The two elements that “compete” in CYPECAD will now be analysed: Shear walls and walls.
These have been created to do what their name indicates: provide vertical support in such a way that they must resist vertical loads and fundamentally, horizontal loads. They are to behave as rigid nuclei.
There is no doubt that they will meet their requirements if they are designed correctly and as long as the architectural distributions allow for this.
Rules for shear wall design in CYPECAD:
- On plan, it is a figure made up of segments or straight sections joined at their vertices. Any shape can be created as long as it consists of a group of interconnected segments, making up an open or closed polygon.
- Its on plan shape is constant throughout its height. A side cannot disappear or become a column.
- It cannot overlap another support, be it column or wall.
- It cannot have any openings. If the user intends on placing a door within it, it is not possible. Its ends can only be joined to the horizontal floor slab elements at each floor level, e.g. a beam.
Shear wall discretisation carried out by CYPECAD:
- The model consists of a thick shell made up of triangular finite elements with six nodes (three vertices plus the three intermediate points) and six degrees of freedom per node. The triangles are always regular and equal at each floor.
- A bar is generated at the intersection with the floor. The bar’s width is the same as the side of the shear wall and has a minimum depth of 20cm or the depth of the slab it touches.
The bar’s nodes consist of the nodes of the triangles that touch it, and the bars that reach it from the floor slab. This bar connects the vertical elements to the horizontal elements, logically transmitting the bending and torsion forces between them. If it is the last job that has been analysed, it may be interesting to click on the Results tab (Envelopes > 3D Model), to view the triangular discretisation and consult the geometric and mechanical properties
Shear wall reinforcement criteria in CYPECAD:
- A shear wall is a surface element submitted to the six corresponding forces. It will have vertical and horizontal reinforcement at either side, which will be constant throughout the height of each floor at which it is built. It may differ at each side and span.
- The reinforcement is obtained from reinforcement tables ordered sequentially. The program checks the reinforcement with the existing forces at each node. The reinforcement is gradually increased until no nodes of either side fail for each reinforcement position.
- The program allows users to establish a percentage of nodes passing all the checks within the program options i.e. the Compliance Factor (Job > General data > By position button > Column options > Required compliance factor for walls and shear walls).
Each node has a tributary area. If the compliance factor is 90% (default value), it means that the reinforcement placed by the program complies with 90% of the surface of the side where the reinforcement has been placed. The rest of the surface may fail a check. In this case, the compliance factor is indicated. For example, if the compliance factor equals 95.6%; 4.4% fails.
By selecting ‘Design’ (Column Introduction tab > Introduction > Columns, shear walls and starts) and clicking on a shear wall, a dialogue box appears displaying the reinforcement laid in the shear wall and any nodes that fail shown in red.
By clicking on the ‘Show reinforcement’ button in the aforementioned dialogue box, the nodes change colour and can be selected. A report is displayed on screen indicating the reinforcement that fails (vertical or horizontal, left or right), the steel area of each reinforcement (ratio between the required steel area and that provided) and an additional reinforcement proposal.
Checks carried out by CYPECAD on shear walls due to them consisting of a thick shell:
The program checks the following for each loadcase:
- Concrete: Its stresses and unit deformations
- Steel: Stresses and deformations of each reinforcement position at each side, be it vertical or horizontal.
Additionally, CYPECAD checks the limits established in the selected code (minimum and maximum diameters, spacing, steel areas, reinforcement criteria and options implemented by CYPE) and indicates if any fail.
The discretisation, forces, displacements and stresses by loadcase can be seen in the Results tab > Envelopes > Forces in columns and shear walls.
The program also provides a report (File > Print > Job report > Forces and reinforcement of columns, shear walls and walls) which displays the forces of the resultant per loadcase applied at the geometrical centre of the shear wall, and the worst case combinations (those which fail the previous entry of the reinforcement table and require the reinforcement which has finally been placed).
