Shear deflection.

[Daven1978]

Hi forum,

I am in the current process of building an quick excel find the BM, SF, defections in beams under simple supports with uniformly/partial distributed load and concentrated load at any point. I need an output on an graph, of length along the beam against defection to find certain results at certain points.

I can find the bending defection equations, for an UDL =(wx/24EI)(l^3 - 2lx^2 + x^3) and Point load (Pbx/6EIl)(l^2 - b^2 - x^2) when x<a.

But I can not shear defection equations of defection against distance along an beam.

Thank you for any help you can give.

 

[rowingengineer]

thread507-20914

http://www.nceng.com.au/

"Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning."

 

[a7x1984]

Daven: Hope you don't mind me asking why you need to include shear deformation. It is usually neglected 99.999% of the time.

In Russia building design you!

 

[Daven1978]

Hi a7x1984, yes in Euro code EC5 C7.2 it asks to for the instantaneous deflection due to permanent actions and variable actions, and can add only an couple of mm's to an lightly loaded beam. But I am looking find an equation, to add to the excel and the bending deflection, so to find the load effects easily.

Thank you again for any help you can give.

 

[Daven1978]

Hi a7x1984,

Yes under Eurocode EN5 CL7.2. Deflection calculations for timber members should include both bending and shear deflection. I am at the moment making an excel to display clearly the load effect ie BM, SF and deflection the change in the across the beam for partial UDL and point loads so that we can quickly cal results live to save time redo the process several times.

Thank you again for any help you can give.

 

[Daven1978]

I am looking to create an excel on the beam design to EN5 and it states that the deflection calculations for timber members should include both bending
and shear deflection.

 

[Daven1978]

Hi a7x1984,

I am in the making of an excel to EN5. To find the final deflection, the easiest way in an excel program to get what I need an graphically (i.e. displacement against distance along an beam) like from the bending equations is to use an formula. That is to add the shear and bending deflection.

Thank again for any help you can give.

 

[kingnero]

I would be interested in that excel file - if you don't mind sharing it?
Been looking into EC5 as well lately due to few metal projects for the time being - could be interesting for the future...

 

[BAretired]

Is your question specifically related to timber beams and rectangular cross sections? Do you have a value of G for timber?

BA

 

[Daven1978]

Hi BAretired,

Yes it only to with timber in EC5 of rectangular mainly solid timber, as all others would be an subcontractor. No I do not have an G as this would be dependant on the timber at the time.

Thank you for your help.

 

[a7x1984]

The National Design Specifications for Wood Construction in the US provide this explanation in their commentary: "For solid rectangular and circular bending members, reference modulus of elasticity values are considered to contain a shear deflection component equivalent to that occurring in a rectangular bending member on a span-depth ratio of between 17 and 21 under uniformly distributed load."

I guess they make it simple for us around these parts.

"permanent actions and variable actions" - Does Eurocode really use those terms for dead and live loads?

In Russia building design you!

 

[BAretired]

I have seen expressions for deflection due to shear in several books, but it is expressed in terms of the Modulus of Rigidity, G. If that value is not known for a specific type of timber, I am not sure how you can take it into account. Does the European code give any idea how to determine shear deflection?

BA

 

[dhengr]

And, do you have all of the other mechanical properties of each and every timber, and the variability of these? Be sure to take that into account as it will affect deflection more than shear deflection does. Our mechanical properties are averages for a species, grade and for a grading method. And, do pay attention to the fact that sawn lumber or timbers are barely homogeneous and certainly not isotropic. And, our design methods generally try to account for these things. You just can’t reliably calc. the deflection for an individual wooden beam to +/- 1mm, nor do you know the loads that well. Does the code requirement really fit this material and problem for most normal beams, or is it explained away, as not significant for most problems. Or, only significant for a few, for example... Take a look at the commentary, or ask a committee person, and see if you can get a feel for the intent of that part of the code.

 

[rb1957]

defelcion due to shear ... is this like the defelction of the beam's web, carrying the shear P/2 (for a point load) ?

would this be shear strain*length (ie L/2) ? and isn't shear strain = shear stress/G

Quando Omni Flunkus Moritati

 

[prex]

Daven1978, coming back to the original question, here you can see the formulae for shear deflection in a simply supported beam under distributed load.
However these are rather numerical formulae than analytical ones, as they are obtained via energy minimization; hope they might inspire you for something though.

prex
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[Daven1978]

@ a7x1984.

Thank you for the info and Yes, there are some minor differences in wording, like dead and live loads are now called permanent and variable action respectively. They are so minor that they are unlikely to present any problem.

@ BAretired

Thank you, yes the equation I think the type of equation you were at looking was on the lines of:

Vs UDL = [Kform*pd*l^2]/(8*G*b*h)

and in Table 1 EN338 the Mean shear modulus (GMean) is given. And yes Eurocode give information what the final deflection should be, it is cover in CL7.2 and CL2.2.3 but not how to calculated the bending or shear deflection.

@ prex

Thank you for your comments, after looking at the possible solutions, I have made the decision to keep, UDL =(wx/24EI)(l^3 - 2lx^2 + x^3) and Point load (Pbx/6EIl)(l^2 - b^2 - x^2) when x<a and add the shear deflection to each point on the graph. Thank you for the link prex, very useful. I did have an look at them, but as I half way through the excel I did not want to rewrite my code.

Thank you for your help.

 

[homostructural]

Daven, have a look also at this article!

All about Eurocodes:

http://www.DesignToEurocodes.com