All about civil construction knowledge- PARAM VISIONS

Punning in civil construction.

 Now, let us go through some of the common questions related to punning in civil construction.

1. What is punning in construction?



Punning is the process of applying a thin layer of the slurry or mortar, made of cement, POP, gypsum, lime, or any such specified binding materials to the wall, ceiling, flooring, or plastering surface. The punning in some regions are also referred to as a skim coating.

They are applied over the existing masonry wall or ceiling surface directly or over the finished plastering surface. In the case of flooring, neat cement punning is done over the fresh mortar surface before fixing the flooring tiles.


2. What are the different types of punning?

The most common types of punning are,

    1. Neat cement punning.

    2. POP  punning.

    3. Gypsum punning. &

   4. Cement plus lime punning.


3. What should be the thickness of punning?




The thickness for different types of punning are,

1.  Neat cement punning πŸ‘‰ between 1 to 3mm.

2. POP punning πŸ‘‰ 3 to 6mm.

3. Gypsum punning πŸ‘‰ between 6mm to 12mm.

4. Cement plus lime punning πŸ‘‰ 2mm to 6mm.


4. What are the advantages & purpose of punning in construction?

Advantages & purpose of punning:

  1. To fill the undulations over the rough surface.

  2. Helps in providing a smooth finished surface.

  3. Punning gives protection against dampness and seepage.

 4. Punning provides an aesthetic appearance and a royal home interior.

 5. It helps in saving time and money.

 6. Punning provides a good base for the painting work.



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What is a rechargeable LED bulb?/ What is LED inverter bulb?

  LED inverter bulbs are rechargeable emergency bulbs that glow with 60 - 75% efficiency when there is a power outage.


When the current is on, these bulbs work as normal LED bulbs and simultaneously get charged. These bulbs fit in the same socket as that of normal bulbs but are larger and heavier when compared to the normal LED bulbs as shown in fig.




These bulbs contain a small rechargeable battery with a charging circuit and have a control circuit with switch sensor technology.

The charging time of these bulbs is approximately 8 - 10 hours and the backup time will be about 4 hrs. if fully charged. They usually contain lithium batteries having a normal life cycle of 400 - 600 times.

Inverter bulbs come in all wattage range like normal LED bulbs at an approximate price of 2.5 times the normal LED bulb of the same wattage.


The system works as follows

When the switch is on πŸ‘‰ the bulb is on.

         The switch is on πŸ‘‰ power outage πŸ‘‰ the bulb is on.

         The switch is off πŸ‘‰ the bulb is off.


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Laser plumb bob.

 1. What is a laser plumb bob?



A laser plumb bob is a tool that transfers a given point in a vertical plane by projecting a laser pointer beam.



 

Suppose, if you want to hang a chandelier light at a certain point in the ceiling, say at the center of the room, then by using the laser plumb bob you can transfer that point from the marking made over the floor to the ceiling by using laser plumb bobs.


Advantages of laser plumb bob over traditional string plumb bobs.

1. String plumb bob consumes much of your time in settling down but the laser plumb bob does this work in a lesser time, as it uses a laser pointer beam.

2. The work done will be more accurate with a minimal tolerance limit.

3. It makes your work much easier and hassle-free.

4. There is less room for the degree of error  

Nowadays, line laser levels are replacing laser plumb bobs as they are multifunctional to use in almost all kinds of constructional works.

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Laser levels for civil construction.

 Now, let us go through some of the FAQ's and let us gain some basic understanding of laser levels.

1. What is a  laser level?


A laser level is a device that projects a laser beam horizontally and vertically and also a point beam, to enable the person using it to level and align the works that he is doing.


2. What is the different types of laser levels?

Laser levels can be broadly divided into three main types.

They are,

1. Line laser level.

2. Rotary laser level &







3. Plumb, dot, or point laser level.


Based on the leveling capability, they are categorized as

1. Manual leveling laser 

2. Self-leveling laser.


Based on the beams emitted through laser levels, they are subdivided as

 1. Horizontal beam or single-beam lasers.

 2. Dual-beam or split beam lasers.


3. What are the applications of laser levels?

Laser levels can be used for both indoor and outdoor construction works. Let us check out the indoor & outdoor applications separately.

