1. Hi, Would appreciate if someone can enlighten me.

"A hollow steel shaft has an internal diameter to external diameter ratio of 0.6".

so how do we determine the diameter of both the internal and external diameter?

thanks and regards,
Daijiro  2. Originally Posted by Daijiro Hi, Would appreciate if someone can enlighten me.

"A hollow steel shaft has an internal diameter to external diameter ratio of 0.6".

so how do we determine the diameter of both the internal and external diameter?

thanks and regards,
Daijiro
Draw a cross-section of the shaft. Do you notice two circles? I think you should be able to figure out the "inner" and "outer" part from there.   3. Originally Posted by Daijiro Hi, Would appreciate if someone can enlighten me.

"A hollow steel shaft has an internal diameter to external diameter ratio of 0.6".

so how do we determine the diameter of both the internal and external diameter?

thanks and regards,
Daijiro
By finding a specification of one of the diameters and using the specified ratio to calculate the other diameter.  4. Originally Posted by tk421 Draw a cross-section of the shaft. Do you notice two circles? I think you should be able to figure out the "inner" and "outer" part from there. Hi tk421, thanks for the reply.

i understand how the hollow shaft looks like, but its only i need to know how to get the diameter as the question only indicate the internal diameter to external diameter ratio of 0.6.

Here's the full question;

A hollow steel shaft has an internal diameter (d) to external diameter (D) ratio of 0.6. It is required to transmit 5MW when rotating at a constant speed of 300rpm. If the shear stress is not to exceed 55MN/m2,

a) Show that the polar second of area J = 0.0855 D4
and hence calculate the value of D.

b) calculate the minimum shear stress in the shaft.  5. okay, lets narrow down to this;

2nd polar moment of area equals to J = pie(D4-d4)/32

any expert knows how to resolve it till J = 0.0855 D4 ?  6. Originally Posted by Daijiro okay, lets narrow down to this;

2nd polar moment of area equals to J = pie(D4-d4)/32
Since this is a homework problem, I will give you only minimum hints. I'll also suggest that you double-check your maths.

That said. you are given the ratio of d to D. You have your equation of J in terms of D and d. Now you have two equations, two unknowns (not counting J).

Where are you stuck?  7. Originally Posted by pikpobedy First you wrote it incorrectly.
0.6 cannot be an OD to ID ratio because 0.6 less than 1 and an OD is necessarily larger than an ID.
However, 0.6 can be a ID to OD ratio.
Which is what he said it was.

And now that he has finally figured out that he needs to divulge additional key information, it is possible to solve directly for the diameters separately.  Posting Permissions
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