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 100 mm pulley at 1450 rpm drives a 200 mm pulley by a belt with 1.5 % slip. Find the speed n₂ of the driven pulley.

n₁ = 1450 rpmd₁ = 100 mmd₂ = 200 mms = 1.5 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 1450 rpm to a 450 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 800 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 80 mmn₁ = 1450 rpmd₂ = 450 mmC = 500 mmμ = 0.3T₁ = 800 N

P = kW

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 = 500 N·md = 40 mmL = 3 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 15 kW at 1800 rpm. Find the maximum shear stress τ_max in the shaft.

P = 15 kWN = 1800 rpmd_o = 40 mmd_i = 28 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = −80 MPa, σᵧ = 70 MPa and τₓᵧ = −20 MPa (tension positive). The element is rotated 25° 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 = −80 MPaσ_y = 70 MPaτ_xy = −20 MPaθ = 25°

σ_y′ = MPa

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = −60 MPa, σᵧ = 0 MPa and τₓᵧ = −10 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 40° counter-clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = −60 MPaσ_y = 0 MPaτ_xy = −10 MPaθ = 40°

τ_θ = MPa

7.Axial stress & strainMedium

Find the elongation δ of the copper bar under the axial load.

P = 55 kNd = 16 mmL = 2.5 mE = 117 GPa

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 20 mm, E = 200 GPa, allowable stress 160 MPa) sits inside a tube made of brass (outer diameter 80 mm, inner diameter 50 mm, E = 100 GPa, allowable stress 80 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø20 mm, E = 200 GPa, σ_allow = 160 MPatube: Ø80/Ø50 mm, E = 100 GPa, σ_allow = 80 MPa

P_max = kN

9.Column bucklingMedium

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

L = 5 mE = 200 GPa (steel)b × h = 25 × 50 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium 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 = 1.5 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

An 80 kg mass on a spring with a stiffness of 80 kN/m has a viscous damper giving a damping ratio of 0.4. Find the damped natural frequency f_d of its free vibration in Hz.

m = 80 kgk = 80 kN/mζ = 0.4

f_d = Hz

12.Mechanical vibrationsHard

A 25 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 60 kN/m and k₂ = 20 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 = 25 kgk₁ = 60 kN/mk₂ = 20 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 714.1 rpm

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.

Speed without slip

1450·100/200 = 725 rpm

With slip

n₂ = 725·0.985 = 714.1 rpm

Answer

n₂ = 714.1 rpm

2.Gears & gear trains

Answer: P = 2.482 kW

Angle of wrap

θ = 2.384 rad

Slack-side tension

e^μθ = e^0.7151 = 2.044, T₂ = 800/2.044 = 391.3 N

Power

P = (800 − 391.3)·6.074 = 2.482 kW

Answer

P = 2.482 kW

3.Torsion of shafts

Answer: φ = 4.274 °

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) = 500·1000·3000/(80000·251327) = 0.0746 rad

φ = 0.0746·180/π = 4.274°

Answer

φ = 4.274°

4.Torsion of shafts

Answer: τ_max = 8.333 MPa

Torque

ω = 2πN/60 = 2π·1800/60 = 188.5 rad/s

T = P/ω = 15000/188.5 = 79.58 N·m

Polar second moment of area

J = π(40⁴ − 28⁴)/32 = 190984 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 79577·20/190984 = 8.333 MPa

Answer

τ_max = 8.333 MPa

5.Mohr’s circle & stress transformation

Answer: σ_y′ = 58.53 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′ = −5 − (−75)·0.6428 − (−20)·0.766 = 58.53 MPa

Check

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

σ_x′ = −68.53 MPa, σ_x′ + σ_y′ = −10 = σ_x + σ_y

Answer

σ_y′ = 58.53 MPa

6.Mohr’s circle & stress transformation

Answer: τ_θ = 27.81 MPa

Transformation equation

τ_θ = −−30·0.9848 + −10·0.1736 = 27.81 MPa

Answer

τ_θ = 27.81 MPa

7.Axial stress & strain

Answer: δ = 5.845 mm

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·16²/4 = 201.1 mm²

σ = P/A = 55·1000/201.1 = 273.5 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 273.5/117000 = 0.002338 = 2338 µε

Elongation

δ = ε·L = 0.002338·2500 mm = 5.845 mm

Answer

δ = 5.845 mm

8.Axial stress & strain

Answer: P_max = 295.3 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

A_r = π·20²/4 = 314.2 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·314.2 + 100·10³·3063 = 3.691 × 10⁸ N

Governing strain

ε = min(0.0008, 0.0008) = 0.0008 (brass)

Largest load

P_max = 0.0008·3.691 × 10⁸ = 295310 N = 295.3 kN

Answer

P_max = 295.3 kN

9.Column buckling

Answer: P_cr = 20.56 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 = 50·25³/12 = 65104 mm⁴

Effective length

KL = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·65104/2500² = 20562 N = 20.56 kN

Answer

P_cr = 20.56 kN

10.Column buckling

Answer: σ_cr = 35.25 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 = 0.7·1500 = 1050 mm

KL/r = 1050/7.5 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/140² = 35.25 MPa

Answer

σ_cr = 35.25 MPa

11.Mechanical vibrations

Answer: f_d = 4.613 Hz

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.

Undamped natural frequency

ω_n = √(80000/80) = 31.62 rad/s

f_n = 31.62/(2π) = 5.033 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 5.033·√(1 − 0.4²) = 4.613 Hz

Answer

f_d = 4.613 Hz

12.Mechanical vibrations

Answer: f_n = 9.003 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 + 20 = 80 kN/m

Natural frequency

ω_n = √(80000/25) = 56.57 rad/s

f_n = 56.57/(2π) = 9.003 Hz

Answer

f_n = 9.003 Hz

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