, σᵧ = 40 MPa and τₓᵧ = 60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −30 MPaσy = 40 MPaτxy = 60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 20 mm carries an axial compressive force of 100 kN together with a torque of 200 N·m. At the surface of the shaft, find the maximum in-plane shear stress τmax.
d = 20 mmF = −100 kNT = 200 N·m
τmax = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 10 kNd = 10 mmL = 2.5 mE = 200 GPa
δ = mm
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 = 115 kNsegment 1: Ø35 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 0.5 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 8 kg mass is supported by a spring with a stiffness of 1 kN/m and a viscous damper with a damping coefficient of 40 N·s/m. Find the damping ratio ζ.
m = 8 kgk = 1 kN/mc = 40 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 15 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 20 kN/mk2 = 15 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: N2 = 30
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 = 2·125/5 = 50
Split by the ratio
N1 = 50/(1 + 1.5) = 20
N2 = 1.5·20 = 30
Answer
N2 = 30
2.Gears & gear trains
Answer: nout = 928 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
First stage
nB = 2900·18/25 = 2088 rpm = nC
Second stage
nout = 2088·16/36 = 928 rpm
Answer
nout = 928 rpm
total ratio = 3.125 : 1
3.Torsion of shafts
Answer: φ = 1.1 °
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 − 30⁴)/32 = 534071 mm⁴
Angle of twist
T = 400 N·m = 400000 N·mm
L = 1 m = 1000 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 400000·1000/(39000·534071) = 0.0192 rad
φ = 0.0192·180/π = 1.1°
Answer
φ = 1.1°
4.Torsion of shafts
Answer: d = 12.93 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·75)) = 12.93 mm
Answer
d = 12.93 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = 60.13 °
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(2·60, −30 − 40) = 120.3°
Angle on the element
θp = 60.13°
Answer
θp = 60.13°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τmax = 203.8 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.
Stresses at the surface
σ = F/A = −100000/(π·20²/4) = −100000/314.2 = −318.3 MPa
τ = 16T/(πd³) = 16·200000/(π·20³) = 127.3 MPa
Mohr’s circle
C = σ/2 = −159.2 MPa
R = √((−159.2)² + 127.3²) = 203.8 MPa
Answer
τmax = 203.8 MPa
7.Axial stress & strain
Answer: δ = 1.592 mm
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 = 10·1000/78.54 = 127.3 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 127.3/200000 = 0.0006366 = 636.6 µε
Elongation
δ = ε·L = 0.0006366·2500 mm = 1.592 mm
Answer
δ = 1.592 mm
8.Axial stress & strain
Answer: σmax = 1017 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·35²/4 = 962.1 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 119.5 MPa
σ2 = 1017 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 1017 MPa
Answer
σmax = 1017 MPa
9.Column buckling
Answer: Pcr = 77.24 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·306796/2800² = 77244 N = 77.24 kN
Answer
Pcr = 77.24 kN
10.Column buckling
Answer: σcr = 19.28 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 2·1000 = 2000 mm
KL/r = 2000/6.25 = 320
Critical stress
σcr = π²E/(KL/r)² = π²·200000/320² = 19.28 MPa
Answer
σcr = 19.28 MPa
11.Mechanical vibrations
Answer: ζ = 0.2236
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·√(1000·8) = 178.9 N·s/m
Damping ratio
ζ = c/cc = 40/178.9 = 0.2236
Answer
ζ = 0.2236
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 8.571 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/20 + 1/15 → keq = 20·15/(20 + 15) = 8.571 kN/m
Answer
keq = 8.571 kN/m
ExamProblems · examproblems.com/courses/machine-design