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 a 150 mm driving pulley at 1440 rpm. The tension is 1200 N on the tight side and 240 N on the slack side. Find the power P transmitted, in kW.

d₁ = 150 mmn₁ = 1440 rpmT₁ = 1200 NT₂ = 240 N

P = kW

2.Gears & gear trainsHard

An open belt connects a 125 mm driving pulley at 2880 rpm to a 200 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 125 mmn₁ = 2880 rpmd₂ = 200 mmC = 1200 mmμ = 0.25T₁ = 600 N

P = kW

3.Torsion of shaftsMedium

A solid brass 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 = 250 N·md = 50 mmL = 0.8 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 300 rpmd_o = 75 mmd_i = 38 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 110 MPa and σᵧ = 70 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 125 MPa?

σ_x = 110 MPaσ_y = 70 MPaσ₁ = 125 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 40 MPa, σᵧ = 10 MPa and τₓᵧ = 20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 50° counter-clockwise from the x-axis.

σ_x = 40 MPaσ_y = 10 MPaτ_xy = 20 MPaθ = 50°

σ_θ = MPa

7.Axial stress & strainMedium

A steel rod (diameter 25 mm, E = 200 GPa) sits inside a tube made of aluminium (outer diameter 100 mm, inner diameter 60 mm, E = 70 GPa). Both are 600 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø25 mm, E = 200 GPatube: Ø100/Ø60 mm, E = 70 GPaL = 600 mm

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 95 kNsegment 1: Ø25 mm, 2 m, brass (E = 100 GPa)segment 2: Ø12 mm, 2 m, copper (E = 117 GPa)

σ_max = MPa

9.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and free at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 3 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is fixed at both ends. It must carry 60 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 60 kNn = 2L = 5 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 12 kg mass is supported by a spring with a stiffness of 4 kN/m and a viscous damper with a damping coefficient of 180 N·s/m. Find the damping ratio ζ.

m = 12 kgk = 4 kN/mc = 180 N·s/m

ζ =

12.Mechanical vibrationsHard

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 10 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 = 50 kgk₁ = 60 kN/mk₂ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 10.86 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 = 11.31 m/s

Power

P = (1200 − 240)·11.31 = 10857 W = 10.86 kW

Answer

P = 10.86 kW

2.Gears & gear trains

Answer: P = 6.072 kW

Angle of wrap

θ = 3.079 rad

Slack-side tension

e^μθ = e^0.7698 = 2.159, T₂ = 600/2.159 = 277.9 N

Power

P = (600 − 277.9)·18.85 = 6.072 kW

Answer

P = 6.072 kW

3.Torsion of shafts

Answer: φ = 0.4789 °

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) = 250·1000·800/(39000·613592) = 0.008358 rad

φ = 0.008358·180/π = 0.4789°

Answer

φ = 0.4789°

4.Torsion of shafts

Answer: τ_max = 41.14 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 100000/31.42 = 3183 N·m

Polar second moment of area

J = π(75⁴ − 38⁴)/32 = 2.902 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 3.183 × 10⁶·37.5/2.902 × 10⁶ = 41.14 MPa

Answer

τ_max = 41.14 MPa

5.Mohr’s circle & stress transformation

Answer: τ_xy = 28.72 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 = (110 + 70)/2 = 90 MPa

R = σ₁ − C = 125 − 90 = 35 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(35² − 20²) = 28.72 MPa

Answer

τ_xy = 28.72 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 42.09 MPa

Transformation equation

σ_θ = 25 + 15·−0.1736 + 20·0.9848 = 42.09 MPa

Answer

σ_θ = 42.09 MPa

7.Axial stress & strain

Answer: δ = 0.3333 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(100² − 60²)/4 = 5027 mm²

Combined axial stiffness

ΣEA = 200·10³·490.9 + 70·10³·5027 = 4.5 × 10⁸ N

Shortening

δ = 250000·600/4.5 × 10⁸ = 0.3333 mm

Answer

δ = 0.3333 mm

8.Axial stress & strain

Answer: σ_max = 840 MPa

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·25²/4 = 490.9 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 193.5 MPa

σ₂ = 840 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 840 MPa

Answer

σ_max = 840 MPa

9.Column buckling

Answer: P_cr = 38.59 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 = 2·3000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/6000² = 38586 N = 38.59 kN

Answer

P_cr = 38.59 kN

10.Column buckling

Answer: d = 52.75 mm

Required critical load

P_cr = n·P = 2·60000 = 120000 N

KL = 0.5·5000 = 2500 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·120000·2500²/(π³·200000))^(1/4) = 52.75 mm

Answer

d = 52.75 mm

Check: the slenderness KL/r is 189.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.4108

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.

Critical damping

c_c = 2·√(k·m) = 2·√(4000·12) = 438.2 N·s/m

Damping ratio

ζ = c/c_c = 180/438.2 = 0.4108

Answer

ζ = 0.4108

ζ < 1, so the system is underdamped and oscillates as it decays.

12.Mechanical vibrations

Answer: f_n = 5.955 Hz

Equivalent stiffness

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

k_eq = 60 + 10 = 70 kN/m

Natural frequency

ω_n = √(70000/50) = 37.42 rad/s

f_n = 37.42/(2π) = 5.955 Hz

Answer

f_n = 5.955 Hz

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