# Engineering Science: Torsion (Hollow Shaft replaced by Solid Shaft)

## Homework Statement

Question

Assuming the torque and maximum shear stress values to be the same for both shafts, determine the size of a suitable soild shaft which could be used instead of the hollow shaft.

Information/Data Known

Firstly, a previous question - A ship's propellar shaft transmits 7.5MW at 440rev/min. The shaft has an external diameter of 230mm. Calculate the maximum permissable bore diameter if the shearing stress in the shaft is limeted to 150MN/m^2. The modulus of rigidity for the shaft material is 79GN/m^2

Ok, the following is known:

Bore Diameter = 108mm = 0.108m
DH = Hollow shaft External Diameter = 0.23m
touH = touS
TH=TS

## Homework Equations

T/J = tou/r = Gθ/l

We're using T/J = tou/r

J=pi(D^4-d^4)/32 [Hollow Shaft]

J=pi(D^4)/32 [Soid Shaft]

## The Attempt at a Solution

Ok so to show understanding, the question says that the torque and stress values are the same for both shafts therefore:

touH = touS
TH=TS

Which means we need to find JS, rS, JH and rH

JS/rS = JH/rH

We can find our H values but not our solid so:

JH = pi(DH^4 - dH^4)/32

DH = 0.23m
dH = 0.108m

JH = 2.6138x10^-4m^4

rH = D/2 = 0.115m

JS = pi/32 [because we can only assume DS atm is equal to 1]

JS = 0.0982m^4

rS = D/2 = 0.5Dm

Back to the equation: JS/rS = JH/rH

Muliply over:

JH(rS)/JS(rH)

Which looks like so:

(2.6138x10^4)x0.5 / 0.0982 x 0.115 = d^3 = 11.5727x10^-3

to get the diamter, we cube route the answer:

(cube root)11.5727x10^-3 = D

D = 0.226m

Therefore the diameter of the solid shaft use to replace the hollow shaft is: 226mm

What i'm asking for here is clarification that what i've just done here is correct. I've been told by my collegues that i may have gotten the bore diamter wrong, someone said they got 198mm instead of 108mm.

So from the second question i have typed out p above, could you find the time to check if i have the correct or incorrect bore diamter and the second request is simply to read the above (my workings out) to see if it looks / is valid.

Thankyou for your time, i really do appreciate your help even if it's a simple word.

nvn
Homework Helper
MathsRetard09: Your dH value is currently wrong. Try again. Your colleague's dH is rounded too much. You need dH to at least four (preferably five) significant digits.

1. By the way, always leave a space between a numeric value and its following unit symbol. E.g., 7.5 MW, not 7.5MW. See the international standard for writing units (ISO 31-0).

2. MN/m^2 is called MPa. GN/m^2 is called GPa. Always use the correct, special name for a unit. E.g., 150 MPa, not 150 MN/m^2. See the above links in item 1.

3. Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits, unless the first significant digit of the final answer is 1, in which case round the final answer to four significant digits.

nvn: Thankyou for your reply. If my dH value is wrong then, as i was regreting, my previous calculations were wrong.

Thanks for this, i'll re-do it all again and see what i get.

P.S. just to clarify - do i have the correct method?

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Okay i've got everything planned out, i know where i've gone wrong, and its at the the start, my d value was wrong which meant it affected everything else.

All i finally ask for is help with transposing the following equation to make (d) the subject:

J = pi(D^4 - d^4) / 32 // d is to be the subject

My attempt:

J = pi(D^4 - d^4) / 32

J/pi = D^4 - d^4 / 32

J+d^4/pi = D^4 / 32

d^4/pi = D^4 / (J*32)

D^4 = (D^4)*pi / (J*32)

But this gives me d^4, not d, so how do i get (d) from here??

I tried on my calculator with the values in, when i get d^4 i tried 4(root)ANS on the calculator which gave 2.2026.

Values known are:

D = 0.23m
J = 1.2473x10^-4

THanks again for your time, this is the only bit of help i need, once i have (d) i can finish of the assessment.

nvn
Homework Helper
MathsRetard09: Your algebra is grossly incorrect. You will probably want to take an introductory algebra course.

J = (pi/32)(D^4 - d^4)

32*J/pi = D^4 - d^4

d^4 = D^4 - (32*J/pi)

(d^4)^0.25 = [D^4 - (32*J/pi)]^0.25

d^(4*0.25) = [D^4 - (32*J/pi)]^0.25

d^1 = [D^4 - (32*J/pi)]^0.25

d = [D^4 - (32*J/pi)]^0.25

I do not obtain JH = 1.2473e-4; I get 1.2479e-4 m^4, because I did not round intermediate values. Also, see item 1 in post 2. You should write 0.23 m, not 0.23m.

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MathsRetard09: Your algebra is grossly incorrect. You will probably want to take an introductory algebra course.

J = (pi/32)(D^4 - d^4)

32*J/pi = D^4 - d^4

d^4 = D^4 - (32*J/pi)

(d^4)^0.25 = [D^4 - (32*J/pi)]^0.25

d^(4*0.25) = [D^4 - (32*J/pi)]^0.25

d^1 = [D^4 - (32*J/pi)]^0.25

d = [D^4 - (32*J/pi)]^0.25

I do not obtain JH = 1.2473e-4; I get 1.2479e-4 m^4, because I did not round intermediate values. Also, see item 1 in post 2. You should write 0.23 m, not 0.23m.

Again thankyou, I do appolagise i did read your advice previously about the units being spaced from the digits, i guess i was just rushing with my typing which is what i normally do.

I did manage to use some common sense since my previous post and yesterday did find a method very similar to what you've posted.

All of this is now sorted, I have a peaceful mind now, thankyou for your help and after finding so many posts by you on other topics, thankyou on behalf of this community for your efforts to give advice and help to complete strangers on the internet, it is appreciated very much.