Presentation is loading. Please wait.

Presentation is loading. Please wait.

PRESENTED BY : TAILOR SHIVOM R. 130180106117 GOVERNMENT ENGINEERING COLLEGE, DAHOD CIVIL ENGINEERING DEPARTMENT ACTIVE LEARNING ASSIGMENT SURVEYING (2130601)

Similar presentations


Presentation on theme: "PRESENTED BY : TAILOR SHIVOM R. 130180106117 GOVERNMENT ENGINEERING COLLEGE, DAHOD CIVIL ENGINEERING DEPARTMENT ACTIVE LEARNING ASSIGMENT SURVEYING (2130601)"— Presentation transcript:

1 PRESENTED BY : TAILOR SHIVOM R. 130180106117 GOVERNMENT ENGINEERING COLLEGE, DAHOD CIVIL ENGINEERING DEPARTMENT ACTIVE LEARNING ASSIGMENT SURVEYING (2130601) GUIDED BY:- PROF. H. D. GOLAKIYA

2 This is the most widely used method. The total volume is divided into a series of solids by the plan of cross section s. The fundamental solids on which measurement is based are the prism, wedge and prismoid. The spacing of the sections depends upon the character of the ground and the accuracy required in the measurement. The area of the cross-sections taken along the line are first calculated by standard formulae developed below, and the volume

3  of the prismoids between successive cross section are then calculated by either trapezoidal formula or by prismoidal formula.  The various cross-sections may be classed as 1. Level section 2. Two-level section’ 3. Side hill two level section 4. Three level section 5. Multi level section

4  Let  b = The constant formation (or sub grade) width  h = The depth on cutting on the centre line  w 1 and w 2 = The side width,or half breadths, i.e., the horizontal Distance from the inter section of the side slopes with original ground level.  h 1 and h 2 = the side heights, i.e., the vertical distance from formation level to the intersection of the slopes with the original surface

5 FIG.1 CROSS SECTION

6 n horizontal to 1 vertical = inclination of the side slopes. m horizontal to 1 vertical = the transverse slope of the original ground A = area of the cross-section

7  In this case the ground slopes crosses the formation level so that one portion of the area is in cutting and the other in filing.  Now, BJ= nh 1  Also ∴..........(1

8 But w 1 = ( h 1 – h )m.......(2 Solving (1) and (2) as before we get h 1 = And w 1 =

9 FIG.1 SIDE HILL TWO- LEVEL SECTION

10  Let us now derive expression for w 2 and h 2 IA=nh 2  Also IA= IH – AH = ∴......(3 Also w 2 =(h + h 2 )m ∴ nh 2 = (h + h 2 )m......(4 And Or h2h2 h2h2

11  Hence  By inspection, it is clear that the expression for w 1 and w 2 are similar ; also expression of h 1 and h 2 are similar, except –h in place of +h  Now area of filing = ∆PBE = A 1 (say),  And, area of cutting = ∆PAD = A 2 (say),

12 FIG.3 THREE- LEVEL SECTION

13  Let 1 in m 1 be the transverse slop of the ground to one side and 1 in m 2 be the slop to the other side of the centre line of the cross section.  The expression for w 1, w 2, h 1 and h 2 can be derived in the similar way as for case (2)  Thus,

14  The area ABECD =

15  In the multi level section the co-ordinate system provides the most general method of calculating the area. The cross-section notes provide with x and y co- ordinate for each vertex of the section, the origin being at the central point (H). The x co-ordinates measure positive to the right and negative to the left of H. Similarly, the y co-ordinates are measured positive for cuts and negatives for fills. In usual form, the notes are recorded as below:

16 FIG.3 MULTI-LEVEL SECTION

17  If the co-ordinates are given proper sign and if the co- ordinates of formation points A and B are also include (one at extreme left and other at extreme right),they appear as follow:  There are several method to calculate the area. In one of the methods the opposite algebraic sign is placed on the opposite side of each lower term. The co-ordinate than appear as :

18  The area can now be computed by multiplying each upper term by the algebraic sum of the two adjacent lower terms, using the sign facing the upper term.The algebraic sum of these product will be double the area of the cross section.  Thus, we get  H =

19


Download ppt "PRESENTED BY : TAILOR SHIVOM R. 130180106117 GOVERNMENT ENGINEERING COLLEGE, DAHOD CIVIL ENGINEERING DEPARTMENT ACTIVE LEARNING ASSIGMENT SURVEYING (2130601)"

Similar presentations


Ads by Google