Below we discuss how to assign a rigidity diagram to a bar end. The same workflow can be applied to support functions .

• Select the relevant bars and click on .
• Select the option “Function” from the drop down menu next to the relevant internal force.
• Either choose one of the existing functions or define a new one using .
  Let’s assume we want to create a new one.
• In the dialogue box that appears, click .
  
• Name the function.
• For each sign of the internal force, choose if:
  • the internal force cannot be transfered = free
  • the internal force can be transfered = fixed
  • the internal force can be transfered linear = value
  • the internal force can be transfered variable = user defined
• You can save a defined support function in a TXT-file using the button . An existing function (TXT format) can be loaded via .

Below, you will find an example of a rigidity diagram possible for an arbitrary connection.

Rigidity diagram of a fixed moment connection on a bar end	Rigidity diagram of a deep foundation that can’t bear tension an only 100kN compression
The diagram shows that the rotation rigidity M/φ is constant as long as the applied moment is less than 45,4kNm. Once the moment is exceeded, the rigidity varies according to M. If the moment exceeds 68,2kNm, a plastic hinge is formed because an additional moment produces an angular deformation infinitely large.	The rigidity diagram should lay in the 1st and 3rd quadrant because in the formula F=k∙x the spring constant k cannot take on a negative value. If the function does not have a valid range, a red message will appear below this window.

(Balconies with thermal breaks – source: Plaka)

A thermal break is a reinforcement system with the main function of eliminating thermal bridges caused by structure elements overhanging reinforced concrete structures (balconies, cornices, canopies, etc.).

To design thermal breaks, manufacturers usually request the moment and shear force at the break. So what should you model and which internal forces should you give to the manufacturer?

• As seen on the image aboven, the reinforcement is not continuous over the total length of the balcony. But in a pre-design, we make an abstraction of this. We assume that the thermal break has the same propeties over the entire length of the balcony.
• The reinforcement present in the thermal break, is drawn on the Diamonds model below. This reinforcement ensures that moment Mxx ( = moment about the local z’-axis = orange arrows) can be transferred.
  Note: determining which forces can (or cannot) be transferred can only be done correctly by making the local coordinate system of the plates visible!

• Select the thermal break together with the plate and click on  .

• The orange arrows (two images above), have the same orientation as the moment framed in red:
  The red framed moment represents M_xx.

• If we then assume that the other moments cannot be transferred, then the two moments at the bottom must be unchecked:

• If we then assume that only the out-of-plane shear can be transmitted, then the normal force and in-plane shear must also be unchecked:

• If the manufacturer then asks what force is present in the thermal break, you draw a cut line over the entire length of the thermal break and request the average value for Mxx and Vx (= the shear force in a plane perpendicular to the x’ axis).

If you also want to assess the cracked deflection of the balcony, you must request the rotational capacity of the thermal break from the manufacturer (the 1kNm/rad/m in the printscreen below is an arbitrary value).

Shape 1 – individual steps are recognizable

Note: the video has no audio

Shape 2 – curved surface is recognizable

To generate a surface:

1. the selected perimeter must be closed
2. the selected perimeter must be in same plane.

For curved surfaces, the difficulty lies in the second condition. The easiest way to walk around around it, is by using triangles. But still this can cause struggles in Diamonds. Alternatively, you could use other software (like AutoCAD, Rhino or Grashopper) which are more intended to model complex shapes and import the curvature using DXF.

Note: the video has no audio

Here are some ways of handling beams with openings.

Take the opening into account in the global analysis in Diamonds

There is no direct & simple solution to deal with openings in concrete beams in Diamonds.

Here is a work-around that you can follow – for small & simple models only(!):

1. Divide the beam in at least 3 parts
2. Position the middle part to be at the location of the opening
3. Make the middle part have the length of the opening (more or less)
4. Edit the middle part with Section Utility so that the opening is present – this involves deleting a part of the section.
5. Calculate the model

Following this method, you will understand that small openings can be ignored, they are not worth modelling.

Detailing of the reinforcement around the opening

The detailing of the reinforcement around the opening cannot be done in Diamonds.

Theories on how to handle this can be found here:

• [EN] Book & Article ‘Concrete Beams with Openings’ by M.A. Mansur
• [NL] Article in Cement magazine

You can either use the above references to do the detailing by hand or use a software to help you, for example MatrixTools – Strut&Tie. In this module, you can model the beam and its opening. The software will determine the tension & compression rods and propose a suited reinforcement scheme.

