Plastic Modulus of shear area. 1

[damnaged]

Hi,
Is there anybody who would be kind to explain how to calculate the plastic modulus of shear area for a pipe @ circular hollow section?

There's only the formula Sv=tD^2/4 but its for rolled section with equal flanges.

Please help. Thanks.

 

[ishvaaag]

The plastic modulus of asection is a value that multiplied by Fy gives the plastic Moment capacity (Mp) of the SECTION. It is a geometrical property. The value you quote is for rectangular sections, not double tee. See p. 68 and following of the following reference

http://books.google.es/books?id=cKs...resnum=1&ved=0CAoQ6AEwAA#v=onepage&q=&f=false

where it is seen that for a relatively thin wall pipe the plastic modulus of section is 1.27 times the elastic modulus of section.

It is a geometrical property. Mp respect a line parallel to some line is a line parallel line that divides the section in two equal areas, all loaded at Fy, so the way of getting Mp is dividing the area in two equal parts and making one work at Fy in compression and the other at Fy in tension. Multiplying such C (or T) by the arm between centers of gravity of the respective (halved) sections gives Mp. Divide by Fy and you get your plastic section modulus.

So since Fy is eliminated on the division of Mp the plastic modulus is just a property of the section.

For a rectangular section

Mp=C·arm=Fy·Acomp·arm=Fy·(b·h/2)·(h/2)=Fy·Z=Fy·b·h^2/4

Z=b·h^2/4 the plastic modulus.

Since the elastic modulus of section is b·h^2/6 you can extoll 1.5 times more strength of the structure if allowed (and can) go to a plastic section behaviour. For a pipe section, that would be 1.27 times.

That is the plastic modulus of a section. Then the Shear Modulus of a Material is entirely another different thing, it is a material property in theory of elasticity akin to what the Young modulus represents for axial deformation, but for shear action... Is someone misleading you? If you can precise your question we may also write a little about the shear modulus.

 

[BAretired]

For a hollow circle,

Z = (d³ - d_i³)/6

where Z = plastic modulus
d = outer diameter
d_i = inside diameter

For a symmetrical flanged section, consider a rectangle of b*d and a negative rectangle of (b - w)(d - 2t) where b is the flange width, d is the overall depth, t is the flange thickness and w is the web thickness.

The plastic modulus, Z for the section is the algebraic sum of the Z for each rectangle. For a rectangle, Z = bd[²]/4.

BA

 

[271828]

explain "of shear area"

 

[damnaged]

thanx all for your kind replies..

i guess ishvaaag gave the approach i needed to understand better how this thing works, i'll need to read further to understand properly.

plastic modulus calculation is not the issue as most formulas have been establshied for most normal sections - I beams, circular hollow section, rectangular hollow section, rectangles..

problem now is plastic modulus of shear area, where for a circular hollow section (pipe), the shear area is taken as 0.6 of section area affected from the main axis.

so plastic modulus of shear area shall require calculations of plastic modulus of whole section minus the plastic modulus of area not affected by shear...or is there a simpler way?

 

[BAretired]

damnaged,

I don't understand what you are asking. You seem to be inventing your own theory. Kindly let us know how it works.

BA

 

[271828]

The plastic section modulus is easily calculated for just about any section, including a round HSS, using the standard methods we have now.

Shear area is a kinda-fictitious quantity that is used to compute deflections from shear deformation.

I don't see how they relate. Actually, I don't have any idea what you're typing about, so some more explanations would help.

 

[damnaged]

BA,i didn't invent the theory (or did i?)..

if you refer to british standard BS5950-1:2000, clause 4.2.5.3 bending moment check against high shear,value Sv (plastic modulus of shear area) is required. So how do i calculate this Sv value?

 

[ishvaaag]

Certainly (have checked) you didn't invent is. It is the Plastic Modulus of Shear Area Av as per 4.2.3 Shear capacity, and Av is taken the rectangular area of the web taken rectangular d·t. So a rectangular section as my derevation showed, and simply reading jointly what must be read separate, then comply with the (most surely an interaction) check.

 

[BAretired]

The term Plastic Moment of Shear Area (Sv) is not used in Canada, so far as I know. The shear area of a WF or I beam is the web, h x w, so I guess Sv = w*h²/4 where h = height and w = web thickness. Perhaps someone else can comment on this as I am not at all sure.

BA

 

[271828]

That explains it. We don't use the concept in AISC steel design. Assuming that the quantity is as its name implies, BA gave the answer. The height of the shear area is up to interpretation and some people think that it should be d-tf, but that's a nit-picky detail.

 

[rowingengineer]

BAretired,
you have it in one, it is the plastic moment of the the shear area.

Arguing with an engineer is like wrestling with a pig in mud. After a while you realize that them like it

 

[damnaged]

ishvaaag & BA, understand both of your comments and that is thus far that i found info on.. but i am looking at how to calculate this based on a pipe section where the shear area is 60% of total area acting at the major axis.

Can somebody who understands how this work can perhaps give a simplified formula on getting the platic modulus of the part section of the pipe?

thanks a lot guys for all your input.

 

[ishvaaag]

The answer to your question is in 4.2.3 where says

"In CHS and RHS sections the shear area should be assumed to be located adjacent to the neutral axis."

CHS it seems stands for circular hollow sections, hence you have C and symm to C 30% of Pi·D at each side for which to calculate the plastic moment. You have an arc of .15·D·t area where Fy is applied for C or T zones (twice that) and we identify the ordinate of such arcs by looking at the cog of the 15% of 360 deg, that is 54 deg arc. I pass this to a Mathcad sheet, printout attached and gives plastic modulus of the shear area for thin wall pipes at Z=.412·D^2·t

 

[damnaged]

thanks ishvaaag, exactly what i needed.. sadly i'm not good enough in this to understand it..

i'll base on that..formula wise i think not a problem, the platic modulus value is decreased by about a half, so seems workable and i'd use same thought to be conservative (decrease value in half).

 

[ishvaaag]

In the post above each half arc covers 15% of the circle. I omitted PI when stating .15·D·t, should be .15·PI·D·t. In the printout I think all goes well.