## Dear all, I have a question for

## Dear all, I have a question for

(OP)

Dear all,

I have a question for weak rock. To ask this, I will illustrate with an example.

Assume that our uniaxial compression strength is 3000 kPa. (Average of many, that do not scatter much.) RQD is %30. We are looking for its bearing capacity.

There are three possibilities for me to calculate it:

First, I will reduce the UCS with a factor. I have one reference, that says for RQD b/w 25-50, reduce the lab result with 0.20. Ok. Now our design UCS is 3000 x 0.2 = 600 kPa.

Now, Canadian Design Manual recommends that qall=KspxUCS where UCS is the sound rock's UCS (in our case, we reduce and can use it.) Ksp changes between 0.1 and 0.4. For this RQD, I will choose 0.2. So qall = 600 x 0.2 = 120 kPa is the allowable bearing capacity.

UCS = 3000 kPa. Desing UCS = 3000 x 0.2 = 600 kPa. qult = 5.14 x 600 = 3000 kPa. Using FS=3, qall=1000 kPa. (Do not mind 5.14, Nc can be calculated for foundation dimension. )

Using Hoek-Brown method, c=75 Ø=17 can be found via RocData. Terzaghi Bearing Capacity equation Nc=12.34, Nq=4.77, Ngama=3.53.

If we do not ignore, Ngama, for depth of foundation = 0, bearing capacity with FS=3 is 600 kPa. If we ignore, 482 kPa.

Now we have reached a conclusion with three different bearing capacity:

WHY ROCK BEARING CAPACITY GIVES RESULT THIS LOW? I should reduce the UCS definetely. I do not know what to do and think.

I have a question for weak rock. To ask this, I will illustrate with an example.

Assume that our uniaxial compression strength is 3000 kPa. (Average of many, that do not scatter much.) RQD is %30. We are looking for its bearing capacity.

There are three possibilities for me to calculate it:

***Rock Bearing Capacity**First, I will reduce the UCS with a factor. I have one reference, that says for RQD b/w 25-50, reduce the lab result with 0.20. Ok. Now our design UCS is 3000 x 0.2 = 600 kPa.

Now, Canadian Design Manual recommends that qall=KspxUCS where UCS is the sound rock's UCS (in our case, we reduce and can use it.) Ksp changes between 0.1 and 0.4. For this RQD, I will choose 0.2. So qall = 600 x 0.2 = 120 kPa is the allowable bearing capacity.

*Clay Like Behavior.*Clay Like Behavior.

UCS = 3000 kPa. Desing UCS = 3000 x 0.2 = 600 kPa. qult = 5.14 x 600 = 3000 kPa. Using FS=3, qall=1000 kPa. (Do not mind 5.14, Nc can be calculated for foundation dimension. )

***Hoek-Brown**Using Hoek-Brown method, c=75 Ø=17 can be found via RocData. Terzaghi Bearing Capacity equation Nc=12.34, Nq=4.77, Ngama=3.53.

If we do not ignore, Ngama, for depth of foundation = 0, bearing capacity with FS=3 is 600 kPa. If we ignore, 482 kPa.

Now we have reached a conclusion with three different bearing capacity:

**Rock Bearing = 120 kPa****Clay Like = 1000 kPa****Hoek Brown = 600 kPa.**

WHY ROCK BEARING CAPACITY GIVES RESULT THIS LOW? I should reduce the UCS definetely. I do not know what to do and think.

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

Please.. My question is if you have these, what would you do? Rock is claystone. At least 70% of the projects are designed without the things you have asked. If you think the question cannot be answered without these info, please pass this topic

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

I admire your experience. And I login everyday to see what others ask and what people answer. Almost at every single topic, your answer is like these. You never answer to question but bring the argument to somewhere else, which may be important in general theory, but totally irrelavant to the answer.

