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​ = 120 mmd2​ = 300 mm960 rpmn2​ = ?

A 120 mm pulley at 960 rpm drives a 300 mm pulley by a belt with 1 % slip. Find the speed n2 of the driven pulley.

n1 = 960 rpmd1 = 120 mmd2 = 300 mms = 1 %
n2 = rpm
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 315 mm1450 rpmC = 800 mmθ = ?

An open belt connects a 100 mm pulley and a 315 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d1 = 100 mmd2 = 315 mmC = 800 mm
θ = °
3.Torsion of shaftsMedium
do​ = 40, di​ = 30 mmT = 500 N·mL = 2.5 m

A hollow brass shaft (outer diameter 40 mm, inner diameter 30 mm, length 2.5 m) is fixed at one end and twisted by a torque of 500 N·m at the other. Find the angle of twist φ in degrees.

T = 500 N·mdo = 40 mmdi = 30 mmL = 2.5 mG = 39 GPa
φ = °
4.Torsion of shaftsHard
do​ = ?20 kW, 1000 rpm

A hollow shaft must transmit 20 kW at 1000 rpm. Its inner diameter is to be 0.7 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter do.

P = 20 kWN = 1000 rpmdi/do = 0.7τallow = 75 MPa
do = mm
5.Mohr’s circle & stress transformationMedium
σx​ = 40σy​ = 60MPaτxy​ = ?

A point has σₓ = 40 MPa and σᵧ = 60 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 115 MPa?

σx = 40 MPaσy = 60 MPaσ1 = 115 MPa
τxy = MPa
6.Mohr’s circle & stress transformationHard
σx​ = −80σy​ = 50τxy​ = 60MPa

A point in a loaded part has σₓ = −80 MPa, σᵧ = 50 MPa and τₓᵧ = 60 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 35° counter-clockwise from the x-axis.

σx = −80 MPaσy = 50 MPaτxy = 60 MPaθ = 35°
σθ = MPa
7.Axial stress & strainMedium
P = ?12400 mm300 mmØ40Ø50δ = 0.3 mm

A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 300 mm, E = 70 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
50 kN12600 mm300 mmØ25Ø30σ2​ = ?

A steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 30 mm, length 300 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 50 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ2 in segment 2.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 300 mm, E = 70 GPaP = 50 kN
σ2 = MPa
9.Column bucklingMedium
PL = 6 md = 50 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 = 50 mm
Pcr = kN
10.Column bucklingHard
PL = 4 md = 60 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 = 4 mE = 70 GPa (aluminium)d = 60 mm
σcr = MPa
11.Mechanical vibrationsMedium
tx10 mm4 cycles3.7 mm

In a free-vibration test of a damped mass–spring system, a displacement peak of 10 mm is followed 4 cycles later by a peak of 3.7 mm. Find the damping ratio ζ.

x0 = 10 mmx4 = 3.7 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 50k2​ = 15k3​ = 12m = 10 kg

A 10 kg mass hangs from spring k₃ = 12 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 = 15 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 = 10 kgk1 = 50 kN/mk2 = 15 kN/mk3 = 12 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: n2 = 380.2 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

960·120/300 = 384 rpm

With slip

n2 = 384·0.99 = 380.2 rpm

Answer

n2 = 380.2 rpm
2.Gears & gear trains

Answer: θ = 164.6 °

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(0.1344) = 7.722°

Angle of wrap

θ = 180 − 2·7.722 = 164.6°

Answer

θ = 164.6°
3.Torsion of shafts

Answer: φ = 10.69 °

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 − 30⁴)/32 = 171806 mm⁴

Angle of twist

T = 500 N·m = 500000 N·mm
L = 2.5 m = 2500 mm
G = 39 GPa = 39000 MPa
φ = TL/(GJ) = 500000·2500/(39000·171806) = 0.1866 rad
φ = 0.1866·180/π = 10.69°

Answer

φ = 10.69°
4.Torsion of shafts

Answer: do = 25.75 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 = 20000/104.7 = 191 N·m

Outer diameter from the allowable stress

T = 191 N·m = 190986 N·mm
1 − k⁴ = 1 − 0.7⁴ = 0.7599
do = ∛(16T/(π·τallow·(1 − k⁴))) = ∛(16·190986/(π·75·0.7599)) = 25.75 mm
di = 0.7·25.75 = 18.02 mm

Answer

do = 25.75 mm

A solid shaft would need d = 23.49 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: τxy = 64.23 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.

Required radius

C = (40 + 60)/2 = 50 MPa
R = σ1 − C = 115 − 50 = 65 MPa
στxyCσ1​σ2​

Shear stress

τxy = √(R² − ((σₓ − σᵧ)/2)²) = √(65² − 10²) = 64.23 MPa

Answer

τxy = 64.23 MPa
6.Mohr’s circle & stress transformation

Answer: σθ = 19.15 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 = −15 + −65·0.342 + 60·0.9397 = 19.15 MPa

Answer

σθ = 19.15 MPa
7.Axial stress & strain

Answer: P = 79.49 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 = 1257 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 300/(70000·1963) = 2.183 × 10⁻⁶ mm/N

Load

P = 0.3/3.774 × 10⁻⁶ = 79486 N = 79.49 kN

Answer

P = 79.49 kN
8.Axial stress & strain

Answer: σ2 = 35.51 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 = 200000·490.9/600 = 163625 N/mm
k₂ = 70000·706.9/300 = 164934 N/mm

Share of segment 2

F2 = 50000·164934/328558 = 25100 N

Stress

σ2 = 25100/706.9 = 35.51 MPa (compression)

Answer

σ2 = 35.51 MPa
9.Column buckling

Answer: Pcr = 23.55 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 0.5·6000 = 3000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·306796/3000² = 23551 N = 23.55 kN

Answer

Pcr = 23.55 kN
10.Column buckling

Answer: σcr = 38.86 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·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²
r = √(I/A) = √(636173/2827) = 15 mm
KL = 0.5·4000 = 2000 mm
KL/r = 2000/15 = 133.3

Critical stress

σcr = π²E/(KL/r)² = π²·70000/133.3² = 38.86 MPa

Answer

σcr = 38.86 MPa
11.Mechanical vibrations

Answer: ζ = 0.03953

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 = (1/4)·ln(10/3.7) = 0.2486

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2486/√(39.48 + 0.06178) = 0.03953

Answer

ζ = 0.03953
12.Mechanical vibrations

Answer: fn = 5.065 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 + 15 = 65 kN/m
keq = 65·12/(65 + 12) = 10.13 kN/m

Natural frequency

ωn = √(10130/10) = 31.83 rad/s
fn = 31.83/(2π) = 5.065 Hz

Answer

fn = 5.065 Hz

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