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​ = 315 mm960 rpmT1​ = 1200 NT2​ = 600 NP = ?

A belt runs on a 160 mm driving pulley at 960 rpm. The tension is 1200 N on the tight side and 600 N on the slack side. Find the power P transmitted, in kW.

d1 = 160 mmn1 = 960 rpmT1 = 1200 NT2 = 600 N
P = kW
2.Gears & gear trainsHard
d1​ = 200 mmd2​ = 300 mm1500 rpmC = 500 mmθ = ?

An open belt connects a 200 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 = 200 mmd2 = 300 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
do​ = ?T = 2000 N·m

A hollow shaft must carry a torque of 2000 N·m. 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.

T = 2000 N·mdi/do = 0.5τallow = 60 MPa
do = mm
4.Torsion of shaftsHard
do​ = ?30 kW, 300 rpm

A hollow shaft must transmit 30 kW at 300 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

P = 30 kWN = 300 rpmdi/do = 0.75τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 120σy​ = 110τxy​ = 30MPa

A point in a loaded part has σₓ = 120 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 120 MPaσy = 110 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 130σy​ = 0τxy​ = −60MPa

A point in a loaded part has σₓ = 130 MPa, σᵧ = 0 MPa and τₓᵧ = −60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σx = 130 MPaσy = 0 MPaτxy = −60 MPaθ = 50°
σθ = MPa
7.Axial stress & strainMedium
300 kNØ2070 GPaØ100/60600 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 300 kN. Find the shortening δ.

P = 300 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 600 mm
δ = mm
8.Axial stress & strainHard
40 kN12400 mm500 mmØ40Ø60σ2​ = ?

A steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) and a segment made of bronze (diameter 60 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 40 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 500 mm, E = 110 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
PL = 1.5 md = 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 = 1.5 mE = 200 GPa (steel)d = 50 mm
Pcr = kN
10.Column bucklingHard
PL = 1 md = 20 mmcross-section

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

L = 1 mE = 200 GPa (steel)d = 20 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx12 mm3 cycles5.6 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 3 cycles later by a peak of 5.6 mm. Find the damping ratio ζ.

x0 = 12 mmx3 = 5.6 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 12k2​ = 30m = 25 kg

A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 12 kN/m and the lower one k₂ = 30 kN/m. Find the equivalent stiffness keq of the spring system in kN/m.

m = 25 kgk1 = 12 kN/mk2 = 30 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 4.825 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 = 8.042 m/s

Power

P = (1200 − 600)·8.042 = 4825 W = 4.825 kW

Answer

P = 4.825 kW
2.Gears & gear trains

Answer: θ = 168.5 °

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(0.1) = 5.739°

Angle of wrap

θ = 180 − 2·5.739 = 168.5°

Answer

θ = 168.5°
3.Torsion of shafts

Answer: do = 56.58 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.

Outer diameter from the allowable stress

T = 2000 N·m = 2 × 10⁶ N·mm
1 − k⁴ = 1 − 0.5⁴ = 0.9375
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·2 × 10⁶/(π·60·0.9375)) = 56.58 mm
di = 0.5·56.58 = 28.29 mm

Answer

do = 56.58 mm

A solid shaft would need d = 55.37 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: do = 56.24 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·300/60 = 31.42 rad/s
T = P/ω = 30000/31.42 = 954.9 N·m

Outer diameter from the allowable stress

T = 954.9 N·m = 954930 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·954930/(π·40·0.6836)) = 56.24 mm
di = 0.75·56.24 = 42.18 mm

Answer

do = 56.24 mm

A solid shaft would need d = 49.54 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τmax = 30.41 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 = √(((120 − 110)/2)² + 30²) = 30.41 MPa

Answer

τmax = 30.41 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = −5.376 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

σθ = 65 + 65·−0.1736 + −60·0.9848 = −5.376 MPa

Answer

σθ = −5.376 MPa
7.Axial stress & strain

Answer: δ = 0.4341 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(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·5027 = 4.147 × 10⁸ N

Shortening

δ = 300000·600/4.147 × 10⁸ = 0.4341 mm

Answer

δ = 0.4341 mm
8.Axial stress & strain

Answer: σ2 = 7.038 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 = 200000·1257/400 = 628319 N/mm
k₂ = 110000·2827/500 = 622035 N/mm

Share of segment 2

F2 = 40000·622035/1.25 × 10⁶ = 19899 N

Stress

σ2 = 19899/2827 = 7.038 MPa (compression)

Answer

σ2 = 7.038 MPa
9.Column buckling

Answer: Pcr = 269.2 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·306796/1500² = 269152 N = 269.2 kN

Answer

Pcr = 269.2 kN
10.Column buckling

Answer: σcr = 197.4 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·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²
r = √(I/A) = √(7854/314.2) = 5 mm
KL = 0.5·1000 = 500 mm
KL/r = 500/5 = 100

Critical stress

σcr = π²E/(KL/r)² = π²·200000/100² = 197.4 MPa

Answer

σcr = 197.4 MPa
11.Mechanical vibrations

Answer: ζ = 0.0404

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(12/5.6) = 0.254

Damping ratio

ζ = δ/√(4π² + δ²) = 0.254/√(39.48 + 0.06454) = 0.0404

Answer

ζ = 0.0404
12.Mechanical vibrations

Answer: keq = 8.571 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 = 1/12 + 1/30 → keq = 12·30/(12 + 30) = 8.571 kN/m

Answer

keq = 8.571 kN/m

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