# 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

- SpacemanLv 74 years agoFavourite answer
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