Objectiv 1

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1.0OBJECTIVEThis experiment deals with determination of rock strength when a certain load implied on the

rocks. Students should be able conducted the experiment, understanding the theory and

recognize the rock strength on different types of rocks in Malaysia.

2.0LEARNING OUTCOMESa) To determine a rock strength on different types of rock formation in Malaysia.b) To evaluate the physical properties of rocks for civil engineering application.c) To understand the theory rock test.

3.0THEORYThis apparatus, also named Franklin Press (1970), is used to obtain quick information

concerning rock strength. Point load measurement represents one of the most widely used in

classification tests for rocks, both in the field and in the laboratory. A rock piece is subjected

to a compression load along its diameter with two apposite conical platens. The index of rock

strength is calculated from the following:

IS= P


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4.0EQUIPMENT AND MATERIALSa) Digital or manual rock strength index apparatus ( Point Load Test )b) Irregular pieces rock samplesc) Clear safety goggles

5.0PROCEDURE

The apparatus consists of a load frame 55kN capacity with hydraulic loading ram actuated by

hand pump. Students should tests on block and irregular lump. Check first on the block and

lumps to be tested respect the shape prescription here indicated:

15 D 85 0.3 D/W 1.0 0.5 D L

Where;

D = Distance of the contact conical points (mm)

W = Average width (Waverage) of the sample perpendicular to the loading direction (mm)

L = Distance between the contact conical points and the nearest free end (mm)


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6.0RESULT AND ANALYSISExample Calculation:

L = 60mm

D = 45mm

W1 = 50mmW2= 50mm

Wave = 50mm

Load, P = 10.17kN

Equivalent diameter of the core sampleA = Waverage x D

= 50mm x 45mm= 2250mm

2

DE2

= 4 A/ = 4 (2250) /

= 2864.79mm2

Point load strength index, IsIs= P (1000)

DE2

= 10.17 (1000)

2864.79mm2

= 3.55 MPa


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7.0QUESTION AND DISCUSSION1. Why point load strength index, Is(50) obtained are not recommended to be used as

design input parameters?

2. Describe generally the differences between index and direct test.


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3. Explain the discontinuities in rock and their effect on strength.Discontinuities are usually categorized according to the manner in which they were

formed. The following are standard definitions of the most commonly encountered types

of discontinuities:

FaultA discontinuity along which there has been an observable amount of

displacement. Faults are rarely single planar units; normally they occur as parallel

or sub-parallel sets of discontinuities along which movement has taken place to agreater or less extent.

Bedding planeThis is surface parallel to the surface of deposition, which may or may not have

physical expression. Note that the original attitude of the bedding plane should

not be assumed to be horizontal.

FoliationFoliation is parallel orientation of platy minerals, or minerals banding in

metamorphic rock.

JointA joint is a discontinuity in which there has been no observable relative

movement. A series of parallel joint is called a joint set; two or more intersecting

sets produced a joint system. Two sets of joint approximately at right angle to one


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8.0CONCLUSION


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Average point load index, Is(50) (MPa) = (3.37 + 39.942 + 5.112)/3

= 4.14

Hence, uniaxial compression strength UCS of rock, c = Is(50)average(MPa)

= 24 x 4.14

= 99.36 MPa

Sample

No.

Length,

L

(mm)

Depth,

D

(mm)

Upper

width,

W1

(mm)

Lower

width,

W2

(mm)

Average

width,

Waverage

(mm)

Load,

P

(kN)

Equivalent

diameter of

core

sample,

DE2

(mm

2)

Point Load

Index

Strength,

Is

(MPa)

FIs(50)

(MPa)

Sketches

sample

diagram

before &

after failure

(use

attachment)

1 60 45 50 50 50 10.17 2864.79 3.55 0.95 3.37

2 60 40 90 90 90 20.06 4583.66 4.38 0.90 3.942

3 90 40 60 50 55 15.92 2801.13 5.68 0.90 5.112

c = Is(50)average

= 20 for soft rock after Broch & Franklin, 1972; Bieniawaski, 1975)

= 24 for hard rock


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Objectiv 1