Rock Mechanics, An Introduction by Sivakugan - During an excavation for a tunnel

Question

Rock Mechanics, An Introduction by Sivakugan - During an excavation for a tunnel, 250 m below ground level, a highly fractured siltstone rock mass with two major joint sets and many random fractures is encountered. The RQD from the rock cores is 40% and the average UCS is 70 MPa. The joints with average spacing of 75 mm are rather continuous with high persistence with apertures of 3-5 mm, and they are filled with silty clay. The joint surface walls are highly weathered, undulating, and slickensided. There is some water inflow into the tunnel estimated as 15 L/min per 10 m of tunnel length, with some outwash of joint fillings. Estimate RMR, and Q. Assume a unit weight of 28 kN/m^3 for the rock mass.

Explanation / Answer

solution:

we know, Rock Mass Rating(RMR) = J1 + J2 + J3 + J4 + J5 + JB

where J1 =rating for UCS(FOR 75MPa) = 7

           J2 = rating for RQD(40%) = 8

              J3 = rating for spacing(75mm) = 8

            J4 = rating for discontinuties(1-5mm) = 10

           J5 = rating for groundwater(10-25mm) = 7

and      JB = discontinuity orientation(fair) = -5

therefore RMR = 7 + 8 + 8 + 10 + 7 - 5 = 35

which can be said as poor rock.

and Q = RQD/JN * JR/JA *JW/SRF

Where JN = rating for no. of joints

           JR = rating for joint roughness

           JA = rating for joint alteration

           JW = rating for groundwater

           SRF =rating for stress situation

therefore Q = 10/4 * 1.5/2 * 0.5/2.5 = 0.375

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