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 7.5 kW motor running at 960 rpm drives a pinion with 20 teeth and a module of 2 mm. Find the tangential force F_t on the pinion teeth.

P = 7.5 kWn = 960 rpmN₁ = 20m = 2 mm

F_t = N

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 960 rpm drives gear B (80 teeth). Gear C (15 teeth) is fixed to the same shaft as B and drives gear D (16 teeth) on the output shaft. Find the output speed n_out.

n_in = 960 rpmA = 16B = 80C = 15D = 16

n_out = rpm

3.Torsion of shaftsMedium

A solid aluminium 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 = 1 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 5 kW at 1500 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 5 kWN = 1500 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 100 MPaσ_y = −50 MPaτ_xy = 30 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A solid shaft of diameter 25 mm carries an axial tensile force of 140 kN together with a torque of 400 N·m. At the surface of the shaft, find the maximum in-plane shear stress τ_max.

d = 25 mmF = 140 kNT = 400 N·m

τ_max = 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 brass (diameter 30 mm, length 400 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.25 mm in total.

segment 1: steel, Ø25 mm, 600 mm, E = 200 GPasegment 2: brass, Ø30 mm, 400 mm, E = 100 GPaδ = 0.25 mm

P = kN

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 = 80 kNsegment 1: Ø40 mm, 1 m, aluminium (E = 70 GPa)segment 2: Ø20 mm, 2 m, copper (E = 117 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 1 mE = 70 GPa (aluminium)b × h = 20 × 60 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel 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 = 5 mE = 200 GPa (steel)d = 80 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 one cycle later by a peak of 13.3 mm. Find the damping ratio ζ.

x₀ = 25 mmx₁ = 13.3 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 50 kN/m and the lower one k₂ = 100 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 50 kN/mk₂ = 100 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: F_t = 3730 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·960/60 = 100.5 rad/s

T = P/ω = 7500/100.5 = 74.6 N·m

Pitch diameter

d = m·N = 2·20 = 40 mm

Tangential force

F_t = 2T/d = 2·74.6/0.04 = 3730 N

Answer

F_t = 3730 N

2.Gears & gear trains

Answer: n_out = 180 rpm

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.

First stage

n_B = 960·16/80 = 192 rpm = n_C

Second stage

n_out = 192·15/16 = 180 rpm

Answer

n_out = 180 rpm

total ratio = 5.333 : 1

3.Torsion of shafts

Answer: φ = 11.08 °

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) = 400·1000·1000/(26000·79522) = 0.1935 rad

φ = 0.1935·180/π = 11.08°

Answer

φ = 11.08°

4.Torsion of shafts

Answer: d = 15.94 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 5000/157.1 = 31.83 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·31831/(π·40)) = 15.94 mm

Answer

d = 15.94 mm

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

5.Mohr’s circle & stress transformation

Answer: θ_s = −34.1 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = 30/80.78 = 0.3714, sin 2θ_s = −75/80.78 = −0.9285

2θ_s = atan2(−75, 30) = −68.2°

Angle on the element

θ_s = −34.1°

check: θ_s − θ_p = −34.1 − 10.9 = −45°

Answer

θ_s = −34.1°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τ_max = 193.2 MPa

Stresses at the surface

σ = F/A = 140000/(π·25²/4) = 140000/490.9 = 285.2 MPa

τ = 16T/(πd³) = 16·400000/(π·25³) = 130.4 MPa

Mohr’s circle

C = σ/2 = 142.6 MPa

R = √(142.6² + 130.4²) = 193.2 MPa

Answer

τ_max = 193.2 MPa

7.Axial stress & strain

Answer: P = 21.24 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 = 706.9 mm²

Flexibilities

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

L₂/(E₂A₂) = 400/(100000·706.9) = 5.659 × 10⁻⁶ mm/N

Load

P = 0.25/1.177 × 10⁻⁵ = 21240 N = 21.24 kN

Answer

P = 21.24 kN

8.Axial stress & strain

Answer: σ_max = 254.6 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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 63.66 MPa

σ₂ = 254.6 MPa

Largest stress

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

σ_max = 254.6 MPa

Answer

σ_max = 254.6 MPa

9.Column buckling

Answer: P_cr = 27.63 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 = 60·20³/12 = 40000 mm⁴

Effective length

KL = 1·1000 = 1000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·40000/1000² = 27635 N = 27.63 kN

Answer

P_cr = 27.63 kN

10.Column buckling

Answer: σ_cr = 64.45 MPa

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

Radius of gyration and slenderness

A = 5027 mm²

r = √(I/A) = √(2.011 × 10⁶/5027) = 20 mm

KL = 0.7·5000 = 3500 mm

KL/r = 3500/20 = 175

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/175² = 64.45 MPa

Answer

σ_cr = 64.45 MPa

11.Mechanical vibrations

Answer: ζ = 0.09994

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 = (1/1)·ln(25/13.3) = 0.6311

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6311/√(39.48 + 0.3983) = 0.09994

Answer

ζ = 0.09994

12.Mechanical vibrations

Answer: k_eq = 33.33 kN/m

Equivalent stiffness

Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k₁ + 1/k₂.

1/k_eq = 1/50 + 1/100 → k_eq = 50·100/(50 + 100) = 33.33 kN/m

Answer

k_eq = 33.33 kN/m

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