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Damm [24]
4 years ago
12

A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The

allowable stress for the hoops is σallos=11.4 ksi. Determine the maximum spacing s along the pipe so that the pipe can resist an internal gauge pressure of 4 psi.
Engineering
1 answer:
Minchanka [31]4 years ago
5 0

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

S = 4560/144

S = 31.66667 inches

S = 31.7 inches (approximated)

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a)

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2.53*10^{-5}T_{AB}=1.3266*10^{-6}T_{BC}\\T_{BC}=19.0717T_{AB}

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Maximum shear stress in AB

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3.     return(volume)

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