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 1.5 kW motor running at 2900 rpm drives a pinion with 18 teeth and a module of 5 mm. Find the tangential force F_t on the pinion teeth.

P = 1.5 kWn = 2900 rpmN₁ = 18m = 5 mm

F_t = N

2.Gears & gear trainsHard

In a planetary gear set the sun gear has 16 teeth and each planet gear has 12 teeth. The planet carrier is held fixed and the sun is driven at 1500 rpm. Find the speed n_r of the ring gear (magnitude).

N_s = 16N_p = 12n_s = 1500 rpm

n_r = 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 = 50 mmL = 1 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 50 kW at 500 rpm without the shear stress exceeding 50 MPa. Find the smallest diameter d that works.

P = 50 kWN = 500 rpmτ_allow = 50 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 50 MPa and σᵧ = 20 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 95 MPa?

σ_x = 50 MPaσ_y = 20 MPaσ₁ = 95 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = −60 MPa, σᵧ = 60 MPa and τₓᵧ = −30 MPa (σ_z = 0). The material has a yield strength of 250 MPa. Find the factor of safety n against yielding by the von Mises (distortion energy) criterion.

σ_x = −60 MPaσ_y = 60 MPaτ_xy = −30 MPaσ_Y = 250 MPa

n =

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) joined end to end with a segment made of brass (diameter 50 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 0.5 mm in total.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaδ = 0.5 mm

P = kN

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the largest normal stress σ_max in the bar.

P = 100 kNsegment 1: Ø40 mm, 1.5 m, steel (E = 200 GPa)segment 2: Ø12 mm, 1.5 m, brass (E = 100 GPa)

σ_max = MPa

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 = 6 mE = 200 GPa (steel)d = 40 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 = 3 mE = 70 GPa (aluminium)d = 20 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 5 kg mass is supported by a spring with a stiffness of 50 kN/m and a viscous damper with a damping coefficient of 200 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 50 kN/mc = 200 N·s/m

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from spring k₃ = 60 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 10 kN/m and k₂ = 80 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 = 20 kgk₁ = 10 kN/mk₂ = 80 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 109.8 N

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.

Torque on the pinion

ω = 2π·2900/60 = 303.7 rad/s

T = P/ω = 1500/303.7 = 4.939 N·m

Pitch diameter

d = m·N = 5·18 = 90 mm

Tangential force

F_t = 2T/d = 2·4.939/0.09 = 109.8 N

Answer

F_t = 109.8 N

2.Gears & gear trains

Answer: n_r = 600 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 = 16 + 2·12 = 40

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 = −1500·16/40 = −600 rpm

|n_r| = 600 rpm

Answer

n_r = 600 rpm

The ring turns opposite to the sun.

3.Torsion of shafts

Answer: φ = 1.437 °

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) = 400·1000·1000/(26000·613592) = 0.02507 rad

φ = 0.02507·180/π = 1.437°

Answer

φ = 1.437°

4.Torsion of shafts

Answer: d = 45.99 mm

Torque

ω = 2πN/60 = 2π·500/60 = 52.36 rad/s

T = P/ω = 50000/52.36 = 954.9 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·954930/(π·50)) = 45.99 mm

Answer

d = 45.99 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: τ_xy = 58.09 MPa

Required radius

C = (50 + 20)/2 = 35 MPa

R = σ₁ − C = 95 − 35 = 60 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(60² − 15²) = 58.09 MPa

Answer

τ_xy = 58.09 MPa

6.Mohr’s circle & stress transformation

Answer: n = 2.152

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 = 0 MPa, R = 67.08 MPa

σ₁ = 67.08 MPa, σ₂ = −67.08 MPa, σ₃ = 0

Von Mises

σ_vM = √(67.08² − 67.08·(−67.08) + (−67.08)²) = 116.2 MPa

n = 250/116.2 = 2.152

Answer

n = 2.152

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

7.Axial stress & strain

Answer: P = 137 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 = 706.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 300/(200000·706.9) = 2.122 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(100000·1963) = 1.528 × 10⁻⁶ mm/N

Load

P = 0.5/3.65 × 10⁻⁶ = 136988 N = 137 kN

Answer

P = 137 kN

8.Axial stress & strain

Answer: σ_max = 884.2 MPa

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·40²/4 = 1257 mm², A₂ = π·12²/4 = 113.1 mm²

σ₁ = 79.58 MPa

σ₂ = 884.2 MPa

Largest stress

The thinner segment has the smaller area, so it carries the larger stress.

σ_max = 884.2 MPa

Answer

σ_max = 884.2 MPa

9.Column buckling

Answer: P_cr = 14.06 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·40⁴/64 = 125664 mm⁴

Effective length

KL = 0.7·6000 = 4200 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·125664/4200² = 14062 N = 14.06 kN

Answer

P_cr = 14.06 kN

10.Column buckling

Answer: σ_cr = 7.676 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·20⁴/64 = 7854 mm⁴

Radius of gyration and slenderness

A = 314.2 mm²

r = √(I/A) = √(7854/314.2) = 5 mm

KL = 0.5·3000 = 1500 mm

KL/r = 1500/5 = 300

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/300² = 7.676 MPa

Answer

σ_cr = 7.676 MPa

11.Mechanical vibrations

Answer: ζ = 0.2

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·√(50000·5) = 1000 N·s/m

Damping ratio

ζ = c/c_c = 200/1000 = 0.2

Answer

ζ = 0.2

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

12.Mechanical vibrations

Answer: f_n = 6.752 Hz

Equivalent stiffness

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

k₁₂ = 10 + 80 = 90 kN/m

k_eq = 90·60/(90 + 60) = 36 kN/m

Natural frequency

ω_n = √(36000/20) = 42.43 rad/s

f_n = 42.43/(2π) = 6.752 Hz

Answer

f_n = 6.752 Hz

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