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 worm with 4 start(s) turns at 1440 rpm and drives a worm wheel with 80 teeth. Find the speed nw of the wheel.
z = 4Nw = 80n = 1440 rpm
nw = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turns at 3000 rpm and drives gear B (48 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 625 rpm. How many teeth ND must gear D have?
nin = 3000 rpmA = 20B = 48C = 12nout = 625 rpm
ND =
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 = 40 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 50 mm is fixed at A. A torque of 500 N·m acts at B, 1.5 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 2.5 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 500 N·mTC = 750 N·mAB = 1.5 mBC = 2.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 60 MPa, σᵧ = 0 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 = 60 MPaσy = 0 MPaτxy = −20 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 30 MPa, σᵧ = 80 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.
σx = 30 MPaσy = 80 MPaτxy = −20 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa). Both are 800 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.
P = 80 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 70 GPaL = 800 mm
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 = 115 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)b × h = 20 × 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 6 mE = 200 GPa (steel)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 8 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 170 N·s/m. Find the damping ratio ζ.
m = 8 kgk = 40 kN/mc = 170 N·s/m
ζ =
12.Mechanical vibrationsHard
A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 12 kN/m and k₂ = 50 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 = 25 kgk1 = 12 kN/mk2 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: nw = 72 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.
Wheel speed
nw = n·z/Nw = 1440·4/80 = 72 rpm
Answer
nw = 72 rpm
2.Gears & gear trains
Answer: ND = 24
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.
Intermediate shaft speed
nB = nC = 3000·20/48 = 1250 rpm
Teeth on gear D
ND = C·nC/nout = 12·1250/625 = 24
Answer
ND = 24
3.Torsion of shafts
Answer: φ = 0.7124 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·500/(80000·251327) = 0.01243 rad
φ = 0.01243·180/π = 0.7124°
Answer
φ = 0.7124°
4.Torsion of shafts
Answer: τmax = 50.93 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC + TB = 750 + 500 = 1250 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 1250 N·m
τmax = 16·1.25 × 10⁶/(π·50³) = 50.93 MPa
Answer
τmax = 50.93 MPa
5.Mohr’s circle & stress transformation
Answer: θp = −16.85 °
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, 60 − 0) = −33.69°
Angle on the element
θp = −16.85°
Answer
θp = −16.85°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 23.35 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 = 55 + −25·0.866 + −20·0.5 = 23.35 MPa
Answer
σθ = 23.35 MPa
7.Axial stress & strain
Answer: δ = 0.1517 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·30²/4 = 706.9 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 70·10³·4006 = 4.218 × 10⁸ N
Shortening
δ = 80000·800/4.218 × 10⁸ = 0.1517 mm
Answer
δ = 0.1517 mm
8.Axial stress & strain
Answer: δ = 4.258 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 119.5 MPa
σ2 = 366.1 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 119.5·500/100000 = 0.5976 mm
δ2 = σ2L2/E2 = 366.1·2000/200000 = 3.661 mm
δ = δ1 + δ2 = 4.258 mm
Answer
δ = 4.258 mm
9.Column buckling
Answer: Pcr = 18.42 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 = 40·20³/12 = 26667 mm⁴
Effective length
KL = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·26667/1000² = 18423 N = 18.42 kN
Answer
Pcr = 18.42 kN
10.Column buckling
Answer: σcr = 34.27 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.5·6000 = 3000 mm
KL/r = 3000/12.5 = 240
Critical stress
σcr = π²E/(KL/r)² = π²·200000/240² = 34.27 MPa
Answer
σcr = 34.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.1503
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·√(40000·8) = 1131 N·s/m
Damping ratio
ζ = c/cc = 170/1131 = 0.1503
Answer
ζ = 0.1503
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 7.926 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 = 12 + 50 = 62 kN/m
Natural frequency
ωn = √(62000/25) = 49.8 rad/s
fn = 49.8/(2π) = 7.926 Hz
Answer
fn = 7.926 Hz
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