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 100 mm driving pulley at 960 rpm. The tension is 400 N on the tight side and 100 N on the slack side. Find the power P transmitted, in kW.
d1 = 100 mmn1 = 960 rpmT1 = 400 NT2 = 100 N
P = kW
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 1450 rpm to a 450 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 100 mmn1 = 1450 rpmd2 = 450 mmC = 500 mmμ = 0.35T1 = 1000 N
P = kW
3.Torsion of shaftsMedium
A hollow shaft must carry a torque of 750 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
T = 750 N·mdi/do = 0.6τallow = 60 MPa
do = mm
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 = 50 mmdi = 40 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = −10 MPa, σᵧ = 80 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = −10 MPaσy = 80 MPaτxy = −20 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = −30 MPa and τₓᵧ = 30 MPa (σz = 0). Find the von Mises equivalent stress σvM.
σx = −20 MPaσy = −30 MPaτxy = 30 MPa
σvM = MPa
7.Axial stress & strainMedium
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
P = 90 kNd = 25 mmL = 1 mE = 70 GPa
ε = µε
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 400 mm, E = 100 GPaP = 100 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 200 GPa (steel)d = 25 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 = 2.5 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 150 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency fd of its free vibration in Hz.
m = 150 kgk = 20 kN/mζ = 0.7
fd = Hz
12.Mechanical vibrationsHard
A 5 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 = 40 kN/m and k₂ = 15 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 = 5 kgk1 = 40 kN/mk2 = 15 kN/mk3 = 50 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: P = 1.508 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 = 5.027 m/s
Power
P = (400 − 100)·5.027 = 1508 W = 1.508 kW
Answer
P = 1.508 kW
2.Gears & gear trains
Answer: P = 4.345 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 = 2.426 rad
Slack-side tension
eμθ = e0.8493 = 2.338, T₂ = 1000/2.338 = 427.7 N
Power
P = (1000 − 427.7)·7.592 = 4.345 kW
Answer
P = 4.345 kW
3.Torsion of shafts
Answer: do = 41.82 mm
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.
Outer diameter from the allowable stress
T = 750 N·m = 750000 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·750000/(π·60·0.8704)) = 41.82 mm
di = 0.6·41.82 = 25.09 mm
Answer
do = 41.82 mm
A solid shaft would need d = 39.93 mm; the hollow one is a little larger but much lighter.
4.Torsion of shafts
Answer: τmax = 87.87 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(50⁴ − 40⁴)/32 = 362265 mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 1.273 × 10⁶·25/362265 = 87.87 MPa
Answer
τmax = 87.87 MPa
5.Mohr’s circle & stress transformation
Answer: θs = 56.98 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = −20/49.24 = −0.4061, sin 2θs = −(−45)/49.24 = 0.9138
2θs = atan2(45, −20) = 114°
Angle on the element
θs = 56.98°
check: θs − θp = 56.98 − (−78.02) = 135°
Answer
θs = 56.98°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σvM = 58.31 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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.
Principal stresses
C = −25 MPa, R = 30.41 MPa
σ1 = 5.414 MPa, σ2 = −55.41 MPa, σ3 = 0
Von Mises
σvM = √(5.414² − 5.414·(−55.41) + (−55.41)²) = 58.31 MPa
Answer
σvM = 58.31 MPa
7.Axial stress & strain
Answer: ε = 2619 µε
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·25²/4 = 490.9 mm²
σ = P/A = 90·1000/490.9 = 183.3 MPa
Strain (Hooke’s law)
E for aluminium = 70 GPa = 70000 MPa.
ε = σ/E = 183.3/70000 = 0.002619 = 2619 µε
Answer
ε = 2619 µε
8.Axial stress & strain
Answer: σ2 = 18.72 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 = 200000·1257/400 = 628319 N/mm
k₂ = 100000·2827/400 = 706858 N/mm
Share of segment 2
F2 = 100000·706858/1.335 × 10⁶ = 52941 N
Stress
σ2 = 52941/2827 = 18.72 MPa (compression)
Answer
σ2 = 18.72 MPa
9.Column buckling
Answer: Pcr = 6.056 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 = 0.5·5000 = 2500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·19175/2500² = 6056 N = 6.056 kN
Answer
Pcr = 6.056 kN
10.Column buckling
Answer: σcr = 35.25 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.7·2500 = 1750 mm
KL/r = 1750/12.5 = 140
Critical stress
σcr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa
Answer
σcr = 35.25 MPa
11.Mechanical vibrations
Answer: fd = 1.312 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.
Undamped natural frequency
ωn = √(20000/150) = 11.55 rad/s
fn = 11.55/(2π) = 1.838 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 1.838·√(1 − 0.7²) = 1.312 Hz
Answer
fd = 1.312 Hz
12.Mechanical vibrations
Answer: fn = 11.52 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 + 15 = 55 kN/m
keq = 55·50/(55 + 50) = 26.19 kN/m
Natural frequency
ωn = √(26190/5) = 72.37 rad/s
fn = 72.37/(2π) = 11.52 Hz
Answer
fn = 11.52 Hz
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