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 belt runs on an 80 mm driving pulley at 2880 rpm. The tension is 600 N on the tight side and 240 N on the slack side. Find the power P transmitted, in kW.

d₁ = 80 mmn₁ = 2880 rpmT₁ = 600 NT₂ = 240 N

P = kW

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 1440 rpm drives gear B (18 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (25 teeth) on the output shaft. Find the output speed n_out.

n_in = 1440 rpmA = 16B = 18C = 20D = 25

n_out = rpm

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 75 mm, inner diameter 56 mm, length 2 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 = 75 mmd_i = 56 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid aluminium shaft of diameter 60 mm is fixed at A. A torque of 100 N·m acts at B, 2.5 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 2 m along. Find the magnitude of the angle of twist φ_C of end C, in degrees.

d = 60 mmT_B = 100 N·mT_C = 750 N·mAB = 2.5 mBC = 2 mG = 26 GPa

φ_C = °

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 70 MPa, σᵧ = −40 MPa and τₓᵧ = 50 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 = 70 MPaσ_y = −40 MPaτ_xy = 50 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 100 MPa, σᵧ = −20 MPa and τₓᵧ = 10 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 100 MPaσ_y = −20 MPaτ_xy = 10 MPaθ = 50°

τ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 500 mm long and are compressed together between rigid plates by 50 kN. Find the shortening δ.

P = 50 kNrod: Ø40 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 20 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: Ø20 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 pinned at both ends. Using Euler’s formula, find its critical buckling load P_cr.

L = 2 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

10.Column bucklingHard

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

L = 5 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 25 mm is followed one cycle later by a peak of 17.1 mm. Find the damping ratio ζ.

x₀ = 25 mmx₁ = 17.1 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 12 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 30 kN/mk₂ = 12 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: P = 4.343 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 = 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 speed

v = 12.06 m/s

Power

P = (600 − 240)·12.06 = 4343 W = 4.343 kW

Answer

P = 4.343 kW

2.Gears & gear trains

Answer: n_out = 1024 rpm

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.

First stage

n_B = 1440·16/18 = 1280 rpm = n_C

Second stage

n_out = 1280·20/25 = 1024 rpm

Answer

n_out = 1024 rpm

total ratio = 1.406 : 1

3.Torsion of shafts

Answer: φ = 1.544 °

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(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴

Angle of twist

T = 750 N·m = 750000 N·mm

L = 2 m = 2000 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 750000·2000/(26000·2.141 × 10⁶) = 0.02695 rad

φ = 0.02695·180/π = 1.544°

Answer

φ = 1.544°

4.Torsion of shafts

Answer: φ_C = 5.412 °

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 750 N·m

T_AB = T_C − T_B = 750 − 100 = 650 N·m

Twist of each segment

J = π·60⁴/32 = 1.272 × 10⁶ mm⁴

G = 26 GPa = 26000 MPa

φ_AB = 650000·2500/(26000·1.272 × 10⁶) = 0.04912 rad

φ_BC = 750000·2000/(26000·1.272 × 10⁶) = 0.04534 rad

Twist of end C

φ_C = 0.04912 + 0.04534 = 0.09447 rad → |φ_C| = 5.412°

Answer

φ_C = 5.412°

5.Mohr’s circle & stress transformation

Answer: θ_p = 21.14 °

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.

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(2·50, 70 − −40) = 42.27°

Angle on the element

θ_p = 21.14°

Answer

θ_p = 21.14°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τ_θ = −60.82 MPa

Transformation equation

τ_θ = −60·0.9848 + 10·−0.1736 = −60.82 MPa

Answer

τ_θ = −60.82 MPa

7.Axial stress & strain

Answer: δ = 0.0218 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·40²/4 = 1257 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·7461 = 1.147 × 10⁹ N

Shortening

δ = 50000·500/1.147 × 10⁹ = 0.0218 mm

Answer

δ = 0.0218 mm

8.Axial stress & strain

Answer: P_max = 468.1 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 70·10³·7461 = 5.851 × 10⁸ N

Governing strain

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

Largest load

P_max = 0.0008·5.851 × 10⁸ = 468097 N = 468.1 kN

Answer

P_max = 468.1 kN

9.Column buckling

Answer: P_cr = 313.9 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/2000² = 313939 N = 313.9 kN

Answer

P_cr = 313.9 kN

10.Column buckling

Answer: σ_cr = 1.554 MPa

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 = 2·5000 = 10000 mm

KL/r = 10000/15 = 666.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/666.7² = 1.554 MPa

Answer

σ_cr = 1.554 MPa

11.Mechanical vibrations

Answer: ζ = 0.06034

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 = (1/1)·ln(25/17.1) = 0.3798

Damping ratio

ζ = δ/√(4π² + δ²) = 0.3798/√(39.48 + 0.1442) = 0.06034

Answer

ζ = 0.06034

12.Mechanical vibrations

Answer: k_eq = 8.571 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 = 1/30 + 1/12 → k_eq = 30·12/(30 + 12) = 8.571 kN/m

Answer

k_eq = 8.571 kN/m

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