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 = 1, 1440 rpm

A worm with 1 start(s) turns at 1440 rpm and drives a worm wheel with 80 teeth. Find the speed nw of the wheel.

z = 1Nw = 80n = 1440 rpm
nw = rpm
2.Gears & gear trainsHard
d1​ = 80 mmd2​ = 400 mm1500 rpmC = 500 mmT1​ = 2000 NT2​P = ?

An open belt connects an 80 mm driving pulley at 1500 rpm to a 400 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.35 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 = 80 mmn1 = 1500 rpmd2 = 400 mmC = 500 mmμ = 0.35T1 = 2000 N
P = kW
3.Torsion of shaftsMedium
T = 1000 N·md = 50 mmL = 1 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 = 1000 N·md = 50 mmL = 1 mG = 26 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?10 kW, 1000 rpm

A hollow shaft must transmit 10 kW at 1000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter do.

P = 10 kWN = 1000 rpmdi/do = 0.6τallow = 40 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 30σy​ = −60τxy​ = −50MPa

A point in a loaded part has σₓ = 30 MPa, σᵧ = −60 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 = 30 MPaσy = −60 MPaτxy = −50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = −40σy​ = 20τxy​ = −40MPa

A point in a thin plate is in plane stress: σₓ = −40 MPa, σᵧ = 20 MPa and τₓᵧ = −40 MPa (σz = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −40 MPaσy = 20 MPaτxy = −40 MPaσY = 350 MPa
n =
7.Axial stress & strainMedium
250 kNØ20110 GPaØ80/50300 mmδ = ?

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 50 mm, E = 110 GPa). Both are 300 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø20 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 110 GPaL = 300 mm
δ = mm
8.Axial stress & strainHard
P = ?Ø25100 GPaØ100/70400 mm

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load Pmax.

rod: Ø25 mm, E = 200 GPa, σallow = 120 MPatube: Ø100/Ø70 mm, E = 100 GPa, σallow = 100 MPa
Pmax = kN
9.Column bucklingMedium
PL = 1.5 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 = 1.5 mE = 200 GPa (steel)b × h = 30 × 80 mm
Pcr = kN
10.Column bucklingHard
PL = 1 md = 25 mmcross-section

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

L = 1 mE = 200 GPa (steel)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 10 kN/mc = 450 N·s/mm = 80 kg

An 80 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 450 N·s/m. Find the damping ratio ζ.

m = 80 kgk = 10 kN/mc = 450 N·s/m
ζ =
12.Mechanical vibrationsHard
k1​ = 40k2​ = 40m = 40 kg

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

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

3Solutions

1.Gears & gear trains

Answer: nw = 18 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 = 1440·1/80 = 18 rpm

Answer

nw = 18 rpm
2.Gears & gear trains

Answer: P = 7.31 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 = 2.49 rad

Slack-side tension

eμθ = e0.8715 = 2.391, T₂ = 2000/2.391 = 836.6 N

Power

P = (2000 − 836.6)·6.283 = 7.31 kW

Answer

P = 7.31 kW
3.Torsion of shafts

Answer: φ = 3.591 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 1000·1000·1000/(26000·613592) = 0.06268 rad
φ = 0.06268·180/π = 3.591°

Answer

φ = 3.591°
4.Torsion of shafts

Answer: do = 24.08 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·1000/60 = 104.7 rad/s
T = P/ω = 10000/104.7 = 95.49 N·m

Outer diameter from the allowable stress

T = 95.49 N·m = 95493 N·mm
1 − k⁴ = 1 − 0.6⁴ = 0.8704
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·95493/(π·40·0.8704)) = 24.08 mm
di = 0.6·24.08 = 14.45 mm

Answer

do = 24.08 mm

A solid shaft would need d = 22.99 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θp = −24.01 °

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, 30 − −60) = −48.01°

Angle on the element

θp = −24.01°

Answer

θp = −24.01°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 4.015

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 = −10 MPa, R = 50 MPa
σ1 = 40 MPa, σ2 = −60 MPa, σ3 = 0

Von Mises

σvM = √(40² − 40·(−60) + (−60)²) = 87.18 MPa
n = 350/87.18 = 4.015

Answer

n = 4.015

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

7.Axial stress & strain

Answer: δ = 0.1876 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 + 110·10³·3063 = 3.998 × 10⁸ N

Shortening

δ = 250000·300/3.998 × 10⁸ = 0.1876 mm

Answer

δ = 0.1876 mm
8.Axial stress & strain

Answer: Pmax = 299.2 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

Ar = π·25²/4 = 490.9 mm²
At = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 100·10³·4006 = 4.987 × 10⁸ N

Governing strain

ε = min(0.0006, 0.001) = 0.0006 (steel)

Largest load

Pmax = 0.0006·4.987 × 10⁸ = 299237 N = 299.2 kN

Answer

Pmax = 299.2 kN
9.Column buckling

Answer: Pcr = 157.9 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 = 1·1500 = 1500 mm

Critical load

Pcr = π²EI/(KL)² = π²·200000·180000/1500² = 157914 N = 157.9 kN

Answer

Pcr = 157.9 kN
10.Column buckling

Answer: σcr = 157.4 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·25⁴/64 = 19175 mm⁴

Radius of gyration and slenderness

A = 490.9 mm²
r = √(I/A) = √(19175/490.9) = 6.25 mm
KL = 0.7·1000 = 700 mm
KL/r = 700/6.25 = 112

Critical stress

σcr = π²E/(KL/r)² = π²·200000/112² = 157.4 MPa

Answer

σcr = 157.4 MPa
11.Mechanical vibrations

Answer: ζ = 0.2516

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.

Critical damping

cc = 2·√(k·m) = 2·√(10000·80) = 1789 N·s/m

Damping ratio

ζ = c/cc = 450/1789 = 0.2516

Answer

ζ = 0.2516

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: fn = 3.559 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 = 1/40 + 1/40 → keq = 40·40/(40 + 40) = 20 kN/m

Natural frequency

ωn = √(20000/40) = 22.36 rad/s
fn = 22.36/(2π) = 3.559 Hz

Answer

fn = 3.559 Hz

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