Reinforcement Detailing of Beams from ETABS output

AIM - i) To design beams from Etabs 2018 in detail.

ii) Provide detailed description of longitudinal and shear reinforcement for two continuous beams.

iii) Calculate the value of the maximum shear force in any of the spans of continuous beams

iv) State the causes of failure of the mean field along grid A.





Introduction -



1) Beams are horizontal structural elements used to support lateral loads.



2) In conventional reinforced concrete structures, the beams usually take the load from the floor slab, but they can also

be exposed to other loads such as wall loads, finishes, service installations, etc.



3) RC beam design and detailing includes selection of proper beam size and number of stringers

and shear reinforcement to meet the limit state requirements for bearing capacity and serviceability.



4) Then it is very important that the beam braces are properly positioned and arranged in place,

so that there is no error in construction.









PROCEDURE-



Step 1)

i) Open the Challenge 7 attachment we received and download it.

ii) After downloading, open the prompt and open it in Etabs software.

Step 2)

i) Now check the load combinations that are already created in the given modeling file.




Step 3)

i) We need to check the model before running the analysis

ii) Click Analyze

iii) Click on Check Model

iv) The structure is verified without any error and the model has been checked.


Step 4)

i) Now we need to start the beam design by selecting 'Start Design/Check'

ii) Once you click on Start/Design Check, the Show Concrete Frame Design Results window will appear.

iii) Where you can choose any design output you want

iv) So click on Longitudinal reinforcement .

Step 5)

i) Once you click on , the longitudinal reinforcement details will appear

Step 6)

1) Reinforcement details along grid A -

i) Max Ast with lower support = 194 mm2

ii) Max Ast with bottom middle = 789mm2

iii) Max Ast with highest support = 2046mm2

iv)Max Ast with top center = 511mm2


Calculation of reinforcement -

1) Up -

i) Total required above is = 2046


So we can provide

4 bars with a diameter of 20 mm and 4 rods with a diameter of 16 mm meet the total Ast

4 x T20 = 1256mm2

4 x T16 = 804mm2



4 x T20 + 4 x T16 = 2060mm2.......................Therefore satisfy .





2) For Bottom -

i) Total required above is = 1945

So we can provide

3 bars with a diameter of 20 mm and 5 rods with a diameter of 16 mm comply with the total Ast

3 x T20 = 942mm2

4 x T16 = 1004mm2



3 x T20 + 5 x T16 = 1946mm2.......................Therefore satisfy .







3) Shear Details -

i) Right click on and click Details

ii) Click on Shear Details







iii) According to Article 6.3.3, we must ensure that the Distance within 2d from the face of the column is 100 .

iv) It should not be more than that



v) Hence provide 10 mm diameter 3 leg stirrups spaced 100 mm along the length of the beam.



Step 7)



1) Details of the reinforcement along the grid 3 -



i) Max Ast at lower support = 1042mm2

ii) Max Ast at bottom middle = 1501mm2

iii) Max Ast at highest support = 1940mm2

iv) Max Ast at top middle = 485mm2



Calculation of reinforcement -

1) for top-

i) Total required above is = 1940mm2

So we can provide

3 bars with a diameter of 20 mm and 5 rods with a diameter of 16 mm comply with the total Ast

3 x T20 = 942mm2

4 x T16 = 1004mm2

4 x T20 + 4 x T16 = 1946mm2......................Therefore satisfy .





2) For lower -

i) Total required above is = 1501mm2

So we can provide

3 bars with a diameter of 20 mm and 3 rods with a diameter of 16 mm meet the total Ast

3 x T20 = 942mm2

3 x T16 = 603mm2

3 x T20 + 5 x T16 = 1545mm2......................Therefore satisfy .







3) shear Details -

i) Right click on and click Details

ii) Click on Shear Details








iii) According to Article 6.3.3, we must ensure that the Distance within 2d from the face of the column is 100 .

iv) It should not be more than that

v) So provide 3-legged stirrups of 8 mm diameter spaced 100 mm along the length of the beam on both sides

vi) And provide 3-legged stirrups of 8 mm diameter with 150 mm spacing at mid span.




Step 8)

i) Check for Sway shear.

A) For the right Sway



formula for the calculation









From the Etabs model results, substitute the value -

i) VD+Lu= 92 Kn

ii) Mas= 176 kn-m

iii) Mbh= 295 kn

iv) Lab = 4.725 m

Paste this value above eqn

we will get ,



Vu,a= 48 Kn

Vu,b = 230 Kn

B) For Left Sway

From the Etabs model results, substitute the value -

Paste this value above eqn

we will get ,

i) VD+Lu= 74 Kn

ii) Mah= 236 kn

iii) Mbs= 150 kn

iv) Lab = 4.725 m

Put this value in above eqn

we get ,

`V_(u,a)`= 190 Kn

Vu,b = 50 Kn

C) Existing shear capacity in beam as per Is 456-2002

i) Known formula -



Insert values -



i) fy = 415 N/mm2

ii) Asv = 151 mm2

iii) Spacing = 100 mm

iv) d = 450 mm

put this value above eqn
we will get

Vus) = 246 kn................................................ ..> 230 kn (Shear rocking)........... (Therefore Safe)




Result -

i) Beam design is successfully done with detailed bracing.
ii) State the reasons for the failure of the mid-span along grid A. What are the possible solutions to this problem?

1) The main reason for the failure of the mean field along grid A is the shear stress exceeding the limit value

2) It may be due to higher shear force acting on this beam and to provide weak shear

reinforcement on the end supports.

3) This problem can be solved -

i) By providing concrete of higher quality in this particular section

ii) Or by reducing the shear force in this section of the beam

iii) Or by providing higher shear reinforcement that can withstand the shear force in the given section


iii) Calculate the value of the maximum shear force in any of the spans of the two continuous beams,

as per clause 6.3.3 (b) in IS 13920 – 2016.



Since we have already calculated the shear force value as per IS 13920 - 2016 Clause 6.3.3 (b)



We get the shear swing value Right and Left support within the limit



246 kn > 230 kn ...................( Safe )