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 1450 rpm drives a pinion with 12 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 1.1 kWn1 = 1450 rpmN1 = 12N2 = 72η = 98%
T1 = N·m
2.Gears & gear trainsHard
A 7.5 kW motor at 2900 rpm drives gear A (25 teeth), which meshes with gear B (30 teeth). Gear C (36 teeth) on the same shaft as B drives gear D (50 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 7.5 kWnin = 2900 rpmA = 25B = 30C = 36D = 50ηmesh = 98%
Tout = N·m
3.Torsion of shaftsMedium
A solid steel shaft 1.5 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 1°. Find the smallest diameter d that meets this limit.
T = 1000 N·mL = 1.5 mG = 80 GPaφallow = 1°
d = mm
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 1800 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 5 kWN = 1800 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 50 MPa, σᵧ = 20 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 50 MPaσy = 20 MPaτxy = −50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 5 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 500 mmt = 5 mmp = 1.5 MPaα = 25°
σn = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 120 kNd = 16 mmL = 2 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 = 70 kNsegment 1: Ø40 mm, 2 m, steel (E = 200 GPa)segment 2: Ø12 mm, 0.5 m, aluminium (E = 70 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 = 5 mE = 200 GPa (steel)b × h = 40 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 10 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 10 kNn = 2.5L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
An 8 kg mass on a spring with a stiffness of 25 kN/m has a viscous damper giving a damping ratio of 0.5. Find the damped natural frequency fd of its free vibration in Hz.
m = 8 kgk = 25 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 25 kN/mk2 = 80 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 7.244 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·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 1100/151.8 = 7.244 N·m
Answer
T1 = 7.244 N·m
2.Gears & gear trains
Answer: Tout = 39.53 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·(25/30)·(36/50) = 1740 rpm
ωout = 2π·1740/60 = 182.2 rad/s
Output power and torque
Pout = 7500·0.98² = 7203 W
Tout = Pout/ωout = 7203/182.2 = 39.53 N·m
Answer
Tout = 39.53 N·m
3.Torsion of shafts
Answer: d = 57.51 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.
Required polar moment
T = 1000 N·m = 1 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = 1·π/180 = 0.01745 rad
J = TL/(Gφ) = 1 × 10⁶·1500/(80000·0.01745) = 1.074 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 57.51 mm
Answer
d = 57.51 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 13.11 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·1800/60 = 188.5 rad/s
T = P/ω = 5000/188.5 = 26.53 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·26526/(π·60)) = 13.11 mm
Answer
d = 13.11 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −36.65 °
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·−50, 50 − 20) = −73.3°
Angle on the element
θp = −36.65°
Answer
θp = −36.65°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σn = 68.3 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 500/2 = 250 mm
σh = pr/t = 1.5·250/5 = 75 MPa
σa = pr/(2t) = 37.5 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 37.5·sin²25° + 75·cos²25° = 68.3 MPa
Answer
σn = 68.3 MPa
7.Axial stress & strain
Answer: ε = 5968 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·16²/4 = 201.1 mm²
σ = P/A = 120·1000/201.1 = 596.8 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 596.8/100000 = 0.005968 = 5968 µε
Answer
ε = 5968 µε
8.Axial stress & strain
Answer: σmax = 618.9 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 55.7 MPa
σ2 = 618.9 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 618.9 MPa
Answer
σmax = 618.9 MPa
9.Column buckling
Answer: Pcr = 168.4 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 = 100·40³/12 = 533333 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·533333/2500² = 168441 N = 168.4 kN
Answer
Pcr = 168.4 kN
10.Column buckling
Answer: d = 55.21 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 = 2.5·10000 = 25000 N
KL = 1·6000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·25000·6000²/(π³·200000))^(1/4) = 55.21 mm
Answer
d = 55.21 mm
Check: the slenderness KL/r is 434.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 7.705 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.
Undamped natural frequency
ωn = √(25000/8) = 55.9 rad/s
fn = 55.9/(2π) = 8.897 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 8.897·√(1 − 0.5²) = 7.705 Hz
Answer
fd = 7.705 Hz
12.Mechanical vibrations
Answer: keq = 105 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 = 25 + 80 = 105 kN/m
Answer
keq = 105 kN/m
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