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 1.1 kW motor running at 960 rpm drives a pinion with 25 teeth, which meshes with a gear with 30 teeth. The gear pair has an efficiency of 97%. Find the torque T1 on the motor shaft.
P = 1.1 kWn1 = 960 rpmN1 = 25N2 = 30η = 97%
T1 = N·m
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 250 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 250 mmC = 500 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 = 400 N·md = 30 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 750 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 50 kWN = 750 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa and σᵧ = 10 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 30 MPa?
σx = −30 MPaσy = 10 MPaσ1 = 30 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = −10 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° clockwise from the x-axis.
σx = 50 MPaσy = −10 MPaτxy = 20 MPaθ = -30°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the steel bar, in microstrain (µε = 10⁻⁶).
P = 70 kNd = 10 mmL = 3 mE = 200 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 60 kNsegment 1: Ø35 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 1 m, copper (E = 117 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)b × h = 40 × 60 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 70 GPa (aluminium)d = 50 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 5 cycles later by a peak of 4.1 mm. Find the damping ratio ζ.
x0 = 20 mmx5 = 4.1 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 30 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 10 kgk1 = 25 kN/mk2 = 30 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 10.94 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 1100/100.5 = 10.94 N·m
Answer
T1 = 10.94 N·m
2.Gears & gear trains
Answer: θ = 168.5 °
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(0.1) = 5.739°
Angle of wrap
θ = 180 − 2·5.739 = 168.5°
Answer
θ = 168.5°
3.Torsion of shafts
Answer: φ = 2.882 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·800/(80000·79522) = 0.0503 rad
φ = 0.0503·180/π = 2.882°
Answer
φ = 2.882°
4.Torsion of shafts
Answer: d = 35.1 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·750/60 = 78.54 rad/s
T = P/ω = 50000/78.54 = 636.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·636620/(π·75)) = 35.1 mm
Answer
d = 35.1 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τxy = 34.64 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 = (−30 + 10)/2 = −10 MPa
R = σ1 − C = 30 − (−10) = 40 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(40² − 20²) = 34.64 MPa
Answer
τxy = 34.64 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 17.68 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 = 20 + 30·0.5 + 20·−0.866 = 17.68 MPa
Answer
σθ = 17.68 MPa
7.Axial stress & strain
Answer: ε = 4456 µε
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 = 70·1000/78.54 = 891.3 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 891.3/200000 = 0.004456 = 4456 µε
Answer
ε = 4456 µε
8.Axial stress & strain
Answer: δ = 3.414 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 = 62.36 MPa
σ2 = 191 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 62.36·2000/70000 = 1.782 mm
δ2 = σ2L2/E2 = 191·1000/117000 = 1.632 mm
δ = δ1 + δ2 = 3.414 mm
Answer
δ = 3.414 mm
9.Column buckling
Answer: Pcr = 13.82 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 = 60·40³/12 = 320000 mm⁴
Effective length
KL = 2·2000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·320000/4000² = 13817 N = 13.82 kN
Answer
Pcr = 13.82 kN
10.Column buckling
Answer: σcr = 47.98 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 = 1·1500 = 1500 mm
KL/r = 1500/12.5 = 120
Critical stress
σcr = π²E/(KL/r)² = π²·70000/120² = 47.98 MPa
Answer
σcr = 47.98 MPa
11.Mechanical vibrations
Answer: ζ = 0.05038
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 = (1/5)·ln(20/4.1) = 0.3169
Damping ratio
ζ = δ/√(4π² + δ²) = 0.3169/√(39.48 + 0.1005) = 0.05038
Answer
ζ = 0.05038
12.Mechanical vibrations
Answer: keq = 13.64 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 = 1/25 + 1/30 → keq = 25·30/(25 + 30) = 13.64 kN/m
Answer
keq = 13.64 kN/m
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