→ More info on MatrixTools

A wet sweater (5kg ≅ 0,05kN) is hanging on a steel string which spans 5m. What would happen if we’d consider the the string as a beam and as a cable?

If we would consider the string as a pinned beam, the bending moment would be

That seems small. But the string is only 5 mm in diameter (section modulus W = 12,27 mm³). This means that in order to carry the weight of the sweater, the string should withstand the stress of

Only 20 times more than S235 steel… We wonder were that gigantic strength is hidden. And yet it works…

The above example is a typical linear approach to a problem. Perhaps you noticed that we avoided the topic ‘deflection’. In the beam-approach the deflection would be

Think how much the string would have to elongate to deform that much (5m string deflecting 20m?!?). This question brings us closer to understanding nonlinear geometry… where large deformations theory is being used.

The string cannot elongate 20m as linear approach suggests, but it will elongate some for sure. This elongation will cause a tensile force in the string. In turn, this tensile forse will stabilize the structure, reducing the deflections. Using literature, we can approach the tensile force and resulting stresses, deformations:

It is more to believe the string will deform 57mm under a tensile stress of 55,4 N/mm², instead 20m under a stress of 5093 N/mm².

Conclusion

• A cable does not behave like a beam.
• Cables cannot be modelled in Diamonds. Not with a beam, nor with a tie rod.
  There is no element in Diamonds supporting the tensile forces needed to describe the cable-behaviour. Diamonds is not suitable for large deformation theory.
• Even though tensile forces arise in tie rods, you cannot load tie rods perpendicular to their axis in Diamonds. Diamonds will simply refuse to calculate the model.

(Source cable formula’s: Roark’s Formulas for Stress and Strain, ISBN 0-07-072542-X)

Using button

• Select the bar(s) you want to define as a tie rod.
• Assign the desired cross-section to them using or .
• Click on .
• Select .

Using function

In some models, tie rods defined with the button can result in interation problems because the algorithm deciding which tie rod is working and which one is not, is too strict. The alternative is using a function in which you can define your own limit.

• Select the bar(s) you want to define as a tie rod.
• Assign the desired cross-section to them using or .
• Click on .
• With M_y’ and M_z’ choose ‘Free’.
  
• With N_x’ choose ‘Function’ and click on .
• Click on and give the function a name.
• Enter a small value with ‘<0’ and set ‘>0’ to ‘Fixed’.
  The definition of ‘small’ depends on the model. For some models it will be 0,1 kN/m, for others it will be 1kN/m.
  Choose a value and calculate the structure. If it calculates, divide the value for example by 2 and try again. If it doesn’t calculate, multiply the value by 2 and try again.
  You’ll see that after a few attemps, the compressive force in the tie rods will be neglectable in comparison to the tensile force.
  
• Assign the function also to the end of the bar.

• Find the parent section in the Diamonds section library and adjust its height H.
• Adjust the properties I_y, W_y (elastic and plastic), I_T and A_Vz (substract opening from the properties of the gross section)
• Adjust the imperfection factor for lateral torsional buckling to
  • curve c (=0,49) for double symmetric castellated beams (circular or hexagonal openings) made from hot rolled parent sections
  • curve d (=0,76) for all others
• Check that the cross section of the tee section (part of the web above and below the opening) is less conservative than the cross section of the entire beam. If not, also adjust the cross section classification.

Next to the usual checks from EN1993-1-1 (implemented in Diamonds), the additional checks mentioned below also apply (not implemented in Diamonds). These additional checks are mainly local and relate to the tee section (part of the web above and below the opening) and the web post (part of the web between two openings).

• Buckling of compressed tee due to global bending
• Resistance of tees in Vierendeel bending
• Web post shear, bending and buckling

Also note that the deformation will be larger than calculated with Diamonds because of the effects for Vierendeel shear deformation along the openings. This can be calculated with a Vierendeel (framework) analogy.

A new Eurocode part is in the making about these beams (EN1993-1-13: Steel beams with large web opening). Current design rules can be found in:

• Lawson, R. M., & Hicks, S. J. (2011). Design of composite beams with large web openings (SCI P355). SCI (Steel Construction Institute).
• CTICM. Arcelor Cellular Beams: Detailed Technical Description.

