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 3 kW motor running at 1450 rpm drives a pinion with 20 teeth, which meshes with a gear with 32 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 1450 rpmN1 = 20N2 = 32η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turns at 1500 rpm and drives gear B (18 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 400 rpm. How many teeth ND must gear D have?
nin = 1500 rpmA = 15B = 18C = 16nout = 400 rpm
ND =
3.Torsion of shaftsMedium
A solid brass 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 = 50 mmL = 0.8 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 100 kWN = 1450 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = 160 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = 160 MPaσy = 110 MPaτxy = 30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 20 MPa, σᵧ = 90 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° clockwise from the x-axis.
σx = 20 MPaσy = 90 MPaτxy = 30 MPaθ = -45°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 30 kNd = 10 mmL = 1 mE = 117 GPa
δ = mm
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 = 85 kNsegment 1: Ø40 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 25 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 = 4 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 15 kg mass on a spring with a stiffness of 15 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
m = 15 kgk = 15 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 15 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 = 50 kN/mk2 = 15 kN/mk3 = 15 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 30.03 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 = 3000/151.8 = 19.76 N·m
Output torque
T2 = 19.76·32/20·0.95 = 30.03 N·m
Answer
T2 = 30.03 N·m
2.Gears & gear trains
Answer: ND = 50
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.
Intermediate shaft speed
nB = nC = 1500·15/18 = 1250 rpm
Teeth on gear D
ND = C·nC/nout = 16·1250/400 = 50
Answer
ND = 50
3.Torsion of shafts
Answer: φ = 0.7662 °
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) = 400·1000·800/(39000·613592) = 0.01337 rad
φ = 0.01337·180/π = 0.7662°
Answer
φ = 0.7662°
4.Torsion of shafts
Answer: d = 38.24 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·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·658572/(π·60)) = 38.24 mm
Answer
d = 38.24 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τabs = 87.03 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = 135 MPa, R = √(25² + 30²) = 39.05 MPa
σ1 = 174.1 MPa, σ2 = 95.95 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(78.1, 174.1, 95.95)/2 = 87.03 MPa
Answer
τabs = 87.03 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
6.Mohr’s circle & stress transformation
Answer: σθ = 25 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 = 55 + −35·0 + 30·−1 = 25 MPa
Answer
σθ = 25 MPa
7.Axial stress & strain
Answer: δ = 3.265 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·10²/4 = 78.54 mm²
σ = P/A = 30·1000/78.54 = 382 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 382/117000 = 0.003265 = 3265 µε
Elongation
δ = ε·L = 0.003265·1000 mm = 3.265 mm
Answer
δ = 3.265 mm
8.Axial stress & strain
Answer: σmax = 422.8 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·16²/4 = 201.1 mm²
σ1 = 67.64 MPa
σ2 = 422.8 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 422.8 MPa
Answer
σmax = 422.8 MPa
9.Column buckling
Answer: Pcr = 4.828 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 round section has the same I about every axis.
I = πd⁴/64 = π·25⁴/64 = 19175 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/2800² = 4828 N = 4.828 kN
Answer
Pcr = 4.828 kN
10.Column buckling
Answer: σcr = 38.86 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/15 = 133.3
Critical stress
σcr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa
Answer
σcr = 38.86 MPa
11.Mechanical vibrations
Answer: fd = 4.613 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 = √(15000/15) = 31.62 rad/s
fn = 31.62/(2π) = 5.033 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 5.033·√(1 − 0.4²) = 4.613 Hz
Answer
fd = 4.613 Hz
12.Mechanical vibrations
Answer: fn = 5.556 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 = 50 + 15 = 65 kN/m
keq = 65·15/(65 + 15) = 12.19 kN/m
Natural frequency
ωn = √(12188/10) = 34.91 rad/s
fn = 34.91/(2π) = 5.556 Hz
Answer
fn = 5.556 Hz
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