Wednesday 10 November 2021

FORMULA FOR CLCULATION OF VOLUME

 

FORMULA FOR CLCULATION OF VOLUME

 


D = Common distance between sections.

 Trapezoidal rule or average end area rule.

 

Volume (Cutting or Filling),

 

                           V = D/2[A₁ + An + 2(A₂ + A₃ + A₄ + … +An-)]

 

i.e.   Volume = Common distance {Area of 1st section + Area of last section

                                  2              

                                                + 2(Sum of area of other sections)}

 

Prismoidal rule.

 

Volume (Cutting or Filling),

 

          V = D/3[(A₁+An) + 4(A+A₄+...+ An-₂) + 2(A₃+A₅+... + An-₁)]

 i.e.   Volume = Common distance {Area of 1st section + Area of last section

                                  3              

                                                 + 4(Sum of area of even sections)

                                                 + 2(Sum of area of odd sections)}

 

This formula is applicable when there are an odd number of sections.

 

PRISMOIDAL CORRECTION FOR TRAPEZOIDAL OR AVERAGE END AREA RULE


 

i.             Prismoidal correction for level section:

 

            Cp = D x s (h – h)²    (Considering, transverse slope= 1 in n side slope= S : 1)

                        6       

 

 

 

 

ii.            Prismoidal correction for two level section:

 

            Cp = D x s          x (h₁ – h₂)²

                        6    n² - s     

 

iii.          Prismoidal correction for three level section:

 

            Cp = D  (h₁ – h₂)    (Whole width of 1st section – Whole width of 2nd section)

                           12

 

iv.          Prismoidal correction for Side-hill two level or Part cut-part fill:

 

(a)        Cp (for cutting) =      D       x n²(h₁ – h₂)²    (side slope= s₁ : 1)

                                              12(n – s₁)

 

(b)        Cp (for filling) =      D       x n²(h₁ – h₂)²    (side slope= s₂ : 1)

                                           12(n – s₂)




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Friday 5 November 2021

COMPUTATION OF VOLUME

 

VOLUME

Earthwork operations involve the determination of volumes of material that is to be excavated or embanked in engineering project to bring the ground surface to a predetermined grade. Volumes can be determined via cross-sections, spot levels or contours. For computation of the volume of earth work, the sectional areas of the cross-section which are taken transverse to the longitudinal section during profile leveling ate first calculated. Again, the cross-sections may be different types, namely: (a) Level (b) Two-level, (c) Three-level, (d) Side-hill two level or Part cut-part fill and (e) Multi-level.

After calculation of cross-sectional areas, the volume of earth work is calculated by:

i.             The trapezoidal rule or average end area rule.

ii.            The prismoidal rule.

 

1)   The prismoidal rule gives the correct volume directly.

2)   The trapezoidal rule does not give the correct volume. Prismoidal correction should be applied for this purpose. This correction is always subtractive.

3)   Cutting is denoted by a positive sign and filling by a negative sign.

A. Level Section

 

When the ground is level along the transverse direction



Cross-sectional area = (b + b + 2sh) x h

                                         2

                              = (b + sh)h

B. Two-Level Section

 

When the ground surface has a transverse slope:

 


PB = b/2

Bx = sh₁

b₁ = b/2 + sh₁                   …………………………………………. (a)

Ee = (h₁ – h)

b₁ = n x Ee = n(h₁ - h)       …………………………………….…….(b)

 

From (a) and (b),       b/2 + sh₁ = n(h₁ - h)

 

                     Or        h₁(n – s) = n(h + b/2n)

  

                     Or        h₁ =     n      x (h + b/2n)       ….…(1)

                                        (n – s)

 

From (2) and (a),       b₁ = b +    ns      x (h + b/2n)  ……(2)

                                        2    (n – s)

 

                                h₂ =     n      x (h - b/2n)        . ……(3)

                                        (n + s)

 

                                b₂ = b +    ns      x (h - b/2n)   ……(4)

                                        2    (n + s)


Area ABCDE = ∆DOE + ∆COE - ∆AOB

                   = 1 OE X Dd + 1 OE x Ce - 1 AB x OP

                       2                 2                2

 

Here,      OE = OP + PE =  b + h

                                      2s

Dd = b₂        Ce = b₁

 

AB = b          OP =  b

                             2s

 Area = ½{( b + h)b₂ + ½( b + h)b₁ - 1b x b}

                     2s                      2s              2     2s

 

        = ½{( b + h)(b₁ + b₂)}       ……(5)

                     2s                           2s            

 

 

 C. Three-Level Section

 

When the transverse slope is not uniform:

 


  Area ABCOD = ∆DOP + ∆COP + ∆DAP + ∆BCP

 

