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 11 kW motor running at 2900 rpm drives a pinion with 25 teeth, which meshes with a gear with 40 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 11 kWn1 = 2900 rpmN1 = 25N2 = 40η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (20 teeth) on the input shaft turning at 1440 rpm drives gear B (45 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (80 teeth) on the output shaft. Find the output speed nout.
nin = 1440 rpmA = 20B = 45C = 15D = 80
nout = rpm
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 = 1000 N·md = 30 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 20 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 20 kWN = 1500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = −10 MPa, σᵧ = 120 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = −10 MPaσy = 120 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 130 MPa, σᵧ = 120 MPa and τₓᵧ = −40 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 40° counter-clockwise from the x-axis.
σx = 130 MPaσy = 120 MPaτxy = −40 MPaθ = 40°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 200 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 200 mm, E = 70 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 40 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, Ø40 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 = 3 mE = 200 GPa (steel)b × h = 25 × 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 25 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 25 kNn = 2L = 6 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 10 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 90 N·s/m. Find the damping ratio ζ.
m = 10 kgk = 10 kN/mc = 90 N·s/m
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 5 kN/m and k₂ = 50 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 5 kN/mk2 = 50 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 36.22 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Motor torque
ω₁ = 2π·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 11000/303.7 = 36.22 N·m
Answer
T1 = 36.22 N·m
2.Gears & gear trains
Answer: nout = 120 rpm
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
First stage
nB = 1440·20/45 = 640 rpm = nC
Second stage
nout = 640·15/80 = 120 rpm
Answer
nout = 120 rpm
total ratio = 12 : 1
3.Torsion of shafts
Answer: φ = 9.006 °
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) = 1000·1000·1000/(80000·79522) = 0.1572 rad
φ = 0.1572·180/π = 9.006°
Answer
φ = 9.006°
4.Torsion of shafts
Answer: d = 20.09 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.
Torque
ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s
T = P/ω = 20000/157.1 = 127.3 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·127324/(π·80)) = 20.09 mm
Answer
d = 20.09 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −68.65 °
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
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·−60, −10 − 120) = −137.3°
Angle on the element
θp = −68.65°
Answer
θp = −68.65°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 86.48 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.
Transformation equation
σθ = 125 + 5·0.1736 + −40·0.9848 = 86.48 MPa
Answer
σθ = 86.48 MPa
7.Axial stress & strain
Answer: P = 39.39 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 = 490.9 mm², A₂ = 706.9 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(70000·706.9) = 4.042 × 10⁻⁶ mm/N
Load
P = 0.4/1.015 × 10⁻⁵ = 39395 N = 39.39 kN
Answer
P = 39.39 kN
8.Axial stress & strain
Answer: σ2 = 26.53 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 = 100000·1257/400 = 314159 N/mm
Share of segment 2
F2 = 100000·314159/942478 = 33333 N
Stress
σ2 = 33333/1257 = 26.53 MPa (compression)
Answer
σ2 = 26.53 MPa
9.Column buckling
Answer: Pcr = 91.39 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 = 80·25³/12 = 104167 mm⁴
Effective length
KL = 0.5·3000 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·104167/1500² = 91385 N = 91.39 kN
Answer
Pcr = 91.39 kN
10.Column buckling
Answer: d = 71.41 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 = 2·25000 = 50000 N
KL = 0.7·6000 = 4200 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·50000·4200²/(π³·70000))^(1/4) = 71.41 mm
Answer
d = 71.41 mm
Check: the slenderness KL/r is 235.3, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1423
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·√(10000·10) = 632.5 N·s/m
Damping ratio
ζ = c/cc = 90/632.5 = 0.1423
Answer
ζ = 0.1423
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 55 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 = 5 + 50 = 55 kN/m
Answer
keq = 55 kN/m
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