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 1450 rpm drives a pinion with 15 teeth, which meshes with a gear with 20 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 15 kWn1 = 1450 rpmN1 = 15N2 = 20η = 96%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 1500 rpm drives gear B (50 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (25 teeth) on the output shaft. Find the output speed nout.
nin = 1500 rpmA = 15B = 50C = 20D = 25
nout = rpm
3.Torsion of shaftsMedium
A stepped solid brass shaft is fixed at A. Segment AB has diameter 60 mm and length 1 m; segment BC has diameter 50 mm and length 2.5 m. A torque of 1500 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.
T = 1500 N·md1 = 60 mm, L1 = 1 md2 = 50 mm, L2 = 2.5 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 1500 N·m is applied at C, 1 m from A and 2 m from B. Find the angle of rotation φC of section C, in degrees.
d = 40 mmT0 = 1500 N·ma = 1 mb = 2 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −60 MPa, σᵧ = 70 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 = −60 MPaσy = 70 MPaτxy = −10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −10 MPa, σᵧ = 30 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −10 MPaσy = 30 MPaτxy = 30 MPaθ = 15°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 25 kNd = 30 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 total elongation δ.
P = 20 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)b × h = 40 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed 2 cycles later by a peak of 8 mm. Find the damping ratio ζ.
x0 = 15 mmx2 = 8 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 60 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 = 12 kN/mk2 = 60 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 98.79 N·m
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.
Motor torque
ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 15000/151.8 = 98.79 N·m
Answer
T1 = 98.79 N·m
2.Gears & gear trains
Answer: nout = 360 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 = 1500·15/50 = 450 rpm = nC
Second stage
nout = 450·20/25 = 360 rpm
Answer
nout = 360 rpm
total ratio = 4.167 : 1
3.Torsion of shafts
Answer: φC = 10.71 °
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·60⁴/32 = 1.272 × 10⁶ mm⁴
J2 = π·50⁴/32 = 613592 mm⁴
Twist of each segment
T = 1500 N·m = 1.5 × 10⁶ N·mm
G = 39 GPa = 39000 MPa
φAB = 1.5 × 10⁶·1000/(39000·1.272 × 10⁶) = 0.03023 rad
φBC = 1.5 × 10⁶·2500/(39000·613592) = 0.1567 rad
Total twist
φC = 0.03023 + 0.1567 = 0.1869 rad = 10.71°
Answer
φC = 10.71°
4.Torsion of shafts
Answer: φC = 2.85 °
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 1500
TA·a = TB·b
TA = 1500·2/3 = 1000 N·m
TB = 1500·1/3 = 500 N·m
Rotation of C
Use segment AC: it is fixed at A and carries TA.
J = π·40⁴/32 = 251327 mm⁴
G = 80 GPa = 80000 MPa
φC = TA·a/(GJ) = 1 × 10⁶·1000/(80000·251327) = 0.04974 rad = 2.85°
Answer
φC = 2.85°
5.Mohr’s circle & stress transformation
Answer: θp = −85.63 °
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, −60 − 70) = −171.3°
Angle on the element
θp = −85.63°
Answer
θp = −85.63°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = 35.98 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 = −−20·0.5 + 30·0.866 = 35.98 MPa
Answer
τθ = 35.98 MPa
7.Axial stress & strain
Answer: δ = 0.4421 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·30²/4 = 706.9 mm²
σ = P/A = 25·1000/706.9 = 35.37 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 35.37/200000 = 0.0001768 = 176.8 µε
Elongation
δ = ε·L = 0.0001768·2500 mm = 0.4421 mm
Answer
δ = 0.4421 mm
8.Axial stress & strain
Answer: δ = 1.063 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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 28.29 MPa
σ2 = 99.47 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 28.29·2000/100000 = 0.5659 mm
δ2 = σ2L2/E2 = 99.47·1000/200000 = 0.4974 mm
δ = δ1 + δ2 = 1.063 mm
Answer
δ = 1.063 mm
9.Column buckling
Answer: Pcr = 14.74 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 = 100·40³/12 = 533333 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·533333/5000² = 14739 N = 14.74 kN
Answer
Pcr = 14.74 kN
10.Column buckling
Answer: σcr = 9.715 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 2·2000 = 4000 mm
KL/r = 4000/15 = 266.7
Critical stress
σcr = π²E/(KL/r)² = π²·70000/266.7² = 9.715 MPa
Answer
σcr = 9.715 MPa
11.Mechanical vibrations
Answer: ζ = 0.04996
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 = (1/2)·ln(15/8) = 0.3143
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3143/√(39.48 + 0.09879) = 0.04996
Answer
ζ = 0.04996
12.Mechanical vibrations
Answer: fn = 6.113 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 = 12 + 60 = 72 kN/m
keq = 72·50/(72 + 50) = 29.51 kN/m
Natural frequency
ωn = √(29508/20) = 38.41 rad/s
fn = 38.41/(2π) = 6.113 Hz
Answer
fn = 6.113 Hz
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