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

Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.

Mohr’s circle & stress transformation

Centre: C = (σ_x + σ_y)/2
Radius: R = √(((σ_x − σ_y)/2)² + τ_xy²)
Principal stresses: σ_1,2 = C ± R
Principal direction: tan 2θ_p = 2τ_xy/(σ_x − σ_y)
Inclined plane: σ_θ = C + (σ_x − σ_y)/2·cos 2θ + τ_xy·sin 2θ

Method

Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
The principal stresses are C ± R and the largest in-plane shear stress is R.
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: ε_T = α·ΔT

Column buckling

Euler load: P_cr = π²EI/(KL)²
Effective length factor: K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
Critical stress: σ_cr = P_cr/A
Slenderness: λ = KL/r, r = √(I/A)

Mechanical vibrations

Natural frequency: ω_n = √(k/m), f_n = ω_n/(2π)
From static deflection: ω_n = √(g/δ_st)
Critical damping: c_c = 2√(km) = 2mω_n
Damping ratio: ζ = c/c_c
Damped frequency: ω_d = ω_n√(1 − ζ²)
Logarithmic decrement: δ = (1/n)·ln(x₀/x_n), 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

Two parallel shafts are 125 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 1.5 : 1. How many teeth N₂ must the gear have?

a = 125 mmm = 5 mmi = 1.5

N₂ =

2.Gears & gear trainsHard

A 4 kW motor at 2900 rpm drives gear A (16 teeth), which meshes with gear B (20 teeth). Gear C (12 teeth) on the same shaft as B drives gear D (24 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque T_out.

P = 4 kWn_in = 2900 rpmA = 16B = 20C = 12D = 24η_mesh = 96%

T_out = N·m

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 60 mm, inner diameter 36 mm, length 2.5 m) is fixed at one end and twisted by a torque of 750 N·m at the other. Find the angle of twist φ in degrees.

T = 750 N·md_o = 60 mmd_i = 36 mmL = 2.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 20 kW at 500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 20 kWN = 500 rpmτ_allow = 50 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 30 MPa, σᵧ = 40 MPa and τₓᵧ = 40 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 30 MPaσ_y = 40 MPaτ_xy = 40 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 8 mm holds gas at a gauge pressure of 1.2 MPa. It is made from plate welded along a helical seam at 60° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 800 mmt = 8 mmp = 1.2 MPaα = 60°

σ_n = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 mm, length 400 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.4 mm in total.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 400 mm, E = 70 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø120/Ø50 mm, E = 70 GPa, σ_allow = 50 MPa

P_max = kN

9.Column bucklingMedium

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

L = 2.5 mE = 70 GPa (aluminium)d = 25 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 115 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 115 kNn = 3L = 6 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 200 kg mass on a spring with a stiffness of 150 kN/m has a viscous damper giving a damping ratio of 0.7. Find the damped natural frequency f_d of its free vibration in Hz.

m = 200 kgk = 150 kN/mζ = 0.7

f_d = Hz

12.Mechanical vibrationsHard

A 10 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 = 5 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 10 kgk₁ = 40 kN/mk₂ = 5 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₂ = 30

Gear geometry: the module m sets the tooth size. A gear with N teeth has a pitch diameter d = m·N, two meshing gears have the same module, and their centre distance is a = (d₁ + d₂)/2 = m·(N₁ + N₂)/2. The pitch-line speed is v = π·d·n/60, and the tangential tooth force follows from the torque: F_t = 2T/d.

Total number of teeth

N₁ + N₂ = 2a/m = 2·125/5 = 50

Split by the ratio

N₁ = 50/(1 + 1.5) = 20

N₂ = 1.5·20 = 30

Answer

N₂ = 30

2.Gears & gear trains

Answer: T_out = 30.35 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.

Output speed

n_out = 2900·(16/20)·(12/24) = 1160 rpm

ω_out = 2π·1160/60 = 121.5 rad/s

Output power and torque

P_out = 4000·0.96² = 3686 W

T_out = P_out/ω_out = 3686/121.5 = 30.35 N·m

Answer

T_out = 30.35 N·m

3.Torsion of shafts

Answer: φ = 3.731 °

Torsion of circular shafts: shear stress grows linearly from zero at the centre to τ_max = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(d_o⁴ − d_i⁴)/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(60⁴ − 36⁴)/32 = 1.107 × 10⁶ mm⁴

Angle of twist

T = 750 N·m = 750000 N·mm

L = 2.5 m = 2500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 750000·2500/(26000·1.107 × 10⁶) = 0.06512 rad

φ = 0.06512·180/π = 3.731°

Answer

φ = 3.731°

4.Torsion of shafts

Answer: d = 33.89 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 20000/52.36 = 382 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·381972/(π·50)) = 33.89 mm

Answer

d = 33.89 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: θ_p = 48.56 °

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.

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.

2θ_p = atan2(2·40, 30 − 40) = 97.13°

Angle on the element

θ_p = 48.56°

Answer

θ_p = 48.56°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_n = 37.5 MPa

Wall stresses

r = 800/2 = 400 mm

σ_h = pr/t = 1.2·400/8 = 60 MPa

σ_a = pr/(2t) = 30 MPa

Transform to the seam

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

σ_n = 30·sin²60° + 60·cos²60° = 37.5 MPa

Answer

σ_n = 37.5 MPa

7.Axial stress & strain

Answer: P = 81.64 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, P_i = P·E_iA_i/ΣEA, and σ_i = E_i·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) = 500/(200000·1257) = 1.989 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(70000·1963) = 2.91 × 10⁻⁶ mm/N

Load

P = 0.4/4.9 × 10⁻⁶ = 81638 N = 81.64 kN

Answer

P = 81.64 kN

8.Axial stress & strain

Answer: P_max = 537.4 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(120² − 50²)/4 = 9346 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·9346 = 7.524 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0007143) = 0.0007143 (aluminium)

Largest load

P_max = 0.0007143·7.524 × 10⁸ = 537437 N = 537.4 kN

Answer

P_max = 537.4 kN

9.Column buckling

Answer: P_cr = 8.478 kN

Euler buckling: a slender column becomes unstable at P_cr = 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

Effective length

KL = 0.5·2500 = 1250 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·19175/1250² = 8478 N = 8.478 kN

Answer

P_cr = 8.478 kN

10.Column buckling

Answer: d = 75.24 mm

Required critical load

P_cr = n·P = 3·115000 = 345000 N

KL = 0.5·6000 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·345000·3000²/(π³·200000))^(1/4) = 75.24 mm

Answer

d = 75.24 mm

Check: the slenderness KL/r is 159.5, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: f_d = 3.113 Hz

Free vibration of a mass–spring system: the undamped natural frequency is ω_n = √(k/m) in rad/s, and f_n = ω_n/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/c_c, with critical damping c_c = 2·√(k·m) = 2·m·ω_n. An underdamped system (ζ < 1) oscillates at ω_d = ω_n·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.

Undamped natural frequency

ω_n = √(150000/200) = 27.39 rad/s

f_n = 27.39/(2π) = 4.359 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 4.359·√(1 − 0.7²) = 3.113 Hz

Answer

f_d = 3.113 Hz

12.Mechanical vibrations

Answer: k_eq = 4.444 kN/m

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/k_eq = 1/40 + 1/5 → k_eq = 40·5/(40 + 5) = 4.444 kN/m

Answer

k_eq = 4.444 kN/m

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