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] [ponder] [ponder]](/data/assets/smilies/ponder.gif)
Thanks everyone
David
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] [ponder] [ponder]](/data/assets/smilies/ponder.gif)
Thanks everyone
David