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 100 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 = 100 mmm = 2 mmi = 4
N2 =
2.Gears & gear trainsHard
In a compound gear train, gear A (36 teeth) on the input shaft turning at 1440 rpm drives gear B (80 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (20 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 36B = 80C = 16D = 20
nout = rpm
3.Torsion of shaftsMedium
A stepped solid steel shaft is fixed at A. Segment AB has diameter 30 mm and length 2.5 m; segment BC has diameter 25 mm and length 0.5 m. A torque of 100 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.
T = 100 N·md1 = 30 mm, L1 = 2.5 md2 = 25 mm, L2 = 0.5 m
τmax = MPa
4.Torsion of shaftsHard
A solid brass shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 1.2 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 0.5 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 50 mmTB = 200 N·mTC = 250 N·mAB = 1.2 mBC = 0.5 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa and σᵧ = 20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 60 MPa?
σx = −40 MPaσy = 20 MPaσ1 = 60 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −80 MPa, σᵧ = 50 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis.
σx = −80 MPaσy = 50 MPaτxy = −30 MPaθ = 60°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.
P = 50 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 110 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø30 mm, 500 mm, E = 100 GPaP = 40 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 = 2 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 75 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 75 kNn = 3L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 12 kg mass on a spring with a stiffness of 50 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency fd of its free vibration in Hz.
m = 12 kgk = 50 kN/mζ = 0.3
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 15 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 = 10 kgk1 = 25 kN/mk2 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: N2 = 80
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 = 2·100/2 = 100
Split by the ratio
N1 = 100/(1 + 4) = 20
N2 = 4·20 = 80
Answer
N2 = 80
2.Gears & gear trains
Answer: nout = 518.4 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 = 1440·36/80 = 648 rpm = nC
Second stage
nout = 648·16/20 = 518.4 rpm
Answer
nout = 518.4 rpm
total ratio = 2.778 : 1
3.Torsion of shafts
Answer: τmax = 32.59 MPa
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.
Stress in each segment
T = 100 N·m = 100000 N·mm
τAB = 16·100000/(π·30³) = 18.86 MPa
τBC = 16·100000/(π·25³) = 32.59 MPa
Answer
τmax = 32.59 MPa
4.Torsion of shafts
Answer: φC = 1.592 °
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 250 N·m
TAB = TC + TB = 250 + 200 = 450 N·m
Twist of each segment
J = π·50⁴/32 = 613592 mm⁴
G = 39 GPa = 39000 MPa
φAB = 450000·1200/(39000·613592) = 0.02257 rad
φBC = 250000·500/(39000·613592) = 0.005224 rad
Twist of end C
φC = 0.02257 + 0.005224 = 0.02779 rad → |φC| = 1.592°
Answer
φC = 1.592°
5.Mohr’s circle & stress transformation
Answer: τxy = 63.25 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.
Required radius
C = (−40 + 20)/2 = −10 MPa
R = σ1 − C = 60 − (−10) = 70 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(70² − 30²) = 63.25 MPa
Answer
τxy = 63.25 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −8.481 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 = −15 + −65·−0.5 + −30·0.866 = −8.481 MPa
Answer
σθ = −8.481 MPa
7.Axial stress & strain
Answer: δ = 0.05568 mm
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
Ar = π·25²/4 = 490.9 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·4006 = 5.388 × 10⁸ N
Shortening
δ = 50000·600/5.388 × 10⁸ = 0.05568 mm
Answer
δ = 0.05568 mm
8.Axial stress & strain
Answer: σ2 = 32.51 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 = 200000·314.2/600 = 104720 N/mm
k₂ = 100000·706.9/500 = 141372 N/mm
Share of segment 2
F2 = 40000·141372/246091 = 22979 N
Stress
σ2 = 22979/706.9 = 32.51 MPa (compression)
Answer
σ2 = 32.51 MPa
9.Column buckling
Answer: Pcr = 44.29 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 0.7·2000 = 1400 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/1400² = 44295 N = 44.29 kN
Answer
Pcr = 44.29 kN
10.Column buckling
Answer: d = 101.5 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 = 3·75000 = 225000 N
KL = 1·4000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·225000·4000²/(π³·70000))^(1/4) = 101.5 mm
Answer
d = 101.5 mm
Check: the slenderness KL/r is 157.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 9.8 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.
Undamped natural frequency
ωn = √(50000/12) = 64.55 rad/s
fn = 64.55/(2π) = 10.27 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 10.27·√(1 − 0.3²) = 9.8 Hz
Answer
fd = 9.8 Hz
12.Mechanical vibrations
Answer: fn = 10.07 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 = 25 + 15 = 40 kN/m
Natural frequency
ωn = √(40000/10) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz
Answer
fn = 10.07 Hz
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