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 5.5 kW motor running at 960 rpm drives a pinion with 40 teeth, which meshes with a gear with 72 teeth. The gear pair has an efficiency of 98%. Find the output torque T2 on the gear shaft.
P = 5.5 kWn1 = 960 rpmN1 = 40N2 = 72η = 98%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (60 teeth) on the input shaft turning at 960 rpm drives gear B (72 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D (36 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 60B = 72C = 12D = 36
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 = 20 mmL = 0.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 25 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 3 m from A and 2 m from B. Find the reaction torque TA at wall A.
d = 25 mmT0 = 400 N·ma = 3 mb = 2 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 160 MPa, σᵧ = 80 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 = 160 MPaσy = 80 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 15 mm holds gas at a gauge pressure of 1.2 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.
D = 1200 mmt = 15 mmp = 1.2 MPaα = 50°
σn = 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 brass (diameter 40 mm, length 300 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, Ø30 mm, 600 mm, E = 200 GPasegment 2: brass, Ø40 mm, 300 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø30 mm, E = 200 GPa, σallow = 200 MPatube: Ø120/Ø60 mm, E = 120 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 5 mE = 70 GPa (aluminium)d = 40 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 = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.
m = 2 kgk = 100 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 50 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 50 kgk1 = 50 kN/mk2 = 12 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 96.51 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·960/60 = 100.5 rad/s
T1 = P/ω₁ = 5500/100.5 = 54.71 N·m
Output torque
T2 = 54.71·72/40·0.98 = 96.51 N·m
Answer
T2 = 96.51 N·m
2.Gears & gear trains
Answer: nout = 266.7 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 = 960·60/72 = 800 rpm = nC
Second stage
nout = 800·12/36 = 266.7 rpm
Answer
nout = 266.7 rpm
total ratio = 3.6 : 1
3.Torsion of shafts
Answer: φ = 22.8 °
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·20⁴/32 = 15708 mm⁴
Angle of twist
φ = TL/(GJ) = 1000·1000·500/(80000·15708) = 0.3979 rad
φ = 0.3979·180/π = 22.8°
Answer
φ = 22.8°
4.Torsion of shafts
Answer: TA = 160 N·m
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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).
Equilibrium and compatibility
TA + TB = 400
TA·a = TB·b
TA = 400·2/5 = 160 N·m
TB = 400·3/5 = 240 N·m
Answer
TA = 160 N·m
5.Mohr’s circle & stress transformation
Answer: θp = −28.15 °
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, 160 − 80) = −56.31°
Angle on the element
θp = −28.15°
Answer
θp = −28.15°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σn = 33.92 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.
Wall stresses
r = 1200/2 = 600 mm
σh = pr/t = 1.2·600/15 = 48 MPa
σa = pr/(2t) = 24 MPa
Transform to the seam
Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.
σn = 24·sin²50° + 48·cos²50° = 33.92 MPa
Answer
σn = 33.92 MPa
7.Axial stress & strain
Answer: P = 45.24 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 = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·706.9) = 4.244 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·1257) = 2.387 × 10⁻⁶ mm/N
Load
P = 0.3/6.631 × 10⁻⁶ = 45239 N = 45.24 kN
Answer
P = 45.24 kN
8.Axial stress & strain
Answer: Pmax = 966 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
Ar = π·30²/4 = 706.9 mm²
At = π(120² − 60²)/4 = 8482 mm²
Combined axial stiffness
ΣEA = 200·10³·706.9 + 120·10³·8482 = 1.159 × 10⁹ N
Governing strain
ε = min(0.001, 0.0008333) = 0.0008333 (copper)
Largest load
Pmax = 0.0008333·1.159 × 10⁹ = 966040 N = 966 kN
Answer
Pmax = 966 kN
9.Column buckling
Answer: Pcr = 3.473 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·40⁴/64 = 125664 mm⁴
Effective length
KL = 1·5000 = 5000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·125664/5000² = 3473 N = 3.473 kN
Answer
Pcr = 3.473 kN
10.Column buckling
Answer: σcr = 50.76 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 = 0.7·2500 = 1750 mm
KL/r = 1750/15 = 116.7
Critical stress
σcr = π²E/(KL/r)² = π²·70000/116.7² = 50.76 MPa
Answer
σcr = 50.76 MPa
11.Mechanical vibrations
Answer: fd = 32.62 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 = √(100000/2) = 223.6 rad/s
fn = 223.6/(2π) = 35.59 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 35.59·√(1 − 0.4²) = 32.62 Hz
Answer
fd = 32.62 Hz
12.Mechanical vibrations
Answer: keq = 9.677 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.
1/keq = 1/50 + 1/12 → keq = 50·12/(50 + 12) = 9.677 kN/m
Answer
keq = 9.677 kN/m
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