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​ = 200 mmd2​ = 450 mm1440 rpmT1​ = 400 NT2​ = 160 NP = ?

A belt runs on a 200 mm driving pulley at 1440 rpm. The tension is 400 N on the tight side and 160 N on the slack side. Find the power P transmitted, in kW.

d1 = 200 mmn1 = 1440 rpmT1 = 400 NT2 = 160 N
P = kW
2.Gears & gear trainsHard
A = 30B = 45C = 15D = 24

In a compound gear train, gear A (30 teeth) on the input shaft turning at 960 rpm drives gear B (45 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (24 teeth) on the output shaft. Find the output speed nout.

nin = 960 rpmA = 30B = 45C = 15D = 24
nout = rpm
3.Torsion of shaftsMedium
T = 250 N·md = 40 mmL = 1.2 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 = 250 N·md = 40 mmL = 1.2 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?30 kW, 750 rpm

A hollow shaft must transmit 30 kW at 750 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 = 30 kWN = 750 rpmdi/do = 0.75τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 140σy​ = 0τxy​ = −10MPa

A point in a loaded part has σₓ = 140 MPa, σᵧ = 0 MPa and τₓᵧ = −10 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σx = 140 MPaσy = 0 MPaτxy = −10 MPa
θp = °
6.Mohr’s circle & stress transformationHard
σx​ = −10σy​ = −10τxy​ = −20MPa

A point in a thin plate is in plane stress: σₓ = −10 MPa, σᵧ = −10 MPa and τₓᵧ = −20 MPa (σz = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σx = −10 MPaσy = −10 MPaτxy = −20 MPaσY = 250 MPa
n =
7.Axial stress & strainMedium
P = ?12500 mm200 mmØ25Ø60δ = 0.3 mm

A bar is made of a steel segment (diameter 25 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 200 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø25 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 200 mm, E = 100 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
60 kN12500 mm400 mmØ40Ø30σ2​ = ?

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of brass (diameter 30 mm, length 400 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaP = 60 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 md = 30 mmcross-section

A column of length L made of aluminium is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.

L = 6 mE = 70 GPa (aluminium)d = 30 mm
Pcr = kN
10.Column bucklingHard
PL = 2 md = 25 mmcross-section

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

L = 2 mE = 70 GPa (aluminium)d = 25 mm
σcr = MPa
11.Mechanical vibrationsMedium
k = 20 kN/mζ = 0.4m = 8 kg

An 8 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency fd of its free vibration in Hz.

m = 8 kgk = 20 kN/mζ = 0.4
fd = Hz
12.Mechanical vibrationsHard
k1​ = 100k2​ = 20m = 40 kg

A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 100 kN/m and k₂ = 20 kN/m. The mass stays level. Find the equivalent stiffness keq of the spring system in kN/m.

m = 40 kgk1 = 100 kN/mk2 = 20 kN/m
keq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 3.619 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 = 15.08 m/s

Power

P = (400 − 160)·15.08 = 3619 W = 3.619 kW

Answer

P = 3.619 kW
2.Gears & gear trains

Answer: nout = 400 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.

First stage

nB = 960·30/45 = 640 rpm = nC

Second stage

nout = 640·15/24 = 400 rpm

Answer

nout = 400 rpm

total ratio = 2.4 : 1

3.Torsion of shafts

Answer: φ = 1.754 °

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) = 250·1000·1200/(39000·251327) = 0.03061 rad
φ = 0.03061·180/π = 1.754°

Answer

φ = 1.754°
4.Torsion of shafts

Answer: do = 33.6 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 = 30000/78.54 = 382 N·m

Outer diameter from the allowable stress

T = 382 N·m = 381972 N·mm
1 − k⁴ = 1 − 0.75⁴ = 0.6836
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·381972/(π·75·0.6836)) = 33.6 mm
di = 0.75·33.6 = 25.2 mm

Answer

do = 33.6 mm

A solid shaft would need d = 29.6 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θp = −4.065 °

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·−10, 140 − 0) = −8.13°

Angle on the element

θp = −4.065°

Answer

θp = −4.065°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 6.934

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 = 20 MPa
σ1 = 10 MPa, σ2 = −30 MPa, σ3 = 0

Von Mises

σvM = √(10² − 10·(−30) + (−30)²) = 36.06 MPa
n = 250/36.06 = 6.934

Answer

n = 6.934

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

7.Axial stress & strain

Answer: P = 51.72 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 = 490.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·490.9) = 5.093 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 200/(100000·2827) = 7.074 × 10⁻⁷ mm/N

Load

P = 0.3/5.8 × 10⁻⁶ = 51721 N = 51.72 kN

Answer

P = 51.72 kN
8.Axial stress & strain

Answer: σ2 = 22.08 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 = 200000·1257/500 = 502655 N/mm
k₂ = 100000·706.9/400 = 176715 N/mm

Share of segment 2

F2 = 60000·176715/679369 = 15607 N

Stress

σ2 = 15607/706.9 = 22.08 MPa (compression)

Answer

σ2 = 22.08 MPa
9.Column buckling

Answer: Pcr = 3.052 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·30⁴/64 = 39761 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·39761/3000² = 3052 N = 3.052 kN

Answer

Pcr = 3.052 kN
10.Column buckling

Answer: σcr = 26.99 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 = 0.5·2000 = 1000 mm
KL/r = 1000/6.25 = 160

Critical stress

σcr = π²E/(KL/r)² = π²·70000/160² = 26.99 MPa

Answer

σcr = 26.99 MPa
11.Mechanical vibrations

Answer: fd = 7.293 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.

Undamped natural frequency

ωn = √(20000/8) = 50 rad/s
fn = 50/(2π) = 7.958 Hz

Damped natural frequency

fd = fn·√(1 − ζ²) = 7.958·√(1 − 0.4²) = 7.293 Hz

Answer

fd = 7.293 Hz
12.Mechanical vibrations

Answer: keq = 120 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 side by side share the deflection and add their forces, so their stiffnesses simply add: k = k1 + k2.

keq = 100 + 20 = 120 kN/m

Answer

keq = 120 kN/m

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