Challenge

Reinforcement Detailing of Beams from ETABS output

AIM:

To provide detail design of beams in Etabs.
To provide reasons for failure of the middle span along grid A and provide solutions.
To 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.

Detailing Rule for any RC member:

Minimum longitudinal reinforcement as per IS-13920-2016 Clause 6.2.1 As=360mm^2

Minimum transverse reinforcement as per IS 456:2000 Clause 26.5.1.6 (Assuming 1000mm spacing in the formula) Asv=332mm^2

Development length as per IS 456:2000 clause 26.2.1 Ld= 41x dia of bar

PROCEDURE:

STEP 1:

STEP 2:

Click on concrete frame design icon.
The longitudinal reinforcements and shear reinforcements for all the members have been designed.
To toggle the view between longitudinal reinforcement and shear reinforcement click on Concrete frame design dropdown-> Display design Info-> Shear reinforcement.

Reasons for Failure of the middle span along grid A and solution:

The stress is above allowable value as per Codal requirement in this particular segment.

Therefore the Shear stress capacity of the segment can be increased by changing the grade of concrete to one with higher Compressive strength for this particular beam.

STEP 3:

Sway shear force to the right:

where,

Mas= 174 kNm

Mbh= 293 kNm

Lab= 4.725 m

Vd+l= 92 kN

Vu,a= 46kN; Vu,b= 230kN

Sway shear force to the left:

where,

Mah= 234 kNm

Mbs= 155 kNm

Lab= 4.725 m

Vd+l= 70 kN

Vu,a= 185kN; Vu,b= 45kN.

Details of longitudinal and shear reinforcement for 3 span continuous beam along grid A:

Details of longitudinal and shear reinforcement for 7 span continuous beam along grid 3:

Ship work like this
in your own course.