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 2880 rpm drives a 450 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.
n1 = 2880 rpmd1 = 150 mmd2 = 450 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 160 mm pulley and a 500 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 160 mmd2 = 500 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 15 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 300 mm rotate 0.86° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 20 N·md = 15 mmL = 300 mmφ = 0.86°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 1000 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 1000 rpmdo = 50 mmdi = 30 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 20 MPa, σᵧ = −70 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 20 MPaσy = −70 MPaτxy = 50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = 90 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 150 MPaσy = 90 MPaτxy = −10 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 500 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 500 mm, E = 70 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 40 kNsegment 1: Ø40 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 6 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 250 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 3000 N·s/m. Find the damping ratio ζ.
m = 250 kgk = 100 kN/mc = 3000 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 80 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 = 100 kgk1 = 60 kN/mk2 = 80 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 921.6 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
2880·150/450 = 960 rpm
With slip
n2 = 960·0.96 = 921.6 rpm
Answer
n2 = 921.6 rpm
2.Gears & gear trains
Answer: θ = 163.7 °
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(0.1417) = 8.144°
Angle of wrap
θ = 180 − 2·8.144 = 163.7°
Answer
θ = 163.7°
3.Torsion of shafts
Answer: G = 80.43 GPa
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 moment and angle
J = π·15⁴/32 = 4970 mm⁴
φ = 0.86·π/180 = 0.01501 rad
Shear modulus
T = 20 N·m = 20000 N·mm
G = TL/(Jφ) = 20000·300/(4970·0.01501) = 80429 MPa
G = 80.43 GPa
Answer
G = 80.43 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 44.7 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·1000/60 = 104.7 rad/s
T = P/ω = 100000/104.7 = 954.9 N·m
Polar second moment of area
J = π(50⁴ − 30⁴)/32 = 534071 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·25/534071 = 44.7 MPa
Answer
τmax = 44.7 MPa
5.Mohr’s circle & stress transformation
Answer: θp = 24.01 °
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(2·50, 20 − −70) = 48.01°
Angle on the element
θp = 24.01°
Answer
θp = 24.01°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 110 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 = 120 + 30·0 + −10·1 = 110 MPa
Answer
σθ = 110 MPa
7.Axial stress & strain
Answer: P = 53.74 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 = 490.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(70000·2827) = 2.526 × 10⁻⁶ mm/N
Load
P = 0.3/5.582 × 10⁻⁶ = 53744 N = 53.74 kN
Answer
P = 53.74 kN
8.Axial stress & strain
Answer: δ = 1.572 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 31.83 MPa
σ2 = 198.9 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 31.83·500/200000 = 0.07958 mm
δ2 = σ2L2/E2 = 198.9·1500/200000 = 1.492 mm
δ = δ1 + δ2 = 1.572 mm
Answer
δ = 1.572 mm
9.Column buckling
Answer: Pcr = 2.18 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 1·6000 = 6000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/6000² = 2180 N = 2.18 kN
Answer
Pcr = 2.18 kN
10.Column buckling
Answer: σcr = 71.06 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·30⁴/64 = 39761 mm⁴
Radius of gyration and slenderness
A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 0.5·2500 = 1250 mm
KL/r = 1250/7.5 = 166.7
Critical stress
σcr = π²E/(KL/r)² = π²·200000/166.7² = 71.06 MPa
Answer
σcr = 71.06 MPa
11.Mechanical vibrations
Answer: ζ = 0.3
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·√(100000·250) = 10000 N·s/m
Damping ratio
ζ = c/cc = 3000/10000 = 0.3
Answer
ζ = 0.3
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 5.955 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 = 60 + 80 = 140 kN/m
Natural frequency
ωn = √(140000/100) = 37.42 rad/s
fn = 37.42/(2π) = 5.955 Hz
Answer
fn = 5.955 Hz
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