                     = 1 x h x b₂ + 1 x h x b₁ + 1 x b x h₂ + 1 x b x h₁

                         2                 2                2    2           2    2

 

                Area = {1/2(b₁ + b₂) + b/4(h₁ + h₂)}

 

Here               h₁ = OP + Oe = h + b₁/n₁

 

                      h₂ = OP – ef = h – b₂/n₂         

Deduction of formula for b₂ and b₁

                       b₂ = AP + AK = b/2  + sh₂

 or,                   h₂ = {b – (b/2)}                        … (1)

                                     s

Also

                       b₂ = ef x n₂ = (h – h₂)n₂

or,                   h₂ = hn₂ – b₂                              … (2)

                                  n₂

From (1) and (2)

 

                      {b – (b/2)} = hn₂ – b₂                     

                              s₂                n₂

 

                     b₂n₂ – bn₂/2 = hn₂s - b₂s

  

                    b₂(n₂ + s) = n₂(sh + b/2) = n₂s(h + b/2s)

 

                    b₂ = n₂s   x (h + b/2s)

                           n₂ + s

Similarly,

 

                    b₁ = n₁s   x (h + b/2s)

                           n₁ - s

 

 

D. Side-hill two level or Part cut-part fill

 

When the ground surface has a transverse slope, but the slope of the ground cuts the formation level partly in cutting and partly in filling, the following method is adopted:

 


Here,          h =  n   x (b/2n + h)

                      n – s

                  b₁ = b +    ns      x (h + b/2n)

                         2    (n – s) 

Then h₂ and b₂ are deduced as follows:

                            b₂ = b + AA' = b + sh₂                     … (i)

                                    2             2

Again                     b = EE' = O'E' x n = (h + h₂)n           … (ii)

 

From (i) and (ii),     b + sh₂ = (h + h₂)n 

                             2

 


Or                          h₂(n – s) = b – hn = n  b – h

                                               2               2n

 

                              h₂ =  n   x (b/2n - h)

                                    n – s

 

 From (i)                               b₂ = b +    ns      x (b/2n - h)

                                                    2    (n – s)

Area in Cutting:

     Area of ∆PBC,

 

                         A₁ = 1 x PB x h₁

                                 2

Here,                 PB = OB + OP = b + nh

                                                  2

 


                       A₁ = 1  x   b + nh     n     x  b + h  

                                2      2            n – s    2n

 

                          

                             = 1   b + nh     1    x b + nh

                                 2   2           n – s   2

 

                             = 1   {(b/2) + nh}²

                                 2        n – s

 

Area in Filling:

     Area of ∆APE,

 

                         A₂ = 1 x PA x h₂

                                 2

 

Here,                 PA =  b - nh

                                  2

 

                        A₂ = 1  x   b - nh      n     x  b - h  

                                2      2            n – s    2n

 

                          

                             = 1   b - nh      1    x  b - nh

                                 2   2           n – s   2

 


                             = 1   {(b/2) - nh}²

                                 2        n – s

In the above case, the side slopes for cutting and filling are assumed to be equal. But in actual practice, the side slope of cutting is different from that of filling. Let the side slop of cutting be s₁ : 1.

   Then,

 

                                 b₁ = b +    ns₁      x (h + b/2n)

                                         2    (n – s₁)

Area in cutting,

 


                                 A₁ = 1   {(b/2) + nh}²

                                         2        n – s

E. Multi-level Section

 The cross-sectional data pertaining to an irregular section are noted in the following form:

 



             Left

           Centre

           Right

   

        ±h₂/2            ±h₁/2 

       

                     ±h/0

    

        ±h₃/b            ±h₄/b₄

        

 

A positive sign in the numerator denotes a cut, and a negative sign indicates a fill.

 

The denominator denotes corresponding horizontal distance from the centre. Starting from the centre (E) and running outwards to the right and left, the coordinates of the vertices are arranged, irrespective of algebraic sign, in determinant form:

 

 

A        G        F        E        D       C         B             

 

0₀      h₁       h₂       h₃       h₄      h₅       0              

b/2     b₁       b₂       0        b₄      b₅      b/2

 

The sum of products of the coordinates joined by the solid line is

 

       ∑ P = h₃ x 0 + h₄ x b₃ + 0 x b₄ + h₁ x 0 + h₂ x b₁ + 0 x b₂          

 

The sum of products of the coordinates joined by the dotted line is

 

       ∑ Q = h x b₃ + h₃ x b₄ + h₄ x (b/2) + h x b₁ + h₁ x b₁ + h₂ x (b/2)

 

Area = ½ (∑ P - ∑ Q)         

 

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