Indoor applications:

1. To align and fix the flooring tiles.

2. To plumb the walls, doors, windows, etc.

3. For the electrification work by aligning the switchboards and ceiling fixtures.

4. For fixing the plumbing items like washbasins, sinks, drainpipes, etc. in required levels.

5. To check the height of doors, windows, walls, easily.

7. In aligning kitchen platforms, cabinets, shelves, etc.

8. For leveling & aligning railing works, drop ceiling works, and many more.


Outdoor applications:

1. For layout and fixing the boundaries of all types of land.

2. For contouring in survey work.

3. For providing slopes of different gradients in infrastructure work.

4. For alignment of infra masonry works.

5. To find out the land elevation.

6. In the estimation of earthwork excavation & filling.

7. Aligning fence posts and drainage system.


4.  What is the working principle of laser levels?



The laser level projects a beam of light in a vertical & horizontal plane through the amplified diodes. This focused beam of light is reflected through the mirrors or prisms in one direction to work with. These lights are red or green in color for easy working visibility.

 Laser levels that emit the beams of lesser width are more precise as the line of level narrows down providing greater accuracy.


5. What is the accuracy of a laser level?

Laser level accuracy usually ranges between ±0.75mm to ±3mm. The accuracy of different types of laser level is denoted along with the emitted beam length. That means a particular type of laser level is specified by saying that it has an accuracy of ±1/16 of an inch for 100ft. The higher-end models are more accurate & vice versa.


        BACK  PART 1πŸ‘ˆ                 CONTINUED πŸ‘‰ PART 3






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Cockroach drain trap / Anti cockroach floor trap drain.

 Now, let us look into some of the questions, regarding the cockroach drain trap to understand its concept.

1. What is a cockroach drain trap?

stainless steel cockroach trap.

A cockroach drain trap is a compact plumbing fixture, that can be installed at the drains of the bathroom, washbasin, kitchen sink, etc. to control the entry of cockroaches and all such insects from the drain pipe. They also help to a larger extent in controlling the foul smell entering your home through the drainage system.


2. What are the different materials, sizes, & shapes of the cockroach drain trap?

plastic drain trap.


Cockroach drain traps are usually made up of stainless steel, polymer plastics like PVC, & ABS materials. When compared with each other, the stainless steel type is more advantageous, as they are aesthetic, easy to clean, and lasts longer.

The outer body may be square or round in shape having 4" & 5" diameter( for round type ). First, you have to check your drain pipe diameter and then you have to select the one, that fits in your drainage system.


3. What is the working principle of a cockroach drain trap?

The cockroach drain trap is usually made up of 4 parts as shown in the image below. 



The outer filter is fixed over the rim and the inner cup-like filter part is inserted into the outer filter. The clogging of the water between the two filters and small openings in them acts as a barrier for the cockroach from entering your home through the drain trap. The water level will be maintained constant, no matter if there is an overflow, or the drain is not in use for a long time.

The grating sits over the top of the rim and you can remove this drain trap easily from the drainpipe for cleaning purposes. All these 4 parts can be separated, cleaned, and reassembled within a few minutes.


4. What are the advantages of a cockroach drain trap?

1. Cockroach drain trap prevents the cockroach & other rodents from entering the house.

2. The outer filter of the trap retains unwanted materials and prevents the drain pipe from clogging.

3. It blocks bad odors, and helps in controlling the foul smell from entering your house.

4. It can be removed easily from the drainage system for the cleaning purpose.


If your bathroom & sink is already fitted with the simple gratings ( jali ), you can still install them in your drain system without much effort at a minimal cost (INR 150 to INR 800 )

You can check the price of the top 2 brands of cockroach drain trap by clicking the image below.

                                                         

                     


Thank you for going through this articleπŸ˜ƒ. Have a good day.







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Laser distance measurer / Laser distance meter.

 Now, let us go through some of the FAQ's related to laser distance measurer (LDM ) / Laser distance meter.

1. What is a laser distance measurer?


laser distance measurer
laser distance meter.
A laser distance measurer or laser distance meter is a device to measure the distance between the two objects without moving an inch as in the case of measuring tapes. The LDM measures the distance more precisely and helps in completing the measuring work in few seconds, saving a lot of your time for doing other works.


