00:04
For 8 .1a, the torque equation for a solid shaft is pi over 16 times the torque, times, well, not the torque, but the sheer stress limit, times the diameter cube.
00:29
And so with substitutions, we get pi over 16 times 60.
00:41
N over m squared would actually be 60 times 10 to the 6th would be our substitution and our diameter being 50 millimeters would be 50 times 10 to the negative third and then that's going to be cubed and this comes out to about 1 .474 times 10 to the 6, or call it kilonutons.
01:22
The power equation is 2 pi nt over 60.
01:29
And so that would be 2 pi times 250 times our 1 .474 kilonitons over 60.
01:39
And that calculates to 38 .6 kilowatts.
01:46
In the second part, if the external diameter is 50 millimeters and the shear stress limit is 75, then we get pi over 16 times 75 times 10 to the 6 times now is the difference between our diameters over the original, and that would be 50 to the fourth minus our other diameter to the 4th over 50 and we have t sold from earlier as 1 .474 times 10 to the 6 and we could divide by pi over 16 times 70 75 times 10 to the 6th and then we can multiply by 50 and then divide by 50 and then divide by 50 to the fourth or excuse me subtract 50 to the fourth and then take the negative so essentially d i to the fourth would be 50 to the fourth minus the left what i have currently on the left hand side and then we would take the fourth root and that comes out to about 33 .4 millimeters for our 8 .2, the power equation with substitutions would be two, well, actually, we would substitute the power would be 750 times 10 to the third equals 2 pi times 400.
04:34
And we're going to solve for the mean.
04:39
So you'd multiply by 60 and then divide by 2 pi times 400.
04:48
And that'll give you about 2 ,860 .6.
04:55
So the max would be 1 .2 times that since it's 20 % more.
05:05
1 .2 multiplying will give us 20 % more.
05:10
And that's 3 ,432 .7 newtons.
05:18
And then given that the d .i.
05:24
Is 0 .8d .0...