Pinned Support for Beam Division Points to simulate Beam on CMU

[HoshangMustafa]

Hi

I wonder how one can simulate beams on CMU (Concrete Masonry Unit) supporting a slab? I tried it following these steps:

1- I Divided beams into n parts.

2- I set pinned support for nodes created in step 1.

I keep getting Detailing condition problem.

I used Panel cuts and the results for moments at the supports (nodes created and set as pinned) are as predicted (there are moments at the supports indicating the slab is supported by beams rather than nodes created and set as pinned).

Consider Beam 7. Why one should get My values and Fz values if the beam is continuously supported by created nodes set to pinned (at the positions of these nodes)?

I wonder whether the approached steps simulate the beams as being supported by the created nodes (which set to pinned support).

This is a common practice here to reduce deflection of slab.

I reviewed the threads on Deflection condition problem, followed the solution. Still getting the problem.

Please find the attached file.

[Simau]

Hi @HoshangMustafa 

I will instead model the CMU with elastic ground.
It is necessary to estimate the vertical stiffness of this wall

[HoshangMustafa]

Hi @Simau 

You wrote:

I will instead model the CMU with elastic ground.
It is necessary to estimate the vertical stiffness of this wall

How CMU (concrete masonry unit) could be modeled with elastic ground (soil coefficient)?

How would you estimate the vertical stiffness (membrane stiffness) of this wall?

Your attachment can't be opened. If possible send it with an earlier version (2019).

[Simau]

@HoshangMustafa 

Robot 2019 is no longer installed here.
Here are some screenshots of the model.

To estimate the vertical stifness, model the CMU with a storey height a, its dimensions and  its material characteristics .

Apply a unit effort N and look at the displacement d.

Kz= N/d

[HoshangMustafa]

Hi @Simau 

you wrote:

To estimate the vertical stifness, model the CMU with a storey height a, its dimensions and  its material characteristics .

Can you elaborate more on this?

[Simau]

Hi @HoshangMustafa 

Make a model with the materials and the appropriate shape of the CMU

[Simau]

Model updated

K=416666 kN/m

[HoshangMustafa]

Hi @Simau 

You wrote:

Make a model with the materials and the appropriate shape of the CMU

What material properties did you use?

[Simau]

Hi  @HoshangMustafa 

For the test I took this material

[HoshangMustafa]

Hi @Simau 

you wrote:

For the test I took this material

Very little information is available in my case:

How would you use these information to model masonry wall in my case?

[Simau]

Hi @HoshangMustafa 

Suppose you use a CMU with dimensions 50x20x2 and resistance 5.3 MPa. The limit load will be 5.3x0.5x0.2 = 0.53 MN.

The concrete part of the CMU is 0.0704 m² and therefore the resistance of the material is 0.53/0.0704 = 7.528 MPa.

For Young's modulus take that of C12/15 concrete, i.e. 27000 MPa if you don't have any reference.

No idea about other characteristics

[HoshangMustafa]

Hi @Simau 

you wrote:

The concrete part of the CMU is 0.0704 m²

How did you get 0.0704 m²?

[Simau]

Hiligthed area

[HoshangMustafa]

Hi @Simau 

500mm*20mm*2+160mm*20mm*3=0.03m^2

[Simau]

My mistake, you are right

[HoshangMustafa]

Hi @Simau 

You wrote:

Suppose you use a CMU with dimensions 50x20x2 and resistance 5.3 MPa. The limit load will be 5.3x0.5x0.2 = 0.53 MN.

The concrete part of the CMU is 0.0704 m² and therefore the resistance of the material is 0.53/0.0704 = 7.528 MPa.

Doesn't 5.3 MPa given in test report correspond the compressive strength? I'm not aware how this test is done, is it done on 0.5x0.2 surface , so you multiplied the compressive strength given within test result by the 0.5x0.2 area? Then you divided the result by the sum of shell areas (0.0704m^2 in your calculations) since the masonry units are layed such way (for reinforced masonry). The common practice here is using unreinforced masonry units with mortar layed on 0.5x0.2 surface (the units are upside down), rather than laying mortar on shell areas.

[Simau]

Hi@HoshangMustafa 

This is just my way of interpreting the test.

You may investigate more deeply on this kind of test on CMUs.