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RE: Hankinson with different LDF's

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But, the resultant load is a combination of the two at some angle to the grain.  Anyway, this is why I ask the question in the first place.   I’ve never seen an example or explanation for this type of combination although it is common.  My gut tells me to agree, but……

 

By your thought process then why, with a beam design with snow load multiply by the LDF of 1.15 no matter the ratio of dl to sl?  The dead load in that case is pretty static also.

 

I thought an interesting question anyway.  In either case my design is O.K.

Joe

 

Joseph R. Grill, PE

Verde Valley Engineering, PLLC

email: VVEng(--nospam--at)cableone.net

 

From: Drew Morris [mailto:dmorris(--nospam--at)bbfm.com]
Sent: Monday, August 30, 2010 3:44 PM
To: seaint(--nospam--at)seaint.org
Subject: Re: Hankinson with different LDF's

 

I agree.  The LDF of 0.9 applies to the force perpendicular to the grain from the soil loading and the LDF of 1.6 applies to only the inplane shear.

On 8/30/2010 2:14 PM, David Topete wrote:

But, the soil loading is "static."  How can a 1.6 LDF increase be justified for that loading direction?  I think it's similar to a steel beam loaded in both the strong and weak axes.  The interaction equation from the green steel manual (9th Ed. ASD).  fbx/Fbx + fby/Fby <= 1.00   Any LDF's, such as short term loading due to wind/seismic/impact and applied left of "<=", and not to the right of it.  Just my thoughts...

On Fri, Aug 27, 2010 at 3:54 PM, Joseph R. Grill <vveng(--nospam--at)cableone.net> wrote:

But the loads are being applied at the same time, so wouldn’t the LDF of 1.6 be used on the final Z theta after the equation is run without adjusting Z perp and Z parl?

 

Joseph R. Grill, PE

Verde Valley Engineering, PLLC

email: VVEng(--nospam--at)cableone.net

 

From: David Topete [mailto:d.topete73(--nospam--at)gmail.com]
Sent: Friday, August 27, 2010 10:31 AM
To: seaint(--nospam--at)seaint.org
Subject: Re: Hankinson with different LDF's

 

LDF = 0.9 for "static loading" from soil...  LDF = 1.6 for wind-induced shear....  I'd applied each to each Z_perp and Z_parl respectively...  The 0.9*Z_perp will be the governing case....

On Wed, Aug 25, 2010 at 2:42 PM, Joseph R. Grill <vveng(--nospam--at)cableone.net> wrote:

If you have an anchor bolt in a sill plate with a top of wall restraint force from soil loading of 200 plf  perpendicular to grain and a parallel to grain load from a shear wall of 250 plf.  Are the LDF’s of .9 and 1.6 applied before input to the Hankinson formula or is the result from the formula multiplied by the larger LDF of 1.6?  Sudden brain blockage

Thanks, Joe

 

Joseph R. Grill, PE

Verde Valley Engineering, PLLC

email: VVEng(--nospam--at)cableone.net

 




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David Topete, SE




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David Topete, SE