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 4 kW motor running at 1450 rpm drives a pinion with 40 teeth, which meshes with a gear with 60 teeth. The gear pair has an efficiency of 98%. Find the output torque T₂ on the gear shaft.

P = 4 kWn₁ = 1450 rpmN₁ = 40N₂ = 60η = 98%

T₂ = N·m

2.Gears & gear trainsHard

In a compound gear train, gear A (12 teeth) on the input shaft turning at 2900 rpm drives gear B (50 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (40 teeth) on the output shaft. Find the output speed n_out.

n_in = 2900 rpmA = 12B = 50C = 18D = 40

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 = 40 mmL = 0.8 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 20 kW at 300 rpm. Find the maximum shear stress τ_max in the shaft.

P = 20 kWN = 300 rpmd_o = 75 mmd_i = 56 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = 100 MPa, σᵧ = 90 MPa and τₓᵧ = −30 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = 100 MPaσ_y = 90 MPaτ_xy = −30 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −20 MPa, σᵧ = −20 MPa and τₓᵧ = −50 MPa (σ_z = 0). The material has a yield strength of 420 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −20 MPaσ_y = −20 MPaτ_xy = −50 MPaσ_Y = 420 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 30 mm, length 300 mm, E = 70 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.2 mm in total.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø30 mm, 300 mm, E = 70 GPaδ = 0.2 mm

P = kN

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 65 kNsegment 1: Ø35 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø12 mm, 1 m, aluminium (E = 70 GPa)

δ = mm

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 = 3 mE = 70 GPa (aluminium)d = 80 mm

P_cr = kN

10.Column bucklingHard

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

L = 6 mE = 200 GPa (steel)d = 80 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 2 kg mass on a spring with a stiffness of 4 kN/m has a viscous damper giving a damping ratio of 0.35. Find the damped natural frequency f_d of its free vibration in Hz.

m = 2 kgk = 4 kN/mζ = 0.35

f_d = Hz

12.Mechanical vibrationsHard

A 40 kg mass hangs from spring k₃ = 12 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 40 kN/m and k₂ = 5 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency f_n of vertical vibration in Hz.

m = 40 kgk₁ = 40 kN/mk₂ = 5 kN/mk₃ = 12 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: T₂ = 38.72 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 = 4000/151.8 = 26.34 N·m

Output torque

T₂ = 26.34·60/40·0.98 = 38.72 N·m

Answer

T₂ = 38.72 N·m

2.Gears & gear trains

Answer: n_out = 313.2 rpm

First stage

n_B = 2900·12/50 = 696 rpm = n_C

Second stage

n_out = 696·18/40 = 313.2 rpm

Answer

n_out = 313.2 rpm

total ratio = 9.259 : 1

3.Torsion of shafts

Answer: φ = 2.806 °

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) = 400·1000·800/(26000·251327) = 0.04897 rad

φ = 0.04897·180/π = 2.806°

Answer

φ = 2.806°

4.Torsion of shafts

Answer: τ_max = 11.15 MPa

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 20000/31.42 = 636.6 N·m

Polar second moment of area

J = π(75⁴ − 56⁴)/32 = 2.141 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 636620·37.5/2.141 × 10⁶ = 11.15 MPa

Answer

τ_max = 11.15 MPa

5.Mohr’s circle & stress transformation

Answer: τ_abs = 62.71 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = 95 MPa, R = √(5² + (−30)²) = 30.41 MPa

σ₁ = 125.4 MPa, σ₂ = 64.59 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(60.83, 125.4, 64.59)/2 = 62.71 MPa

Answer

τ_abs = 62.71 MPa

Here the out-of-plane circle governs: the absolute maximum is larger than the in-plane maximum R.

6.Mohr’s circle & stress transformation

Answer: n = 4.725

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 = −20 MPa, R = 50 MPa

σ₁ = 30 MPa, σ₂ = −70 MPa, σ₃ = 0

Von Mises

σ_vM = √(30² − 30·(−70) + (−70)²) = 88.88 MPa

n = 420/88.88 = 4.725

Answer

n = 4.725

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

7.Axial stress & strain

Answer: P = 23.67 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) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(70000·706.9) = 6.063 × 10⁻⁶ mm/N

Load

P = 0.2/8.45 × 10⁻⁶ = 23668 N = 23.67 kN

Answer

P = 23.67 kN

8.Axial stress & strain

Answer: δ = 8.548 mm

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·12²/4 = 113.1 mm²

σ₁ = 67.56 MPa

σ₂ = 574.7 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 67.56·500/100000 = 0.3378 mm

δ₂ = σ₂L₂/E₂ = 574.7·1000/70000 = 8.21 mm

δ = δ₁ + δ₂ = 8.548 mm

Answer

δ = 8.548 mm

9.Column buckling

Answer: P_cr = 154.3 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·80⁴/64 = 2.011 × 10⁶ mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/3000² = 154342 N = 154.3 kN

Answer

P_cr = 154.3 kN

10.Column buckling

Answer: σ_cr = 5.483 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 = 2·6000 = 12000 mm

KL/r = 12000/20 = 600

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/600² = 5.483 MPa

Answer

σ_cr = 5.483 MPa

11.Mechanical vibrations

Answer: f_d = 6.667 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 = √(4000/2) = 44.72 rad/s

f_n = 44.72/(2π) = 7.118 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 7.118·√(1 − 0.35²) = 6.667 Hz

Answer

f_d = 6.667 Hz

12.Mechanical vibrations

Answer: f_n = 2.449 Hz

Equivalent stiffness

Only k₁ and k₂ are in parallel. That pair is in series with k₃, so combine them as reciprocals.

k₁₂ = 40 + 5 = 45 kN/m

k_eq = 45·12/(45 + 12) = 9.474 kN/m

Natural frequency

ω_n = √(9474/40) = 15.39 rad/s

f_n = 15.39/(2π) = 2.449 Hz

Answer

f_n = 2.449 Hz

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