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 960 rpm drives a 400 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 150 mmd2 = 400 mms = 2.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 125 mm driving pulley at 960 rpm to a 300 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 125 mmn1 = 960 rpmd2 = 300 mmC = 600 mmμ = 0.3T1 = 500 N
P = kW
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 = 750 N·md = 50 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft must transmit 30 kW at 300 rpm. Its inner diameter is to be 0.8 times its outer diameter, and the shear stress must not exceed 80 MPa. Find the smallest outer diameter do.
P = 30 kWN = 300 rpmdi/do = 0.8τallow = 80 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 90 MPa, σᵧ = 40 MPa and τₓᵧ = 10 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 90 MPaσy = 40 MPaτxy = 10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 70 MPa, σᵧ = 40 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 70 MPaσy = 40 MPaτxy = 40 MPaθ = 45°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 100 kNd = 12 mmL = 1.5 mE = 100 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.
P = 45 kNsegment 1: Ø30 mm, 2 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 0.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 200 GPa (steel)b × h = 40 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 5 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 5 kNn = 3L = 1 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 20 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 270 N·s/m. Find the damping ratio ζ.
m = 20 kgk = 40 kN/mc = 270 N·s/m
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 60 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: n2 = 351 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
960·150/400 = 360 rpm
With slip
n2 = 360·0.975 = 351 rpm
Answer
n2 = 351 rpm
2.Gears & gear trains
Answer: P = 1.805 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.
Angle of wrap
θ = 2.849 rad
Slack-side tension
eμθ = e0.8547 = 2.351, T₂ = 500/2.351 = 212.7 N
Power
P = (500 − 212.7)·6.283 = 1.805 kW
Answer
P = 1.805 kW
3.Torsion of shafts
Answer: φ = 1.751 °
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) = 750·1000·2000/(80000·613592) = 0.03056 rad
φ = 0.03056·180/π = 1.751°
Answer
φ = 1.751°
4.Torsion of shafts
Answer: do = 46.87 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·300/60 = 31.42 rad/s
T = P/ω = 30000/31.42 = 954.9 N·m
Outer diameter from the allowable stress
T = 954.9 N·m = 954930 N·mm
1 − k⁴ = 1 − 0.8⁴ = 0.5904
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·954930/(π·80·0.5904)) = 46.87 mm
di = 0.8·46.87 = 37.5 mm
Answer
do = 46.87 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: θp = 10.9 °
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(2·10, 90 − 40) = 21.8°
Angle on the element
θp = 10.9°
Answer
θp = 10.9°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = −15 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
τθ = −15·1 + 40·0 = −15 MPa
Answer
τθ = −15 MPa
7.Axial stress & strain
Answer: ε = 8842 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·12²/4 = 113.1 mm²
σ = P/A = 100·1000/113.1 = 884.2 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 884.2/100000 = 0.008842 = 8842 µε
Answer
ε = 8842 µε
8.Axial stress & strain
Answer: σmax = 397.9 MPa
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·30²/4 = 706.9 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 63.66 MPa
σ2 = 397.9 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 397.9 MPa
Answer
σmax = 397.9 MPa
9.Column buckling
Answer: Pcr = 374.3 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 = 80·40³/12 = 426667 mm⁴
Effective length
KL = 0.5·3000 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·426667/1500² = 374314 N = 374.3 kN
Answer
Pcr = 374.3 kN
10.Column buckling
Answer: d = 19.84 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 = 3·5000 = 15000 N
KL = 1·1000 = 1000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·15000·1000²/(π³·200000))^(1/4) = 19.84 mm
Answer
d = 19.84 mm
Check: the slenderness KL/r is 201.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1509
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·√(40000·20) = 1789 N·s/m
Damping ratio
ζ = c/cc = 270/1789 = 0.1509
Answer
ζ = 0.1509
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 140 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 = 60 + 80 = 140 kN/m
Answer
keq = 140 kN/m
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