Presented by: Civil Engineering Academy

Quantity Take-off MethodsPresented by: Civil Engineering Academy

Detailed measurements of materials and labor needed to complete a construction project.

Common Problems are:◦ Soil (hauling, compacting, excavating)

◦ Concrete Formwork (foundations, etc.)

◦ Masonry Work (bricks, mortar, etc.)

◦ Steel (structural beams, rebar in concrete, etc.)

◦ Any material asked to estimate (roofing, etc.)

For earthwork problems use the following:◦ Average End Area Method

◦ Prismoidal Method

Eq. 80.8

Eq. 80.9

Eq. 80.10

Based on the information given for a new

road, find the amount of excavation required.

Solution:

Use the average end area method to solve.

V=L(A1+A2)/2

For station 2 to 2+50: 50(100-25)/2 = 1875 ft³

Station 2+50 to 3+50: 100(-25-85)/2 = -5500 ft³

Station 3+50 to 4+50: 100(-85+0)/2 = -4250 ft³

Total = 1875-5500-4250 = -7875 ft³ = 291.67 yd³ (cut)

Station Cut/Fill

2+00 +100 ft²

2+50 -25 ft²

3+50 -85 ft²

4+50 0 ft²

Earthwork (excavation, hauling, compaction)◦ Understand the difference when transporting soils

and what exactly they are asking for. In place soils, loose soil is during transport, and compacted is the final location.

◦ Become familiar with the equations found in the CERM for BCY, LCY, CCY, shrinkage factor, and swell factor. See CERM Chapter 80 (Table 80.1).

80 yd³ of bank run soil is excavated and stockpiled before being transported and subsequently compacted. The soil has a swell factor of 0.35 and a shrinkage factor of 0.10. The final volume of the compacted soil is most nearly? (CERM PP 80-3)Solution:

Given: 80 yd³ of BCY

SF (swell factor) = 0.35

DF (shrink factor) = 0.10

Use Table 80.1 and find what the CCY equation are:

CCY = (1-DF)BCY = (1-0.10)80 = 72 yd³

Concrete Footings◦ Be aware of what dimensions are given. Are they

the outside perimeter, the centerline (shown), or the

inside perimeter. With any question you need the

length.

If it’s the centerline then the total

length is:

32+31+53+14+10+7 = 147 ft.

If it’s the outside then use the

horizontal in-in and vertical out-out

method:

28+31+49+14+10+7= 139 ft.

28 ft

49 ft

10 ft

Masonry◦ You are typically asked to find the amount of bricks

in a wall or the amount of mortar needed. Either

can be figured by reasoning but here are some tips.

Find the number of bricks needed in a wall:

1) Calculate the net surface area of the wall.

2) Calculate the surface area of one brick including the mortar (1/2

the joint thickness on either side. Brick typical dimensions are 7⅝” L

× 3⅝” H× 2¼” W

3) Divide the net wall area by the surface area of the brick.

4) Multiply the number of rows by bricks required.

5) Add an amount of waste given in the problem.

Masonry

To find the amount of mortar needed:

1) Calculate the total amount of mortar for one brick.

2) Multiply the mortar required per brick by the total number of bricks.

3) If there is more than one row then the volume of the mortar needed

to fill the gap between rows needs to be added, which is the joint

thickness multiplied by the net area of the wall.

4) Add any amount of waste.

Find the number of bricks required for the wall

shown. Length = 125 ft.Height = 12 ft.Size of bricks* = 8”(L)x3-5/8”(D)x2-1/4”(H)Mortar thickness between brick = 0.5”The wall is one brick thick.

Solution:Just follow our steps:1) Net surface area of wall = 125’ x 12’ = 1500 ft² or 216000 in²2) Find surface area of brick with mortar = 8.5”x2.75” = 23.375 in²3) Divide wall area by brick area = 216,000/23.375 = 9240.6 bricks4) Multiply by number of rows (1 brick thick in this case). Total = 9241 bricks needed

*it should be noted that sometimes brick dimension

given are different. Sometimes they are given as

DxHxL. Use common sense – draw it out.

Find the amount of mortar required for the

wall shown. Length = 125 ft.Height = 12 ft.Size of bricks = 8”(L)x3-5/8”(D)x2-1/4”(H)Mortar thickness between brick = 0.5”The wall is one brick thick.

Remember 9240.6 bricks needed.

Solution:Just follow our steps:1) Calculate amount of mortar for 1 brick = (8.5 x 0.25 x 3.625)2+(2.25 x 0.25 x 3.625)2 = 19.484

2) Multiply by number of bricks = 19.48 in³ x 9240.6 = 180047.32 in³ = 3.86 yd³

3) Add any volume to fill gap if more than 1 row (there isn’t more than one).

4) Add any waste given in the problem statement (there wasn’t). Total = 3.86 yd³

Example problems

Cost estimating