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
N1​ = 16N2​ = 50

A 4 kW motor running at 1450 rpm drives a pinion with 16 teeth, which meshes with a gear with 50 teeth. Ignore losses. Find the torque T1 on the motor shaft.

P = 4 kWn1 = 1450 rpmN1 = 16N2 = 50η = 100%
T1 = N·m
2.Gears & gear trainsHard
d1​ = 100 mmd2​ = 400 mm720 rpmC = 600 mmT1​ = 1500 NT2​P = ?

An open belt connects a 100 mm driving pulley at 720 rpm to a 400 mm pulley; the centres are 600 mm apart. The coefficient of friction is 0.35 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d1 = 100 mmn1 = 720 rpmd2 = 400 mmC = 600 mmμ = 0.35T1 = 1500 N
P = kW
3.Torsion of shaftsMedium
d1​ = 50 mmd2​ = 40 mmT = 250 N·mABC1.2 m0.5 m

A stepped solid aluminium shaft is fixed at A. Segment AB has diameter 50 mm and length 1.2 m; segment BC has diameter 40 mm and length 0.5 m. A torque of 250 N·m acts at the free end C. Find the maximum shear stress τmax in the shaft.

T = 250 N·md1 = 50 mm, L1 = 1.2 md2 = 40 mm, L2 = 0.5 m
τmax = MPa
4.Torsion of shaftsHard
d = 30 mmT0​ = 500 N·mACBa = 3 mb = 2 m

A solid steel shaft of diameter 30 mm is fixed to rigid walls at both ends A and B. A torque of 500 N·m is applied at C, 3 m from A and 2 m from B. Find the reaction torque TA at wall A.

d = 30 mmT0 = 500 N·ma = 3 mb = 2 m
TA = N·m
5.Mohr’s circle & stress transformationMedium
σx​ = 0σy​ = −10τxy​ = −50MPa

A point in a loaded part has σₓ = 0 MPa, σᵧ = −10 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 = 0 MPaσy = −10 MPaτxy = −50 MPa
θp = °
6.Mohr’s circle & stress transformationHard
25°D = 1000 mm, t = 6 mm, p = 0.5 MPap

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 6 mm holds gas at a gauge pressure of 0.5 MPa. It is made from plate welded along a helical seam at 25° to the axis of the cylinder. Find the magnitude of the shear stress |τn| acting along the weld seam.

D = 1000 mmt = 6 mmp = 0.5 MPaα = 25°
|τn| = MPa
7.Axial stress & strainMedium
P = ?12500 mm500 mmØ30Ø60δ = 0.2 mm

A bar is made of a steel segment (diameter 30 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø30 mm, 500 mm, E = 200 GPasegment 2: brass, Ø60 mm, 500 mm, E = 100 GPaδ = 0.2 mm
P = kN
8.Axial stress & strainHard
Ø 25 mmBrass0.5 mØ 16 mmSteel1.5 mP = 75 kN

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σmax in the bar.

P = 75 kNsegment 1: Ø25 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)
σmax = MPa
9.Column bucklingMedium
PL = 5 m25 × 100 mmcross-section

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

L = 5 mE = 70 GPa (aluminium)b × h = 25 × 100 mm
Pcr = kN
10.Column bucklingHard
P = 130 kNL = 5 md = ?cross-section

A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 130 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 130 kNn = 2L = 5 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
tx10 mm4 cycles0.5 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 0.5 mm. Find the damping ratio ζ.

x0 = 10 mmx4 = 0.5 mmn = 4
ζ =
12.Mechanical vibrationsHard
k1​ = 40k2​ = 30k3​ = 15m = 50 kg

A 50 kg mass hangs from spring k₃ = 15 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 30 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 = 50 kgk1 = 40 kN/mk2 = 30 kN/mk3 = 15 kN/m
fn = Hz

3Solutions

1.Gears & gear trains

Answer: T1 = 26.34 N·m

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.

Motor torque

ω₁ = 2π·1450/60 = 151.8 rad/s
T1 = P/ω₁ = 4000/151.8 = 26.34 N·m

Answer

T1 = 26.34 N·m
2.Gears & gear trains

Answer: P = 3.407 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 = 2.636 rad

Slack-side tension

eμθ = e0.9227 = 2.516, T₂ = 1500/2.516 = 596.2 N

Power

P = (1500 − 596.2)·3.77 = 3.407 kW

Answer

P = 3.407 kW
3.Torsion of shafts

Answer: τmax = 19.89 MPa

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.

Stress in each segment

T = 250 N·m = 250000 N·mm
τAB = 16·250000/(π·50³) = 10.19 MPa
τBC = 16·250000/(π·40³) = 19.89 MPa

Answer

τmax = 19.89 MPa
4.Torsion of shafts

Answer: TA = 200 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 = 500
TA·a = TB·b
TA = 500·2/5 = 200 N·m
TB = 500·3/5 = 300 N·m

Answer

TA = 200 N·m
5.Mohr’s circle & stress transformation

Answer: θp = −42.14 °

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, 0 − −10) = −84.29°

Angle on the element

θp = −42.14°

Answer

θp = −42.14°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: |τn| = 7.98 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. Thin-walled cylinder: hoop stress σh = pr/t and axial stress σa = pr/(2t); these are the principal stresses in the wall.

Wall stresses

r = 1000/2 = 500 mm
σh = pr/t = 0.5·500/6 = 41.67 MPa
σa = pr/(2t) = 20.83 MPa

Transform to the seam

Take x along the axis (σa) and y around the hoop (σh). The seam’s normal makes 90° − α with the axis.

|τn| = (41.67 − 20.83)·sin 25°·cos 25° = 7.98 MPa

Answer

|τn| = 7.98 MPa
7.Axial stress & strain

Answer: P = 37.7 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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 500/(200000·706.9) = 3.537 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 500/(100000·2827) = 1.768 × 10⁻⁶ mm/N

Load

P = 0.2/5.305 × 10⁻⁶ = 37699 N = 37.7 kN

Answer

P = 37.7 kN
8.Axial stress & strain

Answer: σmax = 373 MPa

Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.

Same force in both segments

The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.

A1 = π·25²/4 = 490.9 mm², A2 = π·16²/4 = 201.1 mm²
σ1 = 152.8 MPa
σ2 = 373 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σmax = 373 MPa

Answer

σmax = 373 MPa
9.Column buckling

Answer: Pcr = 3.598 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 1·5000 = 5000 mm

Critical load

Pcr = π²EI/(KL)² = π²·70000·130208/5000² = 3598 N = 3.598 kN

Answer

Pcr = 3.598 kN
10.Column buckling

Answer: d = 166.4 mm

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.

Required critical load

Pcr = n·P = 2·130000 = 260000 N
KL = 2·5000 = 10000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·260000·10000²/(π³·70000))^(1/4) = 166.4 mm

Answer

d = 166.4 mm

Check: the slenderness KL/r is 240.4, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.1184

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(10/0.5) = 0.7489

Damping ratio

ζ = δ/√(4π² + δ²) = 0.7489/√(39.48 + 0.5609) = 0.1184

Answer

ζ = 0.1184
12.Mechanical vibrations

Answer: fn = 2.502 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 = 40 + 30 = 70 kN/m
keq = 70·15/(70 + 15) = 12.35 kN/m

Natural frequency

ωn = √(12353/50) = 15.72 rad/s
fn = 15.72/(2π) = 2.502 Hz

Answer

fn = 2.502 Hz

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