This Article was Presented by the Author in the Seminar`CMOAT` organised by the ISM-Dhanbad on 15th-16th september 2006

Author: Tikeshwqr Mahto, Dy. Manager, RG OCP-II, SCCL

NEW APPROCH TOWARDS ‘ROCK LOAD `CALCULATION FOR SUPPORT DESIGN IN BORD & PILLAR SYSTEM:

Introduction

Strata control in underground coal mines has always remained a challenging task to practicing mining Engineers and research Institutions. Its all due to mysteries buried inside the earth, which is impossible to understand, that is why all the strata control mechanisms developed are based on the empirical approach. Therefore, any empirical system developed, based on practical experience and studying geotechnical condition of few mines can not be a universal system, which will be applicable for all mines having different geotechnical conditions. In this paper the author is suggesting some modification in the Geomechanical classification system (or RMR system) developed by CMRI-ISM based on Bieniawski`s Geomechanical classification s\system (or RMR system). The author is also suggesting one new method based on mathematical approach for calculating rock load in development as well as depillaring workings.

TOPIC-1Existing CMRI-ISM `S Geomechanics classification system :-

In the Bieniawski`s Geomechanics classification or Rock Mass Rating (RMR) system, CMRI has made some changes for applying in Indian conditions. This modified system is known as CMRI `S RMR system. In this system, five parameters were identified for designing support system in underground roadways, which are listed below- Parameters Max. Rating(a) Layer thickness 30 (b) Structural features 25 (c) Weatherability 20(d) Rock strength 15(e) Ground water 10 --------------------- Total = 100 Based on the geotechnical studies of above five parameters, RMR of immediate roof is calculated by the weighted average method.

R = R1*t1 + R2*t2 + R3*t3 + R4*t4+ ----------+ Rn*tn -------------------------------------------------------------- t1 + t2 + t3 +t4 + -----------------+ tn Where R1, R2,R3,R4,-------- Rn are RMR`S of different layers having thickness t1, t2, t3, t4,--------tn respectively of immediate roof upto 2m heights and R is weighted average RMR of the immediate roof.

Empirical formulae given by CMRI for calculating ‘Rock Load’ are as follows:-

(A) Rock load in the gallery :- RL (t/sq.m) = Wd (1.7 -0.037R +0.0002R2) Where, RL = rock load in the development gallery, W = width of the gallery, D = density of the immediate roof rock, and R = RMR of the immediate roof.

B) Rock load at junction of the galleries:- RL (t/sq.m) = 5(W)1/3*d*[1- RMR/100]2

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Critical studies on rock load calculation formula designed by CMRI for Bord &Pillar development working : -

Since, Rock load (R/L) = W*d* (1.7-0.037R +0.0002R2) t/sq.m

Conditions assumed-

Maximum value of RMR(R) = 100 Minimum value of RMR(R) = 0 Rock load will be maximum when RMR is minimum(0) Rock load will be minimum when RMR is maximum(100 Hence, maximum rock load (R/L) =Wd [1.7-0.037(0) +0.0002(0)] =1.7Wd Minimum rock load (R/L) = Wd [1.7-0.037(100) +0.0002(100)2] = Wd [1.7-3.7 +2] =Wd[0] = 0Where `W` is width of gallery and `d` is density of immediate roof.

Again from rock load equation, R/L =Wd [1.7-0.037R+0.0002R2] = Wd f(R) Where, f(R) =1.7-0.037R+0.0002R2 a function of R(RMR) which is a parabolic equation.

But this equation i.e. f(R) does not fulfill the assumed conditions listed above. Minimum value of rock load(R/L) or f(R) does not occur at R=100, it occurs at R=92.5 and also f(R) has zero value at R=85 and R=100 which indicates the failure of assumed conditions.

