1. 1
Author: Tikeshwqr Mahto,
DDMS, Bilaspur Region
NEW APPROCH TOWARDS ‘ROCK LOAD `CALCULATION FOR SUPPORT DESIGN IN BORD
& PILLAR SYSTEM
: - A MODIFICATION OF INDIAN `RMR` SYSTEM
Abstracts
Strata control in underground coal mines has always remained a challenging task to practicing mining
Engineers and research Institutions. It’s all due to mysteries hidden inside the earth, which is impossible to
understand. Support Design for the roof of the galleries in underground coal mines are being done based on
RMR or Geomechanics classification System, an empirical approach developed by CMRI - ISM. Any
empirical system developed, based on practical experience and studying geotechnical condition of few
mines can not be a universal system 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 system.
Existing CMRI-ISM`S
Geomechanics classification system:-
Indian Rock Mass Rating (RMR) system designed by CMRS-ISM are based on the Z.T.Bieniawski`s RMR
or Geomechanics classification system. In this system, following five parameters were used to classify a
rock mass using the RMR system for designing support system in underground Coal Mines. Each of the
five parameters is assigned a value corresponding to the characteristics of the rock. These values are
derived from field surveys and laboratory tests. The sum of the five parameters is the "RMR value", which
lies between 0 and 100.
Parameters Max. Rating
(a) Layer thickness 30
(b) Structural features 25
(c) Weather ability 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 height and ` R` is weighted average RMR of the immediate roof.

2. 2
Classification of the roof rock on the basis of RMR system.
RMR Rock quality
0 - 20 Very Poor
21 - 40 Poor
41 - 60 Fair
61 - 80 Good
81 - 100 Very good
Empirical formulae given by CMRI for calculating ‘Rock Load’ is as follows:-
(A) Rock load in the gallery :-
RL (t/sq.m) = W*d (1.7 -0.037R +0.0002R2
)
Where, RL = rock load in the development or split 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:- support resistance shall be increased by 25% in the junctions.
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 while designing:-
 Maximum value of RMR(R) = 100
 Minimum value of RMR(R) = 0
 Rock load will be maximum when RMR is minimum (i.e. 0)
 Rock load will be minimum when RMR is maximum (i.e. 100)
Hence, maximum rock load (R/L) = W*d [1.7-0.037(0) +0.0002(0)]
= 1.7*W*d
Minimum rock load (R/L) = W*d [1.7-0.037(100) +0.0002(100)2
]
= W*d [1.7-3.7 +2]
=W*d [0] =0
Where `W` is width of gallery and `d` is density of immediate roof rock.
Again, from rock load equation, R/L = W*d [1.7-0.037R+0.0002R2
]
= W*d* f(R)
Where, f(R) =1.7-0.037R+0.0002R2
, a function of R (RMR), which is a Quadratic equation and the shape
after plotting the equation on X- Y axis is parabolic. As per assumed conditions, this parabola should touch
the X- axis at R=100, means the curve should be symmetrical about the line R=100. 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 Rock Load or f(R) has zero value at two values of RMR
i.e. R=85 and R=100, which indicates the failure of assumed conditions.

3. 3
Justification for the above statements are given below by mathematical calculations and also by plotting
f(R) vs. R in Fig.1, shown below.
Since, f(R) = 1.7-0.037R+0.0002R2
Or, f(R) = a - bR + cR2
, Where, a=1.7, b=0.037 & c=0.0002 are constants……………. (1)
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. minimum value of f(R) or Rock Load occurs at R = 100.
Hence, by applying Maxima- Minima concept of Calculus: first derivative of f(R) at R=100 will be
zero.
i.e. df(R)/dR =0, or, d( a-bR +cR2
)/dR=0
or, -b + 2cR = 0 ……………………………………………………………………..(2)
But, the above condition is not satisfied by putting values of b ,c & R in equation (2)
Here, - b + 2cR = - 0.037 +2*0.0002*100
= -0.037 +0.04
= 0.003
Since, df(R)/dR , or slope of the curve is not equal to zero at R= 100, which means curve not touching the
X- axis at R=100, means equation is not having minimum value at R=100 , which indicates failure of
Conditions assumed.
Again by reverse calculation if, -b +2cR= 0, then, R= b/2c, or R= 0.037/ 2*0.0002 , or, R= 92.5 i.e.
Slope of the curve is zero at R=92.5, means minimum value of f(R) or Rock load occurs at R= 92.5
,which is also failure of assumed conditions. It should be at R= 100.
Since minimum value of f(R) occurs at R= 92.5.
Therefore, 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] ------------------------------(3)
This is not possible, because f(R) or rock load shall never have negative values. This is also a drawback of
the CMRI`S
Geomechanical classification system.
Again, in this system, value of f(R) or Rock load is zero at two values of R, which is also a drawback of
this system. It can be determined mathematically in this way;
When rock load or f(R) curve touches the X- axis, its value becomes `0`
i.e. 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 for which f(R) or Rock load =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
These discrepancies with the original formula are clearly picturised in Fig.1, shown below.

