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
Nw​ = 80z = 4, 1450 rpm

A worm with 4 start(s) turns at 1450 rpm and drives a worm wheel with 80 teeth. Find the speed nw of the wheel.

z = 4Nw = 80n = 1450 rpm
nw = rpm
2.Gears & gear trainsHard
Ns​ = 20ring fixedNp​ = 15nc​ = ?

In a planetary gear set the sun gear has 20 teeth and each planet gear has 15 teeth. The ring gear is held fixed and the sun is driven at 600 rpm. Find the speed nc of the planet carrier.

Ns = 20Np = 15ns = 600 rpm
nc = rpm
3.Torsion of shaftsMedium
do​ = ?T = 1000 N·m

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter do.

T = 1000 N·mdi/do = 0.75τallow = 50 MPa
do = mm
4.Torsion of shaftsHard
d = 60 mmT0​ = 400 N·mACBa = 0.8 mb = 2 m

A solid steel shaft of diameter 60 mm is fixed to rigid walls at both ends A and B. A torque of 400 N·m is applied at C, 0.8 m from A and 2 m from B. Find the reaction torque TA at wall A.

d = 60 mmT0 = 400 N·ma = 0.8 mb = 2 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 0σy​ = −20τxy​ = 30MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = −20 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 0 MPaσy = −20 MPaτxy = 30 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = 90τxy​ = −50MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = 90 MPa and τₓᵧ = −50 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σx = −10 MPaσy = 90 MPaτxy = −50 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
P = ?12500 mm300 mmØ20Ø60δ = 0.3 mm

A bar is made of a steel segment (diameter 20 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 60 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø20 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 300 mm, E = 70 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
40 kN12500 mm300 mmØ25Ø30σ2​ = ?

A steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) and a segment made of copper (diameter 30 mm, length 300 mm, E = 120 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, Ø25 mm, 500 mm, E = 200 GPasegment 2: copper, Ø30 mm, 300 mm, E = 120 GPaP = 40 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 m30 × 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)b × h = 30 × 60 mm
Pcr = kN
10.Column bucklingHard
PL = 6 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 = 6 mE = 70 GPa (aluminium)d = 40 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx25 mm3 cycles5.5 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 3 cycles later by a peak of 5.5 mm. Find the damping ratio ζ.

x0 = 25 mmx3 = 5.5 mmn = 3
ζ =
12.Mechanical vibrationsHard
k1​ = 100k2​ = 10m = 40 kg

A 40 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 10 kN/m. Ignore damping and the masses of the springs. Find the natural frequency fn of vertical vibration in Hz.

m = 40 kgk1 = 100 kN/mk2 = 10 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: nw = 72.5 rpm

Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.

Wheel speed

nw = n·z/Nw = 1450·4/80 = 72.5 rpm

Answer

nw = 72.5 rpm
2.Gears & gear trains

Answer: nc = 171.4 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (ns − nc)/(nr − nc) = −Nr/Ns: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

Nr = Ns + 2·Np = 20 + 2·15 = 50

Willis equation with the ring fixed

Set nr = 0 and solve for the carrier speed.

(600 − nc)/(0 − nc) = −50/20
nc = ns·Ns/(Ns + Nr) = 600·20/70 = 171.4 rpm

Answer

nc = 171.4 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: do = 53.02 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 = 1000 N·m = 1 × 10⁶ N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·50·0.6836)) = 53.02 mm
di = 0.75·53.02 = 39.76 mm

Answer

do = 53.02 mm

A solid shaft would need d = 46.7 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: TA = 285.7 N·m

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. When both ends are fixed, statics alone gives only TA + TB = T₀; the second equation is compatibility (equal twist of both parts at the loaded section).

Equilibrium and compatibility

TA + TB = 400
TA·a = TB·b
TA = 400·2/2.8 = 285.7 N·m
TB = 400·0.8/2.8 = 114.3 N·m

Answer

TA = 285.7 N·m
5.Mohr’s circle & stress transformation

Answer: θp = 35.78 °

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(2·30, 0 − −20) = 71.57°

Angle on the element

θp = 35.78°

Answer

θp = 35.78°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 2.475

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. Yield criteria: von Mises predicts yield when σvM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σY; Tresca when the absolute maximum shear stress reaches σY/2.

στxyCσ1​σ2​

Principal stresses

C = 40 MPa, R = 70.71 MPa
σ1 = 110.7 MPa, σ2 = −30.71 MPa, σ3 = 0

Tresca

σmax − σmin = 141.4 MPa
n = 350/141.4 = 2.475

Answer

n = 2.475

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: P = 31.67 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 = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·314.2) = 7.958 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(70000·2827) = 1.516 × 10⁻⁶ mm/N

Load

P = 0.3/9.474 × 10⁻⁶ = 31667 N = 31.67 kN

Answer

P = 31.67 kN
8.Axial stress & strain

Answer: σ2 = 33.4 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 = 200000·490.9/500 = 196350 N/mm
k₂ = 120000·706.9/300 = 282743 N/mm

Share of segment 2

F2 = 40000·282743/479093 = 23607 N

Stress

σ2 = 23607/706.9 = 33.4 MPa (compression)

Answer

σ2 = 33.4 MPa
9.Column buckling

Answer: Pcr = 5.287 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·135000/4200² = 5287 N = 5.287 kN

Answer

Pcr = 5.287 kN
10.Column buckling

Answer: σcr = 3.917 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 = 0.7·6000 = 4200 mm
KL/r = 4200/10 = 420

Critical stress

σcr = π²E/(KL/r)² = π²·70000/420² = 3.917 MPa

Answer

σcr = 3.917 MPa
11.Mechanical vibrations

Answer: ζ = 0.08007

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(25/5.5) = 0.5047

Damping ratio

ζ = δ/√(4π² + δ²) = 0.5047/√(39.48 + 0.2547) = 0.08007

Answer

ζ = 0.08007
12.Mechanical vibrations

Answer: fn = 2.399 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 = 1/100 + 1/10 → keq = 100·10/(100 + 10) = 9.091 kN/m

Natural frequency

ωn = √(9091/40) = 15.08 rad/s
fn = 15.08/(2π) = 2.399 Hz

Answer

fn = 2.399 Hz

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