# Power transfer of a shaft

## Homework Statement

Calculate the power that can be safely transferred by a medium carbon steel shaft of 100 mm diameter at 150 rpm ?

## Homework Equations

Assumed solid circular shaft.

## The Attempt at a Solution

Material : Medium Carbon steel t=580 MPa y= 50 mm k=6 ( table 9 live load varying)
T= Torque

J= polar moment of inertia for a solid circular section = pid^4/32

J=3.14*1 000 000 00 / 32 = 9 812 500 mm^3

tJ/ky = T = 580 * 9 812 500 / 300/1 000 = 18 970.8333 N m

150 rpm / 60 = 2.5 * 6.28 = 15.7 rad/s

18 970.8333 * 15.7 = Power = 298 kW / 745.7 * 1000 = 400 HP

nvn
Homework Helper
Excellent work, rad10k. Your answer appears correct. It would be better to leave your answer in standard units (kW), instead of converting to an ambiguous, incoherent imperial unit (hp). There are six ambiguous definitions for the name hp. Just stick with standard units (SI). One minor comment. Your units for J should be mm^4, not mm^3.

thanks nvn. I will remove the HP from my final answer . )

this has come as incorrect ?? any inclination as to why ? thanks

nvn