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 1.5 kW motor running at 2900 rpm drives a pinion with 25 teeth and a module of 3 mm. Find the tangential force F_t on the pinion teeth.

P = 1.5 kWn = 2900 rpmN₁ = 25m = 3 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects a 120 mm pulley and a 315 mm pulley whose centres are 600 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 120 mmd₂ = 315 mmC = 600 mm

θ = °

3.Torsion of shaftsMedium

A solid steel 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 = 750 N·md = 50 mmL = 2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 200 N·m acts at B, 0.5 m from A, and a torque of 750 N·m in the same direction acts at the free end C, a further 3 m along. Find the maximum shear stress τ_max in the shaft.

d = 50 mmT_B = 200 N·mT_C = 750 N·mAB = 0.5 mBC = 3 m

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa, σᵧ = −30 MPa and τₓᵧ = −30 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 100 MPaσ_y = −30 MPaτ_xy = −30 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 80 MPa, σᵧ = −40 MPa and τₓᵧ = −10 MPa (σ_z = 0). The material has a yield strength of 275 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 80 MPaσ_y = −40 MPaτ_xy = −10 MPaσ_Y = 275 MPa

n =

7.Axial stress & strainMedium

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 50 mm, E = 70 GPa). Both are 400 mm long and are compressed together between rigid plates by 80 kN. Find the shortening δ.

P = 80 kNrod: Ø25 mm, E = 200 GPatube: Ø120/Ø50 mm, E = 70 GPaL = 400 mm

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 70 kNsegment 1: Ø35 mm, 0.5 m, aluminium (E = 70 GPa)segment 2: Ø12 mm, 2 m, steel (E = 200 GPa)

δ = mm

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 = 4 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at the base and free at the top. It must carry 105 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 105 kNn = 2L = 1.5 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed 2 cycles later by a peak of 15.1 mm. Find the damping ratio ζ.

x₀ = 25 mmx₂ = 15.1 mmn = 2

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 25 kN/m and the lower one k₂ = 30 kN/m. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 100 kgk₁ = 25 kN/mk₂ = 30 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 131.7 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 = 1500/303.7 = 4.939 N·m

Pitch diameter

d = m·N = 3·25 = 75 mm

Tangential force

F_t = 2T/d = 2·4.939/0.075 = 131.7 N

Answer

F_t = 131.7 N

2.Gears & gear trains

Answer: θ = 161.3 °

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 = e^μθ, 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(0.1625) = 9.352°

Angle of wrap

θ = 180 − 2·9.352 = 161.3°

Answer

θ = 161.3°

3.Torsion of shafts

Answer: φ = 1.751 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 750·1000·2000/(80000·613592) = 0.03056 rad

φ = 0.03056·180/π = 1.751°

Answer

φ = 1.751°

4.Torsion of shafts

Answer: τ_max = 38.71 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 750 N·m

T_AB = T_C + T_B = 750 + 200 = 950 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 950 N·m

τ_max = 16·950000/(π·50³) = 38.71 MPa

Answer

τ_max = 38.71 MPa

5.Mohr’s circle & stress transformation

Answer: θ_s = −57.39 °

Angle on the circle

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −30/71.59 = −0.4191, sin 2θ_s = −65/71.59 = −0.908

2θ_s = atan2(−65, −30) = −114.8°

Angle on the element

θ_s = −57.39°

check: θ_s − θ_p = −57.39 − (−12.39) = −45°

Answer

θ_s = −57.39°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: n = 2.26

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 σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 20 MPa, R = 60.83 MPa

σ₁ = 80.83 MPa, σ₂ = −40.83 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 121.7 MPa

n = 275/121.7 = 2.26

Answer

n = 2.26

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

7.Axial stress & strain

Answer: δ = 0.04253 mm

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

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

Shortening

δ = 80000·400/7.524 × 10⁸ = 0.04253 mm

Answer

δ = 0.04253 mm

8.Axial stress & strain

Answer: δ = 6.709 mm

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·35²/4 = 962.1 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 72.76 MPa

σ₂ = 618.9 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 72.76·500/70000 = 0.5197 mm

δ₂ = σ₂L₂/E₂ = 618.9·2000/200000 = 6.189 mm

δ = δ₁ + δ₂ = 6.709 mm

Answer

δ = 6.709 mm

9.Column buckling

Answer: P_cr = 62.01 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 2·4000 = 8000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/8000² = 62013 N = 62.01 kN

Answer

P_cr = 62.01 kN

10.Column buckling

Answer: d = 66.46 mm

Required critical load

P_cr = n·P = 2·105000 = 210000 N

KL = 2·1500 = 3000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·210000·3000²/(π³·200000))^(1/4) = 66.46 mm

Answer

d = 66.46 mm

Check: the slenderness KL/r is 180.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.04009

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.

Logarithmic decrement

δ = (1/2)·ln(25/15.1) = 0.2521

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2521/√(39.48 + 0.06355) = 0.04009

Answer

ζ = 0.04009

12.Mechanical vibrations

Answer: f_n = 1.859 Hz

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 = 1/25 + 1/30 → k_eq = 25·30/(25 + 30) = 13.64 kN/m

Natural frequency

ω_n = √(13636/100) = 11.68 rad/s

f_n = 11.68/(2π) = 1.859 Hz

Answer

f_n = 1.859 Hz

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