numerical modeling of concrete in abaqus

Ad 1st assignment

Open Abaqus

Click on part :-

Click continue :-

Center 0,0 and press Enter

A point on perimeter & press Enter

Now click here

Now here

Enter length of cylinder & Click OK

Obtain a beautiful cylinder

Go here click properties

Click Create Material

Click General - > Density

yield stress inelastic strain

2652974.61 0

3094436.04 0.000103295

13686557.6 0.00048805

21909750.7 0.000954491

26216831.2 0.001555971

26902245.8 0.002282336

25368514.1 0.003085224

22908785.7 0.003920042

20287294 0.004760438

18503462.3 0.00536195

Copy – Paste the Given Values ……….. Getting these values I’ve told in excel sheet attached

yield stess cracking strain

1595000 0.00E+00

981831.7496 0.000151144

798658.4402 0.00028246

697445.153 0.00041095

630420.827 0.000538261

581606.8791 0.000664945

543890.8538 0.000791245

513548.0906 0.000917291

488408.7289 0.001043158

467109.2021 0.001168893

448742.5432 0.001294526

432678.1885 0.00142008

410675.3731 0.001620839

damage parameter cracking strain

0 0.00E+00

0.05 0.000151144

0.1 0.00028246

0.15 0.00041095

0.2 0.000538261

0.3 0.000664945

0.4 0.000791245

0.5 0.000917291

0.6 0.001043158

0.7 0.001168893

0.8 0.001294526

0.9 0.00142008

0.95 0.001620839

Click OK

Select the cylinder and click done

1.For rigid platen put U1=0, U2=0 at both surfaces & U3=0 on the opposite surface of loading. 2. For Soft platen put only U3 = 0 on the opposite surface of loading.

I’m solving only for rigid platen.

Rotate cylinder to put U1=U2=U3=0 on the opposite side of loading for rigid platen. For soft platen it would be only U3=0.

Loading surface…..U1 =

U2=0, U3≠0

Fixed surface…U1=U2=U3=

0.

Now we will apply load on the loading surface.

Rotate & Select this top surface

Remember I told that loading time should be consistence with step time……Here please note that Time span should be “Total time ”

not “Step time” so that in 0.1 sec loading will become 100%

Click OK

Applied pressure > 27Mpa / area

=27e6/(3.14 X 0.15^2/4) = 0.1528 e 9

1. We Have almost done but we have not defined yet damage parameters which will tell us, whether our concrete has failed or not. 2. To plot load vs disp graph we have to plot total reaction at the bottom vs disp of the top point where load was applied. So this point I’m going to mark as reference point.

Reaction here

displacement here

We have to create two sets 1. For top surf disp.

2. For reactions Go to Tools->Set->Create

Marking Surf for disp.

Rename it to “top_surf” and click continue

Select top surf and click done. I’ve created a RP at the center but it’s

useless u can skip it.

Again go to Tools-> set- > Create and create 2nd set and give name as Reactions

Select bottom surface and click “Done”

We have create sets for load vs disp graph but we have not told

Abaqus to give me output for these sets yet.

Let’s 1st use the sets that we have just created and tell abaqus to give me

reactions as well top surf disp.

Output -> History output Request-> create

Let’s name this as “Tip-disp”

Again Output -> History output Request-> create

Let’s set damage failure parameters Output -> Feild output Request-> create

We have to make it independent to mess it. Right click on Part 1-1 and select -> Make Independent

This is quite big mess. You can refine it for convergence study. To do this reduce seed size from 0.021 to 0.01 in

slide no.73

OK Take rest now.

Analysis Got aborted due to errors let’s check error.

This is bcoz I’ve taken taken fcr = 3.65* sqrt(fck) in tensile properties. If you will take it above 4.5 rather 3.65 you will not get this error and also reduce mess size. Lets’s

check results with this data.

DAMAGET :- TENSILE FAILURE DAMAGEC:- Compression failure

Sdef :- Stiffness Degradation

Lets’s plot Load vs disp graph

Select All RF3 and click Save As

We are using ssum bcoz we have to add all reactions

Again for disp.

Select all U3 and click save as

This time we will not use ssum rather we will use avg.

Click OK

Here my disp and rxn data has come

Load vs disp graph

Failure load = 0.4e6 Pressure = 0.4e6 /(area)

= 22.64Mpa < 27 Mpa(given in prob. statement) Prob. is in the properties.