Miscellaneous concerns in ARSA

[HoshangMustafa]

Hi all

Please find the attached file.

I calculated Vu & phiVc for column 31 manually (using ACI code)

d=167.5mm

density of concrete= 23.6kN/m^3

column dimension: 400x400

service DL=2kN/m^2

service LL=3.5kN/m^2

b1=400+167.5=567.5mm

23.6(0.2)=4.7kN/m^2

1.2(4.7+2)+1.6(3.5)=13.64kN/m^2

13.64[(7.1)^2-(0.5675)^2]=683.2kN

Bo=4(567.5)=2270mm

phiVc=0.75(0.33)sqrt(24)(2270)(167.5)(10^-3)=461.12kN

Using ARSA:

From Forces table:

Fx=652.81kN

From Slabs- punching for panel 17:

Qadm=467.67kN (manually 461.12kN)

Q=268.17kN

467.67+268.17=735.84kN (manually 683.2kN)

It seems as there's a discrepancy in Q value, if I'm understanding it correctly.

[HoshangMustafa]

Hi @Krzysztof_Wasik 

Any help would be highly appreciated.

[HoshangMustafa]

Hi all,

I hope one can help with this.

[Krzysztof_Wasik]

Hi @HoshangMustafa 

I understand that you expect different Q value (punching force). I see that from Robot you obtain 268.17 kN

but I do not see in your manual calculations manually calculated Q (or I do not understand which value it is)

Coul you explain?

[napooil]

I think this is picture for question of HoshangMustafa

[Krzysztof_Wasik]

Hi @napooil 

As far as I understand your picture (screenshot is not sharp enough to se all details) you are comparing total vertical load applied to slab with punching force transferred by middle column. Please correct me if I have missed your point.

Please keep in mind that slab is also supported with four columns in the corners, which also transfer vertical load from the slab. It should be considered when punching force for middle column is calculated.

[napooil]

You correct. I forgot to think support with four column. thank you.

[HoshangMustafa]

Hi @Krzysztof_Wasik

You wrote:

but I do not see in your manual calculations manually calculated Q (or I do not understand which value it is)

Maximum punching force:

637.4-372.57=264.83kN

[Krzysztof_Wasik]

Hi @HoshangMustafa 

I understand that you expect explanation why maximum punching force 264.83 kN does not correspond to 268.17 kN

Please correct me if I have missed your point.

[HoshangMustafa]

Hi @Krzysztof_Wasik 

You wrote:

I understand that you expect explanation why maximum punching force 264.83 kN does not correspond to 268.17 kN

maximum punching force 264.83kN corresponds to direct punching force (punching force without unbalanced moment).

268.17kN corresponds to punching force with unbalanced moment.

[Krzysztof_Wasik]

Hi @HoshangMustafa 

Is there anything else which shoud be explained?

[HoshangMustafa]

Hi @Krzysztof_Wasik 

Maximum shear force= 264.83kN

Please find these attachments:

gamma_v=1- {1/ [1+(2/3) *(b1/b2) ^ (0.5)]}

gamma_v=1-{1/ [1+(2/3) *(567.5/567.5) ^ (0.5)]}=0.4

0.4*(0.46+0.35)=0.324kN

264.83kN+0.324kN=265.154 (in ARSA Q=268.17kN)

How Q is calculated from the maximum shear force= 264.83kN and the attached bending moment diagrams?

I need this for edge or corner columns, but took an interior column for instance.

[Krzysztof_Wasik]

Hi @HoshangMustafa 

Equivalent Q force including punching force and moments in column is calcualted iusing the following formula.

Q = t*A for biggest t value obtained from the following formula 

refer to Robot Help

Formula components are displayed in Slab calculation note

[HoshangMustafa]

Hi @Krzysztof_Wasik 

Thanks. I tried it this way:

[HoshangMustafa]

Hi @Krzysztof_Wasik 

I tried calculating deflection for element 36 described in:

Robot Structural Analysis 2019 Help | Plate and Shell Deflections - Calculations | Autodesk

Please find the attached picture:

Am I correct? Because I want to proceed calculating deflection.

[HoshangMustafa]

Hi @Krzysztof_Wasik 

Any help would be highly appreciated.

[HoshangMustafa]

Ix=Iy=b*h^3/12=910*(200)^3/12=606,666,666.67mm^4

E=23249.13MPa

D=EI=14104.472kN*m^2

[HoshangMustafa]

Hi @Krzysztof_Wasik 

I'm sorry for being late.

D/B=(14104.472kN*m^2)/2.98=4733.044

Is it ok?