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uplift capacity of rock anchor based on rock/grout interface? 1

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jenduro

Geotechnical
Nov 7, 2005
5
In computing the rock/grout interface uplift capacity in limestone, I use the assumption that that the uplift capacity is equivalent to the effective weight of a cone- or wedge-shaped failure mechanism within the rock. In this analysis, the shear strength of the rock mass is ignored. The analysis indicates that where the weight of the rock within the contained cone-shaped failure is greater than the design rock anchor load, the anchor is considered adequate. I find this capacity increases as the anchors are spread further apart. What other methods are out there?
 
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You will need to check:
1. the structural capacity of the anchor bar/tendon
2. the bond of the grout to the anchor bar/tendon
3. the bond of the grout to the rock
4. the weight of the rock in the cone as you described
against the uplift load as a "pull-out".

I don't see how the weight increases the further spread out you are. For closely spaced anchors, the intersection of the cones will be, as a limiting case, the end of the anchor bar/tendon. As the installations spread out, you will "lose" the weight of the upward looking triangle from the end of the bar to the point of intersection. Remember that 4 is conservative especially if you have "strong" layers within the rock mass (such as for the Queenston Shale). I have another reference in mind that predates the one jdonville indicates - it may actually be from the NAVFAC series, which if you don't have, download from vulcanhammer site. No geotech should be without it (DM7.1, DM7.2 and DM7.3 with the latter now being not a Navy publication but of another agency.)
 
jenduro,

If you assume that the rock is very fractured (this is possible), even if it is hard rock, the required cone or shaft depth for mass stability will probably be longer than either the rock/grout or the grout/tendon bond length. This is especially true if the rock is below the water table and is buoyant.

If the anchors are close together, you will not develop full cones. The cones will overlap; so you can't count on the same weight of rock twice. If the anchors are closely spaced in a long, single row, you may have almost a continuous, v-shaped rock mass with a little reduction for gaps between the anchor cone points. If the anchors are close together, say in a rectangular grid, the mass of rock per each anchor may be a pointed, straight shaft equal to the grid spacing. If the anchors are in a straight row and are not too close together, for each anchor you will have a rock mass somewhere between a cone and a pointed straight shaft. Don't over-analyze the shape. Go conservative on the anchor length. Drilling a few extra feet of anchor length is not the biggest expense. Setting up the drill and moving from hole to hole is more expensive.
 
PEinc - good! It is what I tried to say on the shape of the block but you said it better. I hadn't thought that he was perhaps, double counting? You are 100% correct in that if you have the machine there and are drilling, the added cost of an extra metre or 2 is not much. [cook]
 
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