horizontal forces on base

[paddyo1]

I'm trying to calculate horizontal force on a portal frame base.
Following the procedure in EC3 manual for design of steelwork and coming up with 1.37 kn. Is this too small a figure for a frame of 18 metre span,6 metre eaves height? I compared it to similar sized frame calculations done by other engineers. I have the remainder of calculations completed except for this!

Thanks

 

[BAretired]

Doesn't sound reasonable to me, but you haven't given many details. Show a sketch of the frame with design loads.

BA

 

[paddyo1]

Try this, thanks

 

[paddyo1]

file:///C:/Users/DELL%20User/Desktop/portal1.pdf

 

[paddyo1]

try this, thanks

 

[paddyo1]

try this

 

[paddyo1]

try this jpeg

file:///C:/Users/DELL%20User/Desktop/portal1.jpg

 

[WinelandV]

paddyo1,

You should use the link below the attachment box that says "...or upload your file to ENGINEERING.com".

 

[paddyo1]

Thanks winelandv,

 

[BAretired]

If the columns are pinned at the base, your frame is indeterminate to the first degree. You can solve for the horizontal reaction quite easily by hand to check the computer output. First, calculate the horizontal deflection when one of the pins is replaced with a roller support. Then calculate the force needed to push it back. If that turns out to be 1.37 kN then all is well. Otherwise, your computer is screwing up.

BA

 

[rb1957]

1) i only looked at the last pic ... BA's 1.37kN seems oddly precise ?
2) BA's method isn't how i remember the unit load method ...
a) solve with one roller support (statically determinate). calc deflection at the roller, D.
b) apply 1 lbs, horizontal, at the roller, solve. calc deflection of roller, d.
c) horizontal reaction for roller location is D/d lbs (in the opposite direction to the unit load).
d) solve for the rest of the frame reactions.

Quando Omni Flunkus Moritati

 

[BAretired]

rb1957,
1.37 kN is oddly precise because that is the value the OP found when he analyzed the frame for a load which he did not stipulate.

The unit load method is one of many which could be used and you offered an excellent description of it.



BA

 

[BAretired]

If you are disinclined to go through the calculations, you could check out this link, particularly Tables for Chapter 13:

http://www.google.ca/url?sa=t&rct=j...=yDmfRTuVyeUBrEkqoSqiaQ&bvm=bv.74649129,d.aWw

BA