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 1440 rpm drives a pinion with 25 teeth, which meshes with a gear with 30 teeth. The gear pair has an efficiency of 97%. Find the output torque T2 on the gear shaft.
P = 3 kWn1 = 1440 rpmN1 = 25N2 = 30η = 97%
T2 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (24 teeth) on the input shaft turning at 1450 rpm drives gear B (25 teeth). Gear C (18 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 = 1450 rpmA = 24B = 25C = 18D = 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 = 500 N·md = 30 mmL = 1 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 30 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 30 kWN = 1500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 60 MPa, σᵧ = 110 MPa and τₓᵧ = 50 MPa (tension positive). Find the angle θs, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).
σx = 60 MPaσy = 110 MPaτxy = 50 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A solid shaft of diameter 50 mm carries an axial compressive force of 100 kN together with a torque of 2000 N·m. At the surface of the shaft, find the largest principal stress σ1.
d = 50 mmF = −100 kNT = 2000 N·m
σ1 = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 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 0.5 mm in total.
segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 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, Ø40 mm, 300 mm, E = 70 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 70 GPa (aluminium)d = 20 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of aluminium is fixed at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 2.5 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 40 kg mass on a spring with a stiffness of 100 kN/m has a viscous damper giving a damping ratio of 0.6. Find the damped natural frequency fd of its free vibration in Hz.
m = 40 kgk = 100 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 100 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 80 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 40 kgk1 = 50 kN/mk2 = 80 kN/mk3 = 100 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T2 = 23.16 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 3000/150.8 = 19.89 N·m
Output torque
T2 = 19.89·30/25·0.97 = 23.16 N·m
Answer
T2 = 23.16 N·m
2.Gears & gear trains
Answer: nout = 348 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 = 1450·24/25 = 1392 rpm = nC
Second stage
nout = 1392·18/72 = 348 rpm
Answer
nout = 348 rpm
total ratio = 4.167 : 1
3.Torsion of shafts
Answer: φ = 9.237 °
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) = 500·1000·1000/(39000·79522) = 0.1612 rad
φ = 0.1612·180/π = 9.237°
Answer
φ = 9.237°
4.Torsion of shafts
Answer: d = 22.99 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 = 30000/157.1 = 191 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·80)) = 22.99 mm
Answer
d = 22.99 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: θs = 13.28 °
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
On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.
cos 2θs = 50/55.9 = 0.8944, sin 2θs = −(−25)/55.9 = 0.4472
2θs = atan2(25, 50) = 26.57°
Angle on the element
θs = 13.28°
check: θs − θp = 13.28 − 58.28 = −45°
Answer
θs = 13.28°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σ1 = 59.91 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.
Stresses at the surface
σ = F/A = −100000/(π·50²/4) = −100000/1963 = −50.93 MPa
τ = 16T/(πd³) = 16·2000000/(π·50³) = 81.49 MPa
Mohr’s circle
C = σ/2 = −25.46 MPa
R = √((−25.46)² + 81.49²) = 85.37 MPa
σ1 = C + R = 59.91 MPa
σ2 = C − R = −110.8 MPa
Answer
σ1 = 59.91 MPa
7.Axial stress & strain
Answer: P = 188.5 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 = 1257 mm², A₂ = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N
Load
P = 0.5/2.653 × 10⁻⁶ = 188496 N = 188.5 kN
Answer
P = 188.5 kN
8.Axial stress & strain
Answer: σ2 = 15.19 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 = 70000·1257/300 = 293215 N/mm
Share of segment 2
F2 = 60000·293215/921534 = 19091 N
Stress
σ2 = 19091/1257 = 15.19 MPa (compression)
Answer
σ2 = 15.19 MPa
9.Column buckling
Answer: Pcr = 5.426 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 = 0.5·2000 = 1000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·7854/1000² = 5426 N = 5.426 kN
Answer
Pcr = 5.426 kN
10.Column buckling
Answer: σcr = 69.09 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·50⁴/64 = 306796 mm⁴
Radius of gyration and slenderness
A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 0.5·2500 = 1250 mm
KL/r = 1250/12.5 = 100
Critical stress
σcr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa
Answer
σcr = 69.09 MPa
11.Mechanical vibrations
Answer: fd = 6.366 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 = √(100000/40) = 50 rad/s
fn = 50/(2π) = 7.958 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 7.958·√(1 − 0.6²) = 6.366 Hz
Answer
fd = 6.366 Hz
12.Mechanical vibrations
Answer: fn = 5.983 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
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 50 + 80 = 130 kN/m
keq = 130·100/(130 + 100) = 56.52 kN/m
Natural frequency
ωn = √(56522/40) = 37.59 rad/s
fn = 37.59/(2π) = 5.983 Hz
Answer
fn = 5.983 Hz
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