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F.L.sd B sd

d

sd B sd

s:1s:1

N.S.L.

F.L.

ds:1s:1

X-Section in Banking/Filling

X-Section in Cutting

sd B sd

N.S.L.

F.L.

s:1

s:1

X-Section in Cut & Fill

N.S.L.

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N.S.L.

F.L.Rise, s:1 Fall, s:1

L-Section in Banking/Filling

N.S.L.

F.L.Rise, s:1

Fall, s:1

L-Section in Cutting

N.S.L.

F.L.

Rise, s:1

Fall, s:1

L-Section in Cut & Fill

Rise, s:1

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LongitudinalSectioning

CrossSectioning

PLAN VIEW

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(a) Mid-Sectional Area Method

(m) (m) (m) (m) (m) (m) (m

2

) (m

2

)- - - -

(b) Mean-Sectional Area Method

(m) (m) (m) (m) (m) (m2) (m

2) (m

2)

-

(c) Prismoidal Formula Method

(m) (m) (m) (m) (m2) (m

2) (m

2) (m

2)

- - - - - -

Area of Side Sloping Surface

Sloping Breadth

d*(s2+1)

(m) (m) (m)

- - -

Total =

(m2)

-

d22 d1d2

Station/

R.D.

Total Area of Side Slopes

2*L*d*(s2+1)

Mean

Height/Depth

dm

Length

b/w

stations,

L

Height/

Depth,

d

Side

Area, As

(sd2)

Total X-

Area, A

(Ac+As)

Station/

R.D.

Length

b/w

stations,

L

Height/

Depth, d(d1+d2)/2

Central Area, Ac

B*(d1+d2)/2d1

2

Station/

R.D.

Length

b/w

stations,

L

F.L. N.S.L.Height/

Depth, d

Central

Area, Ac

(Bd)

Station/

R.D.

Length

b/w

stations,

L

F.L. N.S.L.Height/

Depth,d

Mean

Height/

Depth

(dm)

Central

Area, Ac

(B*dm)

Side

Area, As

(s.dm2)

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Cutting Filling

(m

2

) (m

3

) (m

3

) Area of Mid-Section: Area of rec- - -

= Bdm+1/2sd

= Bdm+sdm2

Quantity = L*(Bdm+sd

Cutting Filling

(m2) (m

3) (m

3)

- - - Mean-Sectional Area: A=(A1+A2)/

A1= Bd1+s

Quantity = L*(A1+A2)/2

Cutting Filling

(m2) (m

3) (m

3) Quantity or Volume = L/6 (A1+A2

- - -

= {B(d1+d2)/2

= (Sectional

Total =

Total =

Total =

Side Area, As

s*{(d12+d2

2+d1d2)/3}

Earthwork

(Ac+As)*L

Mean X-Area, Am

Earthwork

(A*L)

Earthwork

(A*L)Total X-Area, A (Ac+As)

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tangular portion + area of two triangular portions

m2+1/2sdm

2

m2)

12, A2=Bd2+sd2

2

4Am)

+ s(d1 +d2 +d1d2)/3}*L

rea of central portion + sectional area of side slope portions) * Length

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Design Problem-1

Calculate the quantity of earthwork for 200m length for a portion of a road in an unifor

at the two ends being 1.00m and 1.60m. The formation width is 10m and side slopes

Assume that there is no transverse slope.

Design Problem-2

Reduced level (R.L.) of ground along the centre line of a proposed road at 30m interv

The formation level (F.L.) at R.D. of 300m is 107.0m and the road is in downward gra

420m and then the gradient changes to 1 in 100 downward. Formation width of the ro

banking are 2:1 (H:V).i) Draw longitudinal section of the road and a typical cross section. Also, determine th

ii) For estimation of turfing on side slopes, find out the area of side slopes as well.

R.D. (m) 300 330 360 390 420 450 480

R.L. of NSL (m) 105.0 105.6 105.44 105.9 105.42 104.3 105.0

R.L. of F.L. (m) 107.0

Design Problem-3

Estimate the quantities of earthwork for a portion of road for 400m length from the foll

Formation width is 10m. Side slopes are 2:1 and 1.5:1 in banking and cutting, respect

Draw the longitudinal section and typical cross sections.

35 1400 50.6

Fall, 1 in 200

34 1360 51.0

33 1320 51.3

32 1280 51.4

31 1240 51.2

30 1200 50.7

29 1160 50.6

28 1120 50.8

27 1080 50.5

26 1040 50.9

25 1000 51.0

Station No. R.D (m)R.L. of N.S.L.

(m)

R.L. of F.L.

(m)

52.0

Fall, 1 in 150

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Design Problem-

R.D. (m)

m ground, the heights of banks R.L. of NSL (m)

2:1 (H:V). R.L. of F.L. (m)

Height/Depth (m)

al are given below.

ient of 1 in 150 upto R.D. of

ad is 10m and side slopes of

e quantity of earthwork involved.

510 540 570 600

104.1 104.62 104.0 103.3

owing data.

ively.

Fall, 1 in 100

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300 330 360 390 420 450 480 510 540

105.0 105.6 105.44 105.9 105.42 104.3 105.0 104.1 104.62

107.0 106.8 106.6 106.4 106.2 105.9 105.6 105.3 105.0

2.0 1.2 1.2 0.5 0.8 1.6 0.6 1.2 0.4

107.0

106.0

105.0

104.0

103.0

Datum

F.L.

N.S.L.

F.L.

sd 10m sd

d 2:12:1

X-Section in Banking/Filling

N.S.L.

Longitudinal Section

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570 600

104.0 103.3

104.7 104.4

0.7 1.1

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