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 11 kW motor running at 2900 rpm drives a pinion with 30 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 96%. Find the torque T1 on the motor shaft.
P = 11 kWn1 = 2900 rpmN1 = 30N2 = 50η = 96%
T1 = N·m
2.Gears & gear trainsHard
An open belt connects a 100 mm driving pulley at 1440 rpm to a 315 mm pulley; the centres are 1000 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 = 100 mmn1 = 1440 rpmd2 = 315 mmC = 1000 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 = 400 N·md = 30 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid shaft must transmit 30 kW at 500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.
P = 30 kWN = 500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 140 MPa, σᵧ = −50 MPa and τₓᵧ = −10 MPa (tension positive). Find the maximum in-plane shear stress τmax.
σx = 140 MPaσy = −50 MPaτxy = −10 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = −40 MPa, σᵧ = −70 MPa and τₓᵧ = 30 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = −40 MPaσy = −70 MPaτxy = 30 MPaθ = 50°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the steel bar under the axial load.
P = 25 kNd = 16 mmL = 3 mE = 200 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 = 105 kNsegment 1: Ø40 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)
δ = mm
9.Column bucklingMedium
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 3 mE = 70 GPa (aluminium)b × h = 40 × 50 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of steel is fixed at both ends. It must carry 140 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.
P = 140 kNn = 3L = 5 mE = 200 GPa (steel)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 2 cycles later by a peak of 3.6 mm. Find the damping ratio ζ.
x0 = 10 mmx2 = 3.6 mmn = 2
ζ =
12.Mechanical vibrationsHard
A 40 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 15 kN/m and k₂ = 50 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 = 15 kN/mk2 = 50 kN/mk3 = 10 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: T1 = 36.22 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·2900/60 = 303.7 rad/s
T1 = P/ω₁ = 11000/303.7 = 36.22 N·m
Answer
T1 = 36.22 N·m
2.Gears & gear trains
Answer: P = 7.824 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 = 2.926 rad
Slack-side tension
eμθ = e0.7315 = 2.078, T₂ = 2000/2.078 = 962.3 N
Power
P = (2000 − 962.3)·7.54 = 7.824 kW
Answer
P = 7.824 kW
3.Torsion of shafts
Answer: φ = 14.78 °
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) = 400·1000·2000/(39000·79522) = 0.258 rad
φ = 0.258·180/π = 14.78°
Answer
φ = 14.78°
4.Torsion of shafts
Answer: d = 33.16 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·500/60 = 52.36 rad/s
T = P/ω = 30000/52.36 = 573 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·572958/(π·80)) = 33.16 mm
Answer
d = 33.16 mm
In practice you would round up to the next standard size.
5.Mohr’s circle & stress transformation
Answer: τmax = 95.52 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 = √(((140 − −50)/2)² + −10²) = 95.52 MPa
Answer
τmax = 95.52 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −19.98 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
τθ = −15·0.9848 + 30·−0.1736 = −19.98 MPa
Answer
τθ = −19.98 MPa
7.Axial stress & strain
Answer: δ = 1.865 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 = 25·1000/201.1 = 124.3 MPa
Strain (Hooke’s law)
E for steel = 200 GPa = 200000 MPa.
ε = σ/E = 124.3/200000 = 0.0006217 = 621.7 µε
Elongation
δ = ε·L = 0.0006217·3000 mm = 1.865 mm
Answer
δ = 1.865 mm
8.Axial stress & strain
Answer: δ = 14.93 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·40²/4 = 1257 mm², A2 = π·12²/4 = 113.1 mm²
σ1 = 83.56 MPa
σ2 = 928.4 MPa
Add the segment elongations
δ1 = σ1L1/E1 = 83.56·2000/100000 = 1.671 mm
δ2 = σ2L2/E2 = 928.4·1000/70000 = 13.26 mm
δ = δ1 + δ2 = 14.93 mm
Answer
δ = 14.93 mm
9.Column buckling
Answer: Pcr = 41.78 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 = 50·40³/12 = 266667 mm⁴
Effective length
KL = 0.7·3000 = 2100 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·266667/2100² = 41776 N = 41.78 kN
Answer
Pcr = 41.78 kN
10.Column buckling
Answer: d = 72.15 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 = 3·140000 = 420000 N
KL = 0.5·5000 = 2500 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·420000·2500²/(π³·200000))^(1/4) = 72.15 mm
Answer
d = 72.15 mm
Check: the slenderness KL/r is 138.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.08103
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.
Logarithmic decrement
δ = (1/2)·ln(10/3.6) = 0.5108
Damping ratio
ζ = δ/√(4π² + δ²) = 0.5108/√(39.48 + 0.2609) = 0.08103
Answer
ζ = 0.08103
12.Mechanical vibrations
Answer: fn = 2.343 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 = 15 + 50 = 65 kN/m
keq = 65·10/(65 + 10) = 8.667 kN/m
Natural frequency
ωn = √(8667/40) = 14.72 rad/s
fn = 14.72/(2π) = 2.343 Hz
Answer
fn = 2.343 Hz
ExamProblems · examproblems.com/courses/machine-design