2. How does a laser distance meter works? / What is the working principle of laser distance measurer?



The laser distance meter sends a pulse of a laser beam at the targeted object. The emitted laser reflects off the object and comes back to the sending device. This working principle is known as a time of flight or pulse measurement, where the travel time of the laser is calculated accurately and converted into a distance in a required unit.


3. What is the accuracy of measurement in a laser distance meter?

The accuracy of the laser distance meter used for the general purpose usually ranges between 1/8th to 1/16th of an inch i.e. more than enough for the estimation, in civil construction works. The higher-end meters provide a precision of up to ∓ 1mm. in the measurement. 


4. Can we use a laser distance meter outside for measurement?

YES, you can use them anywhere to measure the distance between the two points. The higher range meters have digital viewfinders on the screen to work easily in a broad daylight. 


5. What are the measuring units these LDMs work on?

They show the distance digitally in feet with decimals in inches, & in meters with decimals in cm. You can switch them in any mode as per your convenience.


6. What are the other tasks that you can do over these laser digital meters?



You can calculate the area & volume of the room within a few seconds over these meters. When you measure the length & breadth of the room, the meter will calculate the area of the room for you & if you measure the height of the room, then the volume of the room is displayed when you press the volume button. 

We can measure the horizontal as well as the vertical distance for finding the area of plastering, flooring, size of the windows, beams & slab height, etc., and much more.

 Nowadays, these laser distance meters are becoming a must-have device for civil engineers, contractors, or entrepreneurs, and for all those who are working in the construction field.

These laser distance meters are generally available in measuring ranges of 40m to 250m with different functions like memory, area, volume, etc. for deriving instant results.

The things that we have to look for while buying LDM's are their 

1. Accuracy, 2. Measuring range, 3. Battery life, 4. Value for money, 5. Different operating functions.  6. Warranty and 7. Brand name.

 By all these pros and cons, the best seller in the market is "Bosch laser distance measurer".


  Thank you for going through this articleπŸ˜ƒ. Have a good day.

                                                           CONTINUEDπŸ‘‰

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Calculating the construction cost of a compound wall/ Estimating & costing of a compound wall ( part 8 ).

 BACK  PART 7 πŸ‘ˆ

First, we will sum up the total quantity of materials from all 7 parts series of this compound wall.

 The volume of RCC for compound wall

= for footings+ for plinth beam + for columns

=0.743 cum + 2.75 cum. +1.561 cum

=5.054 cum.i.e. 178.48 cu ft.


Let us make this RCC in M20 grade concrete.

The no. of cement bags required for RCC work

= 8.06 bags ×( 5.054 cum ÷ 1 cum )

= 40.73 bags

The volume of sand required for RCC work

= 0.42 cum ×( 5.054 cum ÷ 1 cum )

=2.122 cum.  i.e. 74.938 cu ft.

The volume of coarse aggregates required for RCC work

= 0.84 ×( 5.054 cum ÷ 1 cum )

= 4.245 cum. i.e. 149.91 cu ft.

Note: The value i.e. directly added above is taken from "Calculating the quantity of materials in different grades of concrete".


Now, we will sum up the total quantity of cement, sand, & aggregates required for the compound wall construction.

Note: The value i.e. mentioned below are taken from all 8 parts of the series.

  The total number of cement bags for compound wall work.

       = for PCC + for masonry + for  RCC + for coping

       = 12.70 +  3.516 + 40.73 + 3.537

       = 60.483 bags 


The total volume of sand required in the compound work

     = for PCC + for masonry + for  RCC + for coping 

     = 31.156 + 25.86 + 74.938 + 8.675

     =  140.629 cu ft.


The total volume of coarse aggregates required in the compound work.

     = for PCC + for RCC + for coping

     = 62.31 + 149.91 + 17.35

     =  229.57cu ft.


The total weight of rebars required for the compound work.

= for footing + for plinth beam + for columns

=   43.98 kgs + 260.69 kgs. + 175.9 kgs.

= 480.57 kgs.


Let us find out the material & labor cost for the construction of a compound wall.

