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 15 kW motor running at 1500 rpm drives a pinion with 48 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 15 kWn1 = 1500 rpmN1 = 48N2 = 50η = 98%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (50 teeth) on the input shaft turning at 2900 rpm drives gear B (72 teeth). Gear C (60 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
nin = 2900 rpmA = 50B = 72C = 60D = 80
nout = rpm
3.Torsion of shaftsMedium
A solid steel 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 = 200 N·md = 40 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 2000 N·m is applied at C, 2 m from A and 1.2 m from B. Find the maximum shear stress τmax in the shaft.
d = 30 mmT0 = 2000 N·ma = 2 mb = 1.2 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 160 MPa, σᵧ = −80 MPa and τₓᵧ = −60 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 = 160 MPaσy = −80 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −30 MPa, σᵧ = 90 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −30 MPaσy = 90 MPaτxy = 40 MPaθ = 60°
τθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 20 kNd = 30 mmL = 1.5 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 20 kNsegment 1: Ø35 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, brass (E = 100 GPa)
δ = mm
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 = 4 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. 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 100 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 1580 N·s/m. Find the damping ratio ζ.
m = 100 kgk = 25 kN/mc = 1580 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from spring k₃ = 25 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 = 60 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 = 5 kgk1 = 50 kN/mk2 = 60 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 97.48 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 = 15000/157.1 = 95.49 N·m
Output torque
T2 = 95.49·50/48·0.98 = 97.48 N·m
Answer
T2 = 97.48 N·m
2.Gears & gear trains
Answer: nout = 1510 rpm
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.
First stage
nB = 2900·50/72 = 2014 rpm = nC
Second stage
nout = 2014·60/80 = 1510 rpm
Answer
nout = 1510 rpm
total ratio = 1.92 : 1
3.Torsion of shafts
Answer: φ = 1.425 °
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) = 200·1000·2500/(80000·251327) = 0.02487 rad
φ = 0.02487·180/π = 1.425°
Answer
φ = 1.425°
4.Torsion of shafts
Answer: τmax = 235.8 MPa
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 2000
TA·a = TB·b
TA = 2000·1.2/3.2 = 750 N·m
TB = 2000·2/3.2 = 1250 N·m
Maximum shear stress
The larger of the two torques governs (same diameter).
τmax = 16·1.25 × 10⁶/(π·30³) = 235.8 MPa
Answer
τmax = 235.8 MPa
5.Mohr’s circle & stress transformation
Answer: θs = −58.28 °
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 = −60/134.2 = −0.4472, sin 2θs = −120/134.2 = −0.8944
2θs = atan2(−120, −60) = −116.6°
Angle on the element
θs = −58.28°
check: θs − θp = −58.28 − (−13.28) = −45°
Answer
θs = −58.28°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: τθ = 31.96 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 = −−60·0.866 + 40·−0.5 = 31.96 MPa
Answer
τθ = 31.96 MPa
7.Axial stress & strain
Answer: ε = 404.2 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·30²/4 = 706.9 mm²
σ = P/A = 20·1000/706.9 = 28.29 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 28.29/70000 = 0.0004042 = 404.2 µε
Answer
ε = 404.2 µε
8.Axial stress & strain
Answer: δ = 2.184 mm
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·12²/4 = 113.1 mm²
σ1 = 20.79 MPa
σ2 = 176.8 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 20.79·2000/100000 = 0.4158 mm
δ2 = σ2L2/E2 = 176.8·1000/100000 = 1.768 mm
δ = δ1 + δ2 = 2.184 mm
Answer
δ = 2.184 mm
9.Column buckling
Answer: Pcr = 151.4 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 round section has the same I about every axis.
I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴
Effective length
KL = 0.5·4000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·306796/2000² = 151398 N = 151.4 kN
Answer
Pcr = 151.4 kN
10.Column buckling
Answer: σcr = 2.429 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 = 2·4000 = 8000 mm
KL/r = 8000/15 = 533.3
Critical stress
σcr = π²E/(KL/r)² = π²·70000/533.3² = 2.429 MPa
Answer
σcr = 2.429 MPa
11.Mechanical vibrations
Answer: ζ = 0.4996
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·√(25000·100) = 3162 N·s/m
Damping ratio
ζ = c/cc = 1580/3162 = 0.4996
Answer
ζ = 0.4996
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 10.16 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 + 60 = 110 kN/m
keq = 110·25/(110 + 25) = 20.37 kN/m
Natural frequency
ωn = √(20370/5) = 63.83 rad/s
fn = 63.83/(2π) = 10.16 Hz
Answer
fn = 10.16 Hz
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