In the column and shear wall reinforcement layout drawings, the layout arrangement of each shear wall at each floor is displayed, with its transverse section, its take-off and take-off summary.
They allow for more flexibility when designing compared to shear walls.
Rules for wall design in CYPECAD:
- Their shape does not have to be constant along their entire height.
- They can go through other supports such as columns. Columns can also start on walls at any floor, or, alternatively, a wall can start on several columns.
- Soil with a water table at a specific elevation can be retained, hence applying at rest horizontal lateral pressures to the wall at one or both of its sides.
- Walls do not only have to be made of reinforced concrete. They may also be generic masonry walls, whereby the user defines its properties, or concrete block walls. If concrete blocks are selected, the selected code will be applied, with the further option to reinforce the wall with ties and vertical bars.
- Openings may be defined at its sides, such as doors or windows.
- The wall may also have associated to it its foundation composed of a strip footing without external fixity.
Wall discretisation carried out by CYPECAD:
- The discretisation in walls is similar to that of shear walls. The model consists of a thick shell made up of triangular finite elements with six nodes (three vertices plus the three intermediate points) and six degrees of freedom per node. On the contrary to shear walls, the finite element mesh in walls is designed by the program. It is composed of irregular triangles with variable dimensions to provide a refined mesh so to adapt itself to the edge nodes and geometrical irregularities. For example, the program makes the bar nodes of the slabs coincide with the nodes of the triangular mesh. This is enough to understand why the results provided by the program are different if a wall is placed instead of a shear wall in the same structure; however, they will be similar.
- Due to the rigid diaphragm hypothesis in slabs, the degrees of freedom at floor level are restricted, as occurs in the case of shear walls. It is important to highlight this in the case of the following example: if a wall is designed to start from two columns (something that cannot be done with shear walls), the start beam at that floor level will be designed for bending, but will not be designed to behave as a tie i.e. designed for axial forces. This case could arise when walls are introduced that behave as beams with great depths.
Wall reinforcement criteria and checks in CYPECAD:
- The user should recall a thick shell is being reinforced with uniform reinforcement on both sides. Reinforcement cannot be concentrated in an arbitrary way be it at the wall edges or corners. This may imply an increase in steel area if compared with other manual solutions where the vertical reinforcement is concentrated at those points or edge zones where it is known that it is required. This situation for walls can be improved in some cases by placing columns at the corners and edges whose side measurements are greater than or equal to the width of the wall. This cannot be done with shear walls. In these cases, the program provides, on one hand, the reinforcement of the sides of the wall with its elevations; and on the other, that of the columns in the column schedule. Therefore, it is best to manually complete the wall drawings with section details which include the reinforcement of the columns contained within the wall to provide a complete reinforcement detail.
- The section design is the same as for shear walls. The only difference being that these are consulted at: Results tab > Beams/Walls > Edit walls.
- The most common cases are described within the Calculations manual of the program, together with the assumptions and limitations to be taken into account to make good use of CYPECAD in accordance with the model used. For example, if lateral pressures have been activated, the minimum steel areas of horizontal reinforcement established by the selected code are applied. If they are not, the program applies half of the minimum steel area, as it consists of a wall providing support, i.e. a shear wall.
Things to bear in mind when using shear walls and walls:
It must not be forgotten that by including shear walls and walls in a structure increases the time taken to run the analysis, due to the enormous increment in degrees of freedom which its discretisation implies. This increment in time is substantially greater in the case of walls which intersect with flat slabs, waffle slabs or large floors; those which have many door and window openings; or those which are exposed to small elevation changes within the same floor; as the discretisation mesh which is created consists of very small triangles.
The user should also recall that a linear first order analysis is being undertaken and that the model is an approximation to reality. Therefore, the degree of precision, although sufficient, cannot be great and so imperfections in the model and method must be accepted. Hence, wanting to rigorously introduce every last detail (bend, curve, small opening) is not going to provide more precise global or local results.