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 3 kW motor running at 1450 rpm drives a pinion with 25 teeth, which meshes with a gear with 50 teeth. The gear pair has an efficiency of 95%. Find the output torque T₂ on the gear shaft.

P = 3 kWn₁ = 1450 rpmN₁ = 25N₂ = 50η = 95%

T₂ = N·m

2.Gears & gear trainsHard

A 15 kW motor at 2900 rpm drives gear A (20 teeth), which meshes with gear B (80 teeth). Gear C (16 teeth) on the same shaft as B drives gear D (25 teeth) on the output shaft. Each gear mesh has an efficiency of 98%. Find the output torque T_out.

P = 15 kWn_in = 2900 rpmA = 20B = 80C = 16D = 25η_mesh = 98%

T_out = N·m

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 = 500 N·md = 40 mmL = 1.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 5 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 5 kWN = 300 rpmd_o = 60 mmd_i = 42 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

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

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −20 MPa, σᵧ = 110 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σ_x = −20 MPaσ_y = 110 MPaτ_xy = 30 MPaθ = 15°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 20 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 60 mm, length 300 mm, E = 100 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø20 mm, 600 mm, E = 200 GPasegment 2: brass, Ø60 mm, 300 mm, E = 100 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of bronze (outer diameter 100 mm, inner diameter 70 mm, E = 110 GPa, allowable stress 60 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø100/Ø70 mm, E = 110 GPa, σ_allow = 60 MPa

P_max = kN

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 = 4 mE = 200 GPa (steel)d = 25 mm

P_cr = kN

10.Column bucklingHard

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

L = 4 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 10 kg mass is supported by a spring with a stiffness of 10 kN/m and a viscous damper with a damping coefficient of 50 N·s/m. Find the damping ratio ζ.

m = 10 kgk = 10 kN/mc = 50 N·s/m

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 10 kN/m and the lower one k₂ = 15 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 100 kgk₁ = 10 kN/mk₂ = 15 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: T₂ = 37.54 N·m

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.

Motor torque

ω₁ = 2π·1450/60 = 151.8 rad/s

T₁ = P/ω₁ = 3000/151.8 = 19.76 N·m

Output torque

T₂ = 19.76·50/25·0.95 = 37.54 N·m

Answer

T₂ = 37.54 N·m

2.Gears & gear trains

Answer: T_out = 296.5 N·m

Output speed

n_out = 2900·(20/80)·(16/25) = 464 rpm

ω_out = 2π·464/60 = 48.59 rad/s

Output power and torque

P_out = 15000·0.98² = 14406 W

T_out = P_out/ω_out = 14406/48.59 = 296.5 N·m

Answer

T_out = 296.5 N·m

3.Torsion of shafts

Answer: φ = 6.576 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·1500/(26000·251327) = 0.1148 rad

φ = 0.1148·180/π = 6.576°

Answer

φ = 6.576°

4.Torsion of shafts

Answer: τ_max = 4.938 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 5000/31.42 = 159.2 N·m

Polar second moment of area

J = π(60⁴ − 42⁴)/32 = 966855 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 159155·30/966855 = 4.938 MPa

Answer

τ_max = 4.938 MPa

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 54.45 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′ = 0 − 30·(−0.1736) − (−50)·0.9848 = 54.45 MPa

Check

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

σ_x′ = −54.45 MPa, σ_x′ + σ_y′ = 0 = σ_x + σ_y

Answer

σ_y′ = 54.45 MPa

6.Mohr’s circle & stress transformation

Answer: σ_θ = 3.708 MPa

Transformation equation

σ_θ = 45 + −65·0.866 + 30·0.5 = 3.708 MPa

Answer

σ_θ = 3.708 MPa

7.Axial stress & strain

Answer: P = 18.85 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 = 314.2 mm², A₂ = 2827 mm²

Flexibilities

L₁/(E₁A₁) = 600/(200000·314.2) = 9.549 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·2827) = 1.061 × 10⁻⁶ mm/N

Load

P = 0.2/1.061 × 10⁻⁵ = 18850 N = 18.85 kN

Answer

P = 18.85 kN

8.Axial stress & strain

Answer: P_max = 274.6 kN

Areas

A_r = π·20²/4 = 314.2 mm²

A_t = π(100² − 70²)/4 = 4006 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 110·10³·4006 = 5.034 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0005455) = 0.0005455 (bronze)

Largest load

P_max = 0.0005455·5.034 × 10⁸ = 274604 N = 274.6 kN

Answer

P_max = 274.6 kN

9.Column buckling

Answer: P_cr = 4.828 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 = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·19175/2800² = 4828 N = 4.828 kN

Answer

P_cr = 4.828 kN

10.Column buckling

Answer: σ_cr = 9.715 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 = 0.5·4000 = 2000 mm

KL/r = 2000/7.5 = 266.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/266.7² = 9.715 MPa

Answer

σ_cr = 9.715 MPa

11.Mechanical vibrations

Answer: ζ = 0.07906

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·√(10000·10) = 632.5 N·s/m

Damping ratio

ζ = c/c_c = 50/632.5 = 0.07906

Answer

ζ = 0.07906

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

12.Mechanical vibrations

Answer: k_eq = 6 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 = 1/10 + 1/15 → k_eq = 10·15/(10 + 15) = 6 kN/m

Answer

k_eq = 6 kN/m

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