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 150 mm apart. They must be connected by a pair of spur gears with a module of 2.5 mm and a speed reduction of 4 : 1. How many teeth N₂ must the gear have?

a = 150 mmm = 2.5 mmi = 4

N₂ =

2.Gears & gear trainsHard

In a compound gear train, gear A (40 teeth) on the input shaft turning at 960 rpm drives gear B (72 teeth). Gear C (16 teeth) is fixed to the same shaft as B and drives gear D (36 teeth) on the output shaft. Find the output speed n_out.

n_in = 960 rpmA = 40B = 72C = 16D = 36

n_out = rpm

3.Torsion of shaftsMedium

A hollow shaft must carry a torque of 1000 N·m. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 75 MPa. Find the smallest outer diameter d_o.

T = 1000 N·md_i/d_o = 0.6τ_allow = 75 MPa

d_o = mm

4.Torsion of shaftsHard

A solid steel shaft transmits 50 kW at 1450 rpm. The shear stress must not exceed 75 MPa and the twist must not exceed 0.75° per m of length. Find the smallest diameter d that satisfies both limits.

P = 50 kWN = 1450 rpmτ_allow = 75 MPaθ_allow = 0.75°/mG = 80 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 140 MPa, σᵧ = 60 MPa and τₓᵧ = −20 MPa (tension positive). Find the angle θ_s, between −90° and 90°, from the x-axis to the normal of the plane on which the shear stress has its maximum positive value +R (shear positive in the same sense as τₓᵧ).

σ_x = 140 MPaσ_y = 60 MPaτ_xy = −20 MPa

θ_s = °

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 400 mm and wall thickness 15 mm holds gas at a gauge pressure of 1.5 MPa. It is made from plate welded along a helical seam at 40° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 400 mmt = 15 mmp = 1.5 MPaα = 40°

σ_n = MPa

7.Axial stress & strainMedium

A bar is made of a steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) joined end to end with a segment made of aluminium (diameter 50 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 0.4 mm in total.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: aluminium, Ø50 mm, 300 mm, E = 70 GPaδ = 0.4 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 = 70 kNsegment 1: Ø35 mm, 1 m, steel (E = 200 GPa)segment 2: Ø16 mm, 2 m, steel (E = 200 GPa)

δ = mm

9.Column bucklingMedium

A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load P_cr.

L = 1 mE = 70 GPa (aluminium)b × h = 20 × 40 mm

P_cr = kN

10.Column bucklingHard

A solid round column of length L made of aluminium is pinned at both ends. It must carry 35 kN with a factor of safety of 3 against buckling. Find the smallest diameter d.

P = 35 kNn = 3L = 6 mE = 70 GPa (aluminium)

d = mm

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed one cycle later by a peak of 21.3 mm. Find the damping ratio ζ.

x₀ = 40 mmx₁ = 21.3 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 50 kN/m and k₂ = 15 kN/m. The mass stays level. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 100 kgk₁ = 50 kN/mk₂ = 15 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: N₂ = 96

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 = 2·150/2.5 = 120

Split by the ratio

N₁ = 120/(1 + 4) = 24

N₂ = 4·24 = 96

Answer

N₂ = 96

2.Gears & gear trains

Answer: n_out = 237 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 = 960·40/72 = 533.3 rpm = n_C

Second stage

n_out = 533.3·16/36 = 237 rpm

Answer

n_out = 237 rpm

total ratio = 4.05 : 1

3.Torsion of shafts

Answer: d_o = 42.73 mm

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.

Outer diameter from the allowable stress

T = 1000 N·m = 1 × 10⁶ N·mm

1 − k⁴ = 1 − 0.6⁴ = 0.8704

d_o = ∛(16T/(π·τ_allow·(1 − k⁴))) = ∛(16·1 × 10⁶/(π·75·0.8704)) = 42.73 mm

d_i = 0.6·42.73 = 25.64 mm

Answer

d_o = 42.73 mm

A solid shaft would need d = 40.8 mm; the hollow one is a little larger but much lighter.