Justification for the above statements are given below by mathematical calculations and also by plotting f(R) vs.R in fig.1

support system designed by CMRI. f(R)=1.7-0.037R+0.0002R2

-50

0

50

100

150

200

250

300

-20 0 20 40 60 80 100 120 140

R(RMR)

f(R)*1

00

f®*100

Fig.1

Since, f(R) =1.7-0.037R+0.0002R2

=a - bR + cR2, Where, a=1.7, b=0.037 & c=0.0002

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For maximum or minimum value of f(R),its first derivative should be zero at R=100 because the parabolic curve touches the X-axis at R=100.

i.e. df(R)/dR =0

or, d( a-bR +cR2 )/dR=0

or, -b + 2cR = 0

By putting values of b ,c & R

- b + 2cR= - 0.037 +2*0.0002*100 = -0.037 +0.04 = 0.003

i.e. df(R)/dR , or slope of the curve is not equal to zero at R= 100, which means failure of conditions assumed.

Again, if -b +2cR= 0

Then, R= b/2c, or R= 0.037/ 2*0.0002

or, R= 92.5i.e. slope of the curve is zero at R=92.5 , which is also failure of assumed conditions. It should be at R= 100.

Again, minimum value of f(R) occurs at R=92.5 .so minimum value of f(R) [at R=92.5] = a –bR + cR2

= 1.7-0.037*92.5 +0.0002*(92.5)2

= - 0.011 [ negative value]Which is impossible, because f(R) or rock load shall never have negative value. This is also a drawback of the CMRI`S Geomechanical classification system.

Again, in this system ,value of f(R) will be zero at two values of R, which is also a drawback of this system. It will be found out mathematically in this way;

Since, f(R) =0or, a- bR +cR2 =0

or, R= [ b_+ (b2- 4ac)1/2]/2c , where a= 1.7, b =0.037 & c= 0.0002

Hence R has two values at which f(R)=0

i.e. R1 =[ b+(b2-4ac)1/2]/2c = [ 0.037+{(0.037)2 - 4*1.7*0.0002}1/2]/2*0.0002 = 100, and

R2 = [ b- (b2-4ac)1/2]/2c

= [0.037- {(0.037)2 – 4*1.7*0.0002}1/2]/2*0.0002

= 85

The parabolic curve f(R) cuts the X- axis at R =85 and extends in negative zone which again reflects back and cuts the X-axis at R= 100. All these details have been shown in fig.1 .In Fig.1 point of reflection is at R= 92.5, but as per assumption it should be at R= 100 and also the value of f(R) or rock load is negative between R=85 & R=100, which is not possible. All these drawbacks have been rectified in modified system suggested by the author in detail , which is mentioned below.

MODIFICATIN SUGGESTED BY AUTHOR

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Modification of values of constants a, b & c for clear representation of previous assumed conditions. Since , f(R) = a – bR +cR2

conditions to be fulfilled :-

f(R) = 0 , at R=100, -------- (1) df(R)/dR = 0 at R= 100 ---------(2)

since , f(R) = 0 at R= 100 or a – bR + cR2 = 0 or R = b+_ (b2- 4ac)1/2

------------------- --------(3) 2c Again ,since df(R)/dR = 0

or -b + 2cR =0 or R= b/ 2c ----------(4) comparing equation (3) & (4) , it can be inferred that b2- 4ac = 0

i.e. b2 =4ac ----------(5)

finally we have two equations (4) & (5) and variables are three , therefore one variable shall be assumed to find out the values of other two variables.

If a= 1.7 has any scientific or practical significance, then it can be assumed as value of one variable. Solving equations (4) & (5) we get,

c = a/R2

or, c =1.7/(100)2 [ where R=100 & a =1.7] = 0.00017 -----------------(6)

Again putting the values of `a` & `c` in equation (5) , we get b = (4ac)1/2

= (4*1.7*0.00017)1/2

= 0.034 ---------------(7)Hence , modified values of constants are: -

a = 1.7, b = 0.034 & c = 0.00017 and modified equations of f(R) & rock load are as follows:f(R) = 1.7 – 0.034R + 0.00017R2, -----------------(8)

Rock load(R/L) = Wd f(R) = Wd (1.7 – 0.034R +0.00017R2) ----------(9) The above equation will fulfill all the pre-assumed conditions i.e.

f(R) =0, at R= 100, and df(R)/dR = 0, at R=100 and also there will be no negative value of f(R) & rock load between the range R= 0

to 100. These are clearly shown in fig.2

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FIG.2

Comparison between the original and modified systems of Geomechanical classification Comparing both the equations i.e. original equation and modified equation , we can clearly distinguish between the original and modified system for support design in development and depillaring working. It is shown below in Fig.3. Rock load calculated by the modified system as per RMR of the roof rock of underground roadways is more in comparison to the original system, which will subsequently improve the support resistance required for strata control.

