RAM Concept Shear Reinforcement [FAQ]
| Applies To | |||
| Product(s): | RAM Concept | ||
| Version(s): | Any | ||
| Environment: | N/A | ||
| Area: | N/A | ||
| Subarea: | N/A | ||
| Original Author: | Bentley Technical Support Group | ||
Why does my flat slab (or flat plate) model have one-way shear reinforcement or fail the shear check?
When engineers design flat slabs by hand, they often IGNORE the one-way shear checks. They decide that punching is all that is appropriate. (This is often decided without much consideration – it just “seems right”).
Concept does not make this decision, as nowhere does the code advise to ignore one-way shear checks in a flat slab or flat plate. In fact, ACI 318-02 rule 11.12.1.1 specifically requires a one-way shear check in flat plates. Nonetheless - you should decide what the possible failure mechanism is and so what is appropriate. It may, or may not, be appropriate to ignore the one-way shear results.
One of the biggest factors in thick slabs (especially in mats) is the fact that the program designs for shear at each cross section, while the design codes generally allow the designer to ignore shear some distance from the face of the support (typically "d"). In cases where shear reinforcement is only required in the last section at the face of the support it can be ignored.
The results have a lot more shear reinforcement than expected.
This is likely to be a shear core issue. Refer to “About shear core” on page 99 and “Shear core in slabs” on page 100 of Chapter 22, “Defining Design Strips”.
For a post-tensioned beam, the reason could be that Concept is deducting a fraction of the (bonded) duct from the web width per the appropriate code rules.
Concept calculates the number of duct by dividing the Strands per tendon by the Max strands per duct (as specified in the Materials) and rounding up to the next integer.
Refer to the following sections for an explanation of Concept’s shear web calculation:
- For AS 3600, “Section 8.2 Shear Design” on page 415
- For BS 8110, “Section 3.4.5 Design shear resistance of beams” on page 432.
- For IS 456, “Section 22.4 Design shear resistance of beams” on page 460.
Note: There is no ACI318 rule concerning deduction of ducts.
What does this audit text mean: “Depth “d” is zero - replacing with “column” effective depth. Depth is still zero - giving up.”?
The is likely a combination of two things:
- there is net compression force and a small moment, and as such the bending designer does not provide any reinforcement
- the minimum designer has been turned off If this is the case, you should consider turning the minimum designer back on.
A similar condition tends to occur near the inflection point of a span. In these locations the moment can be very small and a flexural capacity check can pass with very little top reinforcement, but if there is no tension reinforcement at the tension face, or of there is a lot more reinforcement on the compression face, this skews the calculated depth to the tension reinforcement (d) and drastically reduces the shear capacity.
CanImodeluser shearreinforcement?
Transversereinforcementcannotbemodeledasuserreinforcment.Itisalwaysdesignedbytheprogram.Wearelookingtoadduserdefinedtransversereinforcementinthefuture.
DoesthereinforcingweightintheEstimateReportincludetransversereinforcement?
Yes, transversereinforcementisincludedinthereinforcmentweightintheEstimateReport.
See Also
Structural Product TechNotes And FAQs
External Links
Bentley Technical Support KnowledgeBase
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