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

An 80 mm pulley at 2880 rpm drives a 450 mm pulley by a belt with 4 % slip. Find the speed n₂ of the driven pulley.

n₁ = 2880 rpmd₁ = 80 mmd₂ = 450 mms = 4 %

n₂ = rpm

2.Gears & gear trainsHard

A 1.1 kW motor at 2900 rpm drives gear A (20 teeth), which meshes with gear B (32 teeth). Gear C (15 teeth) on the same shaft as B drives gear D (18 teeth) on the output shaft. Each gear mesh has an efficiency of 95%. Find the output torque T_out.

P = 1.1 kWn_in = 2900 rpmA = 20B = 32C = 15D = 18η_mesh = 95%

T_out = N·m

3.Torsion of shaftsMedium

A solid steel shaft 1.5 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 2°. Find the smallest diameter d that meets this limit.

T = 1000 N·mL = 1.5 mG = 80 GPaφ_allow = 2°

d = mm

4.Torsion of shaftsHard

A hollow shaft transmits 30 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 30 kWN = 300 rpmd_o = 40 mmd_i = 20 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 20 MPa and σᵧ = 70 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 120 MPa?

σ_x = 20 MPaσ_y = 70 MPaσ₁ = 120 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −70 MPa, σᵧ = −80 MPa and τₓᵧ = −30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 25° counter-clockwise from the x-axis.

σ_x = −70 MPaσ_y = −80 MPaτ_xy = −30 MPaθ = 25°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of bronze (diameter 50 mm, length 200 mm, E = 110 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, Ø25 mm, 600 mm, E = 200 GPasegment 2: bronze, Ø50 mm, 200 mm, E = 110 GPaδ = 0.4 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of aluminium (outer diameter 100 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: Ø30 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø100/Ø50 mm, E = 70 GPa, σ_allow = 50 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)b × h = 20 × 80 mm

P_cr = kN

10.Column bucklingHard

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

L = 2.5 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 12 mm is followed 2 cycles later by a peak of 8.2 mm. Find the damping ratio ζ.

x₀ = 12 mmx₂ = 8.2 mmn = 2

ζ =

12.Mechanical vibrationsHard

A 40 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 5 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 40 kgk₁ = 40 kN/mk₂ = 5 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 491.5 rpm

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.

Speed without slip

2880·80/450 = 512 rpm

With slip

n₂ = 512·0.96 = 491.5 rpm

Answer

n₂ = 491.5 rpm

2.Gears & gear trains

Answer: T_out = 6.276 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·(20/32)·(15/18) = 1510 rpm

ω_out = 2π·1510/60 = 158.2 rad/s

Output power and torque

P_out = 1100·0.95² = 992.8 W

T_out = P_out/ω_out = 992.8/158.2 = 6.276 N·m

Answer

T_out = 6.276 N·m

3.Torsion of shafts

Answer: d = 48.36 mm

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.

Required polar moment

T = 1000 N·m = 1 × 10⁶ N·mm

L = 1.5 m = 1500 mm

G = 80 GPa = 80000 MPa

φ = 2·π/180 = 0.03491 rad

J = TL/(Gφ) = 1 × 10⁶·1500/(80000·0.03491) = 537148 mm⁴

Diameter

d = (32J/π)^1/4 = 48.36 mm

Answer

d = 48.36 mm

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

4.Torsion of shafts

Answer: τ_max = 81.06 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 30000/31.42 = 954.9 N·m

Polar second moment of area

J = π(40⁴ − 20⁴)/32 = 235619 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 954930·20/235619 = 81.06 MPa

Answer

τ_max = 81.06 MPa

5.Mohr’s circle & stress transformation

Answer: τ_xy = 70.71 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.

Required radius

C = (20 + 70)/2 = 45 MPa

R = σ₁ − C = 120 − 45 = 75 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(75² − 25²) = 70.71 MPa

Answer

τ_xy = 70.71 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = −94.77 MPa

Transformation equation

σ_θ = −75 + 5·0.6428 + −30·0.766 = −94.77 MPa

Answer

σ_θ = −94.77 MPa

7.Axial stress & strain

Answer: P = 56.84 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 = 490.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·490.9) = 6.112 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 200/(110000·1963) = 9.26 × 10⁻⁷ mm/N

Load

P = 0.4/7.038 × 10⁻⁶ = 56838 N = 56.84 kN

Answer

P = 56.84 kN

8.Axial stress & strain

Answer: P_max = 395.5 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·5890 = 5.537 × 10⁸ N

Governing strain

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

Largest load

P_max = 0.0007143·5.537 × 10⁸ = 395504 N = 395.5 kN

Answer

P_max = 395.5 kN

9.Column buckling

Answer: P_cr = 26.32 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 80·20³/12 = 53333 mm⁴

Effective length

KL = 1·2000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·53333/2000² = 26319 N = 26.32 kN

Answer

P_cr = 26.32 kN

10.Column buckling

Answer: σ_cr = 17.77 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 = 1·2500 = 2500 mm

KL/r = 2500/7.5 = 333.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/333.3² = 17.77 MPa

Answer

σ_cr = 17.77 MPa

11.Mechanical vibrations

Answer: ζ = 0.03029

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(12/8.2) = 0.1904

Damping ratio

ζ = δ/√(4π² + δ²) = 0.1904/√(39.48 + 0.03625) = 0.03029

Answer

ζ = 0.03029

12.Mechanical vibrations

Answer: f_n = 5.338 Hz

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 40 + 5 = 45 kN/m

Natural frequency

ω_n = √(45000/40) = 33.54 rad/s

f_n = 33.54/(2π) = 5.338 Hz

Answer

f_n = 5.338 Hz

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