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 125 mm pulley at 1500 rpm drives a 250 mm pulley by a belt with 4 % slip. Find the speed n₂ of the driven pulley.

n₁ = 1500 rpmd₁ = 125 mmd₂ = 250 mms = 4 %

n₂ = rpm

2.Gears & gear trainsHard

A 2.2 kW motor at 1450 rpm drives gear A (24 teeth), which meshes with gear B (25 teeth). Gear C (50 teeth) on the same shaft as B drives gear D (72 teeth) on the output shaft. Each gear mesh has an efficiency of 96%. Find the output torque T_out.

P = 2.2 kWn_in = 1450 rpmA = 24B = 25C = 50D = 72η_mesh = 96%

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 = 200 N·md = 25 mmL = 2 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft of diameter 25 mm is fixed at A. A torque of 200 N·m acts at B, 3 m from A, and a torque of 200 N·m in the same direction acts at the free end C, a further 1 m along. Find the maximum shear stress τ_max in the shaft.

d = 25 mmT_B = 200 N·mT_C = 200 N·mAB = 3 mBC = 1 m

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 120 MPa, σᵧ = 60 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θ_p from the x-axis to the direction of the largest principal stress, between −90° and 90°.

σ_x = 120 MPaσ_y = 60 MPaτ_xy = −20 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 30 MPa, σᵧ = −40 MPa and τₓᵧ = −10 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 20° clockwise from the x-axis.

σ_x = 30 MPaσ_y = −40 MPaτ_xy = −10 MPaθ = -20°

σ_θ = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 400 mm, E = 70 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 400 mm, E = 70 GPaδ = 0.3 mm

P = kN

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of aluminium (outer diameter 120 mm, inner diameter 70 mm, E = 70 GPa, allowable stress 50 MPa). They are compressed together between rigid plates. Find the largest load P_max.

rod: Ø30 mm, E = 200 GPa, σ_allow = 150 MPatube: Ø120/Ø70 mm, E = 70 GPa, σ_allow = 50 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 = 2 mE = 200 GPa (steel)d = 60 mm

P_cr = kN

10.Column bucklingHard

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

L = 5 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 5 kg mass on a spring with a stiffness of 5 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 = 5 kgk = 5 kN/mζ = 0.4

f_d = Hz

12.Mechanical vibrationsHard

A 50 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 20 kN/m and the lower one k₂ = 80 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 50 kgk₁ = 20 kN/mk₂ = 80 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n₂ = 720 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

1500·125/250 = 750 rpm

With slip

n₂ = 750·0.96 = 720 rpm

Answer

n₂ = 720 rpm

2.Gears & gear trains

Answer: T_out = 20.03 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.

Output speed

n_out = 1450·(24/25)·(50/72) = 966.7 rpm

ω_out = 2π·966.7/60 = 101.2 rad/s

Output power and torque

P_out = 2200·0.96² = 2028 W

T_out = P_out/ω_out = 2028/101.2 = 20.03 N·m

Answer

T_out = 20.03 N·m

3.Torsion of shafts

Answer: φ = 22.99 °

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·25⁴/32 = 38350 mm⁴

Angle of twist

φ = TL/(GJ) = 200·1000·2000/(26000·38350) = 0.4012 rad

φ = 0.4012·180/π = 22.99°

Answer

φ = 22.99°

4.Torsion of shafts

Answer: τ_max = 130.4 MPa

Internal torques

Sense of T_C taken as positive.

T_BC = T_C = 200 N·m

T_AB = T_C + T_B = 200 + 200 = 400 N·m

Maximum shear stress

The segment with the largest internal torque has the largest stress (same diameter).

|T|_max = 400 N·m

τ_max = 16·400000/(π·25³) = 130.4 MPa

Answer

τ_max = 130.4 MPa

5.Mohr’s circle & stress transformation

Answer: θ_p = −16.85 °

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

Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.

2θ_p = atan2(2·−20, 120 − 60) = −33.69°

Angle on the element

θ_p = −16.85°

Answer

θ_p = −16.85°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 28.24 MPa

Transformation equation

σ_θ = −5 + 35·0.766 + −10·−0.6428 = 28.24 MPa

Answer

σ_θ = 28.24 MPa

7.Axial stress & strain

Answer: P = 31.01 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 = 1257 mm², A₂ = 706.9 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·1257) = 1.592 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 400/(70000·706.9) = 8.084 × 10⁻⁶ mm/N

Load

P = 0.3/9.676 × 10⁻⁶ = 31006 N = 31.01 kN

Answer

P = 31.01 kN

8.Axial stress & strain

Answer: P_max = 474 kN

Areas

A_r = π·30²/4 = 706.9 mm²

A_t = π(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 70·10³·7461 = 6.637 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0007143) = 0.0007143 (aluminium)

Largest load

P_max = 0.0007143·6.637 × 10⁸ = 474044 N = 474 kN

Answer

P_max = 474 kN

9.Column buckling

Answer: P_cr = 640.7 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·60⁴/64 = 636173 mm⁴

Effective length

KL = 0.7·2000 = 1400 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·636173/1400² = 640691 N = 640.7 kN

Answer

P_cr = 640.7 kN

10.Column buckling

Answer: σ_cr = 31.58 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·40⁴/64 = 125664 mm⁴

Radius of gyration and slenderness

A = 1257 mm²

r = √(I/A) = √(125664/1257) = 10 mm

KL = 0.5·5000 = 2500 mm

KL/r = 2500/10 = 250

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/250² = 31.58 MPa

Answer

σ_cr = 31.58 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 = √(5000/5) = 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: k_eq = 16 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 = 1/20 + 1/80 → k_eq = 20·80/(20 + 80) = 16 kN/m

Answer

k_eq = 16 kN/m

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