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
An 80 mm pulley at 960 rpm drives a 400 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.
n1 = 960 rpmd1 = 80 mmd2 = 400 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 160 mm driving pulley at 1500 rpm to a 200 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 160 mmn1 = 1500 rpmd2 = 200 mmC = 1200 mmμ = 0.25T1 = 2000 N
P = kW
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 = 0.8 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A hollow shaft transmits 30 kW at 300 rpm. Find the maximum shear stress τmax in the shaft.
P = 30 kWN = 300 rpmdo = 80 mmdi = 48 mm
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 10 MPa, σᵧ = 50 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 10 MPaσy = 50 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 50 MPa, σᵧ = 40 MPa and τₓᵧ = −50 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 45° clockwise from the x-axis.
σx = 50 MPaσy = 40 MPaτxy = −50 MPaθ = -45°
σθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the copper bar under the axial load.
P = 100 kNd = 20 mmL = 3 mE = 117 GPa
δ = mm
8.Axial stress & strainHard
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
P = 70 kNsegment 1: Ø30 mm, 2 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 m, copper (E = 117 GPa)
δ = mm
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 = 5 mE = 70 GPa (aluminium)d = 80 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 120 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 120 kNn = 2.5L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
A 2 kg mass on a spring with a stiffness of 40 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 = 2 kgk = 40 kN/mζ = 0.6
fd = Hz
12.Mechanical vibrationsHard
A 100 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 12 kN/m and k₂ = 15 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 = 100 kgk1 = 12 kN/mk2 = 15 kN/mk3 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 186.2 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
960·80/400 = 192 rpm
With slip
n2 = 192·0.97 = 186.2 rpm
Answer
n2 = 186.2 rpm
2.Gears & gear trains
Answer: P = 13.58 kW
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.
Angle of wrap
θ = 3.108 rad
Slack-side tension
eμθ = e0.7771 = 2.175, T₂ = 2000/2.175 = 919.5 N
Power
P = (2000 − 919.5)·12.57 = 13.58 kW
Answer
P = 13.58 kW
3.Torsion of shafts
Answer: φ = 7.39 °
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) = 500·1000·800/(39000·79522) = 0.129 rad
φ = 0.129·180/π = 7.39°
Answer
φ = 7.39°
4.Torsion of shafts
Answer: τmax = 10.91 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 = 30000/31.42 = 954.9 N·m
Polar second moment of area
J = π(80⁴ − 48⁴)/32 = 3.5 × 10⁶ mm⁴
Maximum shear stress
τmax = T·(do/2)/J = 954930·40/3.5 × 10⁶ = 10.91 MPa
Answer
τmax = 10.91 MPa
5.Mohr’s circle & stress transformation
Answer: τmax = 28.28 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.
Radius of Mohr’s circle
τmax = R = √(((10 − 50)/2)² + −20²) = 28.28 MPa
Answer
τmax = 28.28 MPa
6.Mohr’s circle & stress transformation
Answer: σθ = 95 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 = 45 + 5·0 + −50·−1 = 95 MPa
Answer
σθ = 95 MPa
7.Axial stress & strain
Answer: δ = 8.162 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·20²/4 = 314.2 mm²
σ = P/A = 100·1000/314.2 = 318.3 MPa
Strain (Hooke’s law)
E for copper = 117 GPa = 117000 MPa.
ε = σ/E = 318.3/117000 = 0.002721 = 2721 µε
Elongation
δ = ε·L = 0.002721·3000 mm = 8.162 mm
Answer
δ = 8.162 mm
8.Axial stress & strain
Answer: δ = 5.789 mm
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·30²/4 = 706.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 99.03 MPa
σ2 = 222.8 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 99.03·2000/100000 = 1.981 mm
δ2 = σ2L2/E2 = 222.8·2000/117000 = 3.809 mm
δ = δ1 + δ2 = 5.789 mm
Answer
δ = 5.789 mm
9.Column buckling
Answer: Pcr = 13.89 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·80⁴/64 = 2.011 × 10⁶ mm⁴
Effective length
KL = 2·5000 = 10000 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/10000² = 13891 N = 13.89 kN
Answer
Pcr = 13.89 kN
10.Column buckling
Answer: d = 66.32 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 = 2.5·120000 = 300000 N
KL = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·300000·2500²/(π³·200000))^(1/4) = 66.32 mm
Answer
d = 66.32 mm
Check: the slenderness KL/r is 150.8, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: fd = 18.01 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 = √(40000/2) = 141.4 rad/s
fn = 141.4/(2π) = 22.51 Hz
Damped natural frequency
fd = fn·√(1 − ζ²) = 22.51·√(1 − 0.6²) = 18.01 Hz
Answer
fd = 18.01 Hz
12.Mechanical vibrations
Answer: fn = 2.172 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 = 12 + 15 = 27 kN/m
keq = 27·60/(27 + 60) = 18.62 kN/m
Natural frequency
ωn = √(18621/100) = 13.65 rad/s
fn = 13.65/(2π) = 2.172 Hz
Answer
fn = 2.172 Hz
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