To be honest, I format my questions just thinking of you. Everytime I ask something, I think to myself, oldestguy now will come and ask these, comment on this, and argument will go somewhere else, I will be left without a reply.

Do you understand? I asked something about SPT correlations and you gave a speech about corrections and how important they are. I ask a question in an exam format so for you to make best estimate. Answer is as always.

Anyway, I definetely do not ask you these questions. There are some people here that can improve me, but not you.

## RE: Dear all, I have a question for

I will not post them here though as your comments above (and on other threads you have started) go completely against the ethos of this forum. This forum is intended to provide general guidance on every day problems engineers encounter. An engineer stating that they need more information to provide input is completely fine. If you dont have the information all you have to do is state that. This forum is not a last resort are somewhere that people will jump at the chance to do your work for you.

There is an old saying 'manners cost nothing'.....

## RE: Dear all, I have a question for

What commonly is found at these sites are questions containing limited data. What I have found, through experience working, is that one never knows ahead of time how important some trivial item may become. Thus, as complete a set of data in resolving any problem is always required. Take this post, the title alone says noting of the question. Included should have been, at the least: location of the job,local geology information, local building codes applicable as to bearing, a complete boring log which describes all things found, including percent rock recovery, method and equipment details of sampling rock, elevation of proposed foundation with respect to rock surface and load to be applied.

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

There are some simple and conservative approachs to estimate allowable bearing capacity by RQD or UCS. Anyway i just checked the BS8084 (refered at EC7)and have seen some charts to estimate bearing capacity referring to dıscontinuite of rock mass, UCS and rock type. According to data you have given, the charts show the allowable bearing capacity about 0.5 MPa.

In my opinion, the using of terzaghi's formulas or considering clay like bahaviour to estimate capacity is not realistic since these equations were derived in consideration of general shear failure behaviour. However the bearing capacity failure on rock material is more relavent to compresibilty of rock material.

Dear oldestguy,

i would also like to know what is the effect of bedding direction to the capacity of ground. Is there any avaliable equations which refers to these datas?

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

That said - if you are talking claystone/shale, etc., unless you use triple tube core barrels and have a good driller, you are very unlikely in the upper reaches of the formation to get RQD's greater than about 40%. You would like find some grinding of the core - which implies that the core was better than you retrieved (grinding -you can see little indents on one core and a little point on the other side). Given experience in, say, Queenston shale, you should be able, easily, to have a bearing pressure permitted in the range of 500 kPa at the worse. Bedding shouldn't make much difference in a horizontally bedded claystone. As bearing is typically compressive, the beds wouldn't "delaminate". I have recently seen vertically bedded shales and this might be a bit more along the lines of being a bit more conservative in approach.

In harder rocks, assume, as a worse case, the rock formation is a gravel/cobble model - would you accept 120 kPa? No, unless there were large clays seams. Many harder rocks might have surface "damage" due to glaciation movement, etc - but again, I would surmise, as a minimum, a bearing pressure in excess of 500 kPa would be possible - unless of course, you find fracture zones that could slide on each other (the infilling of angled fractures, etc. You could pick up Coates' mining book (Canadian) that he wrote back in the 60s on rock.

As to other points, be respectful, tactful . . . the purpose as I see in many forums (or forii??) is to provide guidance on your exploration of a problem . . . to get you (or any one of us) a little tidbit to ponder as we make up our own minds. A person is a prisoner of his own experience and that is why I fully enjoy seeing posts by oldestguy, Ron, cvg, et. al., because their experience becomes, in part, now my experience. You may not get a "concrete" answer to all or even most of your questions - but if you read very carefully, you will gain some valuable insights.

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

## RE: Dear all, I have a question for

Finally, I have learned A LOT since I started reading/participating in this forum and really consider very valuable each of the answers from the fellow members here, so I am 100% with BigH, please be thankful and respectful. Engineering is a learning process that never stop and you are lucky that experienced engineers can take the time to read and provide input to your questions...