Reinforcement Detailing of Beams from ETABS output

PROJECT:-Reinforcement Detailing of Beams from ETABS output

Introduction:-

• Beams are horizontal structural elements used for the supporting lateral loads
• In conventional reinforced concrete structure beams usually receive load from the floor slab, but may also be subjected to other loads such as wall loads, finishes services installation etc.
• The design and detailing of RC beams involve the selection of the proper beam size and the quality of longitudinal and shear reinforcement that will satisfy ultimate and serviceability limit state requirements.
• Afterwards, it is very important that the beam reinforcements are placed and arranged properly on site, to avoid construction error.

Challenge 7:- Detailed design of concrete buildings -1

Detail design of beams in Etabs 2018.
The ETABS file for a G+4 building is provided. Run the analysis and design the RCC Moment ResistingFrame. The following challenge deals specifically with two continuous beams in the 1st floor.

1) 3 span continuous beam along grid A
2) 7 span continuous beam along grid 3

Address the following issues thereafter.
• Provide details of longitudinal and shear reinforcement for the two continuous beams. Sketch the beam elevation details or draft it using autocad software. A sample of how beam elevation details is prepared in consultancy firms is provided below. Participants can prepare it as per their preference:

• Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved?

• Calculate 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 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.

Aim:-

• To design the beams in detail from the ETABS.
• To provide details of longitudinal and shear reinforcement for the given two continuous beams.
• To calculate the value of maximum shear force in any one of the spans in the continuous beams.

Introduction :

Detailing rule of any Reinforced concrete member :-

• Minimum reinforcement for shrinkage and temperature
• Reinforcement required for limiting crack - width
• Ultimate limit state design rebar requirement
• Ductility rebar requirement

The above 4 conditions whichever is greater, should be drwan in RC beam detailing.

• Minimum longitudinal reinforcement as per IS:13920-2016 clause 6.2.1 As = 360 sq.mm
• Minimum shear or transverse reinforcement as per IS:456-2000 clause 26.5.1.6(Assuming 1000mm spacing in the formula) Asv = 332 sq.mm
• Developement length as per IS:456-2000 clause 26.2.1 L_d = 41xdiameter of bar

PROCEDURE:-

• Open the challange 7 attachment which are provided to us and download it
• After downloading open the Challenge open it in the Etabs software

• Before Run Analysis we have to check the model
• Click on Analyze
• Click on Check Model
• A structure is verified without any error and Model has been checked .

Run Analysis

• Click on run analysis from the ribbon to analyze the model

• Now we have to start the beam design by selecting the "concrete frame design" an below showing;

• iclick on downwards arrow from the concrete frame design, where you can select whatever design output you want
• So Click on Longitudinal Reinforcement .

• Once you click on , Longitudinal reinforcement details will appear

• Longitudnal reinforcement of beam at grid -A from etabs;

• Shear reinforcement of beam at grid-A etabs;

• Longitudnal reinforcement of beam at grid- 3 from etabs;

• Shear reinforcement of beam at grid-3 from etabs;

1) Reinforcement details along Grid A:-

1. Max `A_(st)` at bottom support = 1754 `mm^2`

2. Max `A_(st)` at bottom middle = 702 `mm^2`

3. Max `A_(st)` at Top support = 1889 `mm^2`

4. Max `A_(st)` at Top middle = 898 `mm^2`

Reinforcement Calculation:-

1) For Top:-

i) Total Ast required at top is = 1889 `mm^2`

So we can provide

4 bars of diameter 20 mm and 4 bars of 16 mm satisfies the overall Ast

3 x `pi*(20/2)^2`= 942`mm^2`

5 x`pi*(16/2)^2`= 1004.8 `mm^2`

4 X `pi*(20/2)^2`+ 4 x `pi*(16/2)^2` = 1946 `mm^2`.......................Hence Satisfy .

2) For Bottom:-

i) Total Ast required at top is = 1754

So we can provide

3 bars of diameter 20 mm and 5 bars of 16 mm satisfies the overall Ast

3 x `pi*(20/2)^2` = 942 `mm^2`

4 x `pi*(16/2)^2` = 1004 `mm^2`

3 X `pi*(20/2)^2` + 5 x `pi*(16/2)^2` = 1746 `mm^2`.......................Hence Satisfy .

3) For Shear details:-

• Right Click on beam and Click on Details
• Click on Shear Details

• As per Clause 6.3.3 we have to provide Spacing till 2d from the face of the column is 100 .
• It should not be more than that
• So provide 3 legged stirrups of 10mm diameter with 100mm spacing throught the beam length .

