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​ = 160 mmd2​ = 450 mm1440 rpmT1​ = 1500 NT2​ = 600 NP = ?

A belt runs on a 160 mm driving pulley at 1440 rpm. The tension is 1500 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.

d1 = 160 mmn1 = 1440 rpmT1 = 1500 NT2 = 600 N
P = kW
2.Gears & gear trainsHard
d1​ = 120 mmd2​ = 300 mm720 rpmC = 500 mmθ = ?

An open belt connects a 120 mm pulley and a 300 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 120 mmd2 = 300 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
do​ = 30, di​ = 21 mmT = 1000 N·mL = 2 m

A hollow steel shaft (outer diameter 30 mm, inner diameter 21 mm, length 2 m) is fixed at one end and twisted by a torque of 1000 N·m at the other. Find the angle of twist φ in degrees.

T = 1000 N·mdo = 30 mmdi = 21 mmL = 2 mG = 80 GPa
φ = °
4.Torsion of shaftsHard
Td = ?

A solid shaft must transmit 20 kW at 1500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 20 kWN = 1500 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 130σy​ = 110τxy​ = −50MPa

A point in a loaded part has σₓ = 130 MPa, σᵧ = 110 MPa and τₓᵧ = −50 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 130 MPaσy = 110 MPaτxy = −50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​τxy​top surface of the shaft

A solid shaft of diameter 50 mm carries a bending moment of 1000 N·m together with a torque of 250 N·m. At the top surface, where the bending stress is tensile and largest, find the maximum in-plane shear stress τmax.

d = 50 mmM = 1000 N·mT = 250 N·m
τmax = MPa
7.Axial stress & strainMedium
80 kNØ20100 GPaØ80/50800 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa). Both are 800 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø20 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 100 GPaL = 800 mm
δ = mm
8.Axial stress & strainHard
20 kN12500 mm200 mmØ40Ø40σ2​ = ?

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 40 mm, length 200 mm, E = 120 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: copper, Ø40 mm, 200 mm, E = 120 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 4 md = 30 mmcross-section

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

L = 4 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 40 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. 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
tx30 mm3 cycles14.1 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 3 cycles later by a peak of 14.1 mm. Find the damping ratio ζ.

x0 = 30 mmx3 = 14.1 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 80k2​ = 15k3​ = 30m = 25 kg

A 25 kg mass hangs from spring k₃ = 30 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 = 15 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 = 25 kgk1 = 80 kN/mk2 = 15 kN/mk3 = 30 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: P = 10.86 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 = 12.06 m/s

Power

P = (1500 − 600)·12.06 = 10857 W = 10.86 kW

Answer

P = 10.86 kW
2.Gears & gear trains

Answer: θ = 159.3 °

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(0.18) = 10.37°

Angle of wrap

θ = 180 − 2·10.37 = 159.3°

Answer

θ = 159.3°
3.Torsion of shafts

Answer: φ = 23.7 °

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 − 21⁴)/32 = 60428 mm⁴

Angle of twist

T = 1000 N·m = 1 × 10⁶ N·mm
L = 2 m = 2000 mm
G = 80 GPa = 80000 MPa
φ = TL/(GJ) = 1 × 10⁶·2000/(80000·60428) = 0.4137 rad
φ = 0.4137·180/π = 23.7°

Answer

φ = 23.7°
4.Torsion of shafts

Answer: d = 20.09 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 = 20000/157.1 = 127.3 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·127324/(π·80)) = 20.09 mm

Answer

d = 20.09 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = −39.35 °

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.

στxyCσ1​σ2​

Angle on the circle

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θp = atan2(2·−50, 130 − 110) = −78.69°

Angle on the element

θp = −39.35°

Answer

θp = −39.35°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τmax = 42 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.

Stresses at the surface

σ = 32M/(πd³) = 32·1000000/(π·50³) = 81.49 MPa
τ = 16T/(πd³) = 16·250000/(π·50³) = 10.19 MPa
στxyCτmax​

Mohr’s circle

C = σ/2 = 40.74 MPa
R = √(40.74² + 10.19²) = 42 MPa

Answer

τmax = 42 MPa
7.Axial stress & strain

Answer: δ = 0.1734 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·20²/4 = 314.2 mm²
At = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·3063 = 3.691 × 10⁸ N

Shortening

δ = 80000·800/3.691 × 10⁸ = 0.1734 mm

Answer

δ = 0.1734 mm
8.Axial stress & strain

Answer: σ2 = 9.549 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 = 200000·1257/500 = 502655 N/mm
k₂ = 120000·1257/200 = 753982 N/mm

Share of segment 2

F2 = 20000·753982/1.257 × 10⁶ = 12000 N

Stress

σ2 = 12000/1257 = 9.549 MPa (compression)

Answer

σ2 = 9.549 MPa
9.Column buckling

Answer: Pcr = 6.867 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·4000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·39761/2000² = 6867 N = 6.867 kN

Answer

Pcr = 6.867 kN
10.Column buckling

Answer: σcr = 8.812 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 = 0.7·4000 = 2800 mm
KL/r = 2800/10 = 280

Critical stress

σcr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σcr = 8.812 MPa
11.Mechanical vibrations

Answer: ζ = 0.04002

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(30/14.1) = 0.2517

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2517/√(39.48 + 0.06334) = 0.04002

Answer

ζ = 0.04002
12.Mechanical vibrations

Answer: fn = 4.806 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 + 15 = 95 kN/m
keq = 95·30/(95 + 30) = 22.8 kN/m

Natural frequency

ωn = √(22800/25) = 30.2 rad/s
fn = 30.2/(2π) = 4.806 Hz

Answer

fn = 4.806 Hz

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