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 200 mm driving pulley at 720 rpm. The tension is 1200 N on the tight side and 360 N on the slack side. Find the power P transmitted, in kW.
d1 = 200 mmn1 = 720 rpmT1 = 1200 NT2 = 360 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 160 mm driving pulley at 960 rpm to a 400 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.35 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 = 160 mmn1 = 960 rpmd2 = 400 mmC = 1200 mmμ = 0.35T1 = 500 N
P = kW
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 = 400 N·md = 30 mmL = 0.8 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 100 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 100 kWN = 750 rpmdo = 40 mmdi = 28 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = −70 MPa and τₓᵧ = 10 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).
σx = −50 MPaσy = −70 MPaτxy = 10 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial tensile force of 20 kN together with a torque of 250 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 50 mmF = 20 kNT = 250 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 400 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.
segment 1: steel, Ø30 mm, 600 mm, E = 200 GPasegment 2: copper, Ø60 mm, 400 mm, E = 120 GPaδ = 0.25 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 = 55 kNsegment 1: Ø30 mm, 1 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1 mE = 200 GPa (steel)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
An 80 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 890 N·s/m. Find the damping ratio ζ.
m = 80 kgk = 10 kN/mc = 890 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 25 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 100 kN/mk2 = 25 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: P = 6.333 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 = 7.54 m/s
Power
P = (1200 − 360)·7.54 = 6333 W = 6.333 kW
Answer
P = 6.333 kW
2.Gears & gear trains
Answer: P = 2.585 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 = 2.941 rad
Slack-side tension
eμθ = e1.029 = 2.799, T₂ = 500/2.799 = 178.6 N
Power
P = (500 − 178.6)·8.042 = 2.585 kW
Answer
P = 2.585 kW
3.Torsion of shafts
Answer: φ = 2.882 °
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·30⁴/32 = 79522 mm⁴
Angle of twist
φ = TL/(GJ) = 400·1000·800/(80000·79522) = 0.0503 rad
φ = 0.0503·180/π = 2.882°
Answer
φ = 2.882°
4.Torsion of shafts
Answer: τmax = 133.3 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·750/60 = 78.54 rad/s
T = P/ω = 100000/78.54 = 1273 N·m
Polar second moment of area
J = π(40⁴ − 28⁴)/32 = 190984 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 1.273 × 10⁶·20/190984 = 133.3 MPa
Answer
τmax = 133.3 MPa
5.Mohr’s circle & stress transformation
Answer: τabs = 37.07 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.
In-plane principal stresses
C = −60 MPa, R = √(10² + 10²) = 14.14 MPa
σ1 = −45.86 MPa, σ2 = −74.14 MPa, σz = 0
Largest of the three circles
τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(28.28, 45.86, 74.14)/2 = 37.07 MPa
Answer
τabs = 37.07 MPa
Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.
6.Mohr’s circle & stress transformation
Answer: σ1 = 16.48 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.
Stresses at the surface
σ = F/A = 20000/(π·50²/4) = 20000/1963 = 10.19 MPa
τ = 16T/(πd³) = 16·250000/(π·50³) = 10.19 MPa
Mohr’s circle
C = σ/2 = 5.093 MPa
R = √(5.093² + 10.19²) = 11.39 MPa
σ1 = C + R = 16.48 MPa
σ2 = C − R = −6.295 MPa
Answer
σ1 = 16.48 MPa
7.Axial stress & strain
Answer: P = 46.1 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 = 706.9 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(120000·2827) = 1.179 × 10⁻⁶ mm/N
Load
P = 0.25/5.423 × 10⁻⁶ = 46099 N = 46.1 kN
Answer
P = 46.1 kN
8.Axial stress & strain
Answer: δ = 2.146 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·30²/4 = 706.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 77.81 MPa
σ2 = 273.5 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 77.81·1000/100000 = 0.7781 mm
δ2 = σ2L2/E2 = 273.5·1000/200000 = 1.368 mm
δ = δ1 + δ2 = 2.146 mm
Answer
δ = 2.146 mm
9.Column buckling
Answer: Pcr = 524.5 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 = 100·25³/12 = 130208 mm⁴
Effective length
KL = 0.7·1000 = 700 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·130208/700² = 524533 N = 524.5 kN
Answer
Pcr = 524.5 kN
10.Column buckling
Answer: σcr = 17.27 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·60⁴/64 = 636173 mm⁴
Radius of gyration and slenderness
A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 2·1500 = 3000 mm
KL/r = 3000/15 = 200
Critical stress
σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa
Answer
σcr = 17.27 MPa
11.Mechanical vibrations
Answer: ζ = 0.4975
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·√(10000·80) = 1789 N·s/m
Damping ratio
ζ = c/cc = 890/1789 = 0.4975
Answer
ζ = 0.4975
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 125 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 100 + 25 = 125 kN/m
Answer
keq = 125 kN/m
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