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 belt runs on a 125 mm driving pulley at 2880 rpm. The tension is 400 N on the tight side and 80 N on the slack side. Find the power P transmitted, in kW.
d1 = 125 mmn1 = 2880 rpmT1 = 400 NT2 = 80 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 125 mm pulley and a 300 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 125 mmd2 = 300 mmC = 1200 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 = 250 N·md = 50 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 0.5 m from A, and a torque of 100 N·m in the same direction acts at the free end C, a further 3 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 50 mmTB = 200 N·mTC = 100 N·mAB = 0.5 mBC = 3 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −20 MPaσy = −60 MPaτxy = 20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −20 MPa, σᵧ = −60 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
σx = −20 MPaσy = −60 MPaτxy = 10 MPaθ = 30°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 500 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.
segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: copper, Ø60 mm, 500 mm, E = 120 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 85 kNsegment 1: Ø25 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 0.5 m, copper (E = 117 GPa)
δ = mm
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 = 2 mE = 200 GPa (steel)d = 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 200 GPa (steel)d = 80 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed 4 cycles later by a peak of 7.3 mm. Find the damping ratio ζ.
x0 = 20 mmx4 = 7.3 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 100 kN/m and k₂ = 40 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 = 100 kgk1 = 100 kN/mk2 = 40 kN/mk3 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 6.032 kW
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 speed
v = 18.85 m/s
Power
P = (400 − 80)·18.85 = 6032 W = 6.032 kW
Answer
P = 6.032 kW
2.Gears & gear trains
Answer: θ = 171.6 °
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(0.07292) = 4.182°
Angle of wrap
θ = 180 − 2·4.182 = 171.6°
Answer
θ = 171.6°
3.Torsion of shafts
Answer: φ = 0.5836 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 250·1000·2000/(80000·613592) = 0.01019 rad
φ = 0.01019·180/π = 0.5836°
Answer
φ = 0.5836°
4.Torsion of shafts
Answer: φC = 0.5252 °
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 = 100 N·m
TAB = TC + TB = 100 + 200 = 300 N·m
Twist of each segment
J = π·50⁴/32 = 613592 mm⁴
G = 80 GPa = 80000 MPa
φAB = 300000·500/(80000·613592) = 0.003056 rad
φBC = 100000·3000/(80000·613592) = 0.006112 rad
Twist of end C
φC = 0.003056 + 0.006112 = 0.009167 rad → |φC| = 0.5252°
Answer
φC = 0.5252°
5.Mohr’s circle & stress transformation
Answer: τmax = 28.28 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.
Radius of Mohr’s circle
τmax = R = √(((−20 − −60)/2)² + 20²) = 28.28 MPa
Answer
τmax = 28.28 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −21.34 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 = −40 + 20·0.5 + 10·0.866 = −21.34 MPa
Answer
σθ = −21.34 MPa
7.Axial stress & strain
Answer: P = 55.62 kN
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
A₁ = 706.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(120000·2827) = 1.474 × 10⁻⁶ mm/N
Load
P = 0.2/3.596 × 10⁻⁶ = 55622 N = 55.62 kN
Answer
P = 55.62 kN
8.Axial stress & strain
Answer: δ = 4.867 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·25²/4 = 490.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 173.2 MPa
σ2 = 270.6 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 173.2·1500/70000 = 3.711 mm
δ2 = σ2L2/E2 = 270.6·500/117000 = 1.156 mm
δ = δ1 + δ2 = 4.867 mm
Answer
δ = 4.867 mm
9.Column buckling
Answer: Pcr = 126.6 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 = π²·200000·125664/1400² = 126556 N = 126.6 kN
Answer
Pcr = 126.6 kN
10.Column buckling
Answer: σcr = 64.45 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Radius of gyration and slenderness
A = 5027 mm²
r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm
KL = 0.7·5000 = 3500 mm
KL/r = 3500/20 = 175
Critical stress
σcr = π²E/(KL/r)² = π²·200000/175² = 64.45 MPa
Answer
σcr = 64.45 MPa
11.Mechanical vibrations
Answer: ζ = 0.04007
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 = (1/4)·ln(20/7.3) = 0.252
Damping ratio
ζ = δ/√(4π² + δ²) = 0.252/√(39.48 + 0.06349) = 0.04007
Answer
ζ = 0.04007
12.Mechanical vibrations
Answer: fn = 2.807 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 = 100 + 40 = 140 kN/m
keq = 140·40/(140 + 40) = 31.11 kN/m
Natural frequency
ωn = √(31111/100) = 17.64 rad/s
fn = 17.64/(2π) = 2.807 Hz
Answer
fn = 2.807 Hz
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