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 100 mm pulley at 1450 rpm drives a 315 mm pulley by a belt with 1.5 % slip. Find the speed n2 of the driven pulley.
n1 = 1450 rpmd1 = 100 mmd2 = 315 mms = 1.5 %
n2 = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (25 teeth) on the input shaft turns at 2880 rpm and drives gear B (32 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 1800 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 25B = 32C = 16nout = 1800 rpm
ND =
3.Torsion of shaftsMedium
A solid aluminium 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 = 750 N·md = 40 mmL = 1.2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 1500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.
P = 50 kWN = 1500 rpmτallow = 50 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 70 MPa, σᵧ = 70 MPa and τₓᵧ = −40 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 70 MPaσy = 70 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries a bending moment of 750 N·m together with a torque of 250 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τmax.
d = 50 mmM = 750 N·mT = 250 N·m
τmax = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the brass bar, in microstrain (µε = 10⁻⁶).
P = 110 kNd = 10 mmL = 0.5 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 = 35 kNsegment 1: Ø40 mm, 1.5 m, copper (E = 117 GPa)segment 2: Ø12 mm, 1.5 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 70 GPa (aluminium)b × h = 20 × 40 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 100 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 100 kNn = 2L = 6 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 12 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 80 N·s/m. Find the damping ratio ζ.
m = 12 kgk = 12 kN/mc = 80 N·s/m
ζ =
12.Mechanical vibrationsHard
A 50 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 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 = 50 kgk1 = 5 kN/mk2 = 15 kN/mk3 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 453.4 rpm
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.
Speed without slip
1450·100/315 = 460.3 rpm
With slip
n2 = 460.3·0.985 = 453.4 rpm
Answer
n2 = 453.4 rpm
2.Gears & gear trains
Answer: ND = 20
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 = 2880·25/32 = 2250 rpm
Teeth on gear D
ND = C·nC/nout = 16·2250/1800 = 20
Answer
ND = 20
3.Torsion of shafts
Answer: φ = 7.891 °
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·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1200/(26000·251327) = 0.1377 rad
φ = 0.1377·180/π = 7.891°
Answer
φ = 7.891°
4.Torsion of shafts
Answer: d = 31.89 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·1500/60 = 157.1 rad/s
T = P/ω = 50000/157.1 = 318.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·318310/(π·50)) = 31.89 mm
Answer
d = 31.89 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −45 °
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·−40, 70 − 70) = −90°
Angle on the element
θp = −45°
Answer
θp = −45°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τmax = 32.21 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) = 32·750000/(π·50³) = 61.12 MPa
τ = 16T/(πd³) = 16·250000/(π·50³) = 10.19 MPa
Mohr’s circle
C = σ/2 = 30.56 MPa
R = √(30.56² + 10.19²) = 32.21 MPa
Answer
τmax = 32.21 MPa
7.Axial stress & strain
Answer: ε = 14006 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·10²/4 = 78.54 mm²
σ = P/A = 110·1000/78.54 = 1401 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 1401/100000 = 0.01401 = 14006 µε
Answer
ε = 14006 µε
8.Axial stress & strain
Answer: σmax = 309.5 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 = 27.85 MPa
σ2 = 309.5 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 309.5 MPa
Answer
σmax = 309.5 MPa
9.Column buckling
Answer: Pcr = 4.606 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 = 40·20³/12 = 26667 mm⁴
Effective length
KL = 2·1000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·26667/2000² = 4606 N = 4.606 kN
Answer
Pcr = 4.606 kN
10.Column buckling
Answer: d = 65.65 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 = 2·100000 = 200000 N
KL = 0.5·6000 = 3000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·200000·3000²/(π³·200000))^(1/4) = 65.65 mm
Answer
d = 65.65 mm
Check: the slenderness KL/r is 182.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1054
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·√(12000·12) = 758.9 N·s/m
Damping ratio
ζ = c/cc = 80/758.9 = 0.1054
Answer
ζ = 0.1054
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.599 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 = 5 + 15 = 20 kN/m
keq = 20·40/(20 + 40) = 13.33 kN/m
Natural frequency
ωn = √(13333/50) = 16.33 rad/s
fn = 16.33/(2π) = 2.599 Hz
Answer
fn = 2.599 Hz
ExamProblems · examproblems.com/courses/machine-design