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​ = 30, m = 3 mm

A 7.5 kW motor running at 1440 rpm drives a pinion with 30 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.

P = 7.5 kWn = 1440 rpmN1 = 30m = 3 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 315 mm720 rpmC = 1000 mmT1​ = 2000 NT2​P = ?

An open belt connects a 150 mm driving pulley at 720 rpm to a 315 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.3 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 = 150 mmn1 = 720 rpmd2 = 315 mmC = 1000 mmμ = 0.3T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
T = 750 N·md = 40 mmL = 1.2 mφ = ?

A solid aluminium 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 = 750 N·md = 40 mmL = 1.2 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
d = 60 mmT0​ = 400 N·mACBa = 1 mb = 1.5 m

A solid steel shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 1 m from A and 1.5 m from B. Find the reaction torque TA at wall A.

d = 60 mmT0 = 400 N·ma = 1 mb = 1.5 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = −10σy​ = 90MPaτxy​ = ?

A point has σₓ = −10 MPa and σᵧ = 90 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 100 MPa?

σx = −10 MPaσy = 90 MPaσ1 = 100 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −60τxy​ = −60MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −60 MPa and τₓᵧ = −60 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = −10 MPaσy = −60 MPaτxy = −60 MPa
σvM = MPa
7.Axial stress & strainMedium
300 kNØ25110 GPaØ120/60600 mmδ = ?

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 120 mm, inner diameter 60 mm, E = 110 GPa). Both are 600 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø25 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 110 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
50 kN12400 mm300 mmØ25Ø30σ2​ = ?

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 300 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 300 mm, E = 100 GPaP = 50 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2 m20 × 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 = 20 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 40 mmcross-section

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σcr from Euler’s formula.

L = 4 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm3 cycles6.9 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 3 cycles later by a peak of 6.9 mm. Find the damping ratio ζ.

x0 = 10 mmx3 = 6.9 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 40k2​ = 5m = 20 kg

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 5 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 20 kgk1 = 40 kN/mk2 = 5 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: Ft = 1105 N

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.

Torque on the pinion

ω = 2π·1440/60 = 150.8 rad/s
T = P/ω = 7500/150.8 = 49.74 N·m

Pitch diameter

d = m·N = 3·30 = 90 mm

Tangential force

Ft = 2T/d = 2·49.74/0.09 = 1105 N

Answer

Ft = 1105 N
2.Gears & gear trains

Answer: P = 6.679 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 = 2.976 rad

Slack-side tension

eμθ = e0.8929 = 2.442, T₂ = 2000/2.442 = 818.9 N

Power

P = (2000 − 818.9)·5.655 = 6.679 kW

Answer

P = 6.679 kW
3.Torsion of shafts

Answer: φ = 7.891 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·1200/(26000·251327) = 0.1377 rad
φ = 0.1377·180/π = 7.891°

Answer

φ = 7.891°
4.Torsion of shafts

Answer: TA = 240 N·m

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 400
TA·a = TB·b
TA = 400·1.5/2.5 = 240 N·m
TB = 400·1/2.5 = 160 N·m

Answer

TA = 240 N·m
5.Mohr’s circle & stress transformation

Answer: τxy = 33.17 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 = (−10 + 90)/2 = 40 MPa
R = σ1 − C = 100 − 40 = 60 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(60² − 50²) = 33.17 MPa

Answer

τxy = 33.17 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 117.9 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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

στxyCσ1​σ2​

Principal stresses

C = −35 MPa, R = 65 MPa
σ1 = 30 MPa, σ2 = −100 MPa, σ3 = 0

Von Mises

σvM = √(30² − 30·(−100) + (−100)²) = 117.9 MPa

Answer

σvM = 117.9 MPa
7.Axial stress & strain

Answer: δ = 0.1745 mm

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

Ar = π·25²/4 = 490.9 mm²
At = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 110·10³·8482 = 1.031 × 10⁹ N

Shortening

δ = 300000·600/1.031 × 10⁹ = 0.1745 mm

Answer

δ = 0.1745 mm
8.Axial stress & strain

Answer: σ2 = 34.65 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 = 200000·490.9/400 = 245437 N/mm
k₂ = 100000·706.9/300 = 235619 N/mm

Share of segment 2

F2 = 50000·235619/481056 = 24490 N

Stress

σ2 = 24490/706.9 = 34.65 MPa (compression)

Answer

σ2 = 34.65 MPa
9.Column buckling

Answer: Pcr = 6.58 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 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·53333/4000² = 6580 N = 6.58 kN

Answer

Pcr = 6.58 kN
10.Column buckling

Answer: σcr = 4.318 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 = 1·4000 = 4000 mm
KL/r = 4000/10 = 400

Critical stress

σcr = π²E/(KL/r)² = π²·70000/400² = 4.318 MPa

Answer

σcr = 4.318 MPa
11.Mechanical vibrations

Answer: ζ = 0.01968

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 = (1/3)·ln(10/6.9) = 0.1237

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1237/√(39.48 + 0.0153) = 0.01968

Answer

ζ = 0.01968
12.Mechanical vibrations

Answer: keq = 4.444 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 in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k1 + 1/k2.

1/keq = 1/40 + 1/5 → keq = 40·5/(40 + 5) = 4.444 kN/m

Answer

keq = 4.444 kN/m

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