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 belt runs on a 160 mm driving pulley at 2880 rpm. The tension is 1500 N on the tight side and 375 N on the slack side. Find the power P transmitted, in kW.
d1 = 160 mmn1 = 2880 rpmT1 = 1500 NT2 = 375 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 160 mm pulley and a 200 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 160 mmd2 = 200 mmC = 600 mm
θ = °
3.Torsion of shaftsMedium
A solid brass 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 = 1000 N·md = 40 mmL = 3 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 50 kW at 300 rpm without the shear stress exceeding 60 MPa. Find the smallest diameter d that works.
P = 50 kWN = 300 rpmτallow = 60 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −70 MPa, σᵧ = −10 MPa and τₓᵧ = −40 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 = −70 MPaσy = −10 MPaτxy = −40 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = −80 MPaσy = 90 MPaτxy = −10 MPa
σvM = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 500 mm, E = 100 GPaδ = 0.4 mm
P = kN
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 = 85 kNsegment 1: Ø40 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2.5 mE = 70 GPa (aluminium)d = 40 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 4 cycles later by a peak of 6.7 mm. Find the damping ratio ζ.
x0 = 50 mmx4 = 6.7 mmn = 4
ζ =
12.Mechanical vibrationsHard
A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 50 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 20 kgk1 = 50 kN/mk2 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 27.14 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.
Belt speed
v = 24.13 m/s
Power
P = (1500 − 375)·24.13 = 27143 W = 27.14 kW
Answer
P = 27.14 kW
2.Gears & gear trains
Answer: θ = 176.2 °
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.
Belt angle
asin(0.03333) = 1.91°
Angle of wrap
θ = 180 − 2·1.91 = 176.2°
Answer
θ = 176.2°
3.Torsion of shafts
Answer: φ = 17.54 °
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) = 1000·1000·3000/(39000·251327) = 0.3061 rad
φ = 0.3061·180/π = 17.54°
Answer
φ = 17.54°
4.Torsion of shafts
Answer: d = 51.31 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·300/60 = 31.42 rad/s
T = P/ω = 50000/31.42 = 1592 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·1.592 × 10⁶/(π·60)) = 51.31 mm
Answer
d = 51.31 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = 71.57 °
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 = −40/50 = −0.8, sin 2θs = −(−30)/50 = 0.6
2θs = atan2(30, −40) = 143.1°
Angle on the element
θs = 71.57°
check: θs − θp = 71.57 − (−63.43) = 135°
Answer
θs = 71.57°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σvM = 148.3 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. 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 = 5 MPa, R = 85.59 MPa
σ1 = 90.59 MPa, σ2 = −80.59 MPa, σ3 = 0
Von Mises
σvM = √(90.59² − 90.59·(−80.59) + (−80.59)²) = 148.3 MPa
Answer
σvM = 148.3 MPa
7.Axial stress & strain
Answer: P = 96.66 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
A₁ = 1257 mm², A₂ = 1963 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(100000·1963) = 2.546 × 10⁻⁶ mm/N
Load
P = 0.4/4.138 × 10⁻⁶ = 96664 N = 96.66 kN
Answer
P = 96.66 kN
8.Axial stress & strain
Answer: σmax = 751.6 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 = 67.64 MPa
σ2 = 751.6 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 751.6 MPa
Answer
σmax = 751.6 MPa
9.Column buckling
Answer: Pcr = 28.35 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 0.7·2500 = 1750 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/1750² = 28349 N = 28.35 kN
Answer
Pcr = 28.35 kN
10.Column buckling
Answer: σcr = 19.28 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 2·1000 = 2000 mm
KL/r = 2000/6.25 = 320
Critical stress
σcr = π²E/(KL/r)² = π²·200000/320² = 19.28 MPa
Answer
σcr = 19.28 MPa
11.Mechanical vibrations
Answer: ζ = 0.07972
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 = (1/4)·ln(50/6.7) = 0.5025
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5025/√(39.48 + 0.2525) = 0.07972
Answer
ζ = 0.07972
12.Mechanical vibrations
Answer: fn = 11.25 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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 50 + 50 = 100 kN/m
Natural frequency
ωn = √(100000/20) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz
Answer
fn = 11.25 Hz
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