1) Reinforcement details along Grid 3:-

1. Max Ast at bottom support = 1026 `mm^2`

2. Max Ast at bottom middle = 1493 `mm^2`

3. Max Ast at Top support = 1915 `mm^2`

4. Max Ast at Top middle = 1026 `mm^2`

Reinforcement Calculation:-

1) For Top -

i) Total Ast required at top is = 1915 `mm^2`

So we can provide

3 bars of diameter 20 mm and 5 bars of 16 mm satisfies the overall Ast

3 x`pi*(20/2)^2` = 942 `mm^2`

4 x `pi*(16/2)^2`= 1004 `mm^2`

4 X T20 + 4 x T16 = 1946 `mm^2` .....................Hence Satisfy .

2) For Bottom:-

i) Total Ast required at top is = 1493 `mm^2`

So we can provide

3 bars of diameter 20 mm and 3 bars of 16 mm satisfies the overall Ast

3 x `pi*(20/2)^2` = 942 `mm^2`

3 x `pi*(16/2)^2` = 603 `mm^2`

3 X T20 + 5 x T16 = 1545 `mm^2`.......................Hence Satisfy .

3) For Shear details:-

• Right Click on and Click on Details
• Click on Shear Details

• As per Clause 6.3.3 we have to provide Spacing till 2d from the face of the column is 100 .
• It should not be more than that
• So provide 3 legged stirrups of 8mm diameter with 100mm spacing throught the beam length on both side
• And provide 3 legged stirrups of 8mm diameter with 150mm spacing at mid span .

Now we have to do Check for Sway shear .

A) For Right Sway:-

We know that the formula to calcuate ;

`V_(u,a)=V_(u,a)^(D+L)-1.4((M_u^(As)+M_u^(Bh))/L_(AB))` AND

`V_(u,b)=V_(u,b)^(D+L)+1.4((M_u^(As)+M_u^(Bh))/L_(AB))`

From Etabs Model results substitute the value -

i) `V_u^(D+L)`= 92 KN

ii) `M_(as)`= 176 KN-m

iii) `M_(bh)`= 295 KN

iv)`L_(ab)`= 4.725 m

Put this value in above formula

We get ,

`V_(u,a)`= 48 KN`

`V_(u,b)`= 230 KN

B) For Left Sway:-

i) Follow the same process

ii) We know the formula

`V_(u,a)=V_(u,a)^(D+L)-1.4((M_u^(Ab)+M_u^(Bs))/L_(AB))` AND

`V_(u,b)=V_(u,b)^(D+L)+1.4((M_u^(Ab)+M_u^(Bs))/L_(AB))`

From Etabs Model results substitute the value -

i)`V_u^(D+L)`= 74 Kn

ii)`M_(ah)`= 236 kn

iii) `M_(bs)`= 150 kn

iv) `L_(ab)`= 4.725 m

Put this value in above formula

We get ,

`V_(u,a)`= 190 Kn

`V_(u,b)`= 50 Kn

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

i) We know the formula -

`V_(us)=(0.87*f_y*A_(sv)*d)/S_v`

Put the values in the formula;

i)`f_y`= 415 N/mm2

ii) `A_(sv)` = 151 mm2

iii) Spacing = 100 mm

iv) d = 450 mm

Put this value in above formula;

We get ;

`V_(us)` = 245 KN...............> 230 KN ( Shear sway)........... ( Hence Safe )

Hence shear rebars in the drawing is okay as per as per IS-13920 code guideline

Result:-

1. Beam Design is done successfully with detailed reinforcement .

2. Provide reasons for failure of the middle span along grid A. What are the possible ways this issue can be resolved ?

• The main reason of failure of middle span along grid A is shear stress is exceeded the limiting value
• This may be beacause of the higher Shear force acting on that beam and due to provision of weak shear reinforcement at the end supports .
• This Issue can be Solved by -

1. By providing Higher grade of concrete in that particular section
2. Or by reducing the Shear force in that beam section
3. Or by providing higher shear reinforcement which can resist the shear force in that section

iii) Calculate 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.

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

We get the Shear sway value Right and Left support within limit

245 kN > 230 KN ...Hence shear rebars in the drawing is okay as per as per IS-13920 code guideline