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 5.5 kW motor running at 960 rpm drives a pinion with 50 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 95%. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 960 rpmN1 = 50N2 = 60η = 95%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (30 teeth) on the input shaft turning at 1440 rpm drives gear B (40 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (60 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 30B = 40C = 16D = 60
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 = 250 N·md = 40 mmL = 2.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 75 kW at 500 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 75 kWN = 500 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 0 MPa, σᵧ = 40 MPa and τₓᵧ = 10 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 = 0 MPaσy = 40 MPaτxy = 10 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 500 mm and wall thickness 6 mm holds gas at a gauge pressure of 3 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 500 mmt = 6 mmp = 3 MPaα = 55°
σn = MPa
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 300 mm long and are compressed together between rigid plates by 100 kN. Find the shortening δ.
P = 100 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø30 mm, E = 200 GPa, σallow = 200 MPatube: Ø100/Ø70 mm, E = 100 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 70 GPa (aluminium)b × h = 50 × 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 15 kNn = 2.5L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 3 cycles later by a peak of 11 mm. Find the damping ratio ζ.
x0 = 50 mmx3 = 11 mmn = 3
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 40 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 = 10 kgk1 = 10 kN/mk2 = 40 kN/mk3 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 62.37 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 = 5500/100.5 = 54.71 N·m
Output torque
T2 = 54.71·60/50·0.95 = 62.37 N·m
Answer
T2 = 62.37 N·m
2.Gears & gear trains
Answer: nout = 288 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 = 1440·30/40 = 1080 rpm = nC
Second stage
nout = 1080·16/60 = 288 rpm
Answer
nout = 288 rpm
total ratio = 5 : 1
3.Torsion of shafts
Answer: φ = 1.781 °
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) = 250·1000·2500/(80000·251327) = 0.03108 rad
φ = 0.03108·180/π = 1.781°
Answer
φ = 1.781°
4.Torsion of shafts
Answer: d = 45.99 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·500/60 = 52.36 rad/s
T = P/ω = 75000/52.36 = 1432 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·1.432 × 10⁶/(π·75)) = 45.99 mm
Answer
d = 45.99 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = 31.72 °
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 = 10/22.36 = 0.4472, sin 2θs = −(−20)/22.36 = 0.8944
2θs = atan2(20, 10) = 63.43°
Angle on the element
θs = 31.72°
check: θs − θp = 31.72 − 76.72 = −45°
Answer
θs = 31.72°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σn = 83.06 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 500/2 = 250 mm
σh = pr/t = 3·250/6 = 125 MPa
σa = pr/(2t) = 62.5 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 62.5·sin²55° + 125·cos²55° = 83.06 MPa
Answer
σn = 83.06 MPa
7.Axial stress & strain
Answer: δ = 0.02616 mm
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·40²/4 = 1257 mm²
At = π(120² − 70²)/4 = 7461 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 120·10³·7461 = 1.147 × 10⁹ N
Shortening
δ = 100000·300/1.147 × 10⁹ = 0.02616 mm
Answer
δ = 0.02616 mm
8.Axial stress & strain
Answer: Pmax = 379.3 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·30²/4 = 706.9 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 100·10³·4006 = 5.419 × 10⁸ N
Governing strain
ε = min(0.001, 0.0007) = 0.0007 (brass)
Largest load
Pmax = 0.0007·5.419 × 10⁸ = 379347 N = 379.3 kN
Answer
Pmax = 379.3 kN
9.Column buckling
Answer: Pcr = 26.99 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 = 60·50³/12 = 625000 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·625000/4000² = 26987 N = 26.99 kN
Answer
Pcr = 26.99 kN
10.Column buckling
Answer: d = 49.88 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 = 2.5·15000 = 37500 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·4000²/(π³·200000))^(1/4) = 49.88 mm
Answer
d = 49.88 mm
Check: the slenderness KL/r is 320.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.08007
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(50/11) = 0.5047
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5047/√(39.48 + 0.2547) = 0.08007
Answer
ζ = 0.08007
12.Mechanical vibrations
Answer: fn = 8.312 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 = 10 + 40 = 50 kN/m
keq = 50·60/(50 + 60) = 27.27 kN/m
Natural frequency
ωn = √(27273/10) = 52.22 rad/s
fn = 52.22/(2π) = 8.312 Hz
Answer
fn = 8.312 Hz
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