SEAINT List Archive

Monday, December 6, 2010

RE: Eccentric Loads on Welds

Thanks. You confirmed the direction I was headed.

-----Original Message-----
From: Conrad Harrison [mailto:sch.tectonic@bigpond.com]
Sent: Monday, December 06, 2010 3:48 PM
To: seaint@seaint.org
Subject: RE: Eccentric Loads on Welds

Thor,

Treat the welds as line elements and calculate section modulus for. Section
properties of various weld groups can be found in handbooks: circular, two
vertical lines, 2 horizontal lines etc..

Calculate normal and shear forces, then translate into how the code checks
the combination. Here older codes checked transverse and longitudinal
actions on different shear planes in the weld, current codes largely ignore
and check against a single resistance.

Regards
Conrad Harrison
B.Tech (mfg & mech), MIIE, gradTIEAust
mailto:sch.tectonic@bigpond.com
Adelaide
South Australia

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RE: Eccentric Loads on Welds

Thor,

Treat the welds as line elements and calculate section modulus for. Section
properties of various weld groups can be found in handbooks: circular, two
vertical lines, 2 horizontal lines etc..

Regards
Conrad Harrison
B.Tech (mfg & mech), MIIE, gradTIEAust
mailto:sch.tectonic@bigpond.com
Adelaide
South Australia

soil press at square pad diagonal

I have a situation where a square footing can be loaded at 45 degrees to the axis. I was just sitting down to work out the equation to calculate the footing pressure when the resultant is just outside of the kern. At first glance it looked like the algebra was going to take a bit of time. I don’t have a reference that provides this. I certainly can do it, just thought I would ask if there is a reference someone can give or possibly a copy of pages from a reference or if they just have the equations handy. It would save me some time this week. Of course the way work is nowadays, I can’t really say I don’t have the time.

Thanks,

Joe

Joseph R. Grill, PE

Verde Valley Engineering, PLLC

email: VVEng@cableone.net

Register for SEAOSC Wood Design Webinar

SEAOSC Member Doug Thompson, S.E. will present a Webinar on Wood Design: Significant Changes from 2006 to 2009 IBC Chapter 23-Wood on Wednesday, Dec. 8, 2011. Webinar starts at noon PST and ends at 1 p.m. Register at: http://www.seaosc.org/events_detail_announcement.cfm?pk_announcement=27

Eccentric Loads on Welds

I'm looking at the weldment of an aluminum post to a plate. Apart from the
heat affect on the post alloy I am looking at the weld itself. I find that
for the example I am working on , that a simple flange weld is inadequate
and I am considering the calculation for the weld all around the post
section. That requires consideration of eccentric loading but the handbooks
etc only give examples for a plate welded to a plate.

Can I assume that the "plate thickness", 't' can be the dimension of the
post?

Thanks

Thor A. Tandy P.Eng

Victoria, BC
Canada

TIA 222 G

Hello,

My company need to purchase TIA 222 G for a job. Do I need to buy TIA 222G, TIA 222 G1 as well as TIA 222 G2?.

We haven't used TIA 222G so far.

Thanks for the advise.

Regards

Ashraf

TIA TIA-222-G-2
2009-NOV-24 • Active/Current •
Structural Standard for Antenna Supporting Structures and Antennas-Addendum 2

TIA TIA-222-G-1
2007-APR-11 • Active/Current •
Structural Standards for Steel Antenna Towers and Antenna Supporting Structures – Addendum 1

TIA TIA-222-G
2005-AUG-02 • Active/Current •
Structural Standard for Antenna Supporting Structures and Antennas-EFFECTIVE JANUARY 1, 2006; Includes County Listing 07/20/2005 after TIA-222-G standard

Friday, December 3, 2010

RE: PEMB Foundation Uplift

The load combination of 0.6D+W will always give a factor of safety of
1.5 or greater as long as you include the weight of the footing in the D
and thus satisfy a "1.5 factor of safety" though I am not sure the 1.5
factor of safety is codified in the code I design too but maybe it is
and I don't know it.

I don't think it is the intent of the code to exclude the weight of the
footing from the load combination and separating it without factoring it
by 0.6 is just bad algebra. The expression for required weight of
footing should be expressed as '5/3W-D' with D being the dead load less
footing weight and thus the total load would reduce to 5/3W-D+D or 5/3W
which will be greater than 3/2D mentioned.

Using the expression 'W-0.6D' to determine required footing weight or
even 150% of 'W-0.6D' will not satisfy the required load combinations...
at least that is my interpretation.

Donny Harris SE
Los Angeles, CA

Subject: RE: PEMB Foundation Uplift
From: Long Chen <longfchen@gmail.com>
To: seaint@seaint.org

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Content-Type: text/plain; charset=ISO-8859-1

Hello, Jeff,

I think that the 1.5 factor is needed. Let's say that the wind load is W
and
the dead load beside the footing is D. To make it simple, let's say that
both are given in positive values. In order to provide an uplifting
safety
factor of 1.5 or greater, we will need to have "D + footing weight" be
greater than 1.5W.

When you design the footing with the net uplift load 0.6D+W, you
actually
get a footing weight of W-0.6D since D and W counteract each other. Then
the
total dead load including the footing is D+(W-0.6D) = W+0.4D, which may
or
may not be larger than 1.5W. Then the safety factor achieved may be
smaller
than 1.5.

But if you design the footing with 1.5 times the net uplift load, you
will
get a footing weight of 1.5 * (W-0.6D) = 1.5W-0.9D. Adding this back to
D
you get a total dead load including footing of D + 1.5W - 0.9D =
1.5W+0.1D.
This value will always be greater than 1.5W since D can only be
positive.
And accordingly you will get a safety factor larger than 1.5 as
required,
regardless how small D is.

Hope that can help.

Regards,
Long

--
Long F. Chen, P.E.
Civil/Structural Engineer
Honolulu, Hawaii

From: Jeff Hedman [mailto:jeff_h@lrpope.com]=20
Sent: Wednesday, December 01, 2010 8:27 AM
To: seaint@seaint.org
Subject: PEMB Foundation Uplift

I designed a foundation for a PEMB. I want to know peoples methods for
designing the footings for uplift. I have typically designed the
footing to weigh as much as the net uplift load (0.6D+(W or 0.7E)). The
calculated net uplift force has a factor of safety of 1/0.6 =3D 1.67 in
=
it
from the reduction of the dead load. I received a plan check comment
that states that with my calculations there is only a factor of safety
of 1. But this is a factor of safety of 1 in regards to the net uplift
force. Should we really be increasing the footing weight to maintain a
factor of safety =3D 1.5 over the net uplift force? To me this is more
like a factor of safety of 1.67*1.5 =3D 2.51. Not to mention we know =
what
the actual weight of the footing will be. This is a commercial building
and the city inspectors are very good in requiring construction matches
the plans, especially foundations. Am I wrong? Do others design PEMB
footings that weigh 1.5 times the net uplift force?

Thanks in advance for your replies

Jeff Hedman

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