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
An 80 mm pulley at 960 rpm drives a 250 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 80 mmd2 = 250 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 960 rpm to a 250 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1200 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 100 mmn1 = 960 rpmd2 = 250 mmC = 500 mmμ = 0.3T1 = 1200 N
P = kW
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 = 500 N·md = 25 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 75 kW at 750 rpm. Find the maximum shear stress τmax in the shaft.
P = 75 kWN = 750 rpmdo = 100 mmdi = 60 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −20 MPa, σᵧ = 90 MPa and τₓᵧ = 60 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = −20 MPaσy = 90 MPaτxy = 60 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 160 MPa, σᵧ = 20 MPa and τₓᵧ = 50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 160 MPaσy = 20 MPaτxy = 50 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of bronze (diameter 40 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø40 mm, 500 mm, E = 110 GPaP = 60 kN
σ2 = MPa
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 = 4 mE = 200 GPa (steel)d = 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 15 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 15 kNn = 2L = 1.5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 10 kg mass is supported by a spring with a stiffness of 5 kN/m and a viscous damper with a damping coefficient of 70 N·s/m. Find the damping ratio ζ.
m = 10 kgk = 5 kN/mc = 70 N·s/m
ζ =
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 40 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 5 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 = 40 kN/mk2 = 5 kN/mk3 = 40 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 294.9 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
960·80/250 = 307.2 rpm
With slip
n2 = 307.2·0.96 = 294.9 rpm
Answer
n2 = 294.9 rpm
2.Gears & gear trains
Answer: P = 3.459 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 = 2.84 rad
Slack-side tension
eμθ = e0.8521 = 2.345, T₂ = 1200/2.345 = 511.8 N
Power
P = (1200 − 511.8)·5.027 = 3.459 kW
Answer
P = 3.459 kW
3.Torsion of shafts
Answer: φ = 19.15 °
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) = 500·1000·1000/(39000·38350) = 0.3343 rad
φ = 0.3343·180/π = 19.15°
Answer
φ = 19.15°
4.Torsion of shafts
Answer: τmax = 5.588 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 = 75000/78.54 = 954.9 N·m
Polar second moment of area
J = π(100⁴ − 60⁴)/32 = 8.545 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·50/8.545 × 10⁶ = 5.588 MPa
Answer
τmax = 5.588 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 81.39 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 = √(((−20 − 90)/2)² + 60²) = 81.39 MPa
Answer
τmax = 81.39 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 140 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 = 90 + 70·0 + 50·1 = 140 MPa
Answer
σθ = 140 MPa
7.Axial stress & strain
Answer: P = 37.7 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 = 314.2 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·1257) = 1.592 × 10⁻⁶ mm/N
Load
P = 0.3/7.958 × 10⁻⁶ = 37699 N = 37.7 kN
Answer
P = 37.7 kN
8.Axial stress & strain
Answer: σ2 = 18.98 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 = 200000·1257/600 = 418879 N/mm
k₂ = 110000·1257/500 = 276460 N/mm
Share of segment 2
F2 = 60000·276460/695339 = 23855 N
Stress
σ2 = 23855/1257 = 18.98 MPa (compression)
Answer
σ2 = 18.98 MPa
9.Column buckling
Answer: Pcr = 506.2 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 0.7·4000 = 2800 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/2800² = 506225 N = 506.2 kN
Answer
Pcr = 506.2 kN
10.Column buckling
Answer: d = 31.43 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 = 2·15000 = 30000 N
KL = 0.7·1500 = 1050 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·30000·1050²/(π³·70000))^(1/4) = 31.43 mm
Answer
d = 31.43 mm
Check: the slenderness KL/r is 133.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1565
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·√(5000·10) = 447.2 N·s/m
Damping ratio
ζ = c/cc = 70/447.2 = 0.1565
Answer
ζ = 0.1565
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: fn = 2.316 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 = 40 + 5 = 45 kN/m
keq = 45·40/(45 + 40) = 21.18 kN/m
Natural frequency
ωn = √(21176/100) = 14.55 rad/s
fn = 14.55/(2π) = 2.316 Hz
Answer
fn = 2.316 Hz
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