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BANNER-1-9-PRONTUBEAM

Deep beam – Linear VS Non-linear calculation

November , 2th 2023 | Author: Prontubeam (@Prontubeam_en) Read: 2216 times

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As it is well known, as explained in “Deep beams – Strut and tie VS Non-linear models” article, the deep beams are particular elements that generally needs to be studied using a strut and tie model. As presented in the graphic “Stress distribution variation in beams and deep beams” the stress distribution varies in function on the ratio span/height of the deep beam (L/H). As presented below, the lesser is the ratio L/H, the bigger is the difference between the linear plane stress distribution and the real one.

Stress distribution in Deep beams

Figure 1. Stress distribution in Deep beams

As these elements, the deep beams, does not follow the classic Navier-Bernoulli stress distribution they cannot be studied using the classical formulas and a strut and tie model has to be applied. In some cases these elements are studied using a linear finite element model with linear material laws and stress distributions. However, in the ULS combinations, the materials reach their elastic limits, concrete is cracked, and stress redistribution is produced.

 

Linear stress distribution in deep beams – Principal directions

Figure 2. Linear stress distribution in deep beams – Principal directions

The aim of this article is to perform a deep beam calculation using linear hypothesis and compare it against a full nonlinear calculation with nonlinear concrete properties (cracking and crushing) and nonlinear reinforcement law (bi-linear law considering reinforcement plasticity). For this purpose:

·         An Ansys model is used, using SOLID65 elements, representing a 10m x 7m x 0.5m deep beam. The supports are 1m x 1m x 0.5m, used to avoid concentration of stresses that appears when supports are directly applied to the model. It is considered that, for the nonlinear calculation, the wall is reinforced with a ratio of 0.00314 (equivalent to C20@400mm (15.7cm2/m) on each face).

·         Three different distributed loads will be applied on the top of the beam to compare both, the linear and the nonlinear models. The load will be increased as to start with low values, where the reinforcement is not yielded and finishes with the maximum allowed load where the reinforcement will be fully yielded and reached it limit strain.

 

Model geometry, materials and loads

The following picture presents the geometry that, as explained above, it is a 10m x 7m x 0.5m deep beam with supports of 1m x 1m x 0.5m, leading to a free span of 8m:

Ansys model deep beam geometry

Figure 3. Ansys model geometry

For the reinforcement, generally, as explained in the following picture extracted from “Dimensionnement des constructions selon l’eurocode 2 à l’aide des modèles beilles et tirants - Jean-Louis BOSC” the lower tie can be supposed placed at a height of 0.15L that, for our studied model, is around 0.15*9m=1.35m. As the deep beam has been reinforcement with a ratio of 0.00314, this means that, in the given height of 1.35m there should be 1.35*0.5*0.00314*10^4= 21cm² as the main principal tie.

Deep beam strut and tie model

Figure 4. Deep beam reinforcement location following “Dimensionnement des constructions selon l’eurocode 2 à l’aide des modèles beilles et tirants”, a book of Jean-Louis BOSC

The linear model is done using a 27GPa module young material. For the non-linear model it has been used the CONCRETE material available in Ansys considering 0.4MPa as the uniaxial tensile cracking stress and 30MPa of uniaxial crashing stress. The steel has been modelled using the following non-linear material law:

ANSYS non-lineal reinforcement steel material

Figure 5. ANSYS non-lineal reinforcement steel material

For the loading it applied 3 different loads, all of them distributed load applied on the top of the deep beam. The results from the linear and the non-linear calculation will be compared for: 100kN/m, 700 kN/m and 1500 kN/m.

 

Case 1 – Total load: 1000kN (100kN/m on the upper side)

For the case where 1000kN are applied:

·         the concrete is still not cracked in the horizontal direction in the middle of the span

·         small cracks have appeared close to the supports but they are closed (able to transmit  shear)

The following picture presents the crack status (left) and the crack pattern (right):

CCrack status in the nonlinear model (left) – Crack pattern (right) – Case1

Figure 6. Crack status in the nonlinear model (left) – Crack pattern (right) – Case1

The following picture presents the nodal force distribution in the middle of the wall in the whole height. As it can be seen, for the case where the concrete is almost not cracked, both models represent almost the same results:

Horizontal nodal force N) distribution in the middle of the span of the deep beam – Not cracked

Figure 7. Horizontal nodal force (*) (N) distribution in the middle of the span – Case1

(*) Note that each value presented is a nodal force. As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture. The distance between nodes is 0.5m, therefore, for example, the maximum force in the nonlinear is: 24,68(kN)*2(nodes)/0.5(element height) = 98,7kN/m.

The following graph compares the nodal forces of the picture above:

Nodal comparison-linear nonlinear deep beam low crack

Figure 8. Horizontal nodal force (N) comparison – Case1

As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture above.

It has also been calculated the total reinforcement that would be required for the nodal forces presented above. For the negative forces, meaning compression, the reinforcement is set to 0cm2 but the minimum reinforcement should be applied. The figure below presents the total required reinforcement, this means for the two nodes present in the thickness of the element (in the picture above the nodal force of only one node on the thickness is provided).

Figure 9. Required reinforcement (cm2) comparison – Case1

The following picture presents the stresses on the reinforcement. As the concrete is not cracked, the stress on the reinforcement is very low (less than 50MPa):

Figure 10. Reinforcement stress (N/m2) in the non-linear model – Case1

 

Case 2 – Total load: 7000kN (700kN/m on the upper side)

For the case where 7000kN are applied:

·         the concrete has cracked in the horizontal direction in general, through the whole wall

·         The stress on the reinforcement is still low as presented below (250MPa)

The following picture presents the crack status (left) and the crack pattern (right):

Figure 11. Crack status in the nonlinear model (left) – Crack pattern (right) – Case2

The following picture presents the nodal force distribution in the middle of the wall in the whole height. As it can be seen, both models start to differ due to the cracking of the concrete and, therefore, the loose of stiffness:

Figure 12. Horizontal nodal force (*) (N) distribution in the middle of the span – Case2

(*) Note that each value presented is a nodal force. As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture.