Material cost:


Sl.No.    Description        Quantity    Rate      Unit       Amount

  

1.          Boulder for                    42           23         cu ft.       966.00
             foundation

 2.    Total cement bags     

        for compound work      60.483     380      bag      22983.54     

 3.    Total sand for 

       compound work             140.629     60      cu ft.      8437.74           

 4.   Total aggregates for 

       compound work               229.57       54      cu ft.   12396.78    

 5.    concrete blocks               712             30        no.      21360.00 

 6.   reinforcement                        

        bars                                 480.57         48        kg.      23067.36

        

          Total material cost for compound work =  INR  89211.42



           Labor cost:

Sl.No.   Description        Quantity      Rate      Unit      Amount

 1.     Earthwork excavation          

        for the compound        285.365        5        cu ft.     1426.82

 2.    Boulder soling               42                6        cu ft.       252.00

 3.    PCC work                      70.81          24      cu ft.        1699.44

 4.    Muroom filling            85.102        3       cu ft.         255.306

 5.   RCC work                       178.48        65      cu ft.       11601.2

 6.   Block masonry           338.78          24      cu ft.         8130.72                 

 7.   coping                          19.717            32     cu ft          630.94

              Total labor cost for compound work = INR  23996.43


    The total cost of the compound work = INR 1,13,207.85

   Add 1.5% for water charges & labor for curing =INR 1,698.11

                         Add 10% for contractors profit =INR 11,320.78

    Total expenditure for the compound wall 

                    =INR 1,26,226.74 

Note: The rate of the material & labor changes according to its availability in different regions.


The cost of the compound wall/ r ft. length

 = total expenditure ÷ compound wall length

   = 126226.74 ÷ 140 ft.

   = INR 901.62 / r ft.


The cost of the compound wall / sq ft.

cost per r ft.÷ compound wall height above GL.

 = 901.62 ÷ 6.416 ft.

 = INR 140.52 / sq ft.

BACK  PART 7 πŸ‘ˆ


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Calculating volume of coping & backfilling for a compound wall. / Estimating & costing of a compound wall ( part 7 ).

 BACK πŸ‘ˆ PART 6  

Let us make coping over block masonry having 4" (0.33 ft.) thickness in M15 grade.



8. The volume of coping for the compound wall

= coping length × width × thickness

= block masonry length × masonry width × coping thickness

= 119.5 ft. × 0.5 ft. × 0.33 ft.

= 19.717 cu ft. i.e. 0.558 cu m.


Let us make this coping in M15 grade concrete.

The no. of cement bags required for the coping work

=17.942 bags × (19.717 ÷ 100 cu ft. )

= 3.537 bags.

The volume of sand required for coping

= 44 cu ft × (19.717 ÷ 100 cu ft. )

= 8.675 cu ft.

The volume of coarse aggregates required

= 88 cu ft. × (19.717 ÷ 100 cu ft. )

= 17.35 cu ft.

Note: The above value for the calculation purpose is taken from " Calculating the quantity of materials in 100 cu ft. & 1 cum. of M15 (1:2:4 ) grade concrete".

9. The volume of backfilling for the compound wall

Backfilling for plinth beam

= Volume of excavation for plinth beam. - beam PCC volume - plinth beam volume up to GL.


= [(145.365 cu ft.) - (52.33 cu ft.) - (97.125cu ft. × 0.583ft ÷ 1 ft.) 

                                    ( By volume ratio with plinth beam πŸ‘†)

= 145.365 cu ft. - 52.33 cu ft - 56.623 cu ft.

= 36.412 cu ft.


Backfilling for footing.



=[ volume of excavation for footing - {footing vol. - soling vol. - PCC vol. - column vol. up to GL.}]

= [140 cu ft.- {26.24 cu ft. + 42 cu ft. + 18.48 cu ft.+( 14 nos × 0.583 × 0.75 × 0.75) } ]

= [140 cu ft - 91.31 cu ft.]

=48.69 cu ft.


The total volume of backfilling 

= backfilling for plinth beam + backfilling for footing 

= 36.412 cu ft. + 48.69 cu ft.

= 85.102 cu ft. i.e. 2.409 cum.

