×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Log In

Come Join Us!

Are you an
Engineering professional?
Join Eng-Tips Forums!
  • Talk With Other Members
  • Be Notified Of Responses
    To Your Posts
  • Keyword Search
  • One-Click Access To Your
    Favorite Forums
  • Automated Signatures
    On Your Posts
  • Best Of All, It's Free!
  • Students Click Here

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

Students Click Here

Jobs

calculation of polar inertia of a half circle crosssection

calculation of polar inertia of a half circle crosssection

calculation of polar inertia of a half circle crosssection

(OP)
sorry i am a bit rusty here. i need to know the J - polar inertia moment of a half circular body with thickness 't' around an axis that crosses at the half circle center (the circle's center.

ta
guy

RE: calculation of polar inertia of a half circle crosssection

Sounds like it's an open section in torsion. Yuk! Polar moments on open sections! For a full circular (closed) cross section it's 2*t*PI*r^3 (t thickness, r radius PI=3.1415927...). If it's open, goodness only knows what it would be - try Blodgett's "Design of Welded Structures", which has an entire section devoted to polar moments on open sections. Blodgett treats these things quite conservatively by assuming that the polar contribution to the stiffness is the sum of the individual cross sectional elements. If this assumption is carried forward for your open section here, then you may be able to use (where J is the polar moment of area):

J = t^3*2r (units L^4)

or

J = t*PI*r^3 (units L^4)

depending on how conservative you want to be.

Cheers,

-- drej --

RE: calculation of polar inertia of a half circle crosssection

guyguy: Since you have specified the thickness, presumably what you are looking for is simply the polar moment of inertia, rather than the polar second moment of area. In which case it's just half what it would be for a completely circular disk. In other words, it's M*r^2/2, where M is the mass of the half disk, given by pi*r^2*t*ro/2, where ro is the mass density and r is the radius. If you are looking for something else, such as what Drej seems to be talking about, please clarify.

RE: calculation of polar inertia of a half circle crosssection

Hi guyguy

According to Roarks 5th edition stress and strain the formula for an open circular section in torsion is given as:-

      J= (2/3)*pi*r*t^3


      where r = mean radius
            t = thickness

regards desertfox [2thumbs up]

RE: calculation of polar inertia of a half circle crosssection

What ever the formula is, traditional torsional deflection calculations will be valid only if cross sections that were planar in the unloaded state remain planar in the loaded state.  Round shafts and pipes inherently behave this way.  Square shafts, square tubes, I-beams, channels, and your half-pipe do not behave this way, unless the ends are constrained planar.  This can be done by welding a heavy plate to each end, for example.

RE: calculation of polar inertia of a half circle crosssection

(OP)
thank you all!
Philrock: i am welding a fitting with a half pipe shape to a round pipe. so i guess we are still in the traditional approach. Thanks.

Guy

Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Reply To This Thread

Posting in the Eng-Tips forums is a member-only feature.

Click Here to join Eng-Tips and talk with other members!


Resources