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 15 kW motor running at 960 rpm drives a pinion with 16 teeth and a module of 4 mm. Find the tangential force F_t on the pinion teeth.

P = 15 kWn = 960 rpmN₁ = 16m = 4 mm

F_t = N

2.Gears & gear trainsHard

In a compound gear train, gear A (15 teeth) on the input shaft turning at 1500 rpm drives gear B (40 teeth). Gear C (20 teeth) is fixed to the same shaft as B and drives gear D (48 teeth) on the output shaft. Find the output speed n_out.

n_in = 1500 rpmA = 15B = 40C = 20D = 48

n_out = rpm

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 40 mm, inner diameter 24 mm, length 1 m) is fixed at one end and twisted by a torque of 400 N·m at the other. Find the angle of twist φ in degrees.

T = 400 N·md_o = 40 mmd_i = 24 mmL = 1 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft must transmit 5 kW at 3000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 50 MPa. Find the smallest outer diameter d_o.

P = 5 kWN = 3000 rpmd_i/d_o = 0.6τ_allow = 50 MPa

d_o = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 20 MPa, σᵧ = 30 MPa and τₓᵧ = −50 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 = 20 MPaσ_y = 30 MPaτ_xy = −50 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 120 MPa, σᵧ = −10 MPa and τₓᵧ = 40 MPa (tension positive). Find the shear stress τ_θ on a plane whose normal is rotated 30° clockwise from the x-axis (positive in the same sense as τₓᵧ).

σ_x = 120 MPaσ_y = −10 MPaτ_xy = 40 MPaθ = -30°

τ_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 bronze (diameter 30 mm, length 400 mm, E = 110 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, Ø40 mm, 400 mm, E = 200 GPasegment 2: bronze, Ø30 mm, 400 mm, E = 110 GPaδ = 0.5 mm

P = kN

8.Axial stress & strainHard

A steel segment (diameter 30 mm, length 300 mm, E = 200 GPa) and a segment made of bronze (diameter 60 mm, length 500 mm, E = 110 GPa) are joined end to end and fixed between two rigid walls. An axial load of 60 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø30 mm, 300 mm, E = 200 GPasegment 2: bronze, Ø60 mm, 500 mm, E = 110 GPaP = 60 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)d = 80 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 = 4 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 150 kg mass is supported by a spring with a stiffness of 40 kN/m and a viscous damper with a damping coefficient of 240 N·s/m. Find the damping ratio ζ.

m = 150 kgk = 40 kN/mc = 240 N·s/m

ζ =

12.Mechanical vibrationsHard

A 25 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 = 12 kN/m and k₂ = 12 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 = 25 kgk₁ = 12 kN/mk₂ = 12 kN/mk₃ = 60 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 4663 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·960/60 = 100.5 rad/s

T = P/ω = 15000/100.5 = 149.2 N·m

Pitch diameter

d = m·N = 4·16 = 64 mm

Tangential force

F_t = 2T/d = 2·149.2/0.064 = 4663 N

Answer

F_t = 4663 N

2.Gears & gear trains

Answer: n_out = 234.4 rpm

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.

First stage

n_B = 1500·15/40 = 562.5 rpm = n_C

Second stage

n_out = 562.5·20/48 = 234.4 rpm

Answer

n_out = 234.4 rpm

total ratio = 6.4 : 1

3.Torsion of shafts

Answer: φ = 4.029 °

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 − 24⁴)/32 = 218755 mm⁴

Angle of twist

T = 400 N·m = 400000 N·mm

L = 1 m = 1000 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 400000·1000/(26000·218755) = 0.07033 rad

φ = 0.07033·180/π = 4.029°

Answer

φ = 4.029°

4.Torsion of shafts

Answer: d_o = 12.3 mm

Torque

ω = 2πN/60 = 2π·3000/60 = 314.2 rad/s

T = P/ω = 5000/314.2 = 15.92 N·m

Outer diameter from the allowable stress

T = 15.92 N·m = 15915 N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·15915/(π·50·0.8704)) = 12.3 mm

d_i = 0.6·12.3 = 7.382 mm

Answer

d_o = 12.3 mm

A solid shaft would need d = 11.75 mm; the hollow one is a little larger but much lighter.

5.Mohr’s circle & stress transformation

Answer: θ_p = −47.86 °

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·−50, 20 − 30) = −95.71°

Angle on the element

θ_p = −47.86°

Answer

θ_p = −47.86°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: τ_θ = 76.29 MPa

Transformation equation

τ_θ = −65·−0.866 + 40·0.5 = 76.29 MPa

Answer

τ_θ = 76.29 MPa

7.Axial stress & strain

Answer: P = 74.23 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/(110000·706.9) = 5.144 × 10⁻⁶ mm/N

Load

P = 0.5/6.736 × 10⁻⁶ = 74229 N = 74.23 kN

Answer

P = 74.23 kN

8.Axial stress & strain

Answer: σ₂ = 12.07 MPa

Stiffnesses

k₁ = 200000·706.9/300 = 471239 N/mm

k₂ = 110000·2827/500 = 622035 N/mm

Share of segment 2

F₂ = 60000·622035/1.093 × 10⁶ = 34138 N

Stress

σ₂ = 34138/2827 = 12.07 MPa (compression)

Answer

σ₂ = 12.07 MPa

9.Column buckling

Answer: P_cr = 222.3 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 = 0.5·5000 = 2500 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·2.011 × 10⁶/2500² = 222253 N = 222.3 kN

Answer

P_cr = 222.3 kN

10.Column buckling

Answer: σ_cr = 8.812 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 = 0.7·4000 = 2800 mm

KL/r = 2800/10 = 280

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/280² = 8.812 MPa

Answer

σ_cr = 8.812 MPa

11.Mechanical vibrations

Answer: ζ = 0.04899

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·√(40000·150) = 4899 N·s/m

Damping ratio

ζ = c/c_c = 240/4899 = 0.04899

Answer

ζ = 0.04899

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

12.Mechanical vibrations

Answer: f_n = 4.168 Hz

Equivalent stiffness

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

k₁₂ = 12 + 12 = 24 kN/m

k_eq = 24·60/(24 + 60) = 17.14 kN/m

Natural frequency

ω_n = √(17143/25) = 26.19 rad/s

f_n = 26.19/(2π) = 4.168 Hz

Answer

f_n = 4.168 Hz

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