BACK πŸ‘ˆ PART 6                  CONTINUED πŸ‘‰PART 8

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Estimating block masonry of compound wall. / Estimating & costing of a compound wall ( part 6 ).

 BACK πŸ‘ˆ PART 5 

Let us build this compound wall of 6" thickness having 6 ft. height as shown in the drawing.




Plan of the compound wall.


The total length of the compound wall 

     = site perimeter - gate length

     = (40ft. × 2 nos.) + (30 ft. × 2 nos.) - 10 ft.

     = 80 ft. + 60 ft. -10 ft.

     = 130 ft. 


The total length of the block masonry

      = compound wall length - (no. of columns × width of a single column.)

     = 130 ft. - (14 nos.× 0.75 ft.)

     = 130 ft. - 10.5 ft.

     = 119.5 ft.


Height of the block masonry

   = compound wall height - coping thickness

   = 6 ft - 0.33 ft.

  = 5.67 ft.


7. The volume of the block masonry

   = length × height × thickness 

   = 119.5 ft. × 5.67 ft. × 0.5 ft.

   = 338.7825 cu ft.


Number of concrete blocks required

   = 210 nos. × ( 338.78 cu ft. ÷ 100 cu ft.)

   = 712 nos.


The number of cement bags required

   = 1.038 bags × ( 338.78 cu ft. ÷ 100 cu ft.)

   =3.516 bags.


The volume of sand required

   = 7.634 cu ft. × ( 338.78 cu ft. ÷ 100 cu ft.)

  = 25.86 cu ft.

Note: The above-given quantity is taken from the article "Calculating the quantity of materials in a 100 cubic ft. block wall."

BACK πŸ‘ˆ PART 5          CONTINUED πŸ‘‰ PART 7

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Calculating the volume & BBS of columns in a compound wall./ Estimating & costing of a compound wall ( part 5 ).

 BACK πŸ‘ˆ PART 4 

Let us make a column of size 9" × 9"  as shown in the drawing.






Given data :

Column height above GL = 6 ft.+ 5"(0.416 ft.)= 6.416 ft.,  below GL = 7" (0.583ft.) 

Column size      = 9" × 9" (228.6mm × 228.6 mm )

Longitudinal bars 12mm (0.03936 ft.) - 4 nos,  cover  - 40mm. 

Lateral ties dia d1 - 6mm @ 6" (150 mm.) c/c 

From part 1, number of columns = 14 nos.


6a. The volume of the column concrete for the compound wall

  = total nos. × height × length× breadth

 = total nos.  × (height below GL + height above GL )  × length  × breadth

= 14 nos.  ×( 6.416 ft. + 0.583 ft. )  × 0.75 ft. × 0.75 ft.

= 14 nos.  × 7 ft.  × 0.75 ft  × 0.75 ft.

= 55.125 cu ft. i.e. 1.561 cum.


      6b.  BBS of a column for the compound wall.

Length of the longitudinal bar

= above GL + GL to footing top + development length ( Ld )

= 6.416 ft. + 0.583 ft. +( 50d )

( we have taken Ld as 50d, where d = bar diameter.)

= 7 ft. + (50 × 0.03936 ft.)

= 7 ft. + 1.968 ft.

= 8.968 ft.   i.e. 2.733 m.


Length of the lateral ties

= perimeter of lateral ties + total hook length - no. of bends

= 2sides × ( a - 2 × cover ) + 2 sides × ( b - 2 × cover ) +( 2nos × hook length) - (3 nos. × bend )

( Here, we have taken  hook length = 10d1 for 135°∠    & bend = 2d1 for 90°∟)

={ [ 2 × (228.6mm - 2× 40mm.) ]+[ 2 × ( 228.6 mm - 2 × 40 mm.) ] } + { 2 × 10 × 6mm }- {3 × 2 × 6mm }

={ [ 2 × 148.6 mm ] + [2 × 148.6 mm ]} + 120 mm - 36 mm.

= {297.2 mm + 297.2 mm} + 84 mm

= 678.4 mm i.e. 0.678 m.


Total number of lateral ties ( stirrups )

  ={ [length of the longitudinal bar - Ld ] ÷ stirrup spacing } + 1

Note:  Ld is deducted from the length, as no stirrups are provided over that length.

