A ship's propeller shaft transmits 7.5 MW at 440rev/min.

[TC13]

1. A ship's propeller shaft transmits 7.5 MW at 440rev/min. The shaft has an external diameter of 230mm. Calculate the maximum permissible bore diameter if the shearing stress in the shaft is to be limited to 150MN/m squared. The modulus of rigidity for the shaft material is 79 GN/m squared



2. Homework Equations : P=omega*T therefore T=P/omega
T/J=tau/r therefore J=Tr/tau
J=pi (D power of 4*d power of 4)/32 therefore d power of 4=D power of 4 -J*32/pi



3. Are my transpositions of the equations above correct? I realize if I want d i will have to find the fourth route of my susequent answer. I came up with 162760.42 Nm for T, 1.248*10-4 m4 for J and 0.1977m for d.
I also need to find the legth of the shaft if the angle of twist does not exceed 10 degrees. I've used the equation: tau/r=G*theta/l which I transposed to get l=G*theta*r/tau. My answer came out as 605.666m which to me seems a tad large. Any help would be very much appriciated. Cheers.

 

[Valkarie]

Yes, your transpositions of the equations are correct.For the maximum permissible bore diameter, you have the right answer.For the length of the shaft, your answer seems to be a bit large. To calculate the angle of twist, you need to use the formula: θ = T * L / (G * J). You can then substitute this angle of twist into the original equation to calculate the required length of the shaft.

 

[Mene]

1. Yes, your transpositions of the equations are correct. To find the maximum permissible bore diameter, you can use the equation J=Tr/tau and solve for d. Plugging in the values for T, J, r, and tau, you should get a maximum permissible bore diameter of approximately 0.1977m, or 197.7mm. This means that the bore diameter should not exceed 197.7mm to limit the shearing stress to 150MN/m squared.

2. As for the length of the shaft, your calculation seems to be correct. However, it is always a good idea to double check your calculations and make sure you are using the correct units for all the values. Also, make sure to convert the angle of twist from degrees to radians before plugging it into the equation. It is possible that the length of the shaft is indeed large, as the power being transmitted is quite high at 7.5 MW.

Overall, your calculations and equations seem to be correct. Keep in mind that these are ideal calculations and in reality, there may be other factors that need to be considered in the design and construction of the shaft. It is always important to double check your work and consult with other experts in the field to ensure accuracy.