retaining wall

T

teknikisler

New Member
The ground survey information came to the conclusion that the internal friction angle of the ground is 0. First of all, is this possible? Can the internal friction angle be 0? 0 cannot be entered in the program anyway. I entered 1 as 0.65 in cohesion, in this case I couldn't find a 225 cm high retaining wall solution.
 
In the retaining wall account, which I have sent the attached project, although 260 cm high is saved from retaining walls with approximately the same cross-section, the 210 cm high one does not. How could this be?
 
In the calculation of retaining walls, there are critical factors even if you enter all the data in the same way. The program calculates the weight with the Swedish slice method. It should be greater than 1.5. In your case, one of the resisting forces is the section of the soil on the retaining foundation on the side of the active soil pressure as the Vetrical force.
 
"teknikisler":2uodsx8r" said:
I came to the ground survey information that the internal friction angle of the ground is 0. First of all, is this possible. Can the internal friction angle be 0, 0 cannot be entered in the program anyway. I entered it as 1, 0.65 in cohesion, in this case 225 cm high I couldn't find a retaining wall solution.
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There may be a problem as a result of the soil internal friction angle being 0. I would suggest you to talk to a friend who did the soil investigation, because the internal friction angle of the soil (Mohr's Circle) is the angle at which shear failure occurs in the shear stress and normal effective stress graph. The internal friction angle can be determined by shear test or triaxial tensile test.According to the SPT test, it can be taken in empirical values. It is likely to be the N NUMBER IN SPT, which is 0. In such a case, a value between 25-30 degrees internal friction Angle is taken, so the ground is a very loose ground It shows that it belongs to the class . I recommend you to review the soil survey again
 
Thank you for the clarifications, but there is something strange about it, playing with the dimensions I posted earlier, when I increase the height of the retaining wall to 7 meters, the slip safety saves, but when I reduce the height to 80 cm, it does not. It is strange that an 80 cm retaining wall could not be saved with this structure. I think the program's retaining account needs to be checked.
 
While the sliding forces are about 7 tf at the 7 meter screen, the sliding forces are 14 tf at the 80 cm screen. Come on, I understand that the preventive forces decrease, but shouldn't the slider forces decrease? Why do the shear forces increase while the height decreases?
 
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