When the effect of shear is considered, at what point does the design bending moment change?

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One of the most common questions/exercises we do in the FAGUS online course involves showing a cross section with applied forces on the Axis Point and in which we also want to consider the effect of shear. The student should know how to translate those forces onto the centre of gravity, with the necessary transformations to make the calculation.

One of our students asked ‘why the design moment My does not increase if a tensile axial force appears running along the reinforcement’. The question is quite interesting and made us write this article, in which we insist again in:

• The phenomenon of shear as a ‘travel of the loads to the supports through the analogy of the truss’.
• The way that FAGUS works with the shear through the shear walls.
• The 3 design models V-T that FAGUS proposes to work with shear and torsion

The shear force on a beam always develops when there is a variation in its bending moment law. In other words, the diagram of shear forces is linked to the diagram of bending moments so one cannot be understood without the other.

Due to the effect of the work in the truss, the axial forces are modified on the head of the beam. The longitudinal reinforcements subjected to bending must be capable of holding an increment in tension with respect to the one resulting from Md, for that purpose the diagram of bending forces is shifted one effective depth "d" to the most unfavourable part, thus taking into account the effect of the oblique crack caused by shear (the value d is there for safety purposes).

 

The internal truss structure, generated from the compressed struts and the tie bars that compose the shear reinforcements (vertical frames) and the longitudinal ones, is the one balanced by all the forces acting on it.

 

Taking all of the above into consideration we are going to explain how FAGUS makes the force transformation focussing them on the shear, which is the subject of the question.

In the example given in the video of the course we have a rectangular cross section that, logically, is symmetric with respect to both axes. The transformation of the forces considering the shear will mainly depend on two things:

 a) The geometry of the cross section

In the example, having a rectangular cross section we have the centre of gravity just at mid height. Additionally, our shear wall supporting the vertical shear Vz has that same height, which means that the shear centre of that wall coincides with the centre of gravity of the cross section. The result of the transformation is an equal shear to the one entered and an axial force on the point we are talking about without any eccentricity with respect to the centre of gravity of the cross section that, consequently, generates no moment.

All this is different if instead of a symmetric cross section with respect to a horizontal axis that goes through the centre of gravity we have an asymmetric one with respect to that axis. In this case there is a distance between the centre of gravity of the section and the shear centre of the shear wall. Let’s consider the following cross section:

We see that the centre of gravity is 0.80 m away from its base.

In this cross section we can define the following shear walls:

 

If we look at the vertical shear wall, the height of which lies between the middle line of each wing (1.1 m), we can see that its shear centre would be in the middle point, that is, 0.55m away from any end. This would result in an eccentricity with respect to the centre of gravity of the cross section of 0.15 m.

If we introduce a 10 kN shear and we consider an angle of 45º on the compressed rods it would be normal to expect the appearance of an axial force on said centre of the vertical shear wall (of the same value as the shear due to those 45º).

 

This axial force of 10 kN multiplied by the eccentricity of the cross section until the centre of gravity results in a moment of 10*0.15 = 1.5 kNm.

 If we look at the force transformation done by FAGUS:

We see that increase in My.

This is explained in the following section of the FAGUS manual:

b) The V-T model selected:

For everything explained in point a) to happen it is very important to choose the V-T model correctly. At this point we would like to recommend the reading of our article on that topic which is available in our help centre and to which you can access with this link:

"How to choose the type of shear-torsion model (V-T) that better fits the case of study?" 

In this case the ideal model is the Type A, since in sections like this the program will decide which wall will support the shear force entered depending on the geometry. The Type B model will also be valid but it will be us who must specify what wall supports each force.

The automatic model will make the existing walls on the wings take some of the vertical shear due to the shear circuit introduced inside each wall. That small part of the vertical shear is not shown by the program and we cannot control it. This is why this model is not the most adequate for this cross sections, where is is clear that it will never be of interest to have the wings working on a Vz shear.