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

Two parallel shafts are 125 mm apart. They must be connected by a pair of spur gears with a module of 2.5 mm and a speed reduction of 3 : 1. How many teeth N₁ must the pinion have?

a = 125 mmm = 2.5 mmi = 3

N₁ =

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 30 teeth and each planet gear has 30 teeth. The planet carrier is held fixed and the sun is driven at 3000 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 30N_p = 30n_s = 3000 rpm

n_r = rpm

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 = 50 mmL = 2.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 100 kW at 3000 rpm. Find the maximum shear stress τ_max in the shaft.

P = 100 kWN = 3000 rpmd_o = 80 mmd_i = 40 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −80 MPa, σᵧ = −70 MPa and τₓᵧ = 10 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −80 MPaσ_y = −70 MPaτ_xy = 10 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 140 MPa, σᵧ = 90 MPa and τₓᵧ = −60 MPa (σ_z = 0). Find the von Mises equivalent stress σ_vM.

σ_x = 140 MPaσ_y = 90 MPaτ_xy = −60 MPa

σ_vM = MPa

7.Axial stress & strainMedium

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of bronze (outer diameter 80 mm, inner diameter 50 mm, E = 110 GPa). Both are 500 mm long and are compressed together between rigid plates by 250 kN. Find the shortening δ.

P = 250 kNrod: Ø30 mm, E = 200 GPatube: Ø80/Ø50 mm, E = 110 GPaL = 500 mm

δ = mm

8.Axial stress & strainHard

A steel rod (diameter 30 mm, E = 200 GPa, allowable stress 150 MPa) sits inside a tube made of brass (outer diameter 120 mm, inner diameter 70 mm, E = 100 GPa, allowable stress 80 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 = 100 GPa, σ_allow = 80 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 = 3 mE = 200 GPa (steel)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 = 60 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 100 kg mass on a spring with a stiffness of 150 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 = 100 kgk = 150 kN/mζ = 0.4

f_d = Hz

12.Mechanical vibrationsHard

A 100 kg mass hangs from spring k₃ = 80 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 60 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 = 100 kgk₁ = 20 kN/mk₂ = 60 kN/mk₃ = 80 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₁ = 25

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.

Total number of teeth

N₁ + N₂ = 2a/m = 2·125/2.5 = 100

Split by the ratio

N₁ = 100/(1 + 3) = 25

Answer

N₁ = 25

2.Gears & gear trains

Answer: n_r = 1000 rpm

Planetary gear sets: the speeds of sun, ring and carrier are linked by the Willis equation (n_s − n_c)/(n_r − n_c) = −N_r/N_s: seen from the carrier, the set is an ordinary train with the planets as idlers. Holding one member fixed turns the set into a single-ratio reducer.

Ring gear teeth

N_r = N_s + 2·N_p = 30 + 2·30 = 90

Willis equation with the carrier fixed

With n_c = 0 the sun drives the ring through the planets, which act as idlers.

n_r = −n_s·N_s/N_r = −3000·30/90 = −1000 rpm

|n_r| = 1000 rpm

Answer

n_r = 1000 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ = 1.459 °

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·50⁴/32 = 613592 mm⁴

Angle of twist

φ = TL/(GJ) = 500·1000·2500/(80000·613592) = 0.02546 rad

φ = 0.02546·180/π = 1.459°

Answer

φ = 1.459°

4.Torsion of shafts

Answer: τ_max = 3.377 MPa

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 100000/314.2 = 318.3 N·m

Polar second moment of area

J = π(80⁴ − 40⁴)/32 = 3.77 × 10⁶ mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 318310·40/3.77 × 10⁶ = 3.377 MPa

Answer

τ_max = 3.377 MPa

5.Mohr’s circle & stress transformation

Answer: τ_abs = 43.09 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 = −75 MPa, R = √((−5)² + 10²) = 11.18 MPa

σ₁ = −63.82 MPa, σ₂ = −86.18 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(22.36, 63.82, 86.18)/2 = 43.09 MPa

Answer

τ_abs = 43.09 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: σ_vM = 160.9 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. 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 = 115 MPa, R = 65 MPa

σ₁ = 180 MPa, σ₂ = 50 MPa, σ₃ = 0

Von Mises

σ_vM = √(180² − 180·50 + 50²) = 160.9 MPa

Answer

σ_vM = 160.9 MPa

7.Axial stress & strain

Answer: δ = 0.2613 mm

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·30²/4 = 706.9 mm²

A_t = π(80² − 50²)/4 = 3063 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 110·10³·3063 = 4.783 × 10⁸ N

Shortening

δ = 250000·500/4.783 × 10⁸ = 0.2613 mm

Answer

δ = 0.2613 mm

8.Axial stress & strain

Answer: P_max = 665.6 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 + 100·10³·7461 = 8.875 × 10⁸ N

Governing strain

ε = min(0.00075, 0.0008) = 0.00075 (steel)

Largest load

P_max = 0.00075·8.875 × 10⁸ = 665625 N = 665.6 kN

Answer

P_max = 665.6 kN

9.Column buckling

Answer: P_cr = 900 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 = 0.7·3000 = 2100 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·2.011 × 10⁶/2100² = 899955 N = 900 kN

Answer

P_cr = 900 kN

10.Column buckling

Answer: σ_cr = 3.084 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴

Radius of gyration and slenderness

A = 2827 mm²

r = √(I/A) = √(636173/2827) = 15 mm

KL = 2·6000 = 12000 mm

KL/r = 12000/15 = 800

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/800² = 3.084 MPa

Answer

σ_cr = 3.084 MPa

11.Mechanical vibrations

Answer: f_d = 5.649 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 = √(150000/100) = 38.73 rad/s

f_n = 38.73/(2π) = 6.164 Hz

Damped natural frequency

f_d = f_n·√(1 − ζ²) = 6.164·√(1 − 0.4²) = 5.649 Hz

Answer

f_d = 5.649 Hz

12.Mechanical vibrations

Answer: f_n = 3.183 Hz

Equivalent stiffness

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

k₁₂ = 20 + 60 = 80 kN/m

k_eq = 80·80/(80 + 80) = 40 kN/m

Natural frequency

ω_n = √(40000/100) = 20 rad/s

f_n = 20/(2π) = 3.183 Hz

Answer

f_n = 3.183 Hz

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