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 7.5 kW motor running at 960 rpm drives a pinion with 36 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 98%. Find the torque T1 on the motor shaft.
P = 7.5 kWn1 = 960 rpmN1 = 36N2 = 50η = 98%
T1 = N·m
2.Gears & gear trainsHard
In a compound gear train, gear A (12 teeth) on the input shaft turning at 1450 rpm drives gear B (32 teeth). Gear C (20 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 = 1450 rpmA = 12B = 32C = 20D = 36
nout = rpm
3.Torsion of shaftsMedium
A hollow steel shaft (outer diameter 50 mm, inner diameter 35 mm, length 1.5 m) is fixed at one end and twisted by a torque of 750 N·m at the other. Find the angle of twist φ in degrees.
T = 750 N·mdo = 50 mmdi = 35 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 40 mm is fixed to rigid walls at both ends A and B. A torque of 250 N·m is applied at C, 2.5 m from A and 1 m from B. Find the maximum shear stress τmax in the shaft.
d = 40 mmT0 = 250 N·ma = 2.5 mb = 1 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 140 MPa, σᵧ = −80 MPa and τₓᵧ = −60 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 = 140 MPaσy = −80 MPaτxy = −60 MPa
θs = °
6.Mohr’s circle & stress transformationHard
A point in a thin plate is in plane stress: σₓ = 10 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (σz = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.
σx = 10 MPaσy = −60 MPaτxy = 30 MPaσY = 460 MPa
n =
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 70 kNd = 16 mmL = 1.5 mE = 100 GPa
δ = mm
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: Ø40 mm, 0.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1.5 m, brass (E = 100 GPa)
σmax = 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 = 2.5 mE = 70 GPa (aluminium)b × h = 20 × 100 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at both ends. It must carry 70 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 70 kNn = 2.5L = 4 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
A 500 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 8660 N·s/m. Find the damping ratio ζ.
m = 500 kgk = 150 kN/mc = 8660 N·s/m
ζ =
12.Mechanical vibrationsHard
A 5 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 5 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.
m = 5 kgk1 = 10 kN/mk2 = 5 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: T1 = 74.6 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 = 7500/100.5 = 74.6 N·m
Answer
T1 = 74.6 N·m
2.Gears & gear trains
Answer: nout = 302.1 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 = 1450·12/32 = 543.8 rpm = nC
Second stage
nout = 543.8·20/36 = 302.1 rpm
Answer
nout = 302.1 rpm
total ratio = 4.8 : 1
3.Torsion of shafts
Answer: φ = 1.728 °
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(50⁴ − 35⁴)/32 = 466269 mm⁴
Angle of twist
T = 750 N·m = 750000 N·mm
L = 1.5 m = 1500 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 750000·1500/(80000·466269) = 0.03016 rad
φ = 0.03016·180/π = 1.728°
Answer
φ = 1.728°
4.Torsion of shafts
Answer: τmax = 14.21 MPa
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 = 250
TA·a = TB·b
TA = 250·1/3.5 = 71.43 N·m
TB = 250·2.5/3.5 = 178.6 N·m
Maximum shear stress
The larger of the two torques governs (same diameter).
τmax = 16·178571/(π·40³) = 14.21 MPa
Answer
τmax = 14.21 MPa
5.Mohr’s circle & stress transformation
Answer: θs = −59.31 °
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 = −60/125.3 = −0.4789, sin 2θs = −110/125.3 = −0.8779
2θs = atan2(−110, −60) = −118.6°
Angle on the element
θs = −59.31°
check: θs − θp = −59.31 − (−14.31) = −45°
Answer
θs = −59.31°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: n = 4.989
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 = −25 MPa, R = 46.1 MPa
σ1 = 21.1 MPa, σ2 = −71.1 MPa, σ3 = 0
Tresca
σmax − σmin = 92.2 MPa
n = 460/92.2 = 4.989
Answer
n = 4.989
n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.
7.Axial stress & strain
Answer: δ = 5.222 mm
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
Stress
A = π·16²/4 = 201.1 mm²
σ = P/A = 70·1000/201.1 = 348.2 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 348.2/100000 = 0.003482 = 3482 µε
Elongation
δ = ε·L = 0.003482·1500 mm = 5.222 mm
Answer
δ = 5.222 mm
8.Axial stress & strain
Answer: σmax = 840 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 75.6 MPa
σ2 = 840 MPa
Largest stress
The thinner segment has the smaller area, so it carries the larger stress.
σmax = 840 MPa
Answer
σmax = 840 MPa
9.Column buckling
Answer: Pcr = 29.48 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 rectangle buckles about its weak axis, using the smaller dimension cubed.
Imin = h·b³/12 = 100·20³/12 = 66667 mm⁴
Effective length
KL = 0.5·2500 = 1250 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·66667/1250² = 29477 N = 29.48 kN
Answer
Pcr = 29.48 kN
10.Column buckling
Answer: d = 67.4 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 = 2.5·70000 = 175000 N
KL = 0.5·4000 = 2000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·175000·2000²/(π³·70000))^(1/4) = 67.4 mm
Answer
d = 67.4 mm
Check: the slenderness KL/r is 118.7, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.5
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.
Critical damping
cc = 2·√(k·m) = 2·√(150000·500) = 17321 N·s/m
Damping ratio
ζ = c/cc = 8660/17321 = 0.5
Answer
ζ = 0.5
ζ < 1, so the system is underdamped and oscillates as it decays.
12.Mechanical vibrations
Answer: keq = 3.333 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 = 1/10 + 1/5 → keq = 10·5/(10 + 5) = 3.333 kN/m
Answer
keq = 3.333 kN/m
ExamProblems · examproblems.com/courses/machine-design