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​ = 80 mmd2​ = 400 mm720 rpmn2​ = ?

An 80 mm pulley at 720 rpm drives a 400 mm pulley by a belt with 3 % slip. Find the speed n2 of the driven pulley.

n1 = 720 rpmd1 = 80 mmd2 = 400 mms = 3 %
n2 = rpm
2.Gears & gear trainsHard
d1​ = 200 mmd2​ = 315 mm1440 rpmC = 500 mmT1​ = 2000 NT2​P = ?

An open belt connects a 200 mm driving pulley at 1440 rpm to a 315 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 2000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 200 mmn1 = 1440 rpmd2 = 315 mmC = 500 mmμ = 0.3T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
T = 200 N·md = 25 mmL = 2 mφ = ?

A solid brass 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 = 200 N·md = 25 mmL = 2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?100 kW, 300 rpm

A hollow shaft must transmit 100 kW at 300 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.

P = 100 kWN = 300 rpmdi/do = 0.75τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 20σy​ = 60τxy​ = 30MPa

A point in a loaded part has σₓ = 20 MPa, σᵧ = 60 MPa and τₓᵧ = 30 MPa (tension positive). Find the maximum in-plane shear stress τmax.

σx = 20 MPaσy = 60 MPaτxy = 30 MPa
τmax = MPa
6.Mohr’s circle & stress transformationHard
σx​ = 50σy​ = 70τxy​ = 60MPa

A point in a loaded part has σₓ = 50 MPa, σᵧ = 70 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 20° clockwise from the x-axis.

σx = 50 MPaσy = 70 MPaτxy = 60 MPaθ = -20°
σθ = MPa
7.Axial stress & strainMedium
P = ?12400 mm400 mmØ20Ø30δ = 0.5 mm

A bar is made of a steel segment (diameter 20 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.5 mm in total.

segment 1: steel, Ø20 mm, 400 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaδ = 0.5 mm
P = kN
8.Axial stress & strainHard
20 kN12300 mm300 mmØ40Ø60σ2​ = ?

A steel segment (diameter 40 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 60 mm, length 300 mm, E = 110 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, 300 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 300 mm, E = 110 GPaP = 20 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 md = 60 mmcross-section

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.

L = 6 mE = 70 GPa (aluminium)d = 60 mm
Pcr = kN
10.Column bucklingHard
PL = 2.5 md = 50 mmcross-section

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

L = 2.5 mE = 70 GPa (aluminium)d = 50 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx50 mm1 cycle26.6 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 50 mm is followed one cycle later by a peak of 26.6 mm. Find the damping ratio ζ.

x0 = 50 mmx1 = 26.6 mmn = 1
ζ =
12.Mechanical vibrationsHard
k1​ = 15k2​ = 80m = 10 kg

A 10 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 15 kN/m and k₂ = 80 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 10 kgk1 = 15 kN/mk2 = 80 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 139.7 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

720·80/400 = 144 rpm

With slip

n2 = 144·0.97 = 139.7 rpm

Answer

n2 = 139.7 rpm
2.Gears & gear trains

Answer: P = 17.57 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.

Angle of wrap

θ = 2.911 rad

Slack-side tension

eμθ = e0.8733 = 2.395, T₂ = 2000/2.395 = 835.1 N

Power

P = (2000 − 835.1)·15.08 = 17.57 kW

Answer

P = 17.57 kW
3.Torsion of shafts

Answer: φ = 15.32 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2000/(39000·38350) = 0.2674 rad
φ = 0.2674·180/π = 15.32°

Answer

φ = 15.32°
4.Torsion of shafts

Answer: do = 68.13 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 = 100000/31.42 = 3183 N·m

Outer diameter from the allowable stress

T = 3183 N·m = 3.183 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·3.183 × 10⁶/(π·75·0.6836)) = 68.13 mm
di = 0.75·68.13 = 51.1 mm

Answer

do = 68.13 mm

A solid shaft would need d = 60.01 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τmax = 36.06 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.

στxyCτmax​

Radius of Mohr’s circle

τmax = R = √(((20 − 60)/2)² + 30²) = 36.06 MPa

Answer

τmax = 36.06 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 13.77 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 = 60 + −10·0.766 + 60·−0.6428 = 13.77 MPa

Answer

σθ = 13.77 MPa
7.Axial stress & strain

Answer: P = 41.58 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 = 314.2 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·314.2) = 6.366 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(100000·706.9) = 5.659 × 10⁻⁶ mm/N

Load

P = 0.5/1.203 × 10⁻⁵ = 41580 N = 41.58 kN

Answer

P = 41.58 kN
8.Axial stress & strain

Answer: σ2 = 3.912 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 = 200000·1257/300 = 837758 N/mm
k₂ = 110000·2827/300 = 1.037 × 10⁶ N/mm

Share of segment 2

F2 = 20000·1.037 × 10⁶/1.874 × 10⁶ = 11061 N

Stress

σ2 = 11061/2827 = 3.912 MPa (compression)

Answer

σ2 = 3.912 MPa
9.Column buckling

Answer: Pcr = 24.92 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·636173/4200² = 24916 N = 24.92 kN

Answer

Pcr = 24.92 kN
10.Column buckling

Answer: σcr = 17.27 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·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²
r = √(I/A) = √(306796/1963) = 12.5 mm
KL = 1·2500 = 2500 mm
KL/r = 2500/12.5 = 200

Critical stress

σcr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa

Answer

σcr = 17.27 MPa
11.Mechanical vibrations

Answer: ζ = 0.09994

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 = (1/1)·ln(50/26.6) = 0.6311

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6311/√(39.48 + 0.3983) = 0.09994

Answer

ζ = 0.09994
12.Mechanical vibrations

Answer: fn = 15.51 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 15 + 80 = 95 kN/m

Natural frequency

ωn = √(95000/10) = 97.47 rad/s
fn = 97.47/(2π) = 15.51 Hz

Answer

fn = 15.51 Hz

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