Flich beam design

[rovi123c]

Hi

I've been following an example of flitch beam design and I just don't get it.
The engineer's example shows e of steel as 210000n/mm2 and e of timber SCIV as 6600n/mm2 x modification factor k9 of 1.14. The plate thickness of the steel is 8mm so he gives an equivalent B of timber as 8x210000/(6600x1.14) plus the original B of timber 100mm. This equates to 324mm.

However, as you would expect, the 'e' of timber increases as the grade of timber improves and vice-versa. Therefore, SCIII, a lower grade timber has an 'e' of 5800n/mm2. So here's my confusion.. If you substitute 6600 with 5800 the equivalent 'B' is actually going to increase! this cannot be correct.

Can any structural engineers out there help me please.

 

[CELinOttawa]

It is correct... The value you are calculating is called "nu" (but usually written with a lower case ETA.... Don't ask) and is simply the ratio of the two Modulli of Elasticity. So, as you have found, the value of nu goes UP the worse a grade of timber is used because steel is even stronger in relation to the now weaker timber.

The trick here is that, if you follow through the example with a couple of different grades of lumber, the beam still winds up being weaker with the crappier liumber. Think about it, the nu factor is just keeping the steel contribution constant while you calculate the strength and stiffness at ever lower values of timber.... When you work it out the strength drops because you've lost strength in the wood.

Does that make sense?

P.S. PLEASE post a copy of the design example you're using; They are pretty rare and I've not seen a metric one...

 

[rovi123c]

Hi and thanks for coming back to me.

OK I understand the ratio and the effect the steel would have on the timber. I get that by calculating this ratio we are in effect, increasing the width 'B' for the timber section, thus finding an equivalent B by sandwiching a steel plate between the two 200x50mm sections of timber. What I don't get is that by using a higher grade timber with a higher value of e, the equivalent 'B' we are calculating should increase, but it doesn't. The higher the value of e for timber the lower the equivalent B becomes. That can't be right?

Ive uploaded a file hopefully you can see it. The flitch design is at the bottom of the page.

Thanks again

 

[CELinOttawa]

Please re-read my post and actually work though two calcs with all variables thee same except for differing timber grades. It is counterintuitive but correct.

 

[bouk715]

The value of "B" is a function of stiffness (based on elastic modulus) between the steel and the timber. In this case, the steel is fixed at 29,000,000 psi (sorry, I don't use metric regularly). The stiffness of the timber changes based on its grade as you point out.

So what the "B" is measuring is how much of the load is being attracted to the stiffer steel element. As the wood on either side of it weakens, more is attracted to the stiffer steel, and B goes up. If the wood is stronger, less goes to the steel, and B goes down.

As CELinOttawa points out, where you will see a reduction in the strength of the section is when you determine the bending stresses in the steel and wood elements. The bending stress in the wood cannot exceed its design value and this is where the section will be limited. The "B" value is just the mechanism to determine how much of the load is going to it vs. the steel.

CELinOttawa: here is a link to an article for flitch beam design and tables that I've used in the past. It's in US units though.

http://www.structuremag.org/article.aspx?articleID=399

Hope this helps.

 

[bouk715]

Also, remember that the "nu" you are caluculating has the wood "E" in the denominator. So as wood stiffness goes up, "nu" goes down.

 

[rovi123c]

Ahh I get it guys. So the B we determine is what is required timber wise as opposed to what is actually produced. And when working the calc through to find Mr (moment of resistance) it works out less if the lesser grade timber is used as you would expect.

Thanks CELinOttawa and bouk715!

 

[tony1851]

You know the udl on the beam, and the E of steel and the timber. As both elements deflect equally, you can use the deflection equation for a udl to find the combined breadth of the timber plates required.

 

[tony1851]

Just realized you can't do it that way; long time since I did flitch beams!