calculation of polar inertia of a half circle crosssection

[guyguy]

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

 

[SlideRuleEra]

Take a look at these on-line calculators:

http://www.engineersedge.com/section_properties_menu.shtml

Best Wishes

 

[Drej]

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 --

 

[EnglishMuffin]

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.

 

[desertfox]

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]

 

[Philrock]

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.

 

[guyguy]

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