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
d1​ = 100 mmd2​ = 500 mm720 rpmT1​ = 1500 NT2​ = 375 NP = ?

A belt runs on a 100 mm driving pulley at 720 rpm. The tension is 1500 N on the tight side and 375 N on the slack side. Find the power P transmitted, in kW.

d1 = 100 mmn1 = 720 rpmT1 = 1500 NT2 = 375 N
P = kW
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 450 mm960 rpmC = 800 mmθ = ?

An open belt connects a 150 mm pulley and a 450 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 150 mmd2 = 450 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
T = 250 N·md = 50 mmL = 1.5 mφ = ?

A solid steel 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 = 250 N·md = 50 mmL = 1.5 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
d = 30 mmT0​ = 100 N·mACBa = 1.2 mb = 1 m

A solid steel shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 100 N·m is applied at C, 1.2 m from A and 1 m from B. Find the maximum shear stress τmax in the shaft.

d = 30 mmT0 = 100 N·ma = 1.2 mb = 1 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
σx​ = −60σy​ = −60τxy​ = −20MPa

A point in a loaded part has σₓ = −60 MPa, σᵧ = −60 MPa and τₓᵧ = −20 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = −60 MPaσy = −60 MPaτxy = −20 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −60σy​ = 120τxy​ = 50MPa

A point in a thin plate is in plane stress: σₓ = −60 MPa, σᵧ = 120 MPa and τₓᵧ = 50 MPa (σz = 0). Find the von Mises equivalent stress σvM.

σx = −60 MPaσy = 120 MPaτxy = 50 MPa
σvM = MPa
7.Axial stress & strainMedium
200 kNØ30100 GPaØ120/60500 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 60 mm, E = 100 GPa). Both are 500 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.

P = 200 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø60 mm, E = 100 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
Ø 40 mmBrass0.5 mØ 20 mmSteel1.5 mP = 75 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 75 kNsegment 1: Ø40 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 1.5 m, steel (E = 200 GPa)
δ = mm
9.Column bucklingMedium
PL = 4 m25 × 50 mmcross-section

A column of length L made of steel is pinned at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 4 mE = 200 GPa (steel)b × h = 25 × 50 mm
Pcr = kN
10.Column bucklingHard
PL = 3 md = 25 mmcross-section

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 = 3 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 1 kN/mζ = 0.45m = 8 kg

An 8 kg mass on a spring with a stiffness of 1 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency fd of its free vibration in Hz.

m = 8 kgk = 1 kN/mζ = 0.45
fd = Hz
12.Mechanical vibrationsHard
k1​ = 5k2​ = 10k3​ = 25m = 10 kg

A 10 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 5 kN/m and k₂ = 10 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 = 10 kgk1 = 5 kN/mk2 = 10 kN/mk3 = 25 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 4.241 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.

Belt speed

v = 3.77 m/s

Power

P = (1500 − 375)·3.77 = 4241 W = 4.241 kW

Answer

P = 4.241 kW
2.Gears & gear trains

Answer: θ = 158.4 °

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(0.1875) = 10.81°

Angle of wrap

θ = 180 − 2·10.81 = 158.4°

Answer

θ = 158.4°
3.Torsion of shafts

Answer: φ = 0.4377 °

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) = 250·1000·1500/(80000·613592) = 0.007639 rad
φ = 0.007639·180/π = 0.4377°

Answer

φ = 0.4377°
4.Torsion of shafts

Answer: τmax = 10.29 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. 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 = 100
TA·a = TB·b
TA = 100·1/2.2 = 45.45 N·m
TB = 100·1.2/2.2 = 54.55 N·m

Maximum shear stress

The larger of the two torques governs (same diameter).

τmax = 16·54545/(π·30³) = 10.29 MPa

Answer

τmax = 10.29 MPa
5.Mohr’s circle & stress transformation

Answer: τmax = 20 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.

σxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((−60 − −60)/2)² + −20²) = 20 MPa

Answer

τmax = 20 MPa
6.Mohr’s circle & stress transformation

Answer: σvM = 180.8 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 = 30 MPa, R = 103 MPa
σ1 = 133 MPa, σ2 = −72.96 MPa, σ3 = 0

Von Mises

σvM = √(133² − 133·(−72.96) + (−72.96)²) = 180.8 MPa

Answer

σvM = 180.8 MPa
7.Axial stress & strain

Answer: δ = 0.1011 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·30²/4 = 706.9 mm²
At = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 100·10³·8482 = 9.896 × 10⁸ N

Shortening

δ = 200000·500/9.896 × 10⁸ = 0.1011 mm

Answer

δ = 0.1011 mm
8.Axial stress & strain

Answer: δ = 2.089 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·20²/4 = 314.2 mm²
σ1 = 59.68 MPa
σ2 = 238.7 MPa

Add the segment elongations

δ1 = σ1L1/E1 = 59.68·500/100000 = 0.2984 mm
δ2 = σ2L2/E2 = 238.7·1500/200000 = 1.79 mm
δ = δ1 + δ2 = 2.089 mm

Answer

δ = 2.089 mm
9.Column buckling

Answer: Pcr = 8.032 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 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 1·4000 = 4000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·65104/4000² = 8032 N = 8.032 kN

Answer

Pcr = 8.032 kN
10.Column buckling

Answer: σcr = 8.567 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 = 1·3000 = 3000 mm
KL/r = 3000/6.25 = 480

Critical stress

σcr = π²E/(KL/r)² = π²·200000/480² = 8.567 MPa

Answer

σcr = 8.567 MPa
11.Mechanical vibrations

Answer: fd = 1.589 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 = √(1000/8) = 11.18 rad/s
fn = 11.18/(2π) = 1.779 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 1.779·√(1 − 0.45²) = 1.589 Hz

Answer

fd = 1.589 Hz
12.Mechanical vibrations

Answer: fn = 4.873 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 = 5 + 10 = 15 kN/m
keq = 15·25/(15 + 25) = 9.375 kN/m

Natural frequency

ωn = √(9375/10) = 30.62 rad/s
fn = 30.62/(2π) = 4.873 Hz

Answer

fn = 4.873 Hz

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