bdbd
Geotechnical
- Sep 17, 2015
- 144
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:
*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.
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:
[ul]
[li]Rock Bearing = 120 kPa[/li]
[li]Clay Like = 1000 kPa[/li]
[li]Hoek Brown = 600 kPa.[/li]
[/ul]
WHAT SHOULD I DO?
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.
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:
[ul]
[li]Rock Bearing = 120 kPa[/li]
[li]Clay Like = 1000 kPa[/li]
[li]Hoek Brown = 600 kPa.[/li]
[/ul]
WHAT SHOULD I DO?
WHY ROCK BEARING CAPACITY GIVES RESULT THIS LOW? I should reduce the UCS definetely. I do not know what to do and think.