﻿ Examples > Uniaxial_Shear_design

# Uniaxial_Shear_design

The above cross section of a beam refers to Example 6.4 in /3/.  C30/37 is the concrete class and S450C is the steel grade of stirrups.

The concrete design strength is:        fcd = αcc fck / γc = 0.85 30 / 1.5 = 17.0 Mpa

The steel design yield strength is:        fyd = fyk/ γs = 450/1.15 = 391.3 Mpa

The coefficient ν1= 0.616 (in eq. (6.9) EC2)

Longitudinal reinforcement for bending resistance are not given but it is known only the lever arm of internal force: z = 67.5 cm

The problem is to design stirrups to satisfy a design shear force VEd = 600 kN.

We want to do the same and compare the results.

INPUT DATA

Once opened a new calculation (by means the command "New" in the menu File) it is important to control if the default Code setting options agree with those in the given example to compare.

In this example we control first the parameters γc, γs , αcc. The EC2 recommended value of the last parameter αcc is  1.00 and it is to change in 0.85 as in the example. If a specific National Annex is selected no change is necessary.

The value ν1= 0.616 is not conform to EC2 recommended value 0.6(1-fck/250) = 0.528; then we entered the assigned value 0.616 in the below window.

To obtain the same diameter for stirrups bars we assigned the same diameter 12 mm to the first and the second choice.

In the General Data windows we selected the "predefined" and Beam as Member Characteristic.

In the below Section Data window we select:

-  The option "Check" as the longitudinal reinforcement is known

-  Materials classes for concrete and steel bars (for longitudinal and stirrups reinforcement)

-  Shape (Rectangular) and Dimension B, H of  the section

-  Cover and data of longitudinal bars: we defined the areas of the two rows of bar such to obtain z=67.5 cm like in the original example.

- To design only shear reinforcement stirrups we set to 0 the diameter of stirrups:

In the Force window we cannot assign only the shear Vy = VEd = 600 kN but also the axial and moment force (not present in the original example).

RESULTS

Below printed results show that the designed stirrups are 2-leg Φ 12 /17.8 cm  very near to  the -leg Φ 12 /17.0 cm in the original example. The angle θ  of inclination of shear struts is 29°.04 (29°.0 is the original result).