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 960 rpm drives a pinion with 25 teeth and a module of 3 mm. Find the tangential force F_t on the pinion teeth.

P = 11 kWn = 960 rpmN₁ = 25m = 3 mm

F_t = N

2.Gears & gear trainsHard

An open belt connects a 160 mm driving pulley at 960 rpm to a 400 mm pulley; the centres are 1200 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 160 mmn₁ = 960 rpmd₂ = 400 mmC = 1200 mmμ = 0.3T₁ = 500 N

P = kW

3.Torsion of shaftsMedium

A hollow steel shaft (outer diameter 30 mm, inner diameter 21 mm, length 1.2 m) is fixed at one end and twisted by a torque of 1000 N·m at the other. Find the angle of twist φ in degrees.

T = 1000 N·md_o = 30 mmd_i = 21 mmL = 1.2 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 15 kW at 500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 15 kWN = 500 rpmd_o = 40 mmd_i = 20 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 20 MPa, σᵧ = −50 MPa and τₓᵧ = 30 MPa (tension positive). The element is rotated 15° 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 = 20 MPaσ_y = −50 MPaτ_xy = 30 MPaθ = 15°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = −50 MPa and τₓᵧ = −60 MPa (σ_z = 0). The material has a yield strength of 460 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 100 MPaσ_y = −50 MPaτ_xy = −60 MPaσ_Y = 460 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 500 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.

segment 1: steel, Ø30 mm, 400 mm, E = 200 GPasegment 2: brass, Ø60 mm, 500 mm, E = 100 GPaδ = 0.3 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 = 65 kNsegment 1: Ø35 mm, 1.5 m, aluminium (E = 70 GPa)segment 2: Ø16 mm, 1.5 m, aluminium (E = 70 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 = 2 mE = 70 GPa (aluminium)d = 50 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 5 mE = 200 GPa (steel)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 30 mm is followed 5 cycles later by a peak of 8.5 mm. Find the damping ratio ζ.

x₀ = 30 mmx₅ = 8.5 mmn = 5

ζ =

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 100 kN/m and the lower one k₂ = 10 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 25 kgk₁ = 100 kN/mk₂ = 10 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: F_t = 2918 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 = 11000/100.5 = 109.4 N·m

Pitch diameter

d = m·N = 3·25 = 75 mm

Tangential force

F_t = 2T/d = 2·109.4/0.075 = 2918 N

Answer

F_t = 2918 N

2.Gears & gear trains

Answer: P = 2.357 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.

Angle of wrap

θ = 2.941 rad

Slack-side tension

e^μθ = e^0.8824 = 2.417, T₂ = 500/2.417 = 206.9 N

Power

P = (500 − 206.9)·8.042 = 2.357 kW

Answer

P = 2.357 kW

3.Torsion of shafts

Answer: φ = 14.22 °

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 − 21⁴)/32 = 60428 mm⁴

Angle of twist

T = 1000 N·m = 1 × 10⁶ N·mm

L = 1.2 m = 1200 mm

G = 80 GPa = 80000 MPa

φ = TL/(GJ) = 1 × 10⁶·1200/(80000·60428) = 0.2482 rad

φ = 0.2482·180/π = 14.22°

Answer

φ = 14.22°

4.Torsion of shafts

Answer: τ_max = 24.32 MPa

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 15000/52.36 = 286.5 N·m

Polar second moment of area

J = π(40⁴ − 20⁴)/32 = 235619 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 286479·20/235619 = 24.32 MPa

Answer

τ_max = 24.32 MPa

5.Mohr’s circle & stress transformation

Answer: σ_y′ = −60.31 MPa

Transformation equation for the y′ face

σ_y′ = −15 − 35·0.866 − 30·0.5 = −60.31 MPa

Check

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

σ_x′ = 30.31 MPa, σ_x′ + σ_y′ = −30 = σ_x + σ_y

Answer

σ_y′ = −60.31 MPa

6.Mohr’s circle & stress transformation

Answer: n = 2.395

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 25 MPa, R = 96.05 MPa

σ₁ = 121 MPa, σ₂ = −71.05 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 192.1 MPa

n = 460/192.1 = 2.395

Answer

n = 2.395

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: P = 65.25 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 = 706.9 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·706.9) = 2.829 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 500/(100000·2827) = 1.768 × 10⁻⁶ mm/N

Load

P = 0.3/4.598 × 10⁻⁶ = 65248 N = 65.25 kN

Answer

P = 65.25 kN

8.Axial stress & strain

Answer: σ_max = 323.3 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·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 67.56 MPa

σ₂ = 323.3 MPa

Largest stress

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

σ_max = 323.3 MPa

Answer

σ_max = 323.3 MPa

9.Column buckling

Answer: P_cr = 13.25 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 2·2000 = 4000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/4000² = 13247 N = 13.25 kN

Answer

P_cr = 13.25 kN

10.Column buckling

Answer: σ_cr = 4.441 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 1·5000 = 5000 mm

KL/r = 5000/7.5 = 666.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/666.7² = 4.441 MPa

Answer

σ_cr = 4.441 MPa

11.Mechanical vibrations

Answer: ζ = 0.04011

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 = (1/5)·ln(30/8.5) = 0.2522

Damping ratio

ζ = δ/√(4π² + δ²) = 0.2522/√(39.48 + 0.06362) = 0.04011

Answer

ζ = 0.04011

12.Mechanical vibrations

Answer: k_eq = 9.091 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 = 1/100 + 1/10 → k_eq = 100·10/(100 + 10) = 9.091 kN/m

Answer

k_eq = 9.091 kN/m

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