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 125 mm driving pulley at 1440 rpm. The tension is 1200 N on the tight side and 480 N on the slack side. Find the power P transmitted, in kW.
d1 = 125 mmn1 = 1440 rpmT1 = 1200 NT2 = 480 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 100 mm pulley and a 450 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 100 mmd2 = 450 mmC = 1200 mm
θ = °
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 = 100 N·md = 40 mmL = 2.5 mG = 80 GPa
φ = °
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 = 75 mmdi = 53 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 140 MPa, σᵧ = −60 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 140 MPaσy = −60 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 120 MPa, σᵧ = 10 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 60° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 120 MPaσy = 10 MPaτxy = 30 MPaθ = 60°
τθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 400 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.
P = 250 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 110 GPaL = 400 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 30 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 500 mm, E = 110 GPaP = 30 kN
σ2 = MPa
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 = 4 mE = 200 GPa (steel)d = 25 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and pinned at the top. It must carry 110 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 110 kNn = 2L = 1.5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 12 kg mass is supported by a spring with a stiffness of 25 kN/m and a viscous damper with a damping coefficient of 110 N·s/m. Find the damping ratio ζ.
m = 12 kgk = 25 kN/mc = 110 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 5 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 5 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 6.786 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 = 9.425 m/s
Power
P = (1200 − 480)·9.425 = 6786 W = 6.786 kW
Answer
P = 6.786 kW
2.Gears & gear trains
Answer: θ = 163.2 °
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(0.1458) = 8.386°
Angle of wrap
θ = 180 − 2·8.386 = 163.2°
Answer
θ = 163.2°
3.Torsion of shafts
Answer: φ = 0.7124 °
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) = 100·1000·2500/(80000·251327) = 0.01243 rad
φ = 0.01243·180/π = 0.7124°
Answer
φ = 0.7124°
4.Torsion of shafts
Answer: τmax = 15.36 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(75⁴ − 53⁴)/32 = 2.332 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·37.5/2.332 × 10⁶ = 15.36 MPa
Answer
τmax = 15.36 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 116.6 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.
Radius of Mohr’s circle
τmax = R = √(((140 − −60)/2)² + 60²) = 116.6 MPa
Answer
τmax = 116.6 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −62.63 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
τθ = −55·0.866 + 30·−0.5 = −62.63 MPa
Answer
τθ = −62.63 MPa
7.Axial stress & strain
Answer: δ = 0.134 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·25²/4 = 490.9 mm²
At = π(100² − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 110·10³·5890 = 7.461 × 10⁸ N
Shortening
δ = 250000·400/7.461 × 10⁸ = 0.134 mm
Answer
δ = 0.134 mm
8.Axial stress & strain
Answer: σ2 = 18.08 MPa
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.
Stiffnesses
k₁ = 200000·314.2/300 = 209440 N/mm
k₂ = 110000·706.9/500 = 155509 N/mm
Share of segment 2
F2 = 30000·155509/364948 = 12783 N
Stress
σ2 = 12783/706.9 = 18.08 MPa (compression)
Answer
σ2 = 18.08 MPa
9.Column buckling
Answer: Pcr = 2.366 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·25⁴/64 = 19175 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/4000² = 2366 N = 2.366 kN
Answer
Pcr = 2.366 kN
10.Column buckling
Answer: d = 39.78 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 = 2·110000 = 220000 N
KL = 0.7·1500 = 1050 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·220000·1050²/(π³·200000))^(1/4) = 39.78 mm
Answer
d = 39.78 mm
Check: the slenderness KL/r is 105.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1004
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·√(25000·12) = 1095 N·s/m
Damping ratio
ζ = c/cc = 110/1095 = 0.1004
Answer
ζ = 0.1004
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 3.529 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/5 + 1/12 → keq = 5·12/(5 + 12) = 3.529 kN/m
Answer
keq = 3.529 kN/m
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