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 150 mm pulley at 1500 rpm drives a 500 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
n1 = 1500 rpmd1 = 150 mmd2 = 500 mms = 2.5 %
n2 = rpm
2.Gears & gear trainsHard
A 2.2 kW motor at 2900 rpm drives gear A (20 teeth), which meshes with gear B (72 teeth). Gear C (30 teeth) on the same shaft as B drives gear D (60 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 2.2 kWnin = 2900 rpmA = 20B = 72C = 30D = 60ηmesh = 98%
Tout = N·m
3.Torsion of shaftsMedium
A solid aluminium 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 = 200 N·md = 30 mmL = 1 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 100 kW at 1000 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter do.
P = 100 kWN = 1000 rpmdi/do = 0.7τallow = 80 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −30 MPa and σᵧ = −20 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 35 MPa?
σx = −30 MPaσy = −20 MPaσ1 = 35 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −30 MPa, σᵧ = 50 MPa and τₓᵧ = 40 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 = 50 MPaτxy = 40 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 40 mm, length 400 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 400 mm, E = 70 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) and a segment made of bronze (diameter 60 mm, length 300 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaP = 50 kN
σ2 = MPa
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 = 1 mE = 70 GPa (aluminium)b × h = 30 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 70 GPa (aluminium)d = 40 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 3 cycles later by a peak of 10.3 mm. Find the damping ratio ζ.
x0 = 15 mmx3 = 10.3 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 40 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 50 kgk1 = 40 kN/mk2 = 40 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: n2 = 438.8 rpm
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.
Speed without slip
1500·150/500 = 450 rpm
With slip
n2 = 450·0.975 = 438.8 rpm
Answer
n2 = 438.8 rpm
2.Gears & gear trains
Answer: Tout = 50.09 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.
Output speed
nout = 2900·(20/72)·(30/60) = 402.8 rpm
ωout = 2π·402.8/60 = 42.18 rad/s
Output power and torque
Pout = 2200·0.98² = 2113 W
Tout = Pout/ωout = 2113/42.18 = 50.09 N·m
Answer
Tout = 50.09 N·m
3.Torsion of shafts
Answer: φ = 5.542 °
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) = 200·1000·1000/(26000·79522) = 0.09673 rad
φ = 0.09673·180/π = 5.542°
Answer
φ = 5.542°
4.Torsion of shafts
Answer: do = 43.09 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·1000/60 = 104.7 rad/s
T = P/ω = 100000/104.7 = 954.9 N·m
Outer diameter from the allowable stress
T = 954.9 N·m = 954930 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·954930/(π·80·0.7599)) = 43.09 mm
di = 0.7·43.09 = 30.16 mm
Answer
do = 43.09 mm
A solid shaft would need d = 39.32 mm; the hollow one is a little larger but much lighter.
5.Mohr’s circle & stress transformation
Answer: τxy = 59.79 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 + (−20))/2 = −25 MPa
R = σ1 − C = 35 − (−25) = 60 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(60² − 5²) = 59.79 MPa
Answer
τxy = 59.79 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = −4.641 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 = 10 + −40·0.866 + 40·0.5 = −4.641 MPa
Answer
σθ = −4.641 MPa
7.Axial stress & strain
Answer: P = 19.99 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 = 314.2 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(70000·1257) = 4.547 × 10⁻⁶ mm/N
Load
P = 0.25/1.251 × 10⁻⁵ = 19992 N = 19.99 kN
Answer
P = 19.99 kN
8.Axial stress & strain
Answer: σ2 = 13.89 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 = 200000·706.9/500 = 282743 N/mm
k₂ = 110000·2827/300 = 1.037 × 10⁶ N/mm
Share of segment 2
F2 = 50000·1.037 × 10⁶/1.319 × 10⁶ = 39286 N
Stress
σ2 = 39286/2827 = 13.89 MPa (compression)
Answer
σ2 = 13.89 MPa
9.Column buckling
Answer: Pcr = 19.43 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 = 50·30³/12 = 112500 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·112500/2000² = 19431 N = 19.43 kN
Answer
Pcr = 19.43 kN
10.Column buckling
Answer: σcr = 11.05 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.5·5000 = 2500 mm
KL/r = 2500/10 = 250
Critical stress
σcr = π²E/(KL/r)² = π²·70000/250² = 11.05 MPa
Answer
σcr = 11.05 MPa
11.Mechanical vibrations
Answer: ζ = 0.01994
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 = (1/3)·ln(15/10.3) = 0.1253
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1253/√(39.48 + 0.0157) = 0.01994
Answer
ζ = 0.01994
12.Mechanical vibrations
Answer: keq = 80 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 40 + 40 = 80 kN/m
Answer
keq = 80 kN/m
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