Waqas Haider

Anique,
 
1. Registration with Pakistan Engineering Council (PEC), as a 'Consulting Engineer' (either as a firm or as an individual), is based primarily on following pre-requisites:
 
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  (Freshly graduated engineers, with a B.Sc. Engg degree, are accepted as 'Registered Engineer' by PEC)
 
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  (Requires a minimum of 5 years of practical experience in relevant field, after being enlisted as 'Registered Engineer', and passing of Engineers' Professional Exam  (EPE), conducted by PEC)
 
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As such, having done masters in 1 yr, should not be a problem, for registering with PEC, provided your BS Engg degree is acceptable.
 
2. Please see the following links for knowing the relevant rules & requirements, and downloading application forms etc.:
 
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     (http://pec.org.pk/downloadables/BYELAWS%201986%20%28APRIL%2023,%202013%29.pdf)
 
c. PEC Registration Forms, for 'Registered', 'Professional' and 'Consulting Engineers
    ((http://pec.org.pk/index.aspx?aspxerrorpath=%2fgetpecno.aspx)
 
 
Regards.

first of all you have to take area as transformed area. I mean A will be (Ac-As) + n As where n = Es/Ec and As is the total area of steel bars.
 
now calculate strain by PL/AE where A is the above area and E is the E of concrete because you have already transformed steel area into concrete above.
 
now this strain is assumed to be same in bars and concrete. and you can back calculate to check how much P steel and concrete is carrying. because it will be different for both. You cannot take the same P for both steel and concrete as you are doing in your post.

I got the solution. For any other who want the solution so they can see this link
 
http://www.ce.memphis.edu/6136/PDF_notes/i1__axial.pdf
 
Someone told me to calculate the stiffness using the concept of spring in series or spring in parallel. and then use this stiffness to find the deflection. This deflection will be the same for concrete and steel. Anyone have any idea what is the procedure for this approach? 

Yes  they both are carrying different load therefore they have different stresses. Thank you

dear shaker,
in case you are having just concrete wall as basement and no columns in wall and the length of wall is long then your wall will bend as one way slab. in such case i prefer to model as line element of unit width. But still you need to provide proper consideration to the corner of the wall where this one way action will be restrained. 
 
But in case we are having columns with in the wall then our slab will be showing a dishing effect which depends on the stiffness of wall and columns. in this case i prefer to model the real model consisting of columns and wall so that i may see actual deformation and i will read forces from this model. because at interface of column and wall behavior will be changed than mid span.
 
I need opinion of seniors over this. Thanks. 

dear shaker,
in case you are having just concrete wall as basement and no columns in wall and the length of wall is long then your wall will bend as one way slab. in such case i prefer to model as line element of unit width. But still you need to provide proper consideration to the corner of the wall where this one way action will be restrained. 
 
But in case we are having columns with in the wall then our slab will be showing a dishing effect which depends on the stiffness of wall and columns. in this case i prefer to model the real model consisting of columns and wall so that i may see actual deformation and i will read forces from this model. because at interface of column and wall behavior will be changed than mid span.
 
I need opinion of seniors over this. Thanks. 

Hello everyone. I would like to model eccentricity of columns and beam but I have encountered some problems using the insertion point in Etabs. Here are my questions:
 
1. What is the difference between using cardinal point and joint offset? I tried to joint offset a column as if it was offset using the cardinal point (eg. bottom left), but then the analyze results are different. Why? 
 
2. When using the insertion point, we have to transform the stiffness. I did not do any offset but changed the transform stiffness option to "yes" then applied a point load at the top. Comparing to columns without stiffness transform, the results are different. Would anybody explain why? And when we use the insertion point, shall we transform the stiffness of the object with offset? Or shall we transform the stiffness of objects which have connection with the offset object as well?
 
Appreciate!
 

Dear Waqas,
 
Don't be confused. It is very simple.
 
All you have referred to in your post, is about and applicable to minimum area of reinforcement, required to resist 'shrinkage and temperature movements' in slabs and mats.
 
On the other hand, I have discussed the distribution of minimum reinforcement, required to resist 'shrinkage and temperature movements' only, reinforcement distribution shall be as mentioned in ACI section 7.12.2 (referred by you).
 
However, when the slab or mat has to resist 'primarily, distribution of reinforcement shall be as per flexural requirements. In this case, total amount of 'As,min' has to be provided at each face of slab or mat subjected to '

yes, you have to calculate the net tensile strain in extreme layer. Approx it will be somewhere near to 0.02

A maximum reinforcement ratio for beams and slabs is not directly given in ACI 318-05. Instead, Section 10.3.5 requires that non-prestressed flexural members must be designed such that the net tensile strain in the extreme layer of longitudinal tension steel at nominal strength t is greater than or equal to 0.004. In essence, this requirement limits the amount of flexural reinforcement that can be provided at a section. Using a strain compatibility analysis for 4 ksi concrete and Grade 60 reinforcement, the maximum reinforcement ratio is 0.0206.
Thanks.

