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 worm with 1 start(s) turns at 960 rpm and drives a worm wheel with 48 teeth. Find the speed nw of the wheel.
z = 1Nw = 48n = 960 rpm
nw = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (15 teeth) on the input shaft turning at 1450 rpm drives gear B (50 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (24 teeth) on the output shaft. Find the output speed nout.
nin = 1450 rpmA = 15B = 50C = 12D = 24
nout = rpm
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 400 N·m. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
T = 400 N·mdi/do = 0.7τallow = 60 MPa
do = mm
4.Torsion of shaftsHard
A hollow shaft transmits 10 kW at 500 rpm. Find the maximum shear stress τmax in the shaft.
P = 10 kWN = 500 rpmdo = 50 mmdi = 25 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −40 MPa, σᵧ = −10 MPa and τₓᵧ = 50 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 = −40 MPaσy = −10 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −60 MPa, σᵧ = −60 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = −60 MPaσy = −60 MPaτxy = 20 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).
P = 90 kNd = 10 mmL = 1.5 mE = 117 GPa
ε = µε
8.Axial stress & strainHard
A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 60 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø25 mm, E = 200 GPa, σallow = 160 MPatube: Ø100/Ø60 mm, E = 100 GPa, σallow = 70 MPa
Pmax = kN
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 = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is pinned at both ends. It must carry 40 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 40 kNn = 2L = 2.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed 3 cycles later by a peak of 2 mm. Find the damping ratio ζ.
x0 = 20 mmx3 = 2 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 20 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 20 kgk1 = 20 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: nw = 20 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.
Wheel speed
nw = n·z/Nw = 960·1/48 = 20 rpm
Answer
nw = 20 rpm
2.Gears & gear trains
Answer: nout = 217.5 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 = 1450·15/50 = 435 rpm = nC
Second stage
nout = 435·12/24 = 217.5 rpm
Answer
nout = 217.5 rpm
total ratio = 6.667 : 1
3.Torsion of shafts
Answer: do = 35.48 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.
Outer diameter from the allowable stress
T = 400 N·m = 400000 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·400000/(π·60·0.7599)) = 35.48 mm
di = 0.7·35.48 = 24.84 mm
Answer
do = 35.48 mm
A solid shaft would need d = 32.38 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 8.3 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.
Torque
ω = 2πN/60 = 2π·500/60 = 52.36 rad/s
T = P/ω = 10000/52.36 = 191 N·m
Polar second moment of area
J = π(50⁴ − 25⁴)/32 = 575243 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 190986·25/575243 = 8.3 MPa
Answer
τmax = 8.3 MPa
5.Mohr’s circle & stress transformation
Answer: θs = 8.35 °
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 = 50/52.2 = 0.9578, sin 2θs = −(−15)/52.2 = 0.2873
2θs = atan2(15, 50) = 16.7°
Angle on the element
θs = 8.35°
check: θs − θp = 8.35 − 53.35 = −45°
Answer
θs = 8.35°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −41.21 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.
Transformation equation
σθ = −60 + 0·0.342 + 20·0.9397 = −41.21 MPa
Answer
σθ = −41.21 MPa
7.Axial stress & strain
Answer: ε = 9794 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 = 90·1000/78.54 = 1146 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 1146/117000 = 0.009794 = 9794 µε
Answer
ε = 9794 µε
8.Axial stress & strain
Answer: Pmax = 420.6 kN
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.
Areas
Ar = π·25²/4 = 490.9 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 100·10³·5027 = 6.008 × 10⁸ N
Governing strain
ε = min(0.0008, 0.0007) = 0.0007 (brass)
Largest load
Pmax = 0.0007·6.008 × 10⁸ = 420581 N = 420.6 kN
Answer
Pmax = 420.6 kN
9.Column buckling
Answer: Pcr = 160.2 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/2800² = 160173 N = 160.2 kN
Answer
Pcr = 160.2 kN
10.Column buckling
Answer: d = 47.66 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 = 2·40000 = 80000 N
KL = 1·2500 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·80000·2500²/(π³·200000))^(1/4) = 47.66 mm
Answer
d = 47.66 mm
Check: the slenderness KL/r is 209.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1213
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.
Logarithmic decrement
δ = (1/3)·ln(20/2) = 0.7675
Damping ratio
ζ = δ/√(4π² + δ²) = 0.7675/√(39.48 + 0.5891) = 0.1213
Answer
ζ = 0.1213
12.Mechanical vibrations
Answer: keq = 120 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 20 + 100 = 120 kN/m
Answer
keq = 120 kN/m
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