Original equation, f(R)1 =1.7 -0.037R +0.0002R2 , and Modified equation,f(R)2= 1.7-0.034R+0.00017R2

Comparison between old and new systems for support design. f(R)1= 1.7-0.037R+0.0002R2 and f(R)2 = 1.7-0.034R+0.00017R2

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0

50

100

150

200

250

300

-20 0 20 40 60 80 100 120 140

R(RMR)

f(R

)*10

0

f(R)2

f( R)1

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Fig.3 Comparative studies:

Assumptions :- Depth of working = 300m,

Gallery width (W) = 4m, Rock density (d) = 1.5 t/m3 RMR of the roof (coal) = 60

(a) Rock load by original equation:-R/L = Wd (1.7 – 0.037R + 0.0002R2) =4*1.5[1.7- 0.037*60 +0.0002(60)2] t/m2

=1.2t/m2.

(b) Rock load by modified equation:-

R/L = Wd (1.7 – 0.034R +0.00017R2) = 4*1.5[1.7 -0.034*60 +0.00017(60)2] = 1.632t/m2

Support Pattern;-

Let the method of support system is roof bolting, and bearing capacity of one roof bolt is about 8t.

(1) Support pattern of original system:-

Rock load = 1.2t/m2

F.O.S. = 1.8 Hence, support resistance offered by a roof bolt = bearing capacity of roof bolt ---------------------------------- Area supported by a roof bolt

= 8/A t/m2

Again, support resistance = rock load * F.O.S. = 1.2 *1.8 =2.16t/m2

. Thus comparing the above two equations we get,

8/A = 2.16, Or, A= 8/2.16 = 3.70 m2, or 1.90m * 1.90m in square pattern

i.e. area supported by one roof bolt is 3.70m2 .

(2) Support pattern of modified system:- Rock load = 1.632t/m2

F.O.S. = 1.8 Support resistance = 8/A

Again, support resistance = rock load * F.O.S. = 1.632 * 1.8 = 2.938t/m2

Comparing the above two equations we get;

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8/A = 2.938

Or, A = 8/2.938 =2.723m2, or 1.65m *1.65m in square pattern.

In the modified system of support, one roof bolt covers the area of 2.723m2, which is less than the area supported by one roof bolt (3.7m2) in original system, which shows over estimation.

CONCLUSION;- Comparing both the support pattern, it can be concluded that the modified pattern of support system is much safer.

TOPIC -2

New approach for calculating `Rock load` suggested by author

Mechanism of stress distribution :-

When a drivage is made in rock mass, there is a re-distribution of stresses around the opening and the rock adjusts itself by moving into the opening. In layered strata ,like coal measure rocks, bed separation and subsequent roof sagging takes place in the immediate roof . simultaneously ,the load originally carried by virgin deposit is transferred on both sides of the solid pillars. These abutment stresses are much larger than the average stress on surrounding area. These are clearly understood in the fig.1 given below.

Stress field

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` Stress field Fig.4

After the opening is made ,all the stresses, which exist above the roof of the opening, are transferred on the surrounding pillars . only a small portion of dead rock mass as shown in Fig.4 , plays an important role in failure of roof rock,which is required to be controlled by artificial mechanisms. The transference of stress (or rock load) on surrounding pillars starts from a certain height `h` above the roof of the opening ,as shown in fig.4 ,which depends on height of working `m` and a factor `k`. Rock load on the immediate roof is nothing ,but dead weight of the trapezoid portion of the rock mass in triangular section as shown in fig.4

Section of rock load on the roof of the gallery :-

Where , h = height of the rock load above the roof, m = working height, and W= width of the gallery.

