Transversal reinforcement calculation - Example

 

We consider a column with a 0.3 x 0.4 m rectangular section and a 3.10 m height. The concrete class is 25/30. The loads on the column are the following:

 

 

Top loads

 

Bottom loads

 

Combinations

 

 

 

The shear verification is done according to article 6.2 of EN1992-1-1.

The sizing combination for the shear verification on the X direction is:

Combination 104 : +1.35x[1 G]+1.5x[2 Q]

Design shear force VEd = 135 kN

The shear verification is accomplished if the following relation is satisfied:

VRd ≥ VEd

VRd = min (VRd,max; VRd,s)

Vrdmax

 

αcw = 1.2

bw = 0.40 m

z = 0.9 × d

 

model

 

 

Since, in the design phase, the real d’ is not known, the program will use in the calculation of the concrete cover from the Concrete cover dialog:

 

Concrete cover

 

 

In the initial stage, the calculation will be done using the value from the viewport. After a certain number of iterations, a longitudinal reinforcement area and a transversal reinforcement area with specific diameters will result.

 

Diameters

 

 

In the final stage, the calculation is done again with the real d’ obtained as:

diameter

The report will reflect the final stage, with the real value of d’.

the real value of d

 

The value of Ɵ can be defined in the Design Assumptions dialog, Transversal Reinforcement section (Strut slope).

 

Compression strut angle

 

 

formula

 

The program verifies if:

Aswreal

 

If not, the spacing sreal is decreased or the transversal bars diameter is increased until the relation is verified.

 

Main reinforcement

 

Top section

 

Asw formula