TUTORIALS

Abstract of quantities for Manhole

Comprises excavation, RCC work, Concrete work (PCC), Brickwork, Plaster work within or outer, Iron steps / stairs.

Let’s assume a sketch of the manhole [all the data is supposed]

The base of PCC which is 6” thick. The brick wall with 9” thickness

RCC slab that is 6” thick

The slab has a manhole that has 18” diameter

Assume 3” thick PCC in floor

Assume iron steps below the man hole

Pipe is at the middle of the triangular portion

The dimension of the manhole is 3’ x 2.5’

Compute the subsequent items quantities:

a) Excavation with regard to manhole construction
b) Concrete work in the foundation and superstructure
c) Brickwork in the walls
d) Inside plaster work
e) Iron steps

Summary of quantities:

a) Direct calculation method
b) To utilize table for inputs

Excavation = Excavation depth x Area in plane

Excavation = Breadth x Length x Depth

Excavation = 5 x 5.5 x 3.5 = 96.25 cft

Length = 3’ + 9” + 9” (2 times the wall thickness) + 6” + 6” (bearing portion) = 5.5’

Breadth = 2.5’ + 9” + 9” (2 times the wall thickness for one side and another side) + 6” + 6” = 5’

Depth = 3’ (wall depth below the ground level) + 6” thickness of PCC work in the bed = 3.5’

PCC in bed = Excavation depth x Area in plane

PCC in bed = length x breadth x depth

PCC in bed = 5.5 x 5 x 0.5 = 13.75 cft

a) Overall excavation = 96.25 cft
b) PCC in bed layer = 13.75 cft

PCC [concrete] in floor = Length x breadth x average depth

PCC [concrete] in floor = 3 x 2.5 x [0.25 + 0]/2 = 0.9375 cft

RCC in slab = Length x breadth x thickness – π D²/4 x T [because there is a manhole in the slab]

RCC in slab = (4.5 x 4 x 0.5) – π (1.5)²/4 x 0.5

RCC in slab = 9 – 0.884 = 8.12 cft

PCC [concrete] in floor = 0.9375 cft

RCC in slab = 9 – 0.884 = 8.12 cft

Brickwork for walls [long wall method, short wall method, centreline method, etc]

Brickwork for walls = external box volume – internal box volume

Brickwork for walls = [4.5 x 4 x 4] – [3 x 2.5 x 4] = 72 – 30 = 42 cft

Brickwork = 42 cft

Compute inside plaster

Inside plaster = inside parameter x height = [(3 x 2) + (2.5 x 2)] x 4 = 44 ft²

Inside plaster = 44 ft²

Iron steps = 3 @ 12” c/c

Manhole covers = 1 cover having 18” diameter

Get through the subsequent video lesson to obtain additional information.

Lecturer: SL Khan