More extensive calculations are possible with:

• ArcelorMittal software (ACB+, developped by CTICM: https://constructalia.arcelormittal.com/en/products/acb)
• Westok software (Westok Cellbeam, developped by SCI: https://www.kloecknermetalsuk.com/westok/design-service-software/cellular-beam-software/).

• The first method is used for vessels with vertical walls
• The second method is mainly used for vessels with curved walls

Method 1

• Enter the perimeter of the vessel in a top view.
  • You can use the drawing function .
  • You can use the tool for parametric shapes (circle and rectangle) .
  • Or you can import a dxf-file.

  No matter with tool you use, make sure the perimeter is not too detailed. Avoid lines smaller than 10 cm.

  For example, a circle with radius 3m and 16 divisions.

  

• Remove the center point.
• The next step is copying the perimeter.
  • Select the perimeter.
  • Click on . Ask one copy. Specify the distance. Here 3m. And don’t forget to generate the connecting surfaces. Click ‘OK’.
  • Change the section and material of the walls as desired with .
• The final step is generating the baseplate.
  • Select the lower perimeter and click on  to generate a plate.
  • Change the section and material of the baseplate as desired.
  • After assigning supports and loads, this model can be calculated.

Method 2

• Enter the perimeter of the vessel in a front view.
  • You can use the drawing function .
  • You can use the tool for parametric shapes (circle and rectangle) .
  • Or you can import a dxf-file.

  No matter with tool you use, make sure the perimeter is not too detailed. Avoid lines smaller than 10 cm.

  For example, this curve imported using a dxf file.

• The next step is copying the perimeter.
  • Select the wall shape.
  • Click on . Ask 16 copys, select the rotation plane (XZ), and specify the angle (22.5° = 360°/16).Don’t forget to generate the connecting surfaces. Click ‘OK’.
  • Change the section and material of the walls as desired.
• The final step is generating the baseplate.
  • Select the lower perimeter and click on to generate a plate.
  • Change the section and material of the baseplate as desired.
  • After assigning supports and loads, this model can be calculated.

Click on . This will set t he bending moment My’ and Mz’ free.

Select ‘Function’ next to ‘Nx’ and click on . 

Compression is negative, tension is positive. So define a function for which applies: if F<0 (= compression) then the bar can resist it (=fixed) and when if F>0 (= tension) the bar can’t resist is (= free).
But because ‘Free’ is the equivalent of 0kN, it’s recommended to enter a very small value. Calculating with 0kN migth give iteration problems. The definition of ‘small’ depends on the model. For some models it will be 0,1 kN/m, for others it will be 1kN/m. Choose a value and calculate the structure. If it calculates, divide the value for example by 2 and try again. If it doesn’t calculate, multiply the value by 10 and try again. You’ll see that after a few attemps, the tensile force in the compression bars will be neglectable in comparison to the compression force.

Select the function for the other end of the bar.

Note: the countarpart of a compression bar is a tie rod.

For a column

With Hooke’s Law and the formula for stress , a formula for the spring constant can be deducted. This formula assumes that the element is supported at the buttom.

• the modulus of elasticity of the column
• the cross-section area of the column
• the length of the column

This value must be entered in :

For a foundation pile

In case of a foundation pile, there are two springs in series: the one for the pile and one for the soil .

If there is no data available on the soil, but you still want to take the effect into account, you could assume that . The formula for the spring contant becomes:

The "2" is refered to as the stiffenss ratio in the calculator below.

This value must be entered in :

For a wall

Again Hook's law and the formula for stress are used. But since the length the wall will be modelled as a line support, this dimension is no longer in the formula:

• the modulus of elasticity of the wall
• the thickness of the wall
• the height of the wall

This value must be entered in :

For a foundation pit (supported area)

Again, there are two springs in series: the one for the pit and one for the soil .

If there is no data available on the soil, but you still want to take the effect into account, you could assume that . Both dimensions of the surface support will be modelled. Which gives this as the formula for the spring constant :

• the modulus of elasticity of the surface support
• the height of the pit
• The "2" is refered to as the stiffenss ratio in the calculator below.

This value must be entered in :

Replacing surface supports (like foundation pits) with springs, implies that you to model the circumference of the foundation pit. Therefor it is advised to apply this type of spring constant only to smaller models. For example: in a model that only contains the foundation slab and not all floors above it