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 2.2 kW motor running at 960 rpm drives a pinion with 12 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 960 rpmN1 = 12N2 = 50η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (60 teeth) on the input shaft turns at 1500 rpm and drives gear B (72 teeth). Gear C (48 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 1200 rpm. How many teeth ND must gear D have?
nin = 1500 rpmA = 60B = 72C = 48nout = 1200 rpm
ND =
3.Torsion of shaftsMedium
A solid steel shaft 1 m long carries a torque of 1500 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.
T = 1500 N·mL = 1 mG = 80 GPaφallow = 1.5°
d = mm
4.Torsion of shaftsHard
A solid shaft must transmit 10 kW at 750 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 10 kWN = 750 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −70 MPa, σᵧ = −80 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −70 MPaσy = −80 MPaτxy = −20 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 90 MPa, σᵧ = −80 MPa and τₓᵧ = −40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 90 MPaσy = −80 MPaτxy = −40 MPaθ = 15°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the aluminium bar under the axial load.
P = 50 kNd = 12 mmL = 1.5 mE = 70 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 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, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 50 kN
σ2 = 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 = 1.5 mE = 200 GPa (steel)d = 20 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 100 kg mass is supported by a spring with a stiffness of 20 kN/m and a viscous damper with a damping coefficient of 570 N·s/m. Find the damping ratio ζ.
m = 100 kgk = 20 kN/mc = 570 N·s/m
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 10 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 88.45 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 2200/100.5 = 21.88 N·m
Output torque
T2 = 21.88·50/12·0.97 = 88.45 N·m
Answer
T2 = 88.45 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 = 1500·60/72 = 1250 rpm
Teeth on gear D
ND = C·nC/nout = 48·1250/1200 = 50
Answer
ND = 50
3.Torsion of shafts
Answer: d = 51.97 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 = 1500 N·m = 1.5 × 10⁶ N·mm
L = 1 m = 1000 mm
G = 80 GPa = 80000 MPa
φ = 1.5·π/180 = 0.02618 rad
J = TL/(Gφ) = 1.5 × 10⁶·1000/(80000·0.02618) = 716197 mm⁴
Diameter
d = (32J/π)1/4 = 51.97 mm
Answer
d = 51.97 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: d = 20.09 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·750/60 = 78.54 rad/s
T = P/ω = 10000/78.54 = 127.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·127324/(π·80)) = 20.09 mm
Answer
d = 20.09 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = −82.98 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −20/20.62 = −0.9701, sin 2θs = −5/20.62 = −0.2425
2θs = atan2(−5, −20) = −166°
Angle on the element
θs = −82.98°
check: θs − θp = −82.98 − (−37.98) = −45°
Answer
θs = −82.98°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = −77.14 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 = −85·0.5 + −40·0.866 = −77.14 MPa
Answer
τθ = −77.14 MPa
7.Axial stress & strain
Answer: δ = 9.474 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 = 50·1000/113.1 = 442.1 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 442.1/70000 = 0.006316 = 6316 µε
Elongation
δ = ε·L = 0.006316·1500 mm = 9.474 mm
Answer
δ = 9.474 mm
8.Axial stress & strain
Answer: σ2 = 28.61 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 = 200000·490.9/400 = 245437 N/mm
k₂ = 100000·1257/200 = 628319 N/mm
Share of segment 2
F2 = 50000·628319/873755 = 35955 N
Stress
σ2 = 35955/1257 = 28.61 MPa (compression)
Answer
σ2 = 28.61 MPa
9.Column buckling
Answer: Pcr = 27.56 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 0.5·1500 = 750 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·7854/750² = 27561 N = 27.56 kN
Answer
Pcr = 27.56 kN
10.Column buckling
Answer: σcr = 12.34 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 = 2·2000 = 4000 mm
KL/r = 4000/10 = 400
Critical stress
σcr = π²E/(KL/r)² = π²·200000/400² = 12.34 MPa
Answer
σcr = 12.34 MPa
11.Mechanical vibrations
Answer: ζ = 0.2015
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·√(20000·100) = 2828 N·s/m
Damping ratio
ζ = c/cc = 570/2828 = 0.2015
Answer
ζ = 0.2015
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 5.455 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/10 + 1/12 → keq = 10·12/(10 + 12) = 5.455 kN/m
Answer
keq = 5.455 kN/m
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