The shape of the section of the rock mass above the roof is not exact triangular . The exact shape is shown in fig.5 . But for simplicity , the shape of rock load can be assumed as triangle. For accurate rock load calculation ,a factor `R` known as shape factor shall be multiplied in the value of rock load calculated by triangular cross-section to represent the real section.

In fig.4 the curve line represents the rock load curve(stress curve) transferring on the surrounding pillars.

Hence from Fig.5 the actual area of rock load (curve ABC) coming on the roof = area of triangle ABC * a factor R

The value of factor `R` can be in range of 50-70% For coal roof , R = 60-70%, and For sand stone roof , R= 50-60%

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The trapezoidal portion of the rock mass ,which creates load on the roof of the gallery can be shown in Fig.6 along the direction of gallery in the following fashion;

Fig.6

Where, W= width the gallery, m= working height, L= length of gallery & h= rock load height above the gallery roof.

Calculation of rock load in the gallery:-

Taking a small portion of the trapezoid as shown in the fig.6, for calculating rock load.

Here W= width of gallery, h= rock load height, m= working height, x= a small length of trapezoid along the gallery, d= density of roof rock & R= shape factor = area of actual rock load on the roof (curveABC) --------------------------------------------------------- [from fig.5] Area of rock load of triangular portion (triangleABC) Now the volume of small portion of the trapezoid (V) = area (curve ABC) * length of the trapezoid = area of triangleABC *shape factor *length of the trapezoid Or, V = ½ W*h *R*x = ½ WhRx

i.e. V= ½ WhRx ------------------------------(2) Hence , weight of the trapezoid causing rock load on the roof(P) = Volume * density of the rock = V * dOr, P = ½ WhRx *d = ½ WhRxd ------------------------------------------(3)

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Again area of base of the trapezoid = W*x .

Therefore , rock load (R/L) above the roof of the gallery = weight of the trapezoid ----------------------------- Area of the base of trapezoid

B Or , rock load = [ ½ WhRxd]/Wx , Or , Rock load = ½ hRd ----------------------------(4)

Again , h = m/k-1 , where m= height of working & k= bulk factor Putting the value of `h` in equation (4) Rock load = mRd/2(k-1)t/m2

Thus ,rock load = [ mRd]/2(k-1)------------------------------(5)

From equation (5) we finally inferred that the rock load is the function of working height(m), shape factor(R), rock density(d) & bulk factor(k)

Value of k= 1.4 – 2.2

TOPIC-3

Critical study of CMRI ` S geomechanical classification or RMR system for support design and comparison with the NEW SYSTEM for support design.

Assumed conditions:-

Depth of working = 350m, Width of gallery = 4.8m, Method of working = mechanized Bord &Pillar method, Height of gallery = 3m, Immediate roof = coal (6m thick), Seam thickness = 10m, Floor = 1m coal, RMR of the roof = 60, Density of coal = 1.4t/m3

Adjustment of RMR as designed by CMRI in modified Geomechanics classification or RMR system; Parameters Adjustment factor

-------------- ---------------------

Depth (350m) 0.9 Width of gallery(4.8m) 0.9 Method of coal winning (solid blasting ) 0.9 Lateral forces in the seam 0.8

Therefore , adjusted RMR1 = RMR * 0.9 *0.9 *0.9 *0.8 = RMR *0.583 = 60*0.583 =35 Initially, category of the roof was fair (RMR=60) but after adjustment of RMR, it comes under the category of poor roof. this is the drawback of the existing system.

Again , rock load(R/L) = Wd[1.7 -0.037R +0.0002R2] t/m2

Where , R= adjusted RMR(35), W= width of gallery (4.8),&

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d = density of roof rock(1.4t/m2)

hence , Rock load = 4.8 *1.4[1.7 - 0.037*35 +0.0002*(35)2] = 4.368 t/m2 ---------------(6)

Assumptions for support resistance; Method of support = = Roof bolting bearing capacity of bolt = 8t Area covered by one bolt =A F.O.S = 1.8

Therefore , support resistance = rock load * F.O.S.

Or , 8/A = 4.368 * 1.8 [from eq. (6) ] Or , 8/A = 7.86, Or , A = 8/7.86 = 1.02m2

Or , A = 1m * 1m in square pattern -----------------------(7) Spacing between two bolts in a row = 1m, Spacing between two rows of bolts = 1m

Therefore, for 4.8m gallery 5 number of bolts to be required.