 ={[ 2733 mm - (50 × 12 mm )] ÷ 150 mm.} + 1

 = {[ 2733mm - 600mm ] ÷ 150 mm.} + 1

= {2133 mm ÷ 150 mm.} + 1

= 14.22 + 1

= 15.22 nos. 

Rounding off, the number of stirrups required = 15 nos.


Now, let us prepare BBS for a column.


sl.    bar                    dia.     no.     length    total      weight       total   
no.                        ( mm)                 (m.)     length     kg/m      weight

1.  longitudinal      12        4        2.733     10.932     0.89       9.729

2. lateral                   6        15      0.678       10.17       0.22       2.237

                                                     Total weight of bars  =  11.966 kgs.

                                                    Add 5% wastage          = 0.5983 kgs.

                                             Grand total of rebars  =  12.564 kgs.     

The total weight of  rebars for compound wall columns

 = 14 nos.× 12.564 kgs = 175.9 kgs.


BACK πŸ‘ˆ PART 4          CONTINUED πŸ‘‰ PART 6

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Calculating the volume & BBS of plinth beam in a compound wall. / Estimating & costing of a compound wall ( part 4 ).

BACK πŸ‘ˆ PART 3 

 Let us make a 9"×12" plinth beam as shown in the drawing.




Given data :
Plinth beam size 9"× 12",

Main bar dia = 12 mm.,   no. of bars - 4 nos.
Stirrups  8mm @ spacing 150mm c/c , clear cover = 25 mm from all the sides.


5a. The volume of plinth beam concrete

= length × breadth × depth

= [ (perimeter of site - no.of columns × column width) × breadth × depth] 

= [(140ft - 14 nos. × 0.75 ft.) × 0.75 ft × 1 ft.]

= [129.5 ft. × 0.75 ft. × 1ft.]

= 97.125 cu ft. i.e. 2.75 cum.

Note: We have deducted the column width from the plinth beam length, as we have included them in the column volume.


5b. BBS of plinth beam for the compound wall.




The perimeter of site = 140 ft. = 42672 mm.

We will provide Ld at the corner columns of the compound wall, passing the plinth bar through the intermediate columns. 

The cutting length of the main bar 

= [ (perimeter of site)+ (8nos × Ld )+ (4 nos.×lap length)- (8nos × column width) ]


Note: Let us assume one overlapping for each bar & we will provide a lap length of 50d.

 Let us provide development length Ld = 40d for the main bar. 

= [ (42672 mm.) + (8 nos. × 40 × 12mm) + (4nos × 50 × 12 mm)- (8 nos.× 230mm)]

= [42672 mm + 3840mm + 2400mm - 1840mm ]

= 47072 mm. i.e. 47.072 m.


Cutting length of the stirrup

= 2nos. × (a +b ) + hook length - 90° bend - 135° bend




Where a = beam width - 2 × cover, & b = beam depth - 2 × cover

= 2 nos.  ×[ ( 230 mm - 2 × 25mm.) + ( 300 mm - 2  × 25mm ) ]+ (10d ) - (3 nos. × 2d ) - (2 nos. × 3d)

Here, 10d is taken for hook length.

We have deducted 2d for 90° bend -3nos., & 3d for 135° bend -2nos. as shown in the above drawing.

= 2 nos.×[ ( 180 mm ) + ( 250 mm ) ] + (10 × 8mm) - ( 3 nos. × 2 × 8mm ) - ( 2 nos. × 3 × 8 mm.)

= 2 nos. × [ 430 mm ] + 80 mm - 48mm - 48mm.

= 860 mm + 80 mm - 96 mm.

= 844 mm i.e. 0.844 m.


Number of stirrups 

= ( length of the plinth beam ÷ stirrup spacing ) + 1

here, length of the plinth beam = 129.5 ft = 39471 mm

= ( 39471mm. ÷ 150 mm) +1 

= 263.14+ 1

= 264.14 nos.

By rounding off, the no. of stirrups required = 264 nos. 


Now, let us prepare a BBS table for the plinth beam.


sl.    bar                    dia.     no.     length    total      weight       total   
no.                        ( mm)                 (m.)     length     kg/m      weight

1.  main bar              12         4    47.072    188.29         0.89     167.57

2. stirrups                 8        264    0.844    222.816   0.395      88.012

                                                     Total weight of bars  =  255.58 kgs.

