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Strut and tie - AS3600-2009

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dadomago75

Structural
Mar 2, 2011
30
I hope someone had ever though about this:
clause 7.2.4 bursting forces have to be checked for SLS and ULS using two differend divergence angle, 26.5 for SLS, and 11.3 for ULS.
Say that for a strut action of C = 1 KN, the ultime action is (average) C* = 1.35 C, and applying the divergence angle specified by the code, the bursting forces are T = C*tan(26.5) = 0.5*C (SLS) and T* = 1.35*C*tan(11.3)=0.27*C, which means that the bursting force in the servicebility limit state we have a bursting force double than the ultimate!
Furthermode in the ultimate limit state we use the yield stress (500 MPa) whereas in the servicebility limit state it is required to use the limit stress according to the crack control limit (considering minor degree of crack conrol, is 50% the yield stress), which leads to an area of reinforcement of As,ULS=T*/fsy=0.27*C/500=0.54*C, and As,SLS=T/fs=0.5*C/250=2*C, which leads to a gap between the required reo of 4 times!
Is there anybody out there that has ever come up with this? any commentary, worked example or similar to help me to understand what f..k did they think when they wrote this code? [ponder]
Thanks everyone
David
 
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Dadomago,
This is the diversion angle for a bottle strut. The code doesn't set the angle it give you the maximum angles at which you can diverge a bottles strut under a load. I think the angle you maybe thinking about for strut to tie angle this is limited to 20 degrees for both SLS and ULS.

There are a few good papers on the conc institute site for the strut tie method



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Thank you for the reply.
Actually I am talking of the divergence angle of the compression strut; the code sets a limit (minimum) for the divergence angle in clause 7.2.4 saying that for the serviceability limit state it has to be used a min of tan(α)=1/2 and for strength tan(α)=1/5
I will check the concrete institute website, thanks.
 
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