With Figures, the amount of concrete poured, rebar, and form used in continuous

Question

With Figures, the amount of concrete poured, rebar, and form used in continuous footings and foundation wall can be estimated. For wall width and continuous footing width, see figure below.

A) Using a waste factor of 20 percent, determine the number of cubic yards of concrete needed to pour the continuous footings shown in Figure below. Assume that the wall is centered over the footing.

B) Determine the amount of rebar needed for the continuous footing shown in Figure below. Allow for two inches of cover. Add 22 percent for lap and waste to the continuous bars.

#5 Dowel bar (                ) pounds

#4 Rebar @ 12” O.C (                ) pounds

#4 Continuous rebar (               ) pounds (apply waste factor for lap)

Total Pounds( ) Total tons( )

42-0 Continuous footing length 42' Wall wall width = 1 foot Continuous footing width 3 feet Centerline Plan of Wall

Explanation / Answer

For question A

Wall is centred over footing and length of footing along long wall= 42 ft

Length of footing along short wall = 26-3-3= 20 ft

Dimensions of footing are : width = 3 ft ; depth = 1'5"

Thus total volume of wet concrete without wastage = 3*(17/12 ) * 42 *2 + 3*(17/12)*20*2 = 527 cft

Total volume of concrete after20% wastage = 1.2* 527 = 632.4 cft =0.037* 632.4 cubic yard = 23.3988 cubic yard

Dimensions of wall are : thickness = 1 ft , height = 12 ft thus volume of wet concrete including 20% wastage = 1.2*(39*2*12*1)+1.2*(23*2*12*1)= 1785.6 cft= 0.037*1785.6 = 66.0672 cubic yard

For part B

#4 bar at top of bottom mesh of footing @ 12"

Total length of 1 bar = 32"

No. Of bars along long wall = 43*2= 86

No. Of bars along short length = 21*2= 24

Total running length = 86*32"+ 24*32" =3520" =293.33 ft

Thus weight = 0.668* 293.33= 195.94 lbs

#4 bars continuous 5 each side at bottom of bottom mesh of footing assuming each bar length to be equal to average bar length which is equal to length of wall centred on footing = 2*(39+23)= 124'

Thus total length without considering overlap = 2*5*124' =1240' and weight without considering overlap = 1240*0.668= 828.32 lbs

Weight considering overlap = 828.32* 1.22 = 1010.5504 lbs

#5 dowel bar length of 1 bar = 17" +(17" -2") + 13"+13" = 58"

No. Of bars @ 12" spacing = 86+24 = 110

Thus weight = 58*110*1.043/12 = 554.52 lbs

Thus for question B summary

#5 bar = 554.52 lbs = 251.75 kg = 0.252 MT

#4 bar = 195.94 lbs = 88.956kg= 0.089 MT

# 4 continuous bar = 1010.5504 lbs = 458.789 kg = 0.459 MT

Thus total weight = 554.52+195.94+1010.5504= 1761.0104 lbs = 799.498 kg = 0.799 MT

For question C

Formwork is required on sides of footing and linear foot of form needed which is 1'5" high = 42*2+20*2= 124'

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