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COMPLETE STEPWISE PROCEDURE TO DESIGN MULTI-SPAN CONTINUOUS BEAMS IN ETABS Open the ETABS Model for Which you Want to Design the Beams Double click on ETABS or search it up from the start menu and open the software. Now, look for the file that contains the building model that needs to be designed. In our case, the building…
VARSHA MOHAN WAGH
updated on 21 Jun 2022
COMPLETE STEPWISE PROCEDURE TO DESIGN MULTI-SPAN CONTINUOUS BEAMS IN ETABS
View/Revise Reference
Concrete Frame Design Option
Design Completed
Designing the Beams
1. Seven-Span Continuous Beam Along Grid-3
So, following the same standard procedure as discussed above, the remaining spans of the beam were designed and the final drawings are inserted as follows:
FINAL OUTPUT:
2. Three-Span Continuous Beam Along Grid-A
FINAL OUTPUT:
• Provide reasons for the failure of the middle span along grid A. What are the possible ways this issue can be resolved?
ANSWER:
The reason why the middle span along gird A fails is due to the overstressing in shear near the right support. The shear stress given by the ETABS analytical model at that particular span exceeds the corresponding limiting value as stated by the design codes. Accordingly, the ETABS detailed section also gives the reason for the failure of the span which is shown in the below image.
Possible ways to resolve this issue:
As we know that shear stress is the factor of shear force and the sectional dimensions of the beam. Hence, the overstressing of shear can be resolved in two ways as listed below:
Calculate the value of the maximum shear force in any one of the spans in both the continuous beams, as per clause 6.3.3 (b) in IS 13920 – 2016. Also, confirm these values from the shear force demand calculated by ETABS. Please note that if longitudinal reinforcement provided is more than required (as per ETABS results), the shear force demand will vary from what is provided by ETABS.
ANSWER:
As per clause 6.3.3, the shear reinforcement provided in the beam must be able to take up the maximum sway shear induced for the same section. At the same time, part (b) of clause 6.3.3 also mentions the formula to obtain the sway shear in both directions that are left and right which is given as follows:
So, using the above formula, we have calculated the sway shear as given below:
However, we need to first calculate the shear force at supports due to the factored gravity loads in order to arrive at the values for Va(D+L) AND Vb(D+L). So, we have added a new load combinatin with dead and live gravity loads with the safety factor 1.2 as mentioned in the IS-code.
Now we can insert the relevant values and obtain the sway shear values as follows:
Sway to the Right
Sway to the Left
So, the maximum shear among the four values is of 230 KN. Therefore, the stirrups designed for this beam must be able to resist more than 230KN of shear force for a safe beam design. Thus, let's check how much our stirrups can take in terms of the shea force.
Shear Taken by Stirrups, (Vus):
Vus = 0.87fy Asv d / Sv
= 0.87x500x100x570 / 75
Vus = 330 KN
Since, 330KN>230KN. Thus, our shear design is safe.
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