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 1500 rpm drives a pinion with 36 teeth, which meshes with a gear with 48 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 2.2 kWn1 = 1500 rpmN1 = 36N2 = 48η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a planetary gear set the sun gear has 18 teeth and each planet gear has 20 teeth. The planet carrier is held fixed and the sun is driven at 3000 rpm. Find the speed nr of the ring gear (magnitude).
Ns = 18Np = 20ns = 3000 rpm
nr = rpm
3.Torsion of shaftsMedium
A stepped solid brass shaft is fixed at A. Segment AB has diameter 75 mm and length 2.5 m; segment BC has diameter 40 mm and length 1.5 m. A torque of 2000 N·m acts at the free end C. Find the angle of twist φC of end C, in degrees.
T = 2000 N·md1 = 75 mm, L1 = 2.5 md2 = 40 mm, L2 = 1.5 mG = 39 GPa
φC = °
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 1800 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.
P = 50 kWN = 1800 rpmτallow = 40 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 20 MPa, σᵧ = 60 MPa and τₓᵧ = −20 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 = 60 MPaτxy = −20 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −40 MPa, σᵧ = −30 MPa and τₓᵧ = 50 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 = −40 MPaσy = −30 MPaτxy = 50 MPaθ = 15°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 20 kNd = 20 mmL = 3 mE = 100 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 = 115 kNsegment 1: Ø35 mm, 1 m, steel (E = 200 GPa)segment 2: Ø20 mm, 0.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 70 GPa (aluminium)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 20 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 = 2 kgk = 20 kN/mζ = 0.5
fd = Hz
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 40 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 20 kgk1 = 60 kN/mk2 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 18.11 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 2200/157.1 = 14.01 N·m
Output torque
T2 = 14.01·48/36·0.97 = 18.11 N·m
Answer
T2 = 18.11 N·m
2.Gears & gear trains
Answer: nr = 931 rpm
Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.
Ring gear teeth
Nr = Ns + 2·Np = 18 + 2·20 = 58
Willis equation with the carrier fixed
With nc = 0 the sun drives the ring through the planets, which act as idlers.
nr = −ns·Ns/Nr = −3000·18/58 = −931 rpm
|nr| = 931 rpm
Answer
nr = 931 rpm
The ring turns opposite to the sun.
3.Torsion of shafts
Answer: φC = 19.9 °
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 moments
J1 = π·75⁴/32 = 3.106 × 10⁶ mm⁴
J2 = π·40⁴/32 = 251327 mm⁴
Twist of each segment
T = 2000 N·m = 2 × 10⁶ N·mm
G = 39 GPa = 39000 MPa
φAB = 2 × 10⁶·2500/(39000·3.106 × 10⁶) = 0.04127 rad
φBC = 2 × 10⁶·1500/(39000·251327) = 0.3061 rad
Total twist
φC = 0.04127 + 0.3061 = 0.3473 rad = 19.9°
Answer
φC = 19.9°
4.Torsion of shafts
Answer: d = 32.32 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 = 50000/188.5 = 265.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·265258/(π·40)) = 32.32 mm
Answer
d = 32.32 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −67.5 °
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·−20, 20 − 60) = −135°
Angle on the element
θp = −67.5°
Answer
θp = −67.5°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = 45.8 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 = −−5·0.5 + 50·0.866 = 45.8 MPa
Answer
τθ = 45.8 MPa
7.Axial stress & strain
Answer: δ = 1.91 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·20²/4 = 314.2 mm²
σ = P/A = 20·1000/314.2 = 63.66 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 63.66/100000 = 0.0006366 = 636.6 µε
Elongation
δ = ε·L = 0.0006366·3000 mm = 1.91 mm
Answer
δ = 1.91 mm
8.Axial stress & strain
Answer: σmax = 366.1 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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 119.5 MPa
σ2 = 366.1 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 366.1 MPa
Answer
σmax = 366.1 MPa
9.Column buckling
Answer: Pcr = 13.89 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/2500² = 13891 N = 13.89 kN
Answer
Pcr = 13.89 kN
10.Column buckling
Answer: σcr = 3.172 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.7·5000 = 3500 mm
KL/r = 3500/7.5 = 466.7
Critical stress
σcr = π²E/(KL/r)² = π²·70000/466.7² = 3.172 MPa
Answer
σcr = 3.172 MPa
11.Mechanical vibrations
Answer: fd = 13.78 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 = √(20000/2) = 100 rad/s
fn = 100/(2π) = 15.92 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 15.92·√(1 − 0.5²) = 13.78 Hz
Answer
fd = 13.78 Hz
12.Mechanical vibrations
Answer: fn = 11.25 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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 60 + 40 = 100 kN/m
Natural frequency
ωn = √(100000/20) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz
Answer
fn = 11.25 Hz
ExamProblems · examproblems.com/courses/machine-design