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 100 mm pulley at 2880 rpm drives a 250 mm pulley by a belt with 2.5 % slip. Find the speed n₂ of the driven pulley.

n₁ = 2880 rpmd₁ = 100 mmd₂ = 250 mms = 2.5 %

n₂ = rpm

2.Gears & gear trainsHard

An open belt connects an 80 mm driving pulley at 1450 rpm to a 450 mm pulley; the centres are 1000 mm apart. The coefficient of friction is 0.25 and the tight-side tension may not exceed 1000 N. Find the largest power P the belt can transmit without slipping, in kW (ignore centrifugal effects).

d₁ = 80 mmn₁ = 1450 rpmd₂ = 450 mmC = 1000 mmμ = 0.25T₁ = 1000 N

P = kW

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 = 250 N·md = 40 mmL = 1.5 mG = 80 GPa

φ = °

4.Torsion of shaftsHard

A solid shaft must transmit 100 kW at 1450 rpm without the shear stress exceeding 40 MPa. Find the smallest diameter d that works.

P = 100 kWN = 1450 rpmτ_allow = 40 MPa

d = mm

5.Mohr’s circle & stress transformationMedium

A point in a thin plate is in plane stress: σₓ = −60 MPa, σᵧ = 50 MPa and τₓᵧ = −60 MPa, with σ_z = 0. Find the absolute maximum shear stress τ_abs at the point (considering all planes in three dimensions).

σ_x = −60 MPaσ_y = 50 MPaτ_xy = −60 MPaσ_z = 0

τ_abs = MPa

6.Mohr’s circle & stress transformationHard

A point in a thin plate is in plane stress: σₓ = 60 MPa, σᵧ = 70 MPa and τₓᵧ = 50 MPa (σ_z = 0). The material has a yield strength of 350 MPa. Find the factor of safety n against yielding by the Tresca (maximum shear stress) criterion.

σ_x = 60 MPaσ_y = 70 MPaτ_xy = 50 MPaσ_Y = 350 MPa

n =

7.Axial stress & strainMedium

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

P = 95 kNd = 12 mmL = 0.5 mE = 100 GPa

ε = µε

8.Axial stress & strainHard

The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.

P = 65 kNsegment 1: Ø40 mm, 0.5 m, brass (E = 100 GPa)segment 2: Ø20 mm, 2 m, brass (E = 100 GPa)

δ = mm

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 = 1 mE = 70 GPa (aluminium)b × h = 30 × 60 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 = 2.5 mE = 70 GPa (aluminium)d = 60 mm

σ_cr = MPa

11.Mechanical vibrationsMedium

In a free-vibration test of a damped mass–spring system, a displacement peak of 15 mm is followed one cycle later by a peak of 9.1 mm. Find the damping ratio ζ.

x₀ = 15 mmx₁ = 9.1 mmn = 1

ζ =

12.Mechanical vibrationsHard

A 50 kg mass hangs from two springs side by side (in parallel), with stiffnesses k₁ = 40 kN/m and k₂ = 25 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 = 50 kgk₁ = 40 kN/mk₂ = 25 kN/m

f_n = Hz

3Solutions

1.Gears & gear trains

Answer: n₂ = 1123 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

2880·100/250 = 1152 rpm

With slip

n₂ = 1152·0.975 = 1123 rpm

Answer

n₂ = 1123 rpm

2.Gears & gear trains

Answer: P = 3.034 kW

Angle of wrap

θ = 2.769 rad

Slack-side tension

e^μθ = e^0.6924 = 1.998, T₂ = 1000/1.998 = 500.4 N

Power

P = (1000 − 500.4)·6.074 = 3.034 kW

Answer

P = 3.034 kW

3.Torsion of shafts

Answer: φ = 1.069 °

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·40⁴/32 = 251327 mm⁴

Angle of twist

φ = TL/(GJ) = 250·1000·1500/(80000·251327) = 0.01865 rad

φ = 0.01865·180/π = 1.069°

Answer

φ = 1.069°

4.Torsion of shafts

Answer: d = 43.77 mm

Torque

ω = 2πN/60 = 2π·1450/60 = 151.8 rad/s

T = P/ω = 100000/151.8 = 658.6 N·m

Diameter from the allowable stress

d = ∛(16T/(π·τ_allow)) = ∛(16·658572/(π·40)) = 43.77 mm

Answer

d = 43.77 mm

In practice you would round up to the next standard size.

