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​ = 25, m = 3 mm

A 22 kW motor running at 960 rpm drives a pinion with 25 teeth and a module of 3 mm. Find the tangential force Ft on the pinion teeth.

P = 22 kWn = 960 rpmN1 = 25m = 3 mm
Ft = N
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 250 mm1450 rpmC = 1200 mmθ = ?

An open belt connects a 120 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 = 120 mmd2 = 250 mmC = 1200 mm
θ = °
3.Torsion of shaftsMedium
T = 750 N·md = 30 mmL = 1 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 = 750 N·md = 30 mmL = 1 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?5 kW, 1500 rpm

A hollow shaft must transmit 5 kW at 1500 rpm. Its inner diameter is to be 0.5 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.

P = 5 kWN = 1500 rpmdi/do = 0.5τallow = 60 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = 50τxy​ = 30MPa

A point in a thin plate is in plane stress: σₓ = 30 MPa, σᵧ = 50 MPa and τₓᵧ = 30 MPa, with σz = 0. Find the absolute maximum shear stress τabs at the point (considering all planes in three dimensions).

σx = 30 MPaσy = 50 MPaτxy = 30 MPaσz = 0
τabs = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 70σy​ = 70τxy​ = 40MPa

A point in a loaded part has σₓ = 70 MPa, σᵧ = 70 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τθ on a plane whose normal is rotated 20° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σx = 70 MPaσy = 70 MPaτxy = 40 MPaθ = -20°
τθ = MPa
7.Axial stress & strainMedium
P = ?12600 mm400 mmØ25Ø40δ = 0.4 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 copper (diameter 40 mm, length 400 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: copper, Ø40 mm, 400 mm, E = 120 GPaδ = 0.4 mm
P = kN
8.Axial stress & strainHard
Ø 40 mmCopper2 mØ 16 mmCopper0.5 mP = 85 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 = 85 kNsegment 1: Ø40 mm, 2 m, copper (E = 117 GPa)segment 2: Ø16 mm, 0.5 m, copper (E = 117 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 3 m30 × 40 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 = 3 mE = 200 GPa (steel)b × h = 30 × 40 mm
Pcr = kN
10.Column bucklingHard
PL = 5 md = 60 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 = 5 mE = 200 GPa (steel)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 40 kN/mc = 390 N·s/mm = 150 kg

A 150 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 390 N·s/m. Find the damping ratio ζ.

m = 150 kgk = 40 kN/mc = 390 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 50k2​ = 15m = 20 kg

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 50 kN/m and the lower one k₂ = 15 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 20 kgk1 = 50 kN/mk2 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: Ft = 5836 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·960/60 = 100.5 rad/s
T = P/ω = 22000/100.5 = 218.8 N·m

Pitch diameter

d = m·N = 3·25 = 75 mm

Tangential force

Ft = 2T/d = 2·218.8/0.075 = 5836 N

Answer

Ft = 5836 N
2.Gears & gear trains

Answer: θ = 173.8 °

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(0.05417) = 3.105°

Angle of wrap

θ = 180 − 2·3.105 = 173.8°

Answer

θ = 173.8°
3.Torsion of shafts

Answer: φ = 6.755 °

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) = 750·1000·1000/(80000·79522) = 0.1179 rad
φ = 0.1179·180/π = 6.755°

Answer

φ = 6.755°
4.Torsion of shafts

Answer: do = 14.23 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·1500/60 = 157.1 rad/s
T = P/ω = 5000/157.1 = 31.83 N·m

Outer diameter from the allowable stress

T = 31.83 N·m = 31831 N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·31831/(π·60·0.9375)) = 14.23 mm
di = 0.5·14.23 = 7.115 mm

Answer

do = 14.23 mm

A solid shaft would need d = 13.93 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τabs = 35.81 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. In three dimensions the absolute maximum shear stress is (σmax − σmin)/2 over all three principal stresses; in plane stress the third one is zero.

στxyCσ1​σ2​

In-plane principal stresses

C = 40 MPa, R = √((−10)² + 30²) = 31.62 MPa
σ1 = 71.62 MPa, σ2 = 8.377 MPa, σz = 0

Largest of the three circles

τ = max(|σ1 − σ2|, |σ1|, |σ2|)/2 = max(63.25, 71.62, 8.377)/2 = 35.81 MPa

Answer

τabs = 35.81 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: τθ = 30.64 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 = −0·−0.6428 + 40·0.766 = 30.64 MPa

Answer

τθ = 30.64 MPa
7.Axial stress & strain

Answer: P = 45.64 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 = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(120000·1257) = 2.653 × 10⁻⁶ mm/N

Load

P = 0.4/8.764 × 10⁻⁶ = 45641 N = 45.64 kN

Answer

P = 45.64 kN
8.Axial stress & strain

Answer: σmax = 422.8 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·40²/4 = 1257 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 67.64 MPa
σ2 = 422.8 MPa

Largest stress

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

σmax = 422.8 MPa

Answer

σmax = 422.8 MPa
9.Column buckling

Answer: Pcr = 4.935 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 = 40·30³/12 = 90000 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·90000/6000² = 4935 N = 4.935 kN

Answer

Pcr = 4.935 kN
10.Column buckling

Answer: σcr = 4.441 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·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 2·5000 = 10000 mm
KL/r = 10000/15 = 666.7

Critical stress

σcr = π²E/(KL/r)² = π²·200000/666.7² = 4.441 MPa

Answer

σcr = 4.441 MPa
11.Mechanical vibrations

Answer: ζ = 0.07961

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.

Critical damping

cc = 2·√(k·m) = 2·√(40000·150) = 4899 N·s/m

Damping ratio

ζ = c/cc = 390/4899 = 0.07961

Answer

ζ = 0.07961

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: fn = 3.823 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

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

1/keq = 1/50 + 1/15 → keq = 50·15/(50 + 15) = 11.54 kN/m

Natural frequency

ωn = √(11538/20) = 24.02 rad/s
fn = 24.02/(2π) = 3.823 Hz

Answer

fn = 3.823 Hz

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