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 worm with 1 start(s) turns at 1450 rpm and drives a worm wheel with 40 teeth. Find the speed n_w of the wheel.

z = 1N_w = 40n = 1450 rpm

n_w = rpm

2.Gears & gear trainsHard

In a compound gear train, gear A (16 teeth) on the input shaft turning at 1500 rpm drives gear B (60 teeth). Gear C (18 teeth) is fixed to the same shaft as B and drives gear D (72 teeth) on the output shaft. Find the output speed n_out.

n_in = 1500 rpmA = 16B = 60C = 18D = 72

n_out = rpm

3.Torsion of shaftsMedium

A solid steel 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 = 30 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A hollow shaft transmits 75 kW at 1500 rpm. Find the maximum shear stress τ_max in the shaft.

P = 75 kWN = 1500 rpmd_o = 60 mmd_i = 45 mm

τ_max = MPa

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 150 MPa, σᵧ = 0 MPa and τₓᵧ = 10 MPa (tension positive). Find the maximum in-plane shear stress τ_max.

σ_x = 150 MPaσ_y = 0 MPaτ_xy = 10 MPa

τ_max = MPa

6.Mohr’s circle & stress transformationHard

A thin-walled cylindrical tank of inner diameter 800 mm and wall thickness 5 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 = 800 mmt = 5 mmp = 2.5 MPaα = 35°

σ_n = MPa

7.Axial stress & strainMedium

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

P = 90 kNd = 20 mmL = 3 mE = 200 GPa

ε = µε

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 = 30 kNsegment 1: Ø35 mm, 1.5 m, brass (E = 100 GPa)segment 2: Ø16 mm, 2 m, brass (E = 100 GPa)

σ_max = MPa

9.Column bucklingMedium

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

L = 5 mE = 70 GPa (aluminium)d = 50 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is fixed at the base and free at the top. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1 mE = 70 GPa (aluminium)d = 30 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 150 kg mass is supported by a spring with a stiffness of 150 kN/m and a viscous damper with a damping coefficient of 760 N·s/m. Find the damping ratio ζ.

m = 150 kgk = 150 kN/mc = 760 N·s/m

ζ =

12.Mechanical vibrationsHard

A 100 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 30 kN/m and the lower one k₂ = 25 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 100 kgk₁ = 30 kN/mk₂ = 25 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n_w = 36.25 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.

Wheel speed

n_w = n·z/N_w = 1450·1/40 = 36.25 rpm

Answer

n_w = 36.25 rpm

2.Gears & gear trains

Answer: n_out = 100 rpm

First stage

n_B = 1500·16/60 = 400 rpm = n_C

Second stage

n_out = 400·18/72 = 100 rpm

Answer

n_out = 100 rpm

total ratio = 15 : 1

3.Torsion of shafts

Answer: φ = 5.404 °

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·30⁴/32 = 79522 mm⁴

Angle of twist

φ = TL/(GJ) = 400·1000·1500/(80000·79522) = 0.09431 rad

φ = 0.09431·180/π = 5.404°

Answer

φ = 5.404°

4.Torsion of shafts

Answer: τ_max = 16.47 MPa

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 75000/157.1 = 477.5 N·m

Polar second moment of area

J = π(60⁴ − 45⁴)/32 = 869767 mm⁴

Maximum shear stress

τ_max = T·(d_o/2)/J = 477465·30/869767 = 16.47 MPa

Answer

τ_max = 16.47 MPa

5.Mohr’s circle & stress transformation

Answer: τ_max = 75.66 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.

Radius of Mohr’s circle

τ_max = R = √(((150 − 0)/2)² + 10²) = 75.66 MPa

Answer

τ_max = 75.66 MPa

6.Mohr’s circle & stress transformation

Answer: σ_n = 167.1 MPa

Wall stresses

r = 800/2 = 400 mm

σ_h = pr/t = 2.5·400/5 = 200 MPa

σ_a = pr/(2t) = 100 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 = 100·sin²35° + 200·cos²35° = 167.1 MPa

Answer

σ_n = 167.1 MPa

7.Axial stress & strain

Answer: ε = 1432 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·20²/4 = 314.2 mm²

σ = P/A = 90·1000/314.2 = 286.5 MPa

Strain (Hooke’s law)

E for steel = 200 GPa = 200000 MPa.

ε = σ/E = 286.5/200000 = 0.001432 = 1432 µε

Answer

ε = 1432 µε

8.Axial stress & strain

Answer: σ_max = 149.2 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·35²/4 = 962.1 mm², A₂ = π·16²/4 = 201.1 mm²

σ₁ = 31.18 MPa

σ₂ = 149.2 MPa

Largest stress

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

σ_max = 149.2 MPa

Answer

σ_max = 149.2 MPa

9.Column buckling

Answer: P_cr = 2.12 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 = 2·5000 = 10000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·306796/10000² = 2120 N = 2.12 kN

Answer

P_cr = 2.12 kN

10.Column buckling

Answer: σ_cr = 9.715 MPa

Smallest second moment of area

A round section has the same I about every axis.

I = πd⁴/64 = π·30⁴/64 = 39761 mm⁴

Radius of gyration and slenderness

A = 706.9 mm²

r = √(I/A) = √(39761/706.9) = 7.5 mm

KL = 2·1000 = 2000 mm

KL/r = 2000/7.5 = 266.7

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/266.7² = 9.715 MPa

Answer

σ_cr = 9.715 MPa

11.Mechanical vibrations

Answer: ζ = 0.08011

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·√(150000·150) = 9487 N·s/m

Damping ratio

ζ = c/c_c = 760/9487 = 0.08011

Answer

ζ = 0.08011

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

12.Mechanical vibrations

Answer: k_eq = 13.64 kN/m

Equivalent stiffness

Springs in series share the same force and add their deflections, so they combine as reciprocals: 1/k = 1/k₁ + 1/k₂.

1/k_eq = 1/30 + 1/25 → k_eq = 30·25/(30 + 25) = 13.64 kN/m

Answer

k_eq = 13.64 kN/m

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