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 125 mm driving pulley at 2880 rpm. The tension is 500 N on the tight side and 150 N on the slack side. Find the power P transmitted, in kW.

d₁ = 125 mmn₁ = 2880 rpmT₁ = 500 NT₂ = 150 N

P = kW

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 24 teeth and each planet gear has 30 teeth. The ring gear is held fixed and the sun is driven at 1800 rpm. Find the speed n_c of the planet carrier.

N_s = 24N_p = 30n_s = 1800 rpm

n_c = rpm

3.Torsion of shaftsMedium

A solid steel 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 = 400 N·md = 30 mmL = 0.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 15 kW at 1800 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 15 kWN = 1800 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −40 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 30° counter-clockwise. Find the normal stress σ_y′ on the faces of the rotated element that are perpendicular to the rotated y′-axis.

σ_x = −40 MPaσ_y = −50 MPaτ_xy = 30 MPaθ = 30°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 1000 mm and wall thickness 6 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 1000 mmt = 6 mmp = 0.8 MPaα = 50°

σ_n = MPa

7.Axial stress & strainMedium

A steel rod (diameter 40 mm, E = 200 GPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 60 mm, E = 100 GPa). Both are 800 mm long and are compressed together between rigid plates by 150 kN. Find the shortening δ.

P = 150 kNrod: Ø40 mm, E = 200 GPatube: Ø80/Ø60 mm, E = 100 GPaL = 800 mm

δ = mm

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 55 kNsegment 1: Ø25 mm, 1 m, copper (E = 117 GPa)segment 2: Ø16 mm, 1.5 m, steel (E = 200 GPa)

δ = mm

9.Column bucklingMedium

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

L = 1 mE = 200 GPa (steel)d = 25 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of steel is pinned at both ends. It must carry 70 kN with a factor of safety of 2 against buckling. Find the smallest diameter d.

P = 70 kNn = 2L = 4 mE = 200 GPa (steel)

d = mm

11.Mechanical vibrationsMedium

A 50 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 950 N·s/m. Find the damping ratio ζ.

m = 50 kgk = 50 kN/mc = 950 N·s/m

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from spring k₃ = 10 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 20 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 = 100 kgk₁ = 10 kN/mk₂ = 20 kN/mk₃ = 10 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: P = 6.597 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 = 18.85 m/s

Power

P = (500 − 150)·18.85 = 6597 W = 6.597 kW

Answer

P = 6.597 kW

2.Gears & gear trains

Answer: n_c = 400 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 24 + 2·30 = 84

Willis equation with the ring fixed

Set n_r = 0 and solve for the carrier speed.

(1800 − n_c)/(0 − n_c) = −84/24

n_c = n_s·N_s/(N_s + N_r) = 1800·24/108 = 400 rpm

Answer

n_c = 400 rpm

The carrier turns in the same direction as the sun.

3.Torsion of shafts

Answer: φ = 1.801 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·500/(80000·79522) = 0.03144 rad

φ = 0.03144·180/π = 1.801°

Answer

φ = 1.801°

4.Torsion of shafts

Answer: d = 21.64 mm

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 15000/188.5 = 79.58 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·79577/(π·40)) = 21.64 mm

Answer

d = 21.64 mm

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

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −73.48 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′ = −45 − 5·0.5 − 30·0.866 = −73.48 MPa

Check

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

σ_x′ = −16.52 MPa, σ_x′ + σ_y′ = −90 = σ_x + σ_y

Answer

σ_y′ = −73.48 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 47.11 MPa

Wall stresses

r = 1000/2 = 500 mm

σ_h = pr/t = 0.8·500/6 = 66.67 MPa

σ_a = pr/(2t) = 33.33 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 33.33·sin²50° + 66.67·cos²50° = 47.11 MPa

Answer

σ_n = 47.11 MPa

7.Axial stress & strain

Answer: δ = 0.2546 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(80² − 60²)/4 = 2199 mm²

Combined axial stiffness

ΣEA = 200·10³·1257 + 100·10³·2199 = 4.712 × 10⁸ N

Shortening

δ = 150000·800/4.712 × 10⁸ = 0.2546 mm

Answer

δ = 0.2546 mm

8.Axial stress & strain

Answer: δ = 3.009 mm

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·16²/4 = 201.1 mm²

σ₁ = 112 MPa

σ₂ = 273.5 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 112·1000/117000 = 0.9577 mm

δ₂ = σ₂L₂/E₂ = 273.5·1500/200000 = 2.052 mm

δ = δ₁ + δ₂ = 3.009 mm

Answer

δ = 3.009 mm

9.Column buckling

Answer: P_cr = 77.24 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·25⁴/64 = 19175 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/700² = 77244 N = 77.24 kN

Answer

P_cr = 77.24 kN

10.Column buckling

Answer: d = 69.34 mm

Required critical load

P_cr = n·P = 2·70000 = 140000 N

KL = 1·4000 = 4000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·140000·4000²/(π³·200000))^(1/4) = 69.34 mm

Answer

d = 69.34 mm

Check: the slenderness KL/r is 230.7, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.3004

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·√(50000·50) = 3162 N·s/m

Damping ratio

ζ = c/c_c = 950/3162 = 0.3004

Answer

ζ = 0.3004

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

12.Mechanical vibrations

Answer: f_n = 1.378 Hz

Equivalent stiffness

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

k₁₂ = 10 + 20 = 30 kN/m

k_eq = 30·10/(30 + 10) = 7.5 kN/m

Natural frequency

ω_n = √(7500/100) = 8.66 rad/s

f_n = 8.66/(2π) = 1.378 Hz

Answer

f_n = 1.378 Hz

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