Rock load at junction = 5(W) 1/3 * d [1- RMR/100]1/2 = 5(4.8)1/3 *1.4[1-35/100]1/2

= 4.73t/m2

Hence, support resistance = rock load * F.O.S. Or , 8/A = 4.73 * 1.8 Or , A = 8/8.52 = 0.94m2

= 0.9m *0.9m in square pattern

Therefore number of bolts required at junction = 4.8m *4.8m/A = 4.8m *4.8m / 0.9m *0.9m = 28

Support assessment by using new method developed by the author

From previous assumptions; Width of gallery (W) = 4.8m, Thickness of coal seam = 10m, Height of gallery (m) = 3m, Rock mass above the roof = coal (6m), Density of coal = 1.4t/m3

Density of stone = 2.2t/m3

Bulk factor of coal = 1.4.

Now h = m/k-1, = 3/1.4-1 = 7.5m.

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Taking into consideration of local geology, we can calculate rock load upto 7.5m height. 5m Roof rock consists of 6m coal and 1.5m stone. Assuming rock load above the gallery is only due to 6m of coal portion.

Hence , h = 6m , d= 1.4t/m3 & R(shape factor) = 60%(assumed)

Therefore, rock load in the gallery = ½ hRd t/m2

=1/2 *6 * 0.6 *1.4 t/m2

=2.52t/m2

And support resistance required = rock load *F.O.S =2.52 *1.8 =4.54t/m2

Or, 8/A = 4.54,

Or, A =8/4.54 =1.76m2 1.32m *1.32m in square pattern,

i.e. spacing between two bolts in a row =1.32m & spacing between two rows of bolts = 1.32m

Again number of bolts required in a row =width of gallery ------------------- + 1 Spacing = 4.8/ 1.32 +1 = 3.6+1 =5

Again, rock load at junction:- Rock load = rock load in gallery + 25% extra = 2.52 +2.54/4 = 3.15t/m2

Hence, support resistance = rock load * F.O.S. Or, 8/A = 3.15 *1.8 [where, A = area supported by a roof bolt, F.O.S. = 1.8 =5.67 & roof bolt capacity = 8t]

Or, A =8/5.67 = 1.41m2

= 1.20m *1.20m in square pattern

Hence number of bolts required in the junction = area of junction /area supported by a roof bolt = 4.8m *4.8m /1.41 =16+25%extra =20

Conclusion: Comparing both the systems, the new system is handy and site specific. All the above statement and comments are personal and not necessarily to the organization.

Comments made on this article by CMPDI, Ranchi, which has been published in the monthly magazine of`CMTM` vol.11, No. 5-6, May-June, 2006

The formula for calculation of rock load from RMR is an empirical one, designed by CMRI based on the analysis of 46 working mines where detailed investigations were made and 96 more workings carried out earlier.

Any underground roof strata having RMR more than 80 normally do not require any support except for spot bolting at disturbed patches. Therefore, the portion of the curve (rock load vs. RMR) is not relevant for RMR exceeding 80.

Very poor roof conditions (RMR less than 20) were also not part of the above-mentioned study. Therefore, the portion of the curve (rock load vs.RMR) is also not applicable for RMR less than 20.

Bulk factor criteria is used for caving method and for development galleries.

Present CMRI system of RMR based support design is quite effective.

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Author`s View: As according to the comment made by the CMPDI team that there is no need of support for strata having

RMR more than 80 and the CMRI`S Geomechanical classification system is not necessary to apply for RMR more than 80 and less than 20.But one thing I want to say that, what is validity of this system, if the system does not fulfill pre assumed conditions and not applicable for every point within the pre assumed boundary conditions.

In place of bulk factor (K) used by the author for calculating value of `h`(height) in the above equation any other factor can be used.

Empirical formula is developed on the basis of experience and trial & error method, so it can be true for certain, limit but it cannot be universal.

I have made some changes in the equation for rock load calculation, for which the CMPDI experts have made comments.

(Authors signature)

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