No dear, the maximum area of steel aims to limit the area of steel put in the section so as to control the strain in steel to be equal to or larger than 0.005.
This will not differ between slabs and beams.
Regards

Since the length of basement wall would likely to be much longer when compared to a conventional shear wall, lateral shear resisted / unit foot of basement wall would not be significant. Critical case for basement wall (generally speaking) should be lateral load due to earth + any increase in that lateral load due to earthquake. This load would be perpendicular to wall.
 
Also, have a look at the following topic (there are 5 threads) as you may find them interesting:
 
http://www.sepakistan.com/tags/forums/Highrise%2Bwith%2Bbasements/
 
Thanks.

But in reality, you r going to construct this beam. What if after making another shear wall, still it gets fail? I have reservations regarding this approach. Kindly explain.

Hello Goher,
 
You are right most of time wind loads govern but in case of Mezzanines, Seismic forces may govern also.
You are absolutely right. You will have to multiply horizontal forces with the depth of footing to get moment which you will have to apply at centroid of footing. If you want to check bearing pressure and strength design of footing, you can use SAFE but according to my knowledge, SAFE does not offer SLIDING, OVERTURNING check and for uplift, you may check pressure and pressure will tell you whether there is contact between your footing or your footing has been lifted.
 
To apply loads in safe, you can either model a footing slab and manually assign loads at centroid (by drawing a point or by meshing footing) or there is a better way also. You can import the whole base with reactions from SAP/ETABS to SAFE. 
 
Moreover, I m attaching excel sheets which I have made for simple calculations of isolated Footings. You can use it. If you find any error, kindly report me back. I have tested them many times. These are a little rough but i think usable. 
 
 
 
 
footing reinforcement design sheet.xlsx
footing sizing sheet.xlsx

The reinforcement in basement wall is similar to reinforcement in slabs. So design basement wall as a slab. Model an area element resembling your basement wall with suitable edge conditions... apply back fill load and gravity UDL loads. Mess/Divide the wall. Run the model and read moment at critical locations. and Provide reinforcement for these moments. You can also read shear force. Check thickness of wall against this shear force.
 
But if your wall is very long and acts as one way cantilever slab, then you can design it as beam or slab. Basement walls are basically designed as three edge supported slabs. 
 
But in high seismic regions, i think your walls will also behave as shear walls so proper consideration should be given to seismic induced forces. 

Dear  Mr. Waqas,
 
Your study about Non Linear Analysis is quite impressive. I just want to add some more details.
 
There are three types of design.
1- Service Level Earthquake (SLE)
2- Design Basis Earthquake (DBE)
3- Maximum Considered Earthquake (MCE)
 
SLE is at service level. Which means that in EQ there should be no problem for non structural members.
DBE: In code based design, we use DBE level earthquake. This earthquake is frwquent earthquake with return period of 475 years( I think). During this code uses actualy 100% forces and reduce these forces by reduction factor R. Means we design our structures for reduced forces as explained before. Actually in this case code allows the structure to yield at certain locations for certainm members. But code do not specify these locations and members as R is the overall division factor. So during an earthquake it is not necessary that the members will yield which we assumed to be yield. For example if beam at any location yielded then it will not transfer forces to the column. But we are not sure that column will yield first or beam will yield first. Code tried to take this by incorporating the stiffness modifiers but still it is not 100% reliable. So thats why nobody can gurrantee the building to perform 100% in earthquake when designed by Code.
 
MCE: This is the third and extreme level of earthquake.Return period is 2475 (I think). This can only be incorporated by using Non Linear Time History Analysis. In this we can check the reinforcement designed on DBE level for capicity check by considering the actual yielding and energy disspation. We can set a limit for structure for example Life Safety, Immediate occupancy or collapse prevention. We allow the memebrs to yield upto acertain level and beyond that level we ll not allow the structure to yield. We can actually see that which member is yielding and the after yeidling what is the force and design for that force.
 
The difference is that in code based design we aare not sure how much energy is dissipated in the structure. Code just provide 2 or 3 types or R factors depending upon the building type. So ist ios not 100 % accurate. Secondly code reduces the forces for all modes of structure but actually it can not happen. We can not reduce all modes because we can only reduce the modes which are yielding.
 
