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 160 mm apart. They must be connected by a pair of spur gears with a module of 5 mm and a speed reduction of 3 : 1. How many teeth N₁ must the pinion have?

a = 160 mmm = 5 mmi = 3

N₁ =

2.Gears & gear trainsHard

An open belt connects a 120 mm driving pulley at 960 rpm to a 500 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 600 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 120 mmn₁ = 960 rpmd₂ = 500 mmC = 500 mmμ = 0.25T₁ = 600 N

P = kW

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 30 mm, inner diameter 23 mm, length 1 m) is fixed at one end and twisted by a torque of 200 N·m at the other. Find the angle of twist φ in degrees.

T = 200 N·md_o = 30 mmd_i = 23 mmL = 1 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid steel shaft transmits 5 kW at 300 rpm. The shear stress must not exceed 60 MPa and the twist must not exceed 1° per m of length. Find the smallest diameter d that satisfies both limits.

P = 5 kWN = 300 rpmτ_allow = 60 MPaθ_allow = 1°/mG = 80 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 100 MPa and σᵧ = 60 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 130 MPa?

σ_x = 100 MPaσ_y = 60 MPaσ₁ = 130 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 10 mm holds gas at a gauge pressure of 2.5 MPa. It is made from plate welded along a helical seam at 35° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 600 mmt = 10 mmp = 2.5 MPaα = 35°

σ_n = MPa

7.Axial stress & strainMedium

Find the axial strain ε in the copper bar, in microstrain (µε = 10⁻⁶).

P = 120 kNd = 16 mmL = 1 mE = 117 GPa

ε = µε

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 500 mm, E = 200 GPa) and a segment made of aluminium (diameter 60 mm, length 400 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 100 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 500 mm, E = 200 GPasegment 2: aluminium, Ø60 mm, 400 mm, E = 70 GPaP = 100 kN

σ₂ = MPa

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 = 6 mE = 70 GPa (aluminium)d = 50 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel is fixed at both ends. 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 25 kg mass is supported by a spring with a stiffness of 200 kN/m and a viscous damper with a damping coefficient of 1790 N·s/m. Find the damping ratio ζ.

m = 25 kgk = 200 kN/mc = 1790 N·s/m

ζ =

12.Mechanical vibrationsHard

A 40 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 = 15 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 = 20 kN/mk₂ = 15 kN/mk₃ = 80 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: N₁ = 16

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 = 2·160/5 = 64

Split by the ratio

N₁ = 64/(1 + 3) = 16

Answer

N₁ = 16

2.Gears & gear trains

Answer: P = 1.614 kW

Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = e^μθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.

Angle of wrap

θ = 2.362 rad

Slack-side tension

e^μθ = e^0.5905 = 1.805, T₂ = 600/1.805 = 332.4 N

Power

P = (600 − 332.4)·6.032 = 1.614 kW

Answer

P = 1.614 kW

3.Torsion of shafts

Answer: φ = 5.645 °

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(30⁴ − 23⁴)/32 = 52048 mm⁴

Angle of twist

T = 200 N·m = 200000 N·mm

L = 1 m = 1000 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 200000·1000/(39000·52048) = 0.09853 rad

φ = 0.09853·180/π = 5.645°

Answer

φ = 5.645°

4.Torsion of shafts

Answer: d = 32.83 mm

Torque

ω = 2πN/60 = 2π·300/60 = 31.42 rad/s

T = P/ω = 5000/31.42 = 159.2 N·m

Diameter for strength

T = 159.2 N·m = 159155 N·mm

d_τ = ∛(16·159155/(π·60)) = 23.82 mm

Diameter for stiffness

θ = 1·π/180/1000 = 1.745 × 10⁻⁵ rad/mm

G = 80 GPa = 80000 MPa

d_φ = (32·159155/(π·80000·1.745 × 10⁻⁵))^1/4 = 32.83 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(23.82, 32.83) = 32.83 mm

Answer

d = 32.83 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: τ_xy = 45.83 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.

Required radius

C = (100 + 60)/2 = 80 MPa

R = σ₁ − C = 130 − 80 = 50 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(50² − 20²) = 45.83 MPa

Answer

τ_xy = 45.83 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 62.66 MPa

Wall stresses

r = 600/2 = 300 mm

σ_h = pr/t = 2.5·300/10 = 75 MPa

σ_a = pr/(2t) = 37.5 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 = 37.5·sin²35° + 75·cos²35° = 62.66 MPa

Answer

σ_n = 62.66 MPa

7.Axial stress & strain

Answer: ε = 5101 µε

Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).

Stress

A = π·16²/4 = 201.1 mm²

σ = P/A = 120·1000/201.1 = 596.8 MPa

Strain (Hooke’s law)

E for copper = 117 GPa = 117000 MPa.

ε = σ/E = 596.8/117000 = 0.005101 = 5101 µε

Answer

ε = 5101 µε

8.Axial stress & strain

Answer: σ₂ = 17.54 MPa

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.

Stiffnesses

k₁ = 200000·1257/500 = 502655 N/mm

k₂ = 70000·2827/400 = 494801 N/mm

Share of segment 2

F₂ = 100000·494801/997456 = 49606 N

Stress

σ₂ = 49606/2827 = 17.54 MPa (compression)

Answer

σ₂ = 17.54 MPa

9.Column buckling

Answer: P_cr = 5.888 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·50⁴/64 = 306796 mm⁴

Effective length

KL = 1·6000 = 6000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/6000² = 5888 N = 5.888 kN

Answer

P_cr = 5.888 kN

10.Column buckling

Answer: σ_cr = 49.35 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 = 0.5·6000 = 3000 mm

KL/r = 3000/15 = 200

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/200² = 49.35 MPa

Answer

σ_cr = 49.35 MPa

11.Mechanical vibrations

Answer: ζ = 0.4003

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·√(200000·25) = 4472 N·s/m

Damping ratio

ζ = c/c_c = 1790/4472 = 0.4003

Answer

ζ = 0.4003

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

12.Mechanical vibrations

Answer: f_n = 3.927 Hz

Equivalent stiffness

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

k₁₂ = 20 + 15 = 35 kN/m

k_eq = 35·80/(35 + 80) = 24.35 kN/m

Natural frequency

ω_n = √(24348/40) = 24.67 rad/s

f_n = 24.67/(2π) = 3.927 Hz

Answer

f_n = 3.927 Hz

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