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 4 kW motor running at 2900 rpm drives a pinion with 18 teeth and a module of 4 mm. Find the tangential force F_t on the pinion teeth.

P = 4 kWn = 2900 rpmN₁ = 18m = 4 mm

F_t = N

2.Gears & gear trainsHard

In a compound gear train, gear A (12 teeth) on the input shaft turns at 2880 rpm and drives gear B (18 teeth). Gear C (12 teeth) is fixed to the same shaft as B and drives gear D on the output shaft. The output must turn at 768 rpm. How many teeth N_D must gear D have?

n_in = 2880 rpmA = 12B = 18C = 12n_out = 768 rpm

N_D =

3.Torsion of shaftsMedium

A hollow aluminium shaft (outer diameter 80 mm, inner diameter 60 mm, length 1.5 m) is fixed at one end and twisted by a torque of 250 N·m at the other. Find the angle of twist φ in degrees.

T = 250 N·md_o = 80 mmd_i = 60 mmL = 1.5 mG = 26 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 50 kW at 1450 rpm. Find the maximum shear stress τ_max in the shaft.

P = 50 kWN = 1450 rpmd_o = 50 mmd_i = 38 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point has σₓ = 60 MPa and σᵧ = 90 MPa. How large can the shear stress τ_xy (magnitude) become before the largest principal stress reaches 105 MPa?

σ_x = 60 MPaσ_y = 90 MPaσ₁ = 105 MPa

τ_xy = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 600 mm and wall thickness 5 mm holds gas at a gauge pressure of 0.8 MPa. It is made from plate welded along a helical seam at 50° to the axis of the cylinder. Find the normal stress σ_n acting across the weld seam.

D = 600 mmt = 5 mmp = 0.8 MPaα = 50°

σ_n = MPa

7.Axial stress & strainMedium

A steel rod (diameter 30 mm, E = 200 GPa) sits inside a tube made of copper (outer diameter 120 mm, inner diameter 70 mm, E = 120 GPa). Both are 400 mm long and are compressed together between rigid plates by 200 kN. Find the shortening δ.

P = 200 kNrod: Ø30 mm, E = 200 GPatube: Ø120/Ø70 mm, E = 120 GPaL = 400 mm

δ = mm

8.Axial stress & strainHard

A steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) and a segment made of aluminium (diameter 40 mm, length 200 mm, E = 70 GPa) are joined end to end and fixed between two rigid walls. An axial load of 80 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: aluminium, Ø40 mm, 200 mm, E = 70 GPaP = 80 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 3 mE = 200 GPa (steel)b × h = 30 × 100 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of steel 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 = 2 mE = 200 GPa (steel)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 2 kg mass is supported by a spring with a stiffness of 12 kN/m and a viscous damper with a damping coefficient of 20 N·s/m. Find the damping ratio ζ.

m = 2 kgk = 12 kN/mc = 20 N·s/m

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 50 kN/m. The mass stays level. Ignore damping and the masses of the springs. Find the natural frequency f_n of vertical vibration in Hz.

m = 20 kgk₁ = 40 kN/mk₂ = 50 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: F_t = 365.9 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 = 4000/303.7 = 13.17 N·m

Pitch diameter

d = m·N = 4·18 = 72 mm

Tangential force

F_t = 2T/d = 2·13.17/0.072 = 365.9 N

Answer

F_t = 365.9 N

2.Gears & gear trains

Answer: N_D = 30

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.

Intermediate shaft speed

n_B = n_C = 2880·12/18 = 1920 rpm

Teeth on gear D

N_D = C·n_C/n_out = 12·1920/768 = 30

Answer

N_D = 30

3.Torsion of shafts

Answer: φ = 0.3006 °

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(80⁴ − 60⁴)/32 = 2.749 × 10⁶ mm⁴

Angle of twist

T = 250 N·m = 250000 N·mm

L = 1.5 m = 1500 mm

G = 26 GPa = 26000 MPa

φ = TL/(GJ) = 250000·1500/(26000·2.749 × 10⁶) = 0.005247 rad

φ = 0.005247·180/π = 0.3006°

Answer

φ = 0.3006°

4.Torsion of shafts

Answer: τ_max = 20.13 MPa

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 50000/151.8 = 329.3 N·m

Polar second moment of area

J = π(50⁴ − 38⁴)/32 = 408885 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 329286·25/408885 = 20.13 MPa

Answer

τ_max = 20.13 MPa

5.Mohr’s circle & stress transformation

Answer: τ_xy = 25.98 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 = (60 + 90)/2 = 75 MPa

R = σ₁ − C = 105 − 75 = 30 MPa

Shear stress

τ_xy = √(R² − ((σₓ − σᵧ)/2)²) = √(30² − 15²) = 25.98 MPa

Answer

τ_xy = 25.98 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 33.92 MPa

Wall stresses

r = 600/2 = 300 mm

σ_h = pr/t = 0.8·300/5 = 48 MPa

σ_a = pr/(2t) = 24 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 = 24·sin²50° + 48·cos²50° = 33.92 MPa

Answer

σ_n = 33.92 MPa

7.Axial stress & strain

Answer: δ = 0.07717 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(120² − 70²)/4 = 7461 mm²

Combined axial stiffness

ΣEA = 200·10³·706.9 + 120·10³·7461 = 1.037 × 10⁹ N

Shortening

δ = 200000·400/1.037 × 10⁹ = 0.07717 mm

Answer

δ = 0.07717 mm

8.Axial stress & strain

Answer: σ₂ = 32.61 MPa

Stiffnesses

k₁ = 200000·1257/600 = 418879 N/mm

k₂ = 70000·1257/200 = 439823 N/mm

Share of segment 2

F₂ = 80000·439823/858702 = 40976 N

Stress

σ₂ = 40976/1257 = 32.61 MPa (compression)

Answer

σ₂ = 32.61 MPa

9.Column buckling

Answer: P_cr = 49.35 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 = 100·30³/12 = 225000 mm⁴

Effective length

KL = 1·3000 = 3000 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·225000/3000² = 49348 N = 49.35 kN

Answer

P_cr = 49.35 kN

10.Column buckling

Answer: σ_cr = 100.7 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 = 0.7·2000 = 1400 mm

KL/r = 1400/10 = 140

Critical stress

σ_cr = π²E/(KL/r)² = π²·200000/140² = 100.7 MPa

Answer

σ_cr = 100.7 MPa

11.Mechanical vibrations

Answer: ζ = 0.06455

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·√(12000·2) = 309.8 N·s/m

Damping ratio

ζ = c/c_c = 20/309.8 = 0.06455

Answer

ζ = 0.06455

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

12.Mechanical vibrations

Answer: f_n = 10.68 Hz

Equivalent stiffness

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

k_eq = 40 + 50 = 90 kN/m

Natural frequency

ω_n = √(90000/20) = 67.08 rad/s

f_n = 67.08/(2π) = 10.68 Hz

Answer

f_n = 10.68 Hz

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