4.Torsion of shafts

Answer: d = 42.3 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 50000/151.8 = 329.3 N·m

Diameter for strength

T = 329.3 N·m = 329286 N·mm

d_τ = ∛(16·329286/(π·75)) = 28.17 mm

Diameter for stiffness

θ = 0.75·π/180/1000 = 1.309 × 10⁻⁵ rad/mm

G = 80 GPa = 80000 MPa

d_φ = (32·329286/(π·80000·1.309 × 10⁻⁵))^1/4 = 42.3 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(28.17, 42.3) = 42.3 mm

Answer

d = 42.3 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_s = −58.28 °

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

On Mohr’s circle the top point (+R) is 90° from the principal point σ₁; on the element that is 45°.

cos 2θ_s = −20/44.72 = −0.4472, sin 2θ_s = −40/44.72 = −0.8944

2θ_s = atan2(−40, −20) = −116.6°

Angle on the element

θ_s = −58.28°

check: θ_s − θ_p = −58.28 − (−13.28) = −45°

Answer

θ_s = −58.28°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_n = 15.87 MPa

Wall stresses

r = 400/2 = 200 mm

σ_h = pr/t = 1.5·200/15 = 20 MPa

σ_a = pr/(2t) = 10 MPa

Transform to the seam

Take x along the axis (σ_a) and y around the hoop (σ_h). The seam’s normal makes 90° − α with the axis.

σ_n = 10·sin²40° + 20·cos²40° = 15.87 MPa

Answer

σ_n = 15.87 MPa

7.Axial stress & strain

Answer: P = 63.93 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 = 490.9 mm², A₂ = 1963 mm²

Flexibilities

L₁/(E₁A₁) = 400/(200000·490.9) = 4.074 × 10⁻⁶ mm/N

L₂/(E₂A₂) = 300/(70000·1963) = 2.183 × 10⁻⁶ mm/N

Load

P = 0.4/6.257 × 10⁻⁶ = 63928 N = 63.93 kN

Answer

P = 63.93 kN

8.Axial stress & strain

Answer: δ = 3.845 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·16²/4 = 201.1 mm²

σ₁ = 72.76 MPa

σ₂ = 348.2 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 72.76·1000/200000 = 0.3638 mm

δ₂ = σ₂L₂/E₂ = 348.2·2000/200000 = 3.482 mm

δ = δ₁ + δ₂ = 3.845 mm

Answer

δ = 3.845 mm

9.Column buckling

Answer: P_cr = 37.6 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 rectangle buckles about its weak axis, using the smaller dimension cubed.

I_min = h·b³/12 = 40·20³/12 = 26667 mm⁴

Effective length

KL = 0.7·1000 = 700 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·26667/700² = 37598 N = 37.6 kN

Answer

P_cr = 37.6 kN

10.Column buckling

Answer: d = 102.7 mm

Required critical load

P_cr = n·P = 3·35000 = 105000 N

KL = 1·6000 = 6000 mm

Solve for the diameter

d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·105000·6000²/(π³·70000))^(1/4) = 102.7 mm

Answer

d = 102.7 mm

Check: the slenderness KL/r is 233.6, so the column is slender and Euler applies.

11.Mechanical vibrations

Answer: ζ = 0.09979

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.

Logarithmic decrement

δ = (1/1)·ln(40/21.3) = 0.6302

Damping ratio

ζ = δ/√(4π² + δ²) = 0.6302/√(39.48 + 0.3971) = 0.09979

Answer

ζ = 0.09979

12.Mechanical vibrations

Answer: k_eq = 65 kN/m

Equivalent stiffness

Springs side by side share the deflection and add their forces, so their stiffnesses simply add: k = k₁ + k₂.

k_eq = 50 + 15 = 65 kN/m

Answer

k_eq = 65 kN/m

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