                                                    Add 2% wastage          = 5.111  kgs.

                                         Grand total of rebars  =  260.69 kgs.       

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Calculating the quantity of materials in the RCC footings of a compound wall. / Estimating & costing of a compound wall ( part 3 ).

 PART 2    πŸ‘ˆ BACK

Let us make a footing of size 1.5 ft × 1.5 ft. having 10" ( 0.833 ft. ) thickness as shown in the drawing.


 


Given data :

footing length = 1.5 ft., width = 1.5 ft. , thickness = 0.833 ft.

rebar diameter = 10 mm.(0.0328 ft.),  spacing =5" (0.416 ft. ) c/c,  cover = 2"(0.166 ft ) on all the sides.

From part 1, the number of footings = 14nos. 


4a. The volume of footing for the compound wall

  = total nos. × length × breadth × thickness

  = 14 nos. × 1.5 ft. × 1.5 ft. × 0.833 ft.

  =26.24 cu ft.i.e.0.743 cum. 


4b. BBS for the footings:



No. of bars along x-axis

  = [ {( footing length ) - ( 2 × cover )} ÷ spacing ] + 1

  = [ {( 1.5 ft.) - ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1

  = [ { 1.168 ft. } ÷ 0.416 ft. ] +1

  = 2.807 +1 

  = 4 nos.

( by rounding off )


No. of bars along y-axis

  = [ {( footing width ) - ( 2 × cover )} ÷ spacing ] + 1

   = [ {( 1.5 ft.) - ( 2 ×0.166 ft.)} ÷ 0.416 ft.] +1

  = [ { 1.168 ft. } ÷ 0.416 ft. ] +1

  = 2.807 +1 

  = 4 nos.

( by rounding off )


Cutting length of the bar along the x-axis

=  [ {bar length in x - axis } +{ 2 nos. ×( L- bend length)}] - 2nos.× ( 2 times bar dia. for 90° bend.) 

      ( we have deducted 2 times bar dia i.e. 2d for the  90° bend of the bar. )

=[ { footing length - 2 × cover } + 2nos.×{ footing height - 2 × cover}] - 2× ( 2 × bar dia. )

= [ { 1.5 ft. - 2 × 0.166 ft. } + 2 × { 0.833 ft. - 2 × 0.166 ft. } ] - 2 × ( 2 × 0.0328 ft.)

= [ 1.168 ft. + 1.002 ft. ] - 0.131ft.

= 2.17 ft. - 0.131 ft.

= 2.039 ft. i.e. 0.6214 m.


Cutting length of the bar along the y -axis

=  [ {bar length in y - axis } +{ 2 nos. ×( L- bend length)}] - 2nos.× ( 2 times bar dia. for 90° bend.) 

=[ { footing width - 2 × cover } + 2nos.×{ footing height - 2 × cover}] - 2× ( 2 × bar dia. )

= [ { 1.5 ft. - 2 × 0.166 ft. } + 2 × { 0.833 ft. - 2 × 0.166 ft. } ] - 2 × ( 2 × 0.0328 ft.)

= [ 1.168 ft. + 1.002 ft. ] - 0.131ft.

= 2.17 ft. - 0.131 ft.

= 2.039 ft. i.e. 0.6214 m.


Note: The cutting length & number of bars in both ( x & y ) directions will be the same, in the case of square footing having a similar bar diameter. 

Now, we will prepare BBS of the footing, from calculated data.

sl       bar      dia.   no.   length      total     weight         total     
no.    type    mm.            in m.    length    in kg/m       weight       

 1.    x- axis    10    4    0.6214     2.4856     0.62         1.54   

2.   y - axis   10     4     0.6214     2.4856      0.62         1.54

                                            Total weight of the bars = 3.08kgs               

                                             Add 2 % wastage            = 0.0616kgs

                  A grand total of rebar for a footing    =  3.1416 kgs.

Note : Weight of 10mm dia bar /meter is 0.62 kg.


The total weight of the 10mm bar for all the footings 

   = 14 nos. × 3.1416 kgs

  = 43.98 kgs.



