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 160 mm pulley at 1450 rpm drives a 300 mm pulley by a belt with 1.5 % slip. Find the speed n2 of the driven pulley.
n1 = 1450 rpmd1 = 160 mmd2 = 300 mms = 1.5 %
n2 = rpm
2.Gears & gear trainsHard
An open belt connects a 120 mm driving pulley at 1500 rpm to a 250 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).
d1 = 120 mmn1 = 1500 rpmd2 = 250 mmC = 1000 mmμ = 0.25T1 = 500 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 = 100 N·md = 50 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
A solid steel shaft of diameter 60 mm is fixed at A. A torque of 200 N·m acts at B, 1.5 m from A, and a torque of 250 N·m in the same direction acts at the free end C, a further 1.5 m along. Find the maximum shear stress τmax in the shaft.
d = 60 mmTB = 200 N·mTC = 250 N·mAB = 1.5 mBC = 1.5 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point has σₓ = 20 MPa and σᵧ = 100 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 150 MPa?
σx = 20 MPaσy = 100 MPaσ1 = 150 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 100 MPa, σᵧ = 50 MPa and τₓᵧ = 10 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).
σx = 100 MPaσy = 50 MPaτxy = 10 MPaθ = 35°
τθ = MPa
7.Axial stress & strainMedium
Find the elongation δ of the brass bar under the axial load.
P = 45 kNd = 10 mmL = 2.5 mE = 100 GPa
δ = mm
8.Axial stress & strainHard
A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 60 mm, length 200 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 30 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.
segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 200 mm, E = 100 GPaP = 30 kN
σ2 = MPa
9.Column bucklingMedium
A column of length L made of aluminium is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.
L = 1.5 mE = 70 GPa (aluminium)d = 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 = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 5 cycles later by a peak of 9.7 mm. Find the damping ratio ζ.
x0 = 25 mmx5 = 9.7 mmn = 5
ζ =
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 = 80 kN/m and k₂ = 5 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 = 80 kN/mk2 = 5 kN/mk3 = 60 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: n2 = 761.7 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
1450·160/300 = 773.3 rpm
With slip
n2 = 773.3·0.985 = 761.7 rpm
Answer
n2 = 761.7 rpm
2.Gears & gear trains
Answer: P = 2.493 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 = 3.012 rad
Slack-side tension
eμθ = e0.7529 = 2.123, T₂ = 500/2.123 = 235.5 N
Power
P = (500 − 235.5)·9.425 = 2.493 kW
Answer
P = 2.493 kW
3.Torsion of shafts
Answer: φ = 0.4789 °
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·50⁴/32 = 613592 mm⁴
Angle of twist
φ = TL/(GJ) = 100·1000·2000/(39000·613592) = 0.008358 rad
φ = 0.008358·180/π = 0.4789°
Answer
φ = 0.4789°
4.Torsion of shafts
Answer: τmax = 10.61 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.
Internal torques
Sense of TC taken as positive.
TBC = TC = 250 N·m
TAB = TC + TB = 250 + 200 = 450 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 450 N·m
τmax = 16·450000/(π·60³) = 10.61 MPa
Answer
τmax = 10.61 MPa
5.Mohr’s circle & stress transformation
Answer: τxy = 80.62 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.
Required radius
C = (20 + 100)/2 = 60 MPa
R = σ1 − C = 150 − 60 = 90 MPa
Shear stress
τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(90² − 40²) = 80.62 MPa
Answer
τxy = 80.62 MPa
6.Mohr’s circle & stress transformation
Answer: τθ = −20.07 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 = −25·0.9397 + 10·0.342 = −20.07 MPa
Answer
τθ = −20.07 MPa
7.Axial stress & strain
Answer: δ = 14.32 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·10²/4 = 78.54 mm²
σ = P/A = 45·1000/78.54 = 573 MPa
Strain (Hooke’s law)
E for brass = 100 GPa = 100000 MPa.
ε = σ/E = 573/100000 = 0.00573 = 5730 µε
Elongation
δ = ε·L = 0.00573·2500 mm = 14.32 mm
Answer
δ = 14.32 mm
8.Axial stress & strain
Answer: σ2 = 9.316 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 100000·2827/200 = 1.414 × 10⁶ N/mm
Share of segment 2
F2 = 30000·1.414 × 10⁶/1.61 × 10⁶ = 26341 N
Stress
σ2 = 26341/2827 = 9.316 MPa (compression)
Answer
σ2 = 9.316 MPa
9.Column buckling
Answer: Pcr = 94.2 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·50⁴/64 = 306796 mm⁴
Effective length
KL = 1·1500 = 1500 mm
Critical load
Pcr = π²EI/(KL)² = π²·70000·306796/1500² = 94203 N = 94.2 kN
Answer
Pcr = 94.2 kN
10.Column buckling
Answer: σcr = 77.11 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·25⁴/64 = 19175 mm⁴
Radius of gyration and slenderness
A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 1·1000 = 1000 mm
KL/r = 1000/6.25 = 160
Critical stress
σcr = π²E/(KL/r)² = π²·200000/160² = 77.11 MPa
Answer
σcr = 77.11 MPa
11.Mechanical vibrations
Answer: ζ = 0.03012
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 = (1/5)·ln(25/9.7) = 0.1893
Damping ratio
ζ = δ/√(4π² + δ²) = 0.1893/√(39.48 + 0.03585) = 0.03012
Answer
ζ = 0.03012
12.Mechanical vibrations
Answer: fn = 2.985 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 = 80 + 5 = 85 kN/m
keq = 85·60/(85 + 60) = 35.17 kN/m
Natural frequency
ωn = √(35172/100) = 18.75 rad/s
fn = 18.75/(2π) = 2.985 Hz
Answer
fn = 2.985 Hz
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