Problem nº1 from Strenght of Materials by J.P. Den Hartog (1949)

[miloricardo]

Homework Statement

A propeller shaft in the largest and most powerful ships transmits about 50,000 hp. Assume that the propeller transforms this power into a forward push on the ship with an efficiency of 70 per cent and that the ship's speed then is 30 knots (1 knot is 6,080 ft/hr).

The shaft is 300 ft long from the propeller to the thrust bearing in the engine room, where the thrust is transmitted to the ship's hull structure. The diameter of the (solid, circular) shaft is 24 in. Calculate the elastic contraction of the shaft at full power.
(Answer:0.985 in).

Homework Equations

They are attached in my attempt solution.

The Attempt at a Solution

View attachment My fail solution.doc

 

[Valkarie]

Euler's Buckling Formula $$P = \frac{\pi^2 E I}{L^2}$$Where P is the maximum load that can be applied before buckling occurs, E is Young's Modulus, I is the second moment of inertia of the cross section of the shaft and L is the length of the shaft. The elastic contraction formula $$\sigma=\frac{F}{A}$$Where F is the force applied and A is the area of the cross-section of the shaft. The Force needed for buckling $$F = \frac{\pi^2 E I}{L^2}A$$The displacement due to buckling $$\delta = \frac{FL^2}{AEI}$$Given: Diameter of Shaft = 24 in Length of Shaft = 300 ft Efficiency of Propeller = 70%Power transmitted = 50,000 hp Speed of ship = 30 knotsConversion Factors: 1hp = 746 Watts1 knot = 6,080 ft/hrCalculations: Power transmitted (in Watts) = 50,000 x 746 = 37,300,000 WForce applied (in Newtons) = 37,300,000 W x 0.7 (efficiency) = 26,110,000 NArea of Cross-sectional Area of Shaft = $\pi$r2 = 3.14 x 12 x 12 = 452.16 in2 Young's Modulus of Steel = 200 GPa Second Moment of Inertia = $\frac{\pi}{64}$d4 = $\frac{\pi}{64}$ x 24 x 24 x 24 x 24 = 2.47 x 107 mm4 The Force needed for buckling = $\frac{\pi^2 E I}{L^2}A$ = $\frac{(3.14)^2 x 200 x 2.47 x 10^7}{(300)^2 x 452.16}$ = 25,837,719 N The displacement due to