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 125 mm pulley at 960 rpm drives a 450 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 125 mmd2 = 450 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
In a compound gear train, gear A (25 teeth) on the input shaft turning at 960 rpm drives gear B (30 teeth). Gear C (50 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed nout.
nin = 960 rpmA = 25B = 30C = 50D = 72
nout = rpm
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 = 400 N·md = 25 mmL = 0.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 5 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 5 kWN = 1500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 10 MPa, σᵧ = 100 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 = 100 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −20 MPa, σᵧ = 50 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.
σx = −20 MPaσy = 50 MPaτxy = −60 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa). Both are 300 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.
P = 50 kNrod: Ø25 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 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: aluminium, Ø30 mm, 300 mm, E = 70 GPaP = 50 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 = 1 mE = 200 GPa (steel)d = 20 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 75 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.
P = 75 kNn = 2L = 2 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 40 kg mass on a spring with a stiffness of 20 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 = 40 kgk = 20 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 25 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 = 100 kgk1 = 25 kN/mk2 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 258.7 rpm
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.
Speed without slip
960·125/450 = 266.7 rpm
With slip
n2 = 266.7·0.97 = 258.7 rpm
Answer
n2 = 258.7 rpm
2.Gears & gear trains
Answer: nout = 555.6 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 = 960·25/30 = 800 rpm = nC
Second stage
nout = 800·50/72 = 555.6 rpm
Answer
nout = 555.6 rpm
total ratio = 1.728 : 1
3.Torsion of shafts
Answer: φ = 7.662 °
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) = 400·1000·500/(39000·38350) = 0.1337 rad
φ = 0.1337·180/π = 7.662°
Answer
φ = 7.662°
4.Torsion of shafts
Answer: d = 12.65 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 = 5000/157.1 = 31.83 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·31831/(π·80)) = 12.65 mm
Answer
d = 12.65 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θp = −63.43 °
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 − 100) = −126.9°
Angle on the element
θp = −63.43°
Answer
θp = −63.43°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = −53.35 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 = 15 + −35·0.342 + −60·0.9397 = −53.35 MPa
Answer
σθ = −53.35 MPa
7.Axial stress & strain
Answer: δ = 0.03708 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·25²/4 = 490.9 mm²
At = π(80² − 50²)/4 = 3063 mm²
Combined axial stiffness
ΣEA = 200·10³·490.9 + 100·10³·3063 = 4.045 × 10⁸ N
Shortening
δ = 50000·300/4.045 × 10⁸ = 0.03708 mm
Answer
δ = 0.03708 mm
8.Axial stress & strain
Answer: σ2 = 14.71 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 = 70000·706.9/300 = 164934 N/mm
Share of segment 2
F2 = 50000·164934/793252 = 10396 N
Stress
σ2 = 10396/706.9 = 14.71 MPa (compression)
Answer
σ2 = 14.71 MPa
9.Column buckling
Answer: Pcr = 62.01 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·20⁴/64 = 7854 mm⁴
Effective length
KL = 0.5·1000 = 500 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·7854/500² = 62013 N = 62.01 kN
Answer
Pcr = 62.01 kN
10.Column buckling
Answer: d = 35.27 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 = 2·75000 = 150000 N
KL = 0.5·2000 = 1000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·150000·1000²/(π³·200000))^(1/4) = 35.27 mm
Answer
d = 35.27 mm
Check: the slenderness KL/r is 113.4, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 3.262 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 = √(20000/40) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 3.559·√(1 − 0.4²) = 3.262 Hz
Answer
fd = 3.262 Hz
12.Mechanical vibrations
Answer: fn = 5.627 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 = 25 + 100 = 125 kN/m
Natural frequency
ωn = √(125000/100) = 35.36 rad/s
fn = 35.36/(2π) = 5.627 Hz
Answer
fn = 5.627 Hz
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