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 15 kW motor running at 2900 rpm drives a pinion with 15 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 15 kWn1 = 2900 rpmN1 = 15N2 = 50η = 97%
T2 = N·m
2.Gears & gear trainsHard
A 1.1 kW motor at 1440 rpm drives gear A (25 teeth), which meshes with gear B (80 teeth). Gear C (15 teeth) on the same shaft as B drives gear D (40 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque Tout.
P = 1.1 kWnin = 1440 rpmA = 25B = 80C = 15D = 40ηmesh = 98%
Tout = N·m
3.Torsion of shaftsMedium
A solid aluminium 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 = 0.5 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
A solid brass shaft of diameter 30 mm is fixed at A. A torque of 750 N·m acts at B, 1.5 m from A, and a torque of 400 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the magnitude of the angle of twist φC of end C, in degrees.
d = 30 mmTB = 750 N·mTC = 400 N·mAB = 1.5 mBC = 0.8 mG = 39 GPa
φC = °
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 120 MPa, σᵧ = 70 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 = 120 MPaσy = 70 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 140 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° counter-clockwise from the x-axis.
σx = 140 MPaσy = 0 MPaτxy = 10 MPaθ = 45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 25 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 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 0.25 mm in total.
segment 1: steel, Ø25 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 200 mm, E = 70 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 largest normal stress σmax in the bar.
P = 95 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 0.5 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 4 mE = 200 GPa (steel)d = 30 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 150 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 150 kNn = 2L = 4 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 20 mm is followed one cycle later by a peak of 12.1 mm. Find the damping ratio ζ.
x0 = 20 mmx1 = 12.1 mmn = 1
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 100 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 40 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 159.7 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 = 15000/303.7 = 49.39 N·m
Output torque
T2 = 49.39·50/15·0.97 = 159.7 N·m
Answer
T2 = 159.7 N·m
2.Gears & gear trains
Answer: Tout = 59.78 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.
Output speed
nout = 1440·(25/80)·(15/40) = 168.8 rpm
ωout = 2π·168.8/60 = 17.67 rad/s
Output power and torque
Pout = 1100·0.98² = 1056 W
Tout = Pout/ωout = 1056/17.67 = 59.78 N·m
Answer
Tout = 59.78 N·m
3.Torsion of shafts
Answer: φ = 5.542 °
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) = 400·1000·500/(26000·79522) = 0.09673 rad
φ = 0.09673·180/π = 5.542°
Answer
φ = 5.542°
4.Torsion of shafts
Answer: φC = 3.787 °
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 = 400 N·m
TAB = TC − TB = 400 − 750 = −350 N·m
Twist of each segment
J = π·30⁴/32 = 79522 mm⁴
G = 39 GPa = 39000 MPa
φAB = −350000·1500/(39000·79522) = −0.1693 rad
φBC = 400000·800/(39000·79522) = 0.1032 rad
Twist of end C
φC = −0.1693 + 0.1032 = −0.0661 rad → |φC| = 3.787°
Answer
φC = 3.787°
5.Mohr’s circle & stress transformation
Answer: θp = −33.69 °
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, 120 − 70) = −67.38°
Angle on the element
θp = −33.69°
Answer
θp = −33.69°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 80 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 = 70 + 70·0 + 10·1 = 80 MPa
Answer
σθ = 80 MPa
7.Axial stress & strain
Answer: P = 61.48 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 300/(200000·490.9) = 3.056 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(70000·2827) = 1.011 × 10⁻⁶ mm/N
Load
P = 0.25/4.066 × 10⁻⁶ = 61481 N = 61.48 kN
Answer
P = 61.48 kN
8.Axial stress & strain
Answer: σmax = 302.4 MPa
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·35²/4 = 962.1 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 98.74 MPa
σ2 = 302.4 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 302.4 MPa
Answer
σmax = 302.4 MPa
9.Column buckling
Answer: Pcr = 4.905 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·30⁴/64 = 39761 mm⁴
Effective length
KL = 1·4000 = 4000 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·39761/4000² = 4905 N = 4.905 kN
Answer
Pcr = 4.905 kN
10.Column buckling
Answer: d = 59.32 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 = 2·150000 = 300000 N
KL = 0.5·4000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·2000²/(π³·200000))^(1/4) = 59.32 mm
Answer
d = 59.32 mm
Check: the slenderness KL/r is 134.9, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.07973
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.
Logarithmic decrement
δ = (1/1)·ln(20/12.1) = 0.5025
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5025/√(39.48 + 0.2525) = 0.07973
Answer
ζ = 0.07973
12.Mechanical vibrations
Answer: fn = 18.83 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
Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.
keq = 40 + 100 = 140 kN/m
Natural frequency
ωn = √(140000/10) = 118.3 rad/s
fn = 118.3/(2π) = 18.83 Hz
Answer
fn = 18.83 Hz
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