5.Mohr’s circle & stress transformation

Answer: τ_abs = 81.39 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. In three dimensions the absolute maximum shear stress is (σ_max − σ_min)/2 over all three principal stresses; in plane stress the third one is zero.

In-plane principal stresses

C = −5 MPa, R = √((−55)² + (−60)²) = 81.39 MPa

σ₁ = 76.39 MPa, σ₂ = −86.39 MPa, σ_z = 0

Largest of the three circles

τ = max(|σ₁ − σ₂|, |σ₁|, |σ₂|)/2 = max(162.8, 76.39, 86.39)/2 = 81.39 MPa

Answer

τ_abs = 81.39 MPa

Here σ₁ and σ₂ have opposite signs, so the in-plane maximum R is also the absolute maximum.

6.Mohr’s circle & stress transformation

Answer: n = 3.037

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. Yield criteria: von Mises predicts yield when σ_vM = √(σ₁² − σ₁σ₂ + σ₂²) reaches σ_Y; Tresca when the absolute maximum shear stress reaches σ_Y/2.

Principal stresses

C = 65 MPa, R = 50.25 MPa

σ₁ = 115.2 MPa, σ₂ = 14.75 MPa, σ₃ = 0

Tresca

σ_max − σ_min = 115.2 MPa

n = 350/115.2 = 3.037

Answer

n = 3.037

n > 1 means the point does not yield; Tresca is slightly more conservative than von Mises.

7.Axial stress & strain

Answer: ε = 8400 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 = 95·1000/113.1 = 840 MPa

Strain (Hooke’s law)

E for brass = 100 GPa = 100000 MPa.

ε = σ/E = 840/100000 = 0.0084 = 8400 µε

Answer

ε = 8400 µε

8.Axial stress & strain

Answer: δ = 4.397 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·40²/4 = 1257 mm², A₂ = π·20²/4 = 314.2 mm²

σ₁ = 51.73 MPa

σ₂ = 206.9 MPa

Add the segment elongations

δ₁ = σ₁L₁/E₁ = 51.73·500/100000 = 0.2586 mm

δ₂ = σ₂L₂/E₂ = 206.9·2000/100000 = 4.138 mm

δ = δ₁ + δ₂ = 4.397 mm

Answer

δ = 4.397 mm

9.Column buckling

Answer: P_cr = 23.32 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 = 60·30³/12 = 135000 mm⁴

Effective length

KL = 2·1000 = 2000 mm

Critical load

P_cr = π²EI/(KL)² = π²·70000·135000/2000² = 23317 N = 23.32 kN

Answer

P_cr = 23.32 kN

10.Column buckling

Answer: σ_cr = 6.218 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 = 2·2500 = 5000 mm

KL/r = 5000/15 = 333.3

Critical stress

σ_cr = π²E/(KL/r)² = π²·70000/333.3² = 6.218 MPa

Answer

σ_cr = 6.218 MPa

11.Mechanical vibrations

Answer: ζ = 0.07929

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(15/9.1) = 0.4998

Damping ratio

ζ = δ/√(4π² + δ²) = 0.4998/√(39.48 + 0.2498) = 0.07929

Answer

ζ = 0.07929

12.Mechanical vibrations

Answer: f_n = 5.738 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 + 25 = 65 kN/m

Natural frequency

ω_n = √(65000/50) = 36.06 rad/s

f_n = 36.06/(2π) = 5.738 Hz

Answer

f_n = 5.738 Hz

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