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 mm1500 rpmn2​ = ?

A 100 mm pulley at 1500 rpm drives a 500 mm pulley by a belt with 4 % slip. Find the speed n2 of the driven pulley.

n1 = 1500 rpmd1 = 100 mmd2 = 500 mms = 4 %
n2 = rpm
2.Gears & gear trainsHard
d1​ = 150 mmd2​ = 300 mm1500 rpmC = 600 mmθ = ?

An open belt connects a 150 mm pulley and a 300 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 150 mmd2 = 300 mmC = 600 mm
θ = °
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
Td = ?

A solid shaft must transmit 15 kW at 750 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 15 kWN = 750 rpmτallow = 80 MPa
d = mm
5.Mohr’s circle & stress transformationMedium
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 angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 70 MPaσy = −70 MPaτxy = 40 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​τxy​top surface of the shaft

A solid shaft of diameter 40 mm carries a bending moment of 300 N·m together with a torque of 1500 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ1.

d = 40 mmM = 300 N·mT = 1500 N·m
σ1 = MPa
7.Axial stress & strainMedium
150 kNØ3070 GPaØ100/70500 mmδ = ?

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 70 mm, E = 70 GPa). Both are 500 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø30 mm, E = 200 GPatube: Ø100/Ø70 mm, E = 70 GPaL = 500 mm
δ = mm
8.Axial stress & strainHard
80 kN12300 mm500 mmØ30Ø40σ2​ = ?

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 500 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 500 mm, E = 70 GPaP = 80 kN
σ2 = MPa
9.Column bucklingMedium
PL = 2 m30 × 80 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 = 2 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
P = 15 kNL = 2 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and pinned at the top. It must carry 15 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.

P = 15 kNn = 2.5L = 2 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx40 mm3 cycles18.8 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 3 cycles later by a peak of 18.8 mm. Find the damping ratio ζ.

x0 = 40 mmx3 = 18.8 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 50k2​ = 5k3​ = 80m = 25 kg

A 25 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 5 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 = 50 kN/mk2 = 5 kN/mk3 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 288 rpm

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.

Speed without slip

1500·100/500 = 300 rpm

With slip

n2 = 300·0.96 = 288 rpm

Answer

n2 = 288 rpm
2.Gears & gear trains

Answer: θ = 165.6 °

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(0.125) = 7.181°

Angle of wrap

θ = 180 − 2·7.181 = 165.6°

Answer

θ = 165.6°
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: d = 22.99 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·750/60 = 78.54 rad/s
T = P/ω = 15000/78.54 = 191 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τallow)) = ∛(16·190986/(π·80)) = 22.99 mm

Answer

d = 22.99 mm

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

5.Mohr’s circle & stress transformation

Answer: θp = 14.87 °

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(2·40, 70 − −70) = 29.74°

Angle on the element

θp = 14.87°

Answer

θp = 14.87°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ1 = 145.6 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) = 32·300000/(π·40³) = 47.75 MPa
τ = 16T/(πd³) = 16·1500000/(π·40³) = 119.4 MPa
στxyCσ1​σ2​

Mohr’s circle

C = σ/2 = 23.87 MPa
R = √(23.87² + 119.4²) = 121.7 MPa
σ1 = C + R = 145.6 MPa
σ2 = C − R = −97.86 MPa

Answer

σ1 = 145.6 MPa
7.Axial stress & strain

Answer: δ = 0.1778 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(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·4006 = 4.218 × 10⁸ N

Shortening

δ = 150000·500/4.218 × 10⁸ = 0.1778 mm

Answer

δ = 0.1778 mm
8.Axial stress & strain

Answer: σ2 = 17.31 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 = 200000·706.9/300 = 471239 N/mm
k₂ = 70000·1257/500 = 175929 N/mm

Share of segment 2

F2 = 80000·175929/647168 = 21748 N

Stress

σ2 = 21748/1257 = 17.31 MPa (compression)

Answer

σ2 = 17.31 MPa
9.Column buckling

Answer: Pcr = 88.83 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·180000/2000² = 88826 N = 88.83 kN

Answer

Pcr = 88.83 kN
10.Column buckling

Answer: d = 38.37 mm

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.

Required critical load

Pcr = n·P = 2.5·15000 = 37500 N
KL = 0.7·2000 = 1400 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·37500·1400²/(π³·70000))^(1/4) = 38.37 mm

Answer

d = 38.37 mm

Check: the slenderness KL/r is 146, so the column is slender and Euler applies.

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(40/18.8) = 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 = 5.747 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 = 50 + 5 = 55 kN/m
keq = 55·80/(55 + 80) = 32.59 kN/m

Natural frequency

ωn = √(32593/25) = 36.11 rad/s
fn = 36.11/(2π) = 5.747 Hz

Answer

fn = 5.747 Hz

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