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 an 80 mm driving pulley at 720 rpm. The tension is 1200 N on the tight side and 480 N on the slack side. Find the power P transmitted, in kW.
d1 = 80 mmn1 = 720 rpmT1 = 1200 NT2 = 480 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 200 mm pulley and a 500 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 200 mmd2 = 500 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 = 500 N·md = 25 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 50 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.
P = 50 kWN = 300 rpmdo = 100 mmdi = 80 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = 50 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −20 MPaσy = 50 MPaτxy = −30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 25 mm carries a bending moment of 300 N·m together with a torque of 100 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.
d = 25 mmM = 300 N·mT = 100 N·m
σ1 = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 15 kNd = 12 mmL = 0.5 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø20 mm, E = 200 GPa, σallow = 200 MPatube: Ø100/Ø50 mm, E = 120 GPa, σallow = 60 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)b × h = 20 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is pinned at both ends. It must carry 30 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 30 kNn = 2.5L = 1.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 5 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 50 N·s/m. Find the damping ratio ζ.
m = 5 kgk = 50 kN/mc = 50 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 60 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 = 10 kgk1 = 20 kN/mk2 = 60 kN/mk3 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 2.171 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 = 3.016 m/s
Power
P = (1200 − 480)·3.016 = 2171 W = 2.171 kW
Answer
P = 2.171 kW
2.Gears & gear trains
Answer: θ = 145.1 °
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(0.3) = 17.46°
Angle of wrap
θ = 180 − 2·17.46 = 145.1°
Answer
θ = 145.1°
3.Torsion of shafts
Answer: φ = 7.47 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 500·1000·800/(80000·38350) = 0.1304 rad
φ = 0.1304·180/π = 7.47°
Answer
φ = 7.47°
4.Torsion of shafts
Answer: τmax = 13.73 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.
Torque
ω = 2πN/60 = 2π·300/60 = 31.42 rad/s
T = P/ω = 50000/31.42 = 1592 N·m
Polar second moment of area
J = π(100⁴ − 80⁴)/32 = 5.796 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 1.592 × 10⁶·50/5.796 × 10⁶ = 13.73 MPa
Answer
τmax = 13.73 MPa
5.Mohr’s circle & stress transformation
Answer: θp = −69.7 °
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·−30, −20 − 50) = −139.4°
Angle on the element
θp = −69.7°
Answer
θp = −69.7°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 200.9 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.
Stresses at the surface
σ = 32M/(πd³) = 32·300000/(π·25³) = 195.6 MPa
τ = 16T/(πd³) = 16·100000/(π·25³) = 32.59 MPa
Mohr’s circle
C = σ/2 = 97.78 MPa
R = √(97.78² + 32.59²) = 103.1 MPa
σ1 = C + R = 200.9 MPa
σ2 = C − R = −5.289 MPa
Answer
σ1 = 200.9 MPa
7.Axial stress & strain
Answer: δ = 0.6631 mm
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 = 15·1000/113.1 = 132.6 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 132.6/100000 = 0.001326 = 1326 µε
Elongation
δ = ε·L = 0.001326·500 mm = 0.6631 mm
Answer
δ = 0.6631 mm
8.Axial stress & strain
Answer: Pmax = 384.8 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
Ar = π·20²/4 = 314.2 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 120·10³·5890 = 7.697 × 10⁸ N
Governing strain
ε = min(0.001, 0.0005) = 0.0005 (copper)
Largest load
Pmax = 0.0005·7.697 × 10⁸ = 384845 N = 384.8 kN
Answer
Pmax = 384.8 kN
9.Column buckling
Answer: Pcr = 3.158 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 = 60·20³/12 = 40000 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·40000/5000² = 3158 N = 3.158 kN
Answer
Pcr = 3.158 kN
10.Column buckling
Answer: d = 47.23 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 = 2.5·30000 = 75000 N
KL = 1·1500 = 1500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·75000·1500²/(π³·70000))^(1/4) = 47.23 mm
Answer
d = 47.23 mm
Check: the slenderness KL/r is 127, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.05
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·√(50000·5) = 1000 N·s/m
Damping ratio
ζ = c/cc = 50/1000 = 0.05
Answer
ζ = 0.05
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 10.07 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 = 20 + 60 = 80 kN/m
keq = 80·80/(80 + 80) = 40 kN/m
Natural frequency
ωn = √(40000/10) = 63.25 rad/s
fn = 63.25/(2π) = 10.07 Hz
Answer
fn = 10.07 Hz
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