Hi,
I'm trying to get a deeper understanding of why RAM Concept is showing my shear requirement as being larger than the same beam I have modelled in RAPT to verify. RAPT does not require any additional shear reinforcement. I have checked the shear core and that seems to be OK.
As an example, I have been looking at design section 24C-2, section 2 in the model I have uploaded via secure link and have the following questions:
1. Why is the design section minimum demand (-1427kN) significantly larger than the design strip maximum shear in this span (-533kN)? What does the design section demand mean?
2. From the cross section audit Envelope 7 of the Final Design Check for Strength design, phi Vuc = 0.75 x 2043kN = 1530kN > 1427kN. Why is shear reinforcement added when capacity > demand and what determines that they need to be closed ligs?
3. Why is there bottom reinforcement required at the column at this location rather than at mid-span if anywhere? Similar question applies to the beam in the north-eastern corner of the building.
Regards
File "MEL MSCP Ramp PT Design_Rev7.cpt" was uploaded through the Secure File Upload.
Is this the correct file? I do not see 24C-2 in this model. The last span segment is numbered 10-5.
Apologies Karl - I've uploaded the correct file now "MEL MSCP L3 PT Design_Rev7".
Thank you for the model.
1) The section is subject to torsion. The higher shear force is the equivalent factored shear force (Veq*), which accounts for both the applied shear and torsion. The lower value excludes the contribution of torsion.
2) Minimum transverse shear reinforcement is required per 8.2.1.6 because torsion exceeds 0.25*phi*Tcr. Closed ligs are detailed because the section is designed for torsion.
3) The bottom reinforcement is added to resist the longitudinal shear-torsion tension (deltaFtd, see 8.2.7). The torsion demand is a key contributor to the shear-torsion tension requirement.