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 11 kW motor running at 1450 rpm drives a pinion with 18 teeth and a module of 4 mm. Find the tangential force F_t on the pinion teeth.

P = 11 kWn = 1450 rpmN₁ = 18m = 4 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects an 80 mm pulley and a 500 mm pulley whose centres are 1000 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 80 mmd₂ = 500 mmC = 1000 mm

θ = °

3.Torsion of shaftsMedium

In a torsion test, a solid specimen of diameter 12 mm is twisted by a torque of 30 N·m. The two ends of a gauge length of 250 mm rotate 2.64° relative to each other (the material is still elastic). Find the shear modulus G of the material.

T = 30 N·md = 12 mmL = 250 mmφ = 2.64°

G = GPa

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 1800 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 1800 rpmd_o = 60 mmd_i = 42 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 100 MPa, σᵧ = −50 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 = 100 MPaσ_y = −50 MPaτ_xy = 50 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 40 mm carries a bending moment of 300 N·m together with a torque of 750 N·m. At the top surface, where the bending stress is tensile and largest, find the largest principal stress σ₁.

d = 40 mmM = 300 N·mT = 750 N·m

σ₁ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 20 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 100 mm, inner diameter 50 mm, E = 120 GPa). Both are 800 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø20 mm, E = 200 GPatube: Ø100/Ø50 mm, E = 120 GPaL = 800 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 120 MPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 60 mm, E = 120 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø40 mm, E = 200 GPa, σ_allow = 120 MPatube: Ø120/Ø60 mm, E = 120 GPa, σ_allow = 70 MPa

P_max = kN

9.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)b × h = 25 × 100 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 = 4 mE = 70 GPa (aluminium)d = 50 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

An 8 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.3. Find the damped natural frequency f_d of its free vibration in Hz.

m = 8 kgk = 4 kN/mζ = 0.3

f_d = Hz

12.Mechanical vibrationsHard

A 25 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 50 kN/m and k₂ = 50 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 = 50 kN/mk₂ = 50 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 2012 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·1450/60 = 151.8 rad/s

T = P/ω = 11000/151.8 = 72.44 N·m

Pitch diameter

d = m·N = 4·18 = 72 mm

Tangential force

F_t = 2T/d = 2·72.44/0.072 = 2012 N

Answer

F_t = 2012 N

2.Gears & gear trains

Answer: θ = 155.8 °

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(0.21) = 12.12°

Angle of wrap

θ = 180 − 2·12.12 = 155.8°

Answer

θ = 155.8°

3.Torsion of shafts

Answer: G = 79.96 GPa

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 moment and angle

J = π·12⁴/32 = 2036 mm⁴

φ = 2.64·π/180 = 0.04608 rad

Shear modulus

T = 30 N·m = 30000 N·mm

G = TL/(Jφ) = 30000·250/(2036·0.04608) = 79957 MPa

G = 79.96 GPa

Answer

G = 79.96 GPa

Compare: steel ≈ 80 GPa, brass ≈ 39 GPa, aluminium ≈ 26 GPa.

4.Torsion of shafts

Answer: τ_max = 16.46 MPa

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 100000/188.5 = 530.5 N·m

Polar second moment of area

J = π(60⁴ − 42⁴)/32 = 966855 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 530516·30/966855 = 16.46 MPa

Answer

τ_max = 16.46 MPa

5.Mohr’s circle & stress transformation

Answer: θ_p = 16.85 °

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, 100 − −50) = 33.69°

Angle on the element

θ_p = 16.85°

Answer

θ_p = 16.85°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ₁ = 88.15 MPa

Stresses at the surface

σ = 32M/(πd³) = 32·300000/(π·40³) = 47.75 MPa

τ = 16T/(πd³) = 16·750000/(π·40³) = 59.68 MPa

Mohr’s circle

C = σ/2 = 23.87 MPa

R = √(23.87² + 59.68²) = 64.28 MPa

σ₁ = C + R = 88.15 MPa

σ₂ = C − R = −40.41 MPa

Answer

σ₁ = 88.15 MPa

7.Axial stress & strain

Answer: δ = 0.1559 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·20²/4 = 314.2 mm²

A_t = π(100² − 50²)/4 = 5890 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 120·10³·5890 = 7.697 × 10⁸ N

Shortening

δ = 150000·800/7.697 × 10⁸ = 0.1559 mm

Answer

δ = 0.1559 mm

8.Axial stress & strain

Answer: P_max = 740.4 kN

Areas

A_r = π·40²/4 = 1257 mm²

A_t = π(120² − 60²)/4 = 8482 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 120·10³·8482 = 1.269 × 10⁹ N

Governing strain

ε = min(0.0006, 0.0005833) = 0.0005833 (copper)

Largest load

P_max = 0.0005833·1.269 × 10⁹ = 740369 N = 740.4 kN

Answer

P_max = 740.4 kN

9.Column buckling

Answer: P_cr = 41.12 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 = 100·25³/12 = 130208 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·130208/2500² = 41123 N = 41.12 kN

Answer

P_cr = 41.12 kN

10.Column buckling

Answer: σ_cr = 1.687 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·50⁴/64 = 306796 mm⁴

Radius of gyration and slenderness

A = 1963 mm²

r = √(I/A) = √(306796/1963) = 12.5 mm

KL = 2·4000 = 8000 mm

KL/r = 8000/12.5 = 640

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/640² = 1.687 MPa

Answer

σ_cr = 1.687 MPa

11.Mechanical vibrations

Answer: f_d = 3.395 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 = √(4000/8) = 22.36 rad/s

f_n = 22.36/(2π) = 3.559 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 3.559·√(1 − 0.3²) = 3.395 Hz

Answer

f_d = 3.395 Hz

12.Mechanical vibrations

Answer: f_n = 6.164 Hz

Equivalent stiffness

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

k₁₂ = 50 + 50 = 100 kN/m

k_eq = 100·60/(100 + 60) = 37.5 kN/m

Natural frequency

ω_n = √(37500/25) = 38.73 rad/s

f_n = 38.73/(2π) = 6.164 Hz

Answer

f_n = 6.164 Hz

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