Explore IDNLearn.com's extensive Q&A database and find the answers you need. Our experts are available to provide accurate, comprehensive answers to help you make informed decisions about any topic or issue you encounter.

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2490 hp and causes the shaft to

Sagot :

The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is [tex]$\frac{3}{8}$[/tex]  in.

A) Determine the maximum shear stress developed in the shaft.

[tex]$\tau_{max}$[/tex] = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

[tex]$ \phi $[/tex] = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              [tex]$ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$[/tex]

                              [tex]$= 56.67 \pi \text{ rad/s}$[/tex]

Power [tex]$= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$[/tex]

          = 1424500 ft. lb/s

Torque, [tex]$T = \frac{P}{\omega}$[/tex]

                 [tex]$=\frac{1424500}{56.67 \pi}$[/tex]

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

[tex]$\tau_{max} = \frac{T_c}{J}$[/tex]

        [tex]$=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$[/tex]

      = 2.93 ksi

B). Angle of twist :

     [tex]$\phi = \frac{TL}{JG}$[/tex]

         [tex]$=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$[/tex]

         = 0.08002 rad

         = 4.58°