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 120 mm pulley at 960 rpm drives a 315 mm pulley by a belt with 3 % slip. Find the speed n₂ of the driven pulley.

n₁ = 960 rpmd₁ = 120 mmd₂ = 315 mms = 3 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects a 125 mm pulley and a 200 mm pulley whose centres are 800 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.

d₁ = 125 mmd₂ = 200 mmC = 800 mm

θ = °

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.75 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

T = 1000 N·md_i/d_o = 0.75τ_allow = 50 MPa

d_o = mm

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 500 rpm without the shear stress exceeding 80 MPa. Find the smallest diameter d that works.

P = 50 kWN = 500 rpmτ_allow = 80 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −10 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 30° clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −10 MPaσ_y = 110 MPaτ_xy = 30 MPaθ = -30°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = 40 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 = 160 MPaσ_y = −30 MPaτ_xy = 40 MPaθ = 25°

σ_θ = MPa

7.Axial stress & strainMedium

Find the elongation δ of the steel bar under the axial load.

P = 40 kNd = 12 mmL = 2 mE = 200 GPa

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 250 MPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 60 mm, E = 110 GPa, allowable stress 100 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 250 MPatube: Ø80/Ø60 mm, E = 110 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 = 2.5 mE = 200 GPa (steel)d = 80 mm

P_cr = kN

10.Column bucklingHard

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

L = 3 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 2 cycles later by a peak of 5.5 mm. Find the damping ratio ζ.

x₀ = 25 mmx₂ = 5.5 mmn = 2

ζ =

12.Mechanical vibrationsHard

A 10 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 = 60 kN/m and k₂ = 5 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 = 10 kgk₁ = 60 kN/mk₂ = 5 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 354.7 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

960·120/315 = 365.7 rpm

With slip

n₂ = 365.7·0.97 = 354.7 rpm

Answer

n₂ = 354.7 rpm

2.Gears & gear trains

Answer: θ = 174.6 °

Belt angle

asin(0.04688) = 2.687°

Angle of wrap

θ = 180 − 2·2.687 = 174.6°

Answer

θ = 174.6°

3.Torsion of shafts

Answer: d_o = 53.02 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.

Outer diameter from the allowable stress

T = 1000 N·m = 1 × 10⁶ N·mm

1 − k⁴ = 1 − 0.75⁴ = 0.6836

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·50·0.6836)) = 53.02 mm

d_i = 0.75·53.02 = 39.76 mm

Answer

d_o = 53.02 mm

A solid shaft would need d = 46.7 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 39.32 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 50000/52.36 = 954.9 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·954930/(π·80)) = 39.32 mm

Answer

d = 39.32 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 106 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.

Transformation equation for the y′ face

σ_y′ = 50 − (−60)·0.5 − 30·(−0.866) = 106 MPa

Check

The sum of the normal stresses does not change with rotation.

σ_x′ = −5.981 MPa, σ_x′ + σ_y′ = 100 = σ_x + σ_y

Answer

σ_y′ = 106 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 156.7 MPa

Transformation equation

σ_θ = 65 + 95·0.6428 + 40·0.766 = 156.7 MPa

Answer

σ_θ = 156.7 MPa

7.Axial stress & strain

Answer: δ = 3.537 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·12²/4 = 113.1 mm²

σ = P/A = 40·1000/113.1 = 353.7 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 353.7/200000 = 0.001768 = 1768 µε

Elongation

δ = ε·L = 0.001768·2000 mm = 3.537 mm

Answer

δ = 3.537 mm

8.Axial stress & strain

Answer: P_max = 348.4 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

A_r = π·30²/4 = 706.9 mm²

A_t = π(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·2199 = 3.833 × 10⁸ N

Governing strain

ε = min(0.00125, 0.0009091) = 0.0009091 (bronze)

Largest load

P_max = 0.0009091·3.833 × 10⁸ = 348431 N = 348.4 kN

Answer

P_max = 348.4 kN

9.Column buckling

Answer: P_cr = 158.8 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 = 2·2500 = 5000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/5000² = 158752 N = 158.8 kN

Answer

P_cr = 158.8 kN

10.Column buckling

Answer: σ_cr = 35.25 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 = 0.7·3000 = 2100 mm

KL/r = 2100/15 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa

Answer

σ_cr = 35.25 MPa

11.Mechanical vibrations

Answer: ζ = 0.1196

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(25/5.5) = 0.7571

Damping ratio

ζ = δ/√(4π² + δ²) = 0.7571/√(39.48 + 0.5731) = 0.1196

Answer

ζ = 0.1196

12.Mechanical vibrations

Answer: f_n = 8.89 Hz

Equivalent stiffness

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

k₁₂ = 60 + 5 = 65 kN/m

k_eq = 65·60/(65 + 60) = 31.2 kN/m

Natural frequency

ω_n = √(31200/10) = 55.86 rad/s

f_n = 55.86/(2π) = 8.89 Hz

Answer

f_n = 8.89 Hz

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