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 1440 rpm. The tension is 600 N on the tight side and 120 N on the slack side. Find the power P transmitted, in kW.
d1 = 160 mmn1 = 1440 rpmT1 = 600 NT2 = 120 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 160 mm pulley and a 450 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 160 mmd2 = 450 mmC = 500 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 = 750 N·md = 25 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 40 mm is fixed at A. A torque of 1000 N·m acts at B, 0.8 m from A, and a torque of 1500 N·m in the opposite direction acts at the free end C, a further 2.5 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 40 mmTB = 1000 N·mTC = 1500 N·mAB = 0.8 mBC = 2.5 mG = 80 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 50 MPa, σᵧ = 30 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 50 MPaσy = 30 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 120 MPa, σᵧ = 80 MPa and τₓᵧ = 60 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 120 MPaσy = 80 MPaτxy = 60 MPaθ = 40°
τθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 50 mm, E = 110 GPa). Both are 300 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/Ø50 mm, E = 110 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 150 MPatube: Ø100/Ø60 mm, E = 110 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 = 30 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and pinned at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 5 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 500 kg mass is supported by a spring with a stiffness of 100 kN/m and a viscous damper with a damping coefficient of 5660 N·s/m. Find the damping ratio ζ.
m = 500 kgk = 100 kN/mc = 5660 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 50 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 30 kN/m and k₂ = 30 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 = 100 kgk1 = 30 kN/mk2 = 30 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 5.791 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 = 12.06 m/s
Power
P = (600 − 120)·12.06 = 5791 W = 5.791 kW
Answer
P = 5.791 kW
2.Gears & gear trains
Answer: θ = 146.3 °
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(0.29) = 16.86°
Angle of wrap
θ = 180 − 2·16.86 = 146.3°
Answer
θ = 146.3°
3.Torsion of shafts
Answer: φ = 14.01 °
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·25⁴/32 = 38350 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1000/(80000·38350) = 0.2445 rad
φ = 0.2445·180/π = 14.01°
Answer
φ = 14.01°
4.Torsion of shafts
Answer: φC = 11.83 °
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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 1500 N·m
TAB = TC − TB = 1500 − 1000 = 500 N·m
Twist of each segment
J = π·40⁴/32 = 251327 mm⁴
G = 80 GPa = 80000 MPa
φAB = 500000·800/(80000·251327) = 0.01989 rad
φBC = 1.5 × 10⁶·2500/(80000·251327) = 0.1865 rad
Twist of end C
φC = 0.01989 + 0.1865 = 0.2064 rad → |φC| = 11.83°
Answer
φC = 11.83°
5.Mohr’s circle & stress transformation
Answer: τmax = 31.62 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 = √(((50 − 30)/2)² + 30²) = 31.62 MPa
Answer
τmax = 31.62 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −9.277 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 = −20·0.9848 + 60·0.1736 = −9.277 MPa
Answer
τθ = −9.277 MPa
7.Axial stress & strain
Answer: δ = 0.06331 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 − 50²)/4 = 5890 mm²
Combined axial stiffness
ΣEA = 200·10³·314.2 + 110·10³·5890 = 7.108 × 10⁸ N
Shortening
δ = 150000·300/7.108 × 10⁸ = 0.06331 mm
Answer
δ = 0.06331 mm
8.Axial stress & strain
Answer: Pmax = 511.8 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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 110·10³·5027 = 8.042 × 10⁸ N
Governing strain
ε = min(0.00075, 0.0006364) = 0.0006364 (bronze)
Largest load
Pmax = 0.0006364·8.042 × 10⁸ = 511794 N = 511.8 kN
Answer
Pmax = 511.8 kN
9.Column buckling
Answer: Pcr = 9.715 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 = 100·30³/12 = 225000 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·225000/4000² = 9715 N = 9.715 kN
Answer
Pcr = 9.715 kN
10.Column buckling
Answer: σcr = 5.64 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 0.7·5000 = 3500 mm
KL/r = 3500/10 = 350
Critical stress
σcr = π²E/(KL/r)² = π²·70000/350² = 5.64 MPa
Answer
σcr = 5.64 MPa
11.Mechanical vibrations
Answer: ζ = 0.4002
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·√(100000·500) = 14142 N·s/m
Damping ratio
ζ = c/cc = 5660/14142 = 0.4002
Answer
ζ = 0.4002
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.628 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 = 30 + 30 = 60 kN/m
keq = 60·50/(60 + 50) = 27.27 kN/m
Natural frequency
ωn = √(27273/100) = 16.51 rad/s
fn = 16.51/(2π) = 2.628 Hz
Answer
fn = 2.628 Hz
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