Soil Stress for Stapling Verification

A

Angel

New Member
Hello, I have a question about stapling verification on raft foundations. Referring to TS500 (TS500 page 35 figure 8-2); As long as stapling control is Vpd<=Vpr it's fine. Vpd=N1-Fa Fa=qsp(b+d)(h+d) what value does the program use as the qsp (ground tension) value here? As far as I can see there is a different value for each column. So I wonder if the program looks at all combinations separately for each column support and gets the most unfavorable value (smallest?largest?average?)? I would be glad if you could inform me. Best wishes, Engin YÜCE
 
Hi.. For each combination in the formula, the axial forces of the columns N1 and the maximum soil stress of that column are taken as basis. Good work...
 
Thanks for your answer; I repeat for correct understanding, A stapling check is done for each column. The N1 and qsp values to be used in this investigation are; The largest of the N1 values formed as a result of each loading combination of 1-O column and the qsp value formed under the column as a result of the combination that gives the largest N1 value is selected (that is, the N1 and qsp values for that column are selected from the same combination, from the combination that gives the largest N1 value). 2-Or, is the largest N1 value of that column as a result of each loading combination and is the largest of the qsp values of that column selected as a result of each loading combination (that is, the N1 and qsp values for that column are selected from different combinations). Which of these is the program doing, I did not fully understand this point. Best wishes, Engin YÜCE
 
Hello again, the two explanations you have written are not made.. It will be clearer if I explain with an example.. Let's get the largest ground stress value from the combination of let's say g+q+E1 in the punching circumference of a column. This is your qsp. In that case, the Ni normal strength value of the g+q+E1 combination is used in the column. This is your N1. This process is done for each column and its stapling perimeter one by one... You can see the result of each operation with the column name in the column punching report/concrete dialog.
 
Thank you very much for your answers, it was very detailed and explanatory. However, I found it appropriate to share my opinion with you based on your answers. Let me try to explain with an example. Let's be solving a system and our loading combination is five (COMB1, COMB2, COMB3, COMB4, COMB5). As you stated, the program will determine the largest qsp value among these five combinations for a column examined, for example, let the qsp obtained for COMB2 be the largest. Thus, the program will use the qsp and N1 values of the COMB2 installation. Here is what we know. The qsp of the COMB2 installation is the largest qsp value among all combinations. But is the N1 value of the COMB2 we will use the largest N1 value among all combinations? If we look at the formula on the subject in TS500; Referring to TS500 (TS500 page 35 figure 8-2); As long as stapling control is Vpd<=Vpr it's fine. Vpd=N1-Fa Fa=qsp(b+d)(h+d) So for a column we need to find the most unfavorable (largest) Vpd value to be used in our Vpd<=Vpr verification. Vpd is dependent on N1 and qsp variables except b, d and h values. According to the formula, large values of N1 move Vpd to large values, while large values of qsp move Vpd to small values (it enters the formula with a minus sign in front of it). So it seems more important that N1 is large and qsp is small to get the most unfavorable Vpd value. To simplify it further, the most unfavorable value of Vpd depends on the N1-qsp(b+d)(h+d) difference. In this case, what I want to ask is: If the qsp and N1 values of these five combinations are determined for a column examined, Vpd=N1-qsp(b+d)(h+d) values for each combination are calculated (with their own N1 and qsp values of that combination) and for each combination Wouldn't it be a more appropriate approach if the most unfavorable Vpd values obtained were used? Best wishes, Engin YÜCE
 
Hello again; Let me add one more thing: For Vpd to be greatest, N1 must be the greatest and Fa must be zero. Accordingly, Nmax values in the maximum normal force control in the column report in the program can also be the largest Vpd. If the largest Vpd is sought in a project, it is sufficient to look at the Nmax values of the column. In other words, the control of Nmax being smaller than VPr for that column is the control of the most unfavorable state... I observed that Nmax values can be half of Vpd in a 15-storey, wide building with a raft height of 100 cm. Of course, it would be appropriate to make more observations for a definite result... I mean, no matter which combination we take, when the system conforming to the regulations is selected, there does not seem to be much problem in terms of punching resistance. However, for the sake of ideal calculation, it seems most logical to take the qsp and Ni within each load and combination case. In other words, checking them with their own qsp and its Ni's for all combinations, respectively. Maybe our other colleagues can share their own views on this subject. . Good work
 
You must log in or register to reply here.
Back
Top