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

A 3 kW motor running at 2900 rpm drives a pinion with 30 teeth and a module of 3 mm. Find the tangential force F_t on the pinion teeth.

P = 3 kWn = 2900 rpmN₁ = 30m = 3 mm

F_t = N

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 30 teeth and each planet gear has 24 teeth. The planet carrier is held fixed and the sun is driven at 600 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 30N_p = 24n_s = 600 rpm

n_r = rpm

3.Torsion of shaftsMedium

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 = 100 N·md = 30 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 5 kW at 3000 rpm. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 40 MPa. Find the smallest outer diameter d_o.

P = 5 kWN = 3000 rpmd_i/d_o = 0.75τ_allow = 40 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = −10 MPa, σᵧ = 120 MPa and τₓᵧ = 20 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 = −10 MPaσ_y = 120 MPaτ_xy = 20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −70 MPa, σᵧ = 120 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° counter-clockwise from the x-axis.

σ_x = −70 MPaσ_y = 120 MPaτ_xy = −30 MPaθ = 20°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 40 mm, length 400 mm, E = 120 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, Ø20 mm, 300 mm, E = 200 GPasegment 2: copper, Ø40 mm, 400 mm, E = 120 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 25 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø25 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø120/Ø70 mm, E = 70 GPa, σ_allow = 100 MPa

P_max = kN

9.Column bucklingMedium

A column of length L made of steel is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 200 GPa (steel)d = 50 mm

P_cr = kN

10.Column bucklingHard

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

L = 3 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 20 kg mass on a spring with a stiffness of 20 kN/m has a viscous damper giving a damping ratio of 0.45. Find the damped natural frequency f_d of its free vibration in Hz.

m = 20 kgk = 20 kN/mζ = 0.45

f_d = Hz

12.Mechanical vibrationsHard

A 25 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 = 25 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 f_n of vertical vibration in Hz.

m = 25 kgk₁ = 25 kN/mk₂ = 30 kN/mk₃ = 12 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 219.5 N

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.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s

T = P/ω = 3000/303.7 = 9.879 N·m

Pitch diameter

d = m·N = 3·30 = 90 mm

Tangential force

F_t = 2T/d = 2·9.879/0.09 = 219.5 N

Answer

F_t = 219.5 N

2.Gears & gear trains

Answer: n_r = 230.8 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 30 + 2·24 = 78

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −600·30/78 = −230.8 rpm

|n_r| = 230.8 rpm

Answer

n_r = 230.8 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ = 2.771 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 100·1000·1500/(39000·79522) = 0.04837 rad

φ = 0.04837·180/π = 2.771°

Answer

φ = 2.771°

4.Torsion of shafts

Answer: d_o = 14.37 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 5000/314.2 = 15.92 N·m

Outer diameter from the allowable stress

T = 15.92 N·m = 15915 N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·15915/(π·40·0.6836)) = 14.37 mm

d_i = 0.75·14.37 = 10.77 mm

Answer

d_o = 14.37 mm

A solid shaft would need d = 12.65 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θ_p = 81.45 °

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(2·20, −10 − 120) = 162.9°

Angle on the element

θ_p = 81.45°

Answer

θ_p = 81.45°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = −67.06 MPa

Transformation equation

σ_θ = 25 + −95·0.766 + −30·0.6428 = −67.06 MPa

Answer

σ_θ = −67.06 MPa

7.Axial stress & strain

Answer: P = 26.93 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 = 314.2 mm², A₂ = 1257 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·314.2) = 4.775 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(120000·1257) = 2.653 × 10⁻⁶ mm/N

Load

P = 0.2/7.427 × 10⁻⁶ = 26928 N = 26.93 kN

Answer

P = 26.93 kN

8.Axial stress & strain

Answer: P_max = 496.4 kN

Areas

A_r = π·25²/4 = 490.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·7461 = 6.205 × 10⁸ N

Governing strain

ε = min(0.0008, 0.001429) = 0.0008 (steel)

Largest load

P_max = 0.0008·6.205 × 10⁸ = 496372 N = 496.4 kN

Answer

P_max = 496.4 kN

9.Column buckling

Answer: P_cr = 16.82 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·306796/6000² = 16822 N = 16.82 kN

Answer

P_cr = 16.82 kN

10.Column buckling

Answer: σ_cr = 17.27 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/7.5 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/200² = 17.27 MPa

Answer

σ_cr = 17.27 MPa

11.Mechanical vibrations

Answer: f_d = 4.495 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 = √(20000/20) = 31.62 rad/s

f_n = 31.62/(2π) = 5.033 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 5.033·√(1 − 0.45²) = 4.495 Hz

Answer

f_d = 4.495 Hz

12.Mechanical vibrations

Answer: f_n = 3.159 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 25 + 30 = 55 kN/m

k_eq = 55·12/(55 + 12) = 9.851 kN/m

Natural frequency

ω_n = √(9851/25) = 19.85 rad/s

f_n = 19.85/(2π) = 3.159 Hz

Answer

f_n = 3.159 Hz

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