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Calculating boulder soling & PCC for the compound wall. / Estimating & costing of a compound wall ( part 2 ).

 PART 1    πŸ‘ˆ BACK

Let us make a boulder soling of 9" thick ( 0.75 ft.) for the footing as shown below.


From part 1,  the volume of excavation for the footing = 140 cu ft.

From the above drawing, depth of excavation = 2.5 ft.,  depth of soling = 0.75 ft.

2.  Boulder soling for the compound wall

The volume of soling for the footing.

   = ( 0.75 ft ÷ 2.5 ft.) × 140 cu ft.

    (by volume  ratio with excavation)

  = 0.30 × 140 cu ft.

 = 42 cu ft.


 3.  PCC for the compound wall

Let us prepare 4" thick ( 0.33 ft.) PCC bed for the footing & plinth beam.

The volume of PCC for the footings

Vp1 = ( 0.33 ft ÷ 2.5 ft.) × 140 cu ft.

           (by volume  ratio with footing excavation)

      = 0.132 × 140 cu ft.

      = 18.48 cu ft.





From part 1, the volume of excavation for the plinth beam = 145.365 cu ft.

From the above drawing, depth of excavation = 0.9166 ft. 

PCC thickness = 4" ( 0.33ft.).


The volume of PCC for the plinth beam

Vp2 = ( 0.33 ft ÷ 0.9166 ft.) × 145.365 cu ft.

         (by volume  ratio with plinth beam excavation)

      = 0.36 × 145.365 cu ft.

      = 52.33 cu ft.


The total volume of the PCC for the compound wall

   = Vp1 + Vp2

  = 18.48 cu ft.+ 52. 33 cu ft.

 = 70.81 cu ft.


Calculating the quantity of materials in PCC.

Let us make PCC in the 1:2:4 mix.

Quantity of cement bags required for PCC 

17.942 bags × ( 70.81cu ft ÷ 100 cu ft.)

 12.70 bags.


 The volume of sand required for PCC

44 cu ft. × ( 70.81cu ft ÷ 100 cu ft.)

 31.156 cu ft.


The volume of aggregates required for PCC

88 cu ft. × ( 70.81cu ft ÷ 100 cu ft.)

= 62.31cu ft.


Note: The value taken for the calculation purpose is derived in the article "Calculating the quantity of materials in 100 cu ft. & 1 cum. of M15 (1:2:4 ) grade concrete".

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Calculating Earth work excavation for a compound wall./ Estimating & costing of a compound wall ( part 1 ).

 Let us now estimate the cost of a compound wall for a 30' × 40' site with a 10' gate as shown below.



Let us build this wall in a concrete block, with the plinth beam and columns as shown in the drawing.


We have provided the columns at 10 ft. c/c.

The number of columns required 

   = site perimeter ÷ c /c column distance

   =[ ( 40ft.× 2 nos.) +( 30 ft. × 2nos.)] ÷ 10 ft.

   = 140 ft.÷10 ft.

   = 14 nos. ( as shown below.)


1. Earthwork excavation for the compound wall :

    Excavation for the footing

The number of excavation pit required for the column footing = 14 nos.

Let us provide  1.5' × 1.5' size footings, and the dimension of the pit to accommodate these footings shall be  2ft. × 2ft.× 2.5 ft. as shown in the drawing (below ).




The total volume of excavation for footing

   = 2ft. × 2ft. × 2.5 ft. × 14 nos.

   = 140 cu.ft.


Excavation for the plinth beam

Let us excavate 4" extra on either side of the plinth beam for formwork removal as shown in the drawing.





 The volume of excavation for the plinth beam

 =[ perimeter of site  - ( no. of footings × footing excavation width.)] × plinth excavation width × plinth excavation depth 

   = [140 ft. - (14 nos. × 2 ft.)] × 1.416 ft. × 0.9166 ft. 

   =  112 ft. × 1.416 ft. × 0.9166 ft. 

   =  145.365 cu ft.


The total earthwork excavation for compound wall

    = footing excavation + plinth excavation

   = 140 cu ft + 145.365 cu ft.

   =  285.365 cu ft.

                                                CONTINUED πŸ‘‰   PART 2

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