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 3 kW motor running at 1500 rpm drives a pinion with 45 teeth, which meshes with a gear with 80 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 1500 rpmN1 = 45N2 = 80η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (16 teeth) on the input shaft turns at 2880 rpm and drives gear B (72 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 400 rpm. How many teeth ND must gear D have?
nin = 2880 rpmA = 16B = 72C = 15nout = 400 rpm
ND =
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 = 1000 N·md = 25 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 500 N·m is applied at C, 1.5 m from A and 2.5 m from B. Find the reaction torque TA at wall A.
d = 60 mmT0 = 500 N·ma = 1.5 mb = 2.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 110 MPa, σᵧ = −70 MPa and τₓᵧ = −10 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 110 MPaσy = −70 MPaτxy = −10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = −50 MPa, σᵧ = −20 MPa and τₓᵧ = 40 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = −50 MPaσy = −20 MPaτxy = 40 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.
P = 250 kNrod: Ø40 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 500 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø50 mm, 500 mm, E = 100 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 55 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 55 kNn = 3L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
An 8 kg mass on a spring with a stiffness of 40 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.
m = 8 kgk = 40 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 100 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 100 kgk1 = 100 kN/mk2 = 100 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T2 = 32.93 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·1500/60 = 157.1 rad/s
T1 = P/ω₁ = 3000/157.1 = 19.1 N·m
Output torque
T2 = 19.1·80/45·0.97 = 32.93 N·m
Answer
T2 = 32.93 N·m
2.Gears & gear trains
Answer: ND = 24
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.
Intermediate shaft speed
nB = nC = 2880·16/72 = 640 rpm
Teeth on gear D
ND = C·nC/nout = 15·640/400 = 24
Answer
ND = 24
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) = 1000·1000·500/(39000·38350) = 0.3343 rad
φ = 0.3343·180/π = 19.15°
Answer
φ = 19.15°
4.Torsion of shafts
Answer: TA = 312.5 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 = 500
TA·a = TB·b
TA = 500·2.5/4 = 312.5 N·m
TB = 500·1.5/4 = 187.5 N·m
Answer
TA = 312.5 N·m
5.Mohr’s circle & stress transformation
Answer: θp = −3.17 °
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·−10, 110 − −70) = −6.34°
Angle on the element
θp = −3.17°
Answer
θp = −3.17°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 2.926
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 = −35 MPa, R = 42.72 MPa
σ1 = 7.72 MPa, σ2 = −77.72 MPa, σ3 = 0
Tresca
σmax − σmin = 85.44 MPa
n = 250/85.44 = 2.926
Answer
n = 2.926
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 0.2487 mm
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·40²/4 = 1257 mm²
At = π(100² − 60²)/4 = 5027 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 70·10³·5027 = 6.032 × 10⁸ N
Shortening
δ = 250000·600/6.032 × 10⁸ = 0.2487 mm
Answer
δ = 0.2487 mm
8.Axial stress & strain
Answer: σ2 = 16.08 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 = 200000·314.2/600 = 104720 N/mm
k₂ = 100000·1963/500 = 392699 N/mm
Share of segment 2
F2 = 40000·392699/497419 = 31579 N
Stress
σ2 = 31579/1963 = 16.08 MPa (compression)
Answer
σ2 = 16.08 MPa
9.Column buckling
Answer: Pcr = 48.83 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·60⁴/64 = 636173 mm⁴
Effective length
KL = 2·1500 = 3000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·636173/3000² = 48835 N = 48.83 kN
Answer
Pcr = 48.83 kN
10.Column buckling
Answer: d = 72.25 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 = 3·55000 = 165000 N
KL = 2·2000 = 4000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·165000·4000²/(π³·200000))^(1/4) = 72.25 mm
Answer
d = 72.25 mm
Check: the slenderness KL/r is 221.5, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 10.05 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 = √(40000/8) = 70.71 rad/s
fn = 70.71/(2π) = 11.25 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 11.25·√(1 − 0.45²) = 10.05 Hz
Answer
fd = 10.05 Hz
12.Mechanical vibrations
Answer: keq = 200 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 = 100 + 100 = 200 kN/m
Answer
keq = 200 kN/m
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