Materials Question?

A solid circular cylinder is needed as a strut. What is the most lightweight solution required to support a 40 kN load with a factor of safety on yield stress of 2 and a strut length of 10 cm?

1. 6061-T6 Aluminum - 40 ksi yield, 0.0975 lbm/in^3

2. 5-5PH H1025 Stainless Steel - 160 ksi yield, 0.282 lbm/in^3

3. Titanium Ti-6Al-4V - 128 ksi, 0.16 lbm/in^3

4. Inconel 625 Annealed - 70 ksi, 0.305 lbm/in^3

Please show work!

2 Answers

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  • 4 years ago
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    6061-T6 Aluminum

    σ = yield stress = 40 ksi = 276 Mpa = 2.76E+08 Pa

    ρ = mass density = 0.0975 lbm/in³ = 2,699 kg/m³

    5-5PH H1025 Stainless Steel

    σ = yield stress = 160 ksi = 1103 Mpa 1.10E+09 Pa

    ρ = mass density = 0.282 lbm/in³ = 7,806 kg/m³

    Titanium Ti-6Al-4V

    σ = yield stress = 128 ksi = 883 Mpa 8.83E+08 Pa

    ρ = mass density = 0.16 lbm/in³ = 4,429 kg/m³

    Inconel 625 Annealed

    σ = yield stress = 70 ksi = 483 Mpa 4.83E+08 Pa

    ρ = mass density = 0.305 lbm/in³ = 8,442 kg/m³

    F = required load = 40 kN = 4.00E+04 N

    FoS = factor of safety = 2

    σd = design stress = σ/FoS (see below)

    6061-T6 Aluminum design stress = 1.38E+08 Pa

    5-5PH H1025 Stainless Steel design stress = 5.52E+08 Pa

    Titanium Ti-6Al-4V design stress = 4.41E+08 Pa

    Inconel 625 Annealed design stress = 2.41E+08 Pa

    Design Stress

    A = cylinder cross-sectional area

    D = cylinger diameter

    σd = F / A

    σd = F / πD²/4

    σd = 4F/πD²

    Required Cylinder Diameter

    σd = 4F/πD²

    σd(πD²) = 4F

    D² = 4F/σdπ

    D = √(4F/σdπ)

    l = cylinder length = 10 cm = 0.1 m

    V = cylinder volume = A∙l = to be determined

    m = cylinder mass = ρ∙V = to be determined

    g = gravitational acceleration = 9.81 m/s²

    W = weight = m∙g = to be determined

    6061-T6 Aluminum

    D = √(4F/σdπ)

    D = √[4(4.00E+04 N)/(1.38E+08 Pa)(3.14)]

    D = 0.019218101 m

    A = cross-sectional area = πD²/4 = 0.00029 m²

    V = cylinder volume = A∙l = 0.000029 m³

    m = cylinder mass = ρ∙V = 0.08 kg

    g = gravitational acceleration = 9.81 m/s²

    W = weight = m∙g = 0.77 N = 0.17 lb

    5-5PH H1025 Stainless Steel

    D = √(4F/σdπ)

    D = √[4(4.00E+04 N)/(5.52E+08 Pa)(3.14)]

    D = 0.009609051 m

    A = cross-sectional area = πD²/4 = 0.000073 m²

    V = cylinder volume = A∙l = 0.0000073 m³

    m = cylinder mass = ρ∙V = 0.06 kg

    g = gravitational acceleration = 9.81 m/s²

    W = weight = m∙g = 0.56 N = 0.12 lb

    Titanium Ti-6Al-4V

    D = √(4F/σdπ)

    D = √[4(4.00E+04 N)/(4.41E+08 Pa)(3.14)]

    D = 0.010743245 m =

    A = cross-sectional area = πD²/4 = 0.000091 m²

    V = cylinder volume = A∙l = 0.0000091 m³

    m = cylinder mass = ρ∙V = 0.04 kg

    g = gravitational acceleration = 9.81 m/s²

    W = weight = m∙g = 0.39 N = 0.09 lb

    Inconel 625 Annealed

    D = √(4F/σdπ)

    D = √[4(4.00E+04 N)/(2.41E+08 Pa)(3.14)]

    D = 0.014527519 m =

    A = cross-sectional area = πD²/4 = 0.000166 m²

    V = cylinder volume = A∙l = 0.0000166 m³

    m = cylinder mass = ρ∙V = 0.14 kg

    g = gravitational acceleration = 9.81 m/s²

    W = weight = m∙g = 1.37 N = 0.31 lb

    The Titanium Ti-6Al-4V is the most lightweight solution.

  • 4 years ago

    stress = load / cross section area

    cross section area required = load/ working stress

    load is 40 kN you need to convert to lbs

    working stress is half of yield stress (FS= 2) for each material

    weight is cross sectional area required for each x length x density

    this sounds way harder than it is; easy plug and chug; spreadsheet can be very helpful with these

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