4. 4
Fig.1
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. 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.
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)*100R=92.5
↑
f(R)*100

5. 5
MODIFICATIN SUGGESTED BY AUTHOR
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 (minimum), at R=100, -----------------------------------(1)
 f(R) = maximum at R=0 ------------------------------ (2)
 df(R)/dR = 0 at R= 100 --------------------------------(3)
Since, f(R) = 0 at R= 100
Therefore, a – bR + cR2
=0
Or, R = [b ± (b2
- 4ac)1/2
]/ 2c
Or, R= b/2c ± (b2
- 4ac)1/2
/2c -------------------------(4)
Again, since df(R)/dR = 0 at R= 100
Or -b + 2cR =0
Or, R= b/ 2c - ---------------------- (5)
Comparing equation (4) & (5), it can be inferred that
b2
- 4ac = 0
Or, b2
= 4ac - ---------------------- (6 )
Finally we have two equations (5) & (6) 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 (5) & (6), we get, c = a/R2
Or, c = 1.7/(100)2
[ where R=100 & a =1.7]
= 0.00017 -----------------(7)
Again putting the values of `a` & `c` in equation (6) , we get
b = (4ac)1/2
= (4*1.7*0.00017)1/2
= 0.034 ---------------(8)
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
, ----------------------------------(9)
In the modified formula, the author tried to correlate the boundary conditions for maximum and minimum
values of rock loads as evaluated by the empirical approach of CMRI- ISM. Thus, value of constant (a=
1.7) is taken as usual in the modified formula assuming its scientific and technical importance and values of
`b` & `c` are calculated from the above equations, because value of constant `a` decides the maximum
value of rock load for minimum value of RMR (i.e. R=0), as shown in Fjg.1,2 & 3

6. 6
Thus the rectified formula of rock load designed by the author is,
(R/L) = W*d* f(R)
= W*d (1.7 – 0.034R +0.00017R2
) ---------(9)
The above equation will fulfill all the pre-assumed conditions i.e.
 f(R) or Rock load = 0 (minimum value), at R= 100, and
 f(R) or Rock load = maximum at R=0
 df(R)/dR = 0, at R=100 and also there will be no negative value of f(R) or rock load
between the range, R= 0 to 100. These are clearly shown in fig.2
FIG.2
Fig.2
Fig.2
R=100

7. 7
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
COMPARATIVE TABULATION OF `ROCK LOADS` AGAINST `RMR` USING BOTH THE
FORMULAE:
TAKING, W(width of gallery) =4m and, d(density of roof strata) =1.5t/m3
SL.NO. RMR
ROCK LOAD1(t/m2)
= W*d(1.7 – 0.037R + 0.0002R2
)
ROCK LOAD2(t/m2)
= W*d(1.7 – 0.034R + 0.00017R2
)
1 0 10.20 10.20
2 10 8.10 8.26
3 20 6.24 6.53
4 30 4.62 5.00
5 40 3.24 3.67
6 50 2.10 2.55
7 60 1.20 1.63
8 70 0.54 0.92
9 80 0.12 0.41
10 85 0 0.23
11 90 -0.06 0.10
12 95 -0.06 0.03
13 100 0 0

8. 8
Fig.3
In the above figure, it is clear that the curve f(R)1 cutting the X-axis at two points i.e. at R=85 & R=100, but
the curve f(R)2 designed by the author touches the X- axis only at R= 100. Initial boundary conditions for
both the curves is same (i.e. at R=0).
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 = W*d (1.7 – 0.037R + 0.0002R2
)
= 4*1.5[1.7- 0.037*60 +0.0002(60)2
] t/m2
= 1.2t/m2.
Graphical comparison between old and new systems for support design.
f(R)1 =1.7-0.037R+0.0002R
2
and f(R)2 = 1.7-0.034R+0.00017R
2
-50
0
50
100
150
200
250
300
-20 0 20 40 60 80 100 120 140
R(RMR) →
↑
f(R)*100
f(R)2
f( R)1

9. 9
(b) Rock load by modified equation:-
R/L = W*d (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 be 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 for support resistance, we get:
8/A = 2.16,
Or, A = 8/2.16
= 3.70 m2
( i.e. area supported by one roof bolt)
= 1.90m X 1.90m in square pattern or 1.5m X 2.5m in rectangular pattern.
(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;
8/A = 2.938
Or, A = 8/2.938
= 2.723m2
( i.e. area supported by one roof bolt)
= 1.65m X 1.65m in square pattern or 1.5m X 1.8m in rectangular -pattern.
In the modified system, 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, means the modified system is safer than the original system.
CONCLUSION;-
Comparing both the support pattern, it can be concluded that the modified pattern of support system is
much safer, scientific and logistic.
Author
Tikeshwar Mahto,
Dy. Director of Mines Safety,
Bilaspur Region
Cell-08982493361
Email- tikeshwarmahto@yahoo.co.in