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 150 mm pulley at 960 rpm drives a 250 mm pulley by a belt with 2.5 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 150 mmd2 = 250 mms = 2.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 2880 rpm to a 500 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 100 mmn1 = 2880 rpmd2 = 500 mmC = 600 mmμ = 0.3T1 = 500 N
P = kW
3.Torsion of shaftsMedium
A solid steel shaft 1.5 m long carries a torque of 2000 N·m. The angle of twist between its ends must not exceed 3°. Find the smallest diameter d that meets this limit.
T = 2000 N·mL = 1.5 mG = 80 GPaφallow = 3°
d = mm
4.Torsion of shaftsHard
A hollow shaft must transmit 100 kW at 1450 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.
P = 100 kWN = 1450 rpmdi/do = 0.75τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 0 MPa, σᵧ = 0 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 = 0 MPaσy = 0 MPaτxy = −40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 0 MPa, σᵧ = 60 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.
σx = 0 MPaσy = 60 MPaτxy = 30 MPaσY = 275 MPa
n =
7.Axial stress & strainMedium
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa). Both are 500 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.
P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 100 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 115 kNsegment 1: Ø25 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
δ = mm
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 = 2 mE = 70 GPa (aluminium)b × h = 50 × 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 20 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 20 kNn = 2.5L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 20 kg mass is supported by a spring with a stiffness of 2 kN/m and a viscous damper with a damping coefficient of 80 N·s/m. Find the damping ratio ζ.
m = 20 kgk = 2 kN/mc = 80 N·s/m
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 5 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 = 20 kN/mk2 = 5 kN/mk3 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 561.6 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
960·150/250 = 576 rpm
With slip
n2 = 576·0.975 = 561.6 rpm
Answer
n2 = 561.6 rpm
2.Gears & gear trains
Answer: P = 3.937 kW
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.
Angle of wrap
θ = 2.462 rad
Slack-side tension
eμθ = e0.7386 = 2.093, T₂ = 500/2.093 = 238.9 N
Power
P = (500 − 238.9)·15.08 = 3.937 kW
Answer
P = 3.937 kW
3.Torsion of shafts
Answer: d = 51.97 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.
Required polar moment
T = 2000 N·m = 2 × 10⁶ N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = 3·π/180 = 0.05236 rad
J = TL/(Gφ) = 2 × 10⁶·1500/(80000·0.05236) = 716197 mm⁴
Diameter
d = (32J/π)1/4 = 51.97 mm
Answer
d = 51.97 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: do = 40.29 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·1450/60 = 151.8 rad/s
T = P/ω = 100000/151.8 = 658.6 N·m
Outer diameter from the allowable stress
T = 658.6 N·m = 658572 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·658572/(π·75·0.6836)) = 40.29 mm
di = 0.75·40.29 = 30.22 mm
Answer
do = 40.29 mm
A solid shaft would need d = 35.5 mm; the hollow one is a little larger but much lighter.
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, 0 − 0) = −90°
Angle on the element
θp = −45°
Answer
θp = −45°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 3.465
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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = 30 MPa, R = 42.43 MPa
σ1 = 72.43 MPa, σ2 = −12.43 MPa, σ3 = 0
Von Mises
σvM = √(72.43² − 72.43·(−12.43) + (−12.43)²) = 79.37 MPa
n = 275/79.37 = 3.465
Answer
n = 3.465
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.1619 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·20²/4 = 314.2 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 100·10³·4006 = 4.634 × 10⁸ N
Shortening
δ = 150000·500/4.634 × 10⁸ = 0.1619 mm
Answer
δ = 0.1619 mm
8.Axial stress & strain
Answer: δ = 7.637 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·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 234.3 MPa
σ2 = 572 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 234.3·1000/70000 = 3.347 mm
δ2 = σ2L2/E2 = 572·1500/200000 = 4.29 mm
δ = δ1 + δ2 = 7.637 mm
Answer
δ = 7.637 mm
9.Column buckling
Answer: Pcr = 143.9 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 = 80·50³/12 = 833333 mm⁴
Effective length
KL = 1·2000 = 2000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·833333/2000² = 143932 N = 143.9 kN
Answer
Pcr = 143.9 kN
10.Column buckling
Answer: d = 42.38 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 = 2.5·20000 = 50000 N
KL = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·2500²/(π³·200000))^(1/4) = 42.38 mm
Answer
d = 42.38 mm
Check: the slenderness KL/r is 236, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.2
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·√(2000·20) = 400 N·s/m
Damping ratio
ζ = c/cc = 80/400 = 0.2
Answer
ζ = 0.2
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 7.118 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 = 20 + 5 = 25 kN/m
keq = 25·100/(25 + 100) = 20 kN/m
Natural frequency
ωn = √(20000/10) = 44.72 rad/s
fn = 44.72/(2π) = 7.118 Hz
Answer
fn = 7.118 Hz
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