The following graph compares the nodal forces of the picture above:

Nodal-comparison-linear-nonlinear-deep-beam-medium-crack

Figure 13. Horizontal nodal force (N) comparison – Case2

As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture above.

It has also been calculated the total required reinforcement for the nodal forces above. For the negative forces, meaning compression, the reinforcement is set to 0cm2 but the minimum reinforcement should be applied. The figure below presents the total required reinforcement, this means for the two nodes present in the thickness of the element (in the picture above the nodal force of only one node on the thickness is provided).

Figure 14. Required reinforcement (cm2) comparison – Case2

The following picture presents the stress force on the reinforcement. As the concrete is already cracked, the stress on the reinforcement is increased compared to the previous case. It present stresses between 200 and 300MPa, no yield is detected:

Figure 15. Reinforcement stress (N/m2) in the non-linear model – Case2

 

Case 3 – Total load: 15000kN (1500kN/m on the upper side)

For the case where 15000kN are applied:

·         the concrete is full cracked on the height of the wall

·         supports are modelled with linear concrete material to avoid local instabilities where the supports are modelled

·         Reinforcement is fully yielded on the lower tie

The following picture presents the crack status (left) and the crack pattern (right):

Figure 16. Crack status in the nonlinear model (left) – Crack pattern (right) – Case3

The following picture presents the stress force on the reinforcement. The concrete was already cracked, the stress on the reinforcement is increased compared to the previous case presenting local yield on the lower main tie (500MPa) reached:

Figure 17. Reinforcement stress (N/m2) in the non-linear model – Case3

The following picture presents the nodal force distribution in the middle of the wall in the whole height. As it can be seen, the stress distribution is totally different and the reinforcement is yielded on the lower part:

Figure 18. Horizontal nodal force (*) (N) distribution in the middle of the span – Case3

(*) Note that each value presented is a nodal force. As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture.

The following graph compares the nodal forces of the picture above:

Figure 19. Horizontal nodal force (N) comparison – Case3

As there is one SOLID65 in the thickness of the wall, there are 2 nodes, so the total force is twice the one presented in the picture above.

It has also been calculated the total required reinforcement for the nodal forces above. For the negative forces, meaning compression, the reinforcement is set to 0cm2 but the minimum reinforcement should be applied. The figure below presents the total required reinforcement, this means for the two nodes present in the thickness of the element (in the picture above the nodal force of only one node on the thickness is provided).

Figure 20. Required reinforcement (cm2) comparison – Case3

 

CONCLUSION

It is noted that the stress distribution is totally different on the elastic linear case and on the non-linear plastic case with the concrete fully cracked and the reinforcement starting to yield.

For the studied cases the following points are observed:

Nodal distribution:

·         When the concrete is not cracked, the nodal distribution on the model is quite similar

·         When cracking appears, there is a huge loose of stiffness on the non linear model on the cracked area. The tensile height is highly increased and the stress distribution is more constant on the height of the wall compared to the linear case

·         For the Case 3 (maximum resisted load) the linear calculation presents a peak of nodal force of 504kN, concentrated on the lower part. It is appreciated that the high nodal forces appears in approximately 1.0-1.3m on the lower part of the wall. The distribution in the linear case matches quite well with the standard strut and tie models

·         However, the nonlinear model stress distribution is totally different to the linear one and to the strut and tie proposed distribution. The non-linear calculation presents a more uniform distribution of the nodal forces. It is checked that, in the approximately 2 lower meters the reinforcement is yielded. It presents a peak of 196kN matching with the yielding value of the reinforcement.

Reinforcement design:

·         Linear calculation: Using the linear results the reinforcement would have been designed to resist a peak load of 1008kN in 0.5m height on the lower part (1707kN/m if the values in 1m are used – 34cm²/m). This means that the linear calculation ends up with a high concentration of reinforcement on the lower tie

·         Non-linear calculation: The studied wall, reinforced with a uniform reinforcement of 15.7cm²/m would resist the loads. The reinforcement on the lower face would have yield but not achieved the limit rupture strain.

·         It is interesting that the total reinforcement, calculated using the nodal forces, is almost the same for both cases, but distributed differently. However, it is noted that if reinforced in the linear calculation, the reinforcement would be concentrated in the lower tie and the minimum reinforcement should also be placed on the whole height of the deep beam, leading to higher amounts of reinforcement. It is reminded that the nodal forces on the non-linear model are linked to the modelled reinforcement. If it would have been modelled a different reinforcement, different loads would have been obtained

In general, it is observed that the distribution of the required reinforcement is strongly concentrated on the lower part of the deep beam while, in reality, it could be permitted a more uniform distribution thought the height of the wall. However, it is needed to rely in the ductility capacity of the reinforcement distributing the stresses leading to higher strains due to a yielding of the reinforcement.

It is reminded that this article studies one particular case and conclusions might vary if applied on other cases under different conditions. It is the responsibility of the engineer to define and verify the correct hypothesis and reinforcement design.

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Carlos Corral
Carlos Corral . MEng Civil Engineering from the Politécnica university of Madrid. Speciality: Structural engineer. Owner and programer of Prontubeam.com and Prontubeam.com/en.
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