 
So thats why 21.1.1 says that integrity should be maintained. But code do not give any gide line how to maintain that integrity. By using that R factor??? how can we do that????? So code just wrote that thing.
Actually that integrity can only be assured by using the Non Linear Time Histoiry Analysis and stops the stucture to deform beyond which we do not require to deform.
 
Sorry if I write something out of topic.
 
Thanks
 
Muneeb

They are doing it right. It depends on you. It is the beauty of the structures that they will behave the way you designed them. 
 
When reducing the torsion modifier for beams that are failing to a value approx equal to 0 then watch for the increased moments in slabs. If you put the reinforcement in slab for additional moment then it is ok!
 
It depends upon the relative stiffness of beam and slab that how much load beam will take (Torsion, moment etc).

As you know that we allow the yielding in the flexural members. In shearwall, we also allow flexural yielding but do not allow shear yielding. 
 
We also do not want the brittle failure we always like to have ductile yielding. 
 
In case of shearwall we want the flexural yielding at certain locations. We can not allow the yielding at unwanted locations. 
 
When earthquake pushes a shearwall (structure) in one direction the in shearwall there will be Tension on one side (one corner) and Compression on the other side (other corner). This can be reversed. We do not want the shearwall to fail in brittle form and specially at corners. So code required that concrete to be confined by the horizontal reinforcement. In this case the fc' of that corners increased significantly and there will be no yielding at these locations. We need the shearwall to be yielded so we provide minimum flexural reinforcement in the NON BOUNDARY ZONE ELEMENTS. So that the shearwall can yield at the middle locations in case of earthquake. The remaining reinforcement ( Total Flexural Reinforcement - minimum reinforcement) is provided in the BOUNDARY ZONE ELEMENTS. 
 
For this actually we need to check the strain limits obtained by Non Linear Time History Analysis but code also provides a method to calculate boundary zone lengths. So thats the reason for confined zones. The length of Boundary zone also depends u[on 
 
Thanks
 
Muneeb

You can design it as propped cantilever. The presence of top floor and foundation would enforce one way bending (Case of a square panel with two sides having stiff supports)
For areas where you don't have a top floor, you can use the information provided in the link below to design your wall as 2 way slab panel.
 
http://www.sepakistan.com/topic/16-rectangular-concrete-tanks/
 
Thanks.

Assalam o alaikum.
I have just designed a frame structure with SMRF. The out put of shear seams weird to me. Column reports design shear Av/s as 0.045. (Images are attached). but when i right click the member, it must show me the most critical case HIGHLIGHTED AUTOMATICALLY. But it highlights load combo 38 (auto-generated combos have been used) which reports Av/s as 0.038. And 0.045 value is at combo 32. Is their any logic behind it?? More over how to interpret this Av/s?? means 0.045 in kip-in units means what? How can i convert this into spacing?? 

I am also a relatively fresh graduate. I completed my civil Engineering in 2013. The first and foremost thing u MUST know is ur interest. In what field you have Interest, learn about ur field the most u can. Because in student life, you have plenty of time  to study and these BOOKISH KNOWLEDGE is going to benefit you ultimately. In field you see things practically but in student life, you must focus on your concepts etc so that they an benifi you when you are in field. The second thing is that, Try to learn maximum softwares. I am a structural design engineer so softwares for this field you MUST LEARN are, AUTOCAD, ETABS, SAP, STAAD, SAFE, MS OFFICE etc. MS Project and Primavera must also be learnt. There is one also software called TEKLA that you must learn. There are 2 years still left so you can learn these softwares well.. And the most important thing,,,,, GO AND JOIN PART TIME JOB/INTERNSHIP Paid/Unpaid with any of professional you get chance. Big names are not necessary. At start attach with any of professional you get chance.. but this will benefit you because when you will complete your degree, you will be already having professional experience. by the way what is ur university?

sir , you were very much right . I got a normal pattern of my results.
As it is confirmed from CSI ETAB (When nonzero Pllf is assigned to a frame object, positive design moments are calculated assuming a simply supported condition under live-load application. While this does affect design moments, analysis moments are not affected)
 https://wiki.csiamerica.com/display/kb/Pattern+live-load+factor

try making pattern live load factor to 0.

I am working on a building with quite high loading.
live load :300 psf
Super Dead :90 Psf
The problem in my etabs result are quite unusual for me.I got more reinforcement in bottom as compared to negative in  continuous beams.
In the analysis, moment diagram is not showing such irregularity. need help form seniors.

Thank you very much sir for detailed answer. The above attached document presents calculations for a beam but since minimum steel limit differs for slab and beams, i think maximum steel ratio limit must also differ for beam and slab.