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 200 mm pulley at 720 rpm drives a 315 mm pulley by a belt with 4 % slip. Find the speed n₂ of the driven pulley.

n₁ = 720 rpmd₁ = 200 mmd₂ = 315 mms = 4 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects a 150 mm driving pulley at 960 rpm to a 250 mm pulley; the centres are 500 mm apart. The coefficient of friction is 0.3 and the tight-side tension may not exceed 1500 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 150 mmn₁ = 960 rpmd₂ = 250 mmC = 500 mmμ = 0.3T₁ = 1500 N

P = kW

3.Torsion of shaftsMedium

A hollow brass shaft (outer diameter 50 mm, inner diameter 35 mm, length 1.5 m) is fixed at one end and twisted by a torque of 100 N·m at the other. Find the angle of twist φ in degrees.

T = 100 N·md_o = 50 mmd_i = 35 mmL = 1.5 mG = 39 GPa

φ = °

4.Torsion of shaftsHard

A solid brass shaft transmits 50 kW at 1500 rpm. The shear stress must not exceed 80 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 = 1500 rpmτ_allow = 80 MPaθ_allow = 0.75°/mG = 39 GPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a loaded part has σₓ = 100 MPa, σᵧ = 30 MPa and τₓᵧ = 30 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 = 100 MPaσ_y = 30 MPaτ_xy = 30 MPa

θ_p = °

6.Mohr’s circle & stress transformationHard

A point in a loaded part has σₓ = 150 MPa, σᵧ = −70 MPa and τₓᵧ = −20 MPa (tension positive). Find the normal stress σ_θ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.

σ_x = 150 MPaσ_y = −70 MPaτ_xy = −20 MPaθ = 15°

σ_θ = MPa

7.Axial stress & strainMedium

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

P = 25 kNd = 12 mmL = 1.5 mE = 100 GPa

ε = µε

8.Axial stress & strainHard

A steel segment (diameter 25 mm, length 400 mm, E = 200 GPa) and a segment made of brass (diameter 50 mm, length 300 mm, E = 100 GPa) are joined end to end and fixed between two rigid walls. An axial load of 20 kN acts at the joint, towards segment 2. Find the magnitude of the stress σ₂ in segment 2.

segment 1: steel, Ø25 mm, 400 mm, E = 200 GPasegment 2: brass, Ø50 mm, 300 mm, E = 100 GPaP = 20 kN

σ₂ = MPa

9.Column bucklingMedium

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

L = 4 mE = 200 GPa (steel)b × h = 30 × 80 mm

P_cr = kN

10.Column bucklingHard

A column of length L made of aluminium is pinned at both ends. Find its slenderness and the critical buckling stress σ_cr from Euler’s formula.

L = 1 mE = 70 GPa (aluminium)d = 40 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

A 5 kg mass is supported by a spring with a stiffness of 80 kN/m and a viscous damper with a damping coefficient of 130 N·s/m. Find the damping ratio ζ.

m = 5 kgk = 80 kN/mc = 130 N·s/m

ζ =

12.Mechanical vibrationsHard

A 20 kg mass hangs from two springs connected end to end (in series): the upper spring has k₁ = 40 kN/m and the lower one k₂ = 15 kN/m. Find the equivalent stiffness k_eq of the spring system in kN/m.

m = 20 kgk₁ = 40 kN/mk₂ = 15 kN/m

k_eq = kN/m

3Solutions

1.Gears & gear trains

Answer: n₂ = 438.9 rpm

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.

Speed without slip

720·200/315 = 457.1 rpm

With slip

n₂ = 457.1·0.96 = 438.9 rpm

Answer

n₂ = 438.9 rpm

2.Gears & gear trains

Answer: P = 6.63 kW

Angle of wrap

θ = 2.941 rad

Slack-side tension

e^μθ = e^0.8824 = 2.417, T₂ = 1500/2.417 = 620.7 N

Power

P = (1500 − 620.7)·7.54 = 6.63 kW

Answer

P = 6.63 kW

3.Torsion of shafts

Answer: φ = 0.4726 °

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(50⁴ − 35⁴)/32 = 466269 mm⁴

Angle of twist

T = 100 N·m = 100000 N·mm

L = 1.5 m = 1500 mm

G = 39 GPa = 39000 MPa

φ = TL/(GJ) = 100000·1500/(39000·466269) = 0.008249 rad

φ = 0.008249·180/π = 0.4726°

Answer

φ = 0.4726°

4.Torsion of shafts

Answer: d = 50.2 mm

Torque

ω = 2πN/60 = 2π·1500/60 = 157.1 rad/s

T = P/ω = 50000/157.1 = 318.3 N·m

Diameter for strength

T = 318.3 N·m = 318310 N·mm

d_τ = ∛(16·318310/(π·80)) = 27.26 mm

Diameter for stiffness

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

G = 39 GPa = 39000 MPa

d_φ = (32·318310/(π·39000·1.309 × 10⁻⁵))^1/4 = 50.2 mm

Governing diameter

Both limits must hold, so the larger diameter governs.

d = max(27.26, 50.2) = 50.2 mm

Answer

d = 50.2 mm

Here stiffness governs.

5.Mohr’s circle & stress transformation

Answer: θ_p = 20.3 °

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(2·30, 100 − 30) = 40.6°

Angle on the element

θ_p = 20.3°

Answer

θ_p = 20.3°

Positive angles are counter-clockwise.

6.Mohr’s circle & stress transformation

Answer: σ_θ = 125.3 MPa

Transformation equation

σ_θ = 40 + 110·0.866 + −20·0.5 = 125.3 MPa

Answer

σ_θ = 125.3 MPa

7.Axial stress & strain

Answer: ε = 2210 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·12²/4 = 113.1 mm²

σ = P/A = 25·1000/113.1 = 221 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 221/100000 = 0.00221 = 2210 µε

Answer

ε = 2210 µε

8.Axial stress & strain

Answer: σ₂ = 7.408 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 = 200000·490.9/400 = 245437 N/mm

k₂ = 100000·1963/300 = 654498 N/mm

Share of segment 2

F₂ = 20000·654498/899935 = 14545 N

Stress

σ₂ = 14545/1963 = 7.408 MPa (compression)

Answer

σ₂ = 7.408 MPa

9.Column buckling

Answer: P_cr = 45.32 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 = 80·30³/12 = 180000 mm⁴

Effective length

KL = 0.7·4000 = 2800 mm

Critical load

P_cr = π²EI/(KL)² = π²·200000·180000/2800² = 45320 N = 45.32 kN

Answer

P_cr = 45.32 kN

10.Column buckling

Answer: σ_cr = 69.09 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 = 1·1000 = 1000 mm

KL/r = 1000/10 = 100

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/100² = 69.09 MPa

Answer

σ_cr = 69.09 MPa

11.Mechanical vibrations

Answer: ζ = 0.1028

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·√(80000·5) = 1265 N·s/m

Damping ratio

ζ = c/c_c = 130/1265 = 0.1028

Answer

ζ = 0.1028

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

12.Mechanical vibrations

Answer: k_eq = 10.91 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 = 1/40 + 1/15 → k_eq = 40·15/(40 + 15) = 10.91 kN/m

Answer

k_eq = 10.91 kN/m

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