Machine Design: cram pack

Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.

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1Formulas and methods

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

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

  1. Find the polar moment of area: J = πd⁴/32 for a solid shaft.
  2. Shear stress at the surface: τ = T·r/J, with r the outer radius.
  3. Angle of twist: φ = TL/(GJ) in radians.
  4. 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

  1. Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
  2. Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
  3. The principal stresses are C ± R and the largest in-plane shear stress is R.
  4. 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
N1​ = 12N2​ = 40

A 2.2 kW motor running at 1440 rpm drives a pinion with 12 teeth, which meshes with a gear with 40 teeth. Ignore losses. Find the torque T1 on the motor shaft.

P = 2.2 kWn1 = 1440 rpmN1 = 12N2 = 40η = 100%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 125 mmd2​ = 250 mm720 rpmC = 1200 mmθ = ?

An open belt connects a 125 mm pulley and a 250 mm pulley whose centres are 1200 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 125 mmd2 = 250 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
d = 20 mmT = 50 N·m200 mmφ = 1.4°

In a torsion test, a solid specimen of diameter 20 mm is twisted by a torque of 50 N·m. The two ends of a gauge length of 200 mm rotate 1.4° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 50 N·md = 20 mmL = 200 mmφ = 1.4°
G = GPa
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 75 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.

P = 75 kWN = 1000 rpmτallow = 75 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 100σy​ = 100τxy​ = −40MPa

A point in a loaded part has σₓ = 100 MPa, σᵧ = 100 MPa and τₓᵧ = −40 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 100 MPaσy = 100 MPaτxy = −40 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
55°D = 1200 mm, t = 15 mm, p = 2.5 MPap

A thin-walled cylindrical tank of inner diameter 1200 mm and wall thickness 15 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 55° to the axis of the cylinder. Find the normal stress σn acting across the weld seam.

D = 1200 mmt = 15 mmp = 2.5 MPaα = 55°
σn = MPa
7.Axial stress & strainMedium
P = ?12600 mm200 mmØ25Ø50δ = 0.25 mm

A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: brass, Ø50 mm, 200 mm, E = 100 GPaδ = 0.25 mm
P = kN
8.Axial stress & strainHard
Ø 25 mmAluminium1 mØ 20 mmSteel0.5 mP = 95 kN

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: Ø25 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 0.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 2 m25 × 80 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 2 mE = 200 GPa (steel)b × h = 25 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 30 mmcross-section

A column of length L made of steel is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 3 mE = 200 GPa (steel)d = 30 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx50 mm5 cycles2.1 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed 5 cycles later by a peak of 2.1 mm. Find the damping ratio ζ.

x0 = 50 mmx5 = 2.1 mmn = 5
ζ =
12.Mechanical vibrationsHard
k1​ = 40k2​ = 80m = 20 kg

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 80 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 20 kgk1 = 40 kN/mk2 = 80 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: T1 = 14.59 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·1440/60 = 150.8 rad/s
T1 = P/ω₁ = 2200/150.8 = 14.59 N·m

Answer

T1 = 14.59 N·m
2.Gears & gear trains

Answer: θ = 174 °

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.

Belt angle

asin(0.05208) = 2.986°

Angle of wrap

θ = 180 − 2·2.986 = 174°

Answer

θ = 174°
3.Torsion of shafts

Answer: G = 26.05 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·20⁴/32 = 15708 mm⁴
φ = 1.4·π/180 = 0.02443 rad

Shear modulus

T = 50 N·m = 50000 N·mm
G = TL/(Jφ) = 50000·200/(15708·0.02443) = 26054 MPa
G = 26.05 GPa

Answer

G = 26.05 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: d = 36.5 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·1000/60 = 104.7 rad/s
T = P/ω = 75000/104.7 = 716.2 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·716197/(π·75)) = 36.5 mm

Answer

d = 36.5 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: τmax = 40 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((100 − 100)/2)² + −40²) = 40 MPa

Answer

τmax = 40 MPa
6.Mohr’s circle & stress transformation

Answer: σn = 66.45 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1200/2 = 600 mm
σh = pr/t = 2.5·600/15 = 100 MPa
σa = pr/(2t) = 50 MPa

Transform to the seam

Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.

σn = 50·sin²55° + 100·cos²55° = 66.45 MPa

Answer

σn = 66.45 MPa
7.Axial stress & strain

Answer: P = 35.06 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 = 490.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·1963) = 1.019 × 10⁻⁶ mm/N

Load

P = 0.25/7.13 × 10⁻⁶ = 35062 N = 35.06 kN

Answer

P = 35.06 kN
8.Axial stress & strain

Answer: σmax = 302.4 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·25²/4 = 490.9 mm², A2 = π·20²/4 = 314.2 mm²
σ1 = 193.5 MPa
σ2 = 302.4 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 302.4 MPa

Answer

σmax = 302.4 MPa
9.Column buckling

Answer: Pcr = 12.85 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 = 80·25³/12 = 104167 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·104167/4000² = 12851 N = 12.85 kN

Answer

Pcr = 12.85 kN
10.Column buckling

Answer: σcr = 3.084 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·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²
r = √(I/A) = √(39761/706.9) = 7.5 mm
KL = 2·3000 = 6000 mm
KL/r = 6000/7.5 = 800

Critical stress

σcr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa

Answer

σcr = 3.084 MPa
11.Mechanical vibrations

Answer: ζ = 0.1004

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(50/2.1) = 0.634

Damping ratio

ζ = δ/√(4π² + δ²) = 0.634/√(39.48 + 0.402) = 0.1004

Answer

ζ = 0.1004
12.Mechanical vibrations

Answer: keq = 120 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 = 40 + 80 = 120 kN/m

Answer

keq = 120 kN/m

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