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
Two parallel shafts are 135 mm apart. They must be connected by a pair of spur gears with a module of 1.5 mm and a speed reduction of 5 : 1. How many teeth N2 must the gear have?
a = 135 mmm = 1.5 mmi = 5
N2 =
2.Gears & gear trainsHard
A 4 kW motor at 960 rpm drives gear A (60 teeth), which meshes with gear B (72 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (36 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque Tout.
P = 4 kWnin = 960 rpmA = 60B = 72C = 16D = 36ηmesh = 96%
Tout = N·m
3.Torsion of shaftsMedium
In a torsion test, a solid specimen of diameter 10 mm is twisted by a torque of 20 N·m. The two ends of a gauge length of 400 mm rotate 5.84° relative to each other (the material is still elastic). Find the shear modulus G of the material.
T = 20 N·md = 10 mmL = 400 mmφ = 5.84°
G = GPa
4.Torsion of shaftsHard
A hollow shaft transmits 75 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.
P = 75 kWN = 300 rpmdo = 100 mmdi = 50 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 140 MPa, σᵧ = −30 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 = 140 MPaσy = −30 MPaτxy = 10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 140 MPa, σᵧ = 120 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° clockwise from the x-axis.
σx = 140 MPaσy = 120 MPaτxy = −60 MPaθ = -45°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 60 mm, length 400 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.
segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: copper, Ø60 mm, 400 mm, E = 120 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of brass (diameter 40 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: brass, Ø40 mm, 200 mm, E = 100 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.
L = 1 mE = 200 GPa (steel)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
A 20 kg mass on a spring with a stiffness of 5 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 = 20 kgk = 5 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 25 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.
m = 40 kgk1 = 60 kN/mk2 = 25 kN/m
keq = kN/m
3Solutions
1.Gears & gear trains
Answer: N2 = 150
Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: Ft = 2T/d.
Total number of teeth
N1 + N2 = 2a/m = 2·135/1.5 = 180
Split by the ratio
N1 = 180/(1 + 5) = 30
N2 = 5·30 = 150
Answer
N2 = 150
2.Gears & gear trains
Answer: Tout = 99.01 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.
Output speed
nout = 960·(60/72)·(16/36) = 355.6 rpm
ωout = 2π·355.6/60 = 37.23 rad/s
Output power and torque
Pout = 4000·0.96² = 3686 W
Tout = Pout/ωout = 3686/37.23 = 99.01 N·m
Answer
Tout = 99.01 N·m
3.Torsion of shafts
Answer: G = 79.95 GPa
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 moment and angle
J = π·10⁴/32 = 981.7 mm⁴
φ = 5.84·π/180 = 0.1019 rad
Shear modulus
T = 20 N·m = 20000 N·mm
G = TL/(Jφ) = 20000·400/(981.7·0.1019) = 79947 MPa
G = 79.95 GPa
Answer
G = 79.95 GPa
Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.
4.Torsion of shafts
Answer: τmax = 12.97 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.
Torque
ω = 2πN/60 = 2π·300/60 = 31.42 rad/s
T = P/ω = 75000/31.42 = 2387 N·m
Polar second moment of area
J = π(100⁴ − 50⁴)/32 = 9.204 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 2.387 × 10⁶·50/9.204 × 10⁶ = 12.97 MPa
Answer
τmax = 12.97 MPa
5.Mohr’s circle & stress transformation
Answer: θp = 3.355 °
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, 140 − −30) = 6.71°
Angle on the element
θp = 3.355°
Answer
θp = 3.355°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 190 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 = 130 + 10·0 + −60·−1 = 190 MPa
Answer
σθ = 190 MPa
7.Axial stress & strain
Answer: P = 99.79 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 = 2827 mm²
Flexibilities
L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(120000·2827) = 1.179 × 10⁻⁶ mm/N
Load
P = 0.4/4.008 × 10⁻⁶ = 99792 N = 99.79 kN
Answer
P = 99.79 kN
8.Axial stress & strain
Answer: σ2 = 12.63 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 = 200000·490.9/600 = 163625 N/mm
k₂ = 100000·1257/200 = 628319 N/mm
Share of segment 2
F2 = 20000·628319/791943 = 15868 N
Stress
σ2 = 15868/1257 = 12.63 MPa (compression)
Answer
σ2 = 12.63 MPa
9.Column buckling
Answer: Pcr = 29.14 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 = 50·25³/12 = 65104 mm⁴
Effective length
KL = 0.7·3000 = 2100 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·65104/2100² = 29141 N = 29.14 kN
Answer
Pcr = 29.14 kN
10.Column buckling
Answer: σcr = 197.4 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·40⁴/64 = 125664 mm⁴
Radius of gyration and slenderness
A = 1257 mm²
r = √(I/A) = √(125664/1257) = 10 mm
KL = 1·1000 = 1000 mm
KL/r = 1000/10 = 100
Critical stress
σcr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa
Answer
σcr = 197.4 MPa
11.Mechanical vibrations
Answer: fd = 2.013 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 = √(5000/20) = 15.81 rad/s
fn = 15.81/(2π) = 2.516 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 2.516·√(1 − 0.6²) = 2.013 Hz
Answer
fd = 2.013 Hz
12.Mechanical vibrations
Answer: keq = 85 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 = 60 + 25 = 85 kN/m
Answer
keq = 85 kN/m
ExamProblems · examproblems.com/courses/machine-design