Stair Core Footing Design: Bearing Capacity

[mrlm]

I have a situation where my stair base (pad footing) has very large moments and low axial force. The stair core is being used as part of the lateral load resisting system. The stair core is 7900x3100mm external and the footing size is 10900x6100x1000mm. To ensure stability against overturning the rock friction was used as the axial load and self weight was not enough. The rock uplift friction used was 175kPa (class 3 rock)
Design Loads:
N* = 5233kN
M* = 52280kNm
V* = 3311kN

Design Actions:
Mo = overturning moment = 52280+3311*1 = 55591kNm
Mcw = counter weight moment = (5233+(10.9*6.1*1*25))*10.9/2 = 37578kNm
Mfb = back face friction = 175*(6.1*1)*10.9 = 11636kNm
Mfs = side friction (assumed triangular distribution) = 175*(10.9*1*1/2)*(10.9*2/3)*2 = 13861kNm
Mr = restoring moment = 37578+11636+13861 = 63075kNm
55591<63075
Stability OK.

I am unsure of how to check bearing due to the addition of the frictional resistance. Allowable bearing capacity is 3500kPa so I believe it will not be an issue, however, still needs to be checked.
Can someone please guide me as to how I could check this?

 

[hokie66]

I disagree with the concept of using friction between the footing and the rock faces. A better and more reliable solution is to use rock anchors to resist the uplift.

 

[mrlm]

Noted.

Assuming rock anchors/piles are not possible and, to rationalise the footing size, rock friction must be used; how would one check the bearing?

 

[hokie66]

Simple statics. You have the footing mass and mass from core, forces on three faces due to friction, and an overturning moment. The total upward force must equal the total downward force, and location of that triangular resultant can be found by summing the moments about a point. Not much different to any footing subject to overturning.

 

[MIStructE_IRE]

While I agree with hokie’s approach in theory I’m not sure I’d be brave enough to do this

 

[hokie66]

I wasn't advocating the approach of using side face friction to resist overturning, as I indicated in my first post.

 

[mrlm]

Curious as to why the opposition to using the rock friction.
If the core can be stable using it without the need for anchors/piles, then is this not a much more economical option as a single anchor can be in the order of $10-20k?

 

[hokie66]

Friction of this type is unpredictable for a few reasons. It depends on the excavation having vertical or with slightly sloping sides in the right direction. It also depends on the soundness of the rock faces after the excavation. Perfect excavation in rock is difficult to achieve.

I don't know the reasons for the high cost of rock anchors where you are, but if that is the case, you are probably better to increase the size of the footing, in plan or depth or both.

 

[mrlm]

Thanks.

Appreciate the feedback.

 

[milkshakelake]

To add to what hokie66 is saying, rock isn't isotropic. It comes in layers and often times in breaking rock, the rock stratum is pitched diagonally so it's hard to get a flat surface but easy to get a sloped one. So the friction coefficient would be unpredictable since it could either be good or bad depending on the orientation of the stratum. Rock anchors are very expensive where I practice, so I try to avoid them as much as possible, instead trying other ways to reduce overturning moment and uplift like adding shear walls or moment frames.