How to calculate RL & distance in a tacheometric surveying? (Type-1)/ Finding RL of a station by using tacheometer.

Eg:

The following observations were made in a tacheometric surveying. A tacheometer was set up at the intermediate point on line AB.

Multiplying constant (f /i) = 100,  additive constant (f +d) = 0.1

Calculate the horizontal distance AB & RL of station B. RL of station A is 465m.

Calculation:

The formula for calculating the horizontal distance 

D = (f/i ) S cos²θ + (f +d ) cosθ

Now,

D1 =  = (f/i ) S1 cos²θ1 + (f +d ) cosθ1

Here,

S1 = staff intercept at station A = 2.425

θ1 = Vertical angle = 6°30'

D1 = [(100 ✕ 2.425 ✕ cos² 6°30') + ( 0.1  ✕ cos6°30')]

      = [239.39 + 0.0993]

  D1  = 239.49 m.

Similarly,

  D2 = [(100 ✕ 1.950 ✕ cos² 9°15') + ( 0.1  ✕ cos9°15')]

        = [189.962 + 0.0987]

 D2  = 190.06 m.

    

The formula for vertical distance

 V = [(f/i ) S sin2θ /2 +  (f +d ) sinθ]

   Now,

 V1 = [(f/i ) S1 sin2θ1 /2 +  (f +d ) sinθ1]

       = [ {100 ✕ 2.425  ✕ sin(2 ✕ 6°30') /2} + (0.1 ✕ sin6°30')]

      =[ 27.275 + 0.0113]

 V1 = 27.29 m.

Similarly,

V2 = [(f/i ) S2 sin2θ2 /2 +  (f +d ) sinθ2]

       = [ {100  ✕ 1.95 ✕ sin(2 ✕ 9°15') /2} + (0.1 ✕ sin9°15')]

      =[ 30.937 + 0.0161]

 V2 = 30.95 m.

RL of a central axis of the instrument at station P

= [RL at station A + axial hair reading of staff at A - V1]

= [ 465m. + 2.150 - 27.29]

= 439.86m.

Find: 👇

1. Horizontal distance AB

            = [D1 +D2]

            = [239.49m. + 190.06m.]

            = 429.55m.

2. RL of station B

         = [RL of instrument axis at P + V2 - axial hair reading of staff at B]

         = [ 439.86m. + 30.95m. - 1.865]

          = 468.945m.

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How to calculate constants of tacheometer ?/ Finding tacheometer constants in surveying.

 Eg: Calculate tacheometer constants for the following observations on site.

Instrument Station	Staff position	Distance	Staff reading
P	A	50	1.425	0.930
	B	150	2.250	0.765

 

Calculation:

Find:

Multiplying constant f /i (or k) =?

Additive constant f +d ( or C) =?

The formula used in tacheometric survey are

D = KS + C

     or

D = (f/i ) S + (f +d)

For distance PA

D1 = (f/i ) S1 + (f +d)

Here,

D1 = 50m. (given)

S1 = [1.425 - 0.930]

     = 0.495

50m. = (f/i ) 0.495 + (f +d) --------------- ①

For distance PB

D2 = (f/i ) S2 + (f +d)

Here,

D2 = 150m. (given)

S2 = [2.250 - 0.765]

     = 1.485

150m = (f/i ) 1.485 + (f +d) --------------- ②

Subtracting equation ① from equation ②,

[150m - 50m.] = (f/i ) (1.485 - 0.495) + 0

100 = [(f/i ) × 0.99]

(f/i ) = [100 ÷ 0.99]

         = 101.01

Now, substituting the value of  (f/i ) in equation  ①

50m = (101.01 × 0.495) + (f +d)

(f +d) = 50m - 49.99995

          = 0.00005 ≈ 0.0

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How to calculate bearings from included angles in surveying?/ Finding bearings from included angles in compass surveying.

 Eg:

Find the bearings of all the lines in a closed traverse for the following given data.

The bearing of AB = 67°30'

ㄥA= 134°15', ㄥB= 94°45',  ㄥC= 54°30', &ㄥD= 76°30'.

Calculation:

First, let us draw the closed traverse for the above-given data.

Known concept or formula:

 If FB>180°, then BB = FB-180°

If FB < 180°, then BB = FB +180°

1. Line AB:

FB of line AB = 67°30' (given) < 180°

BB of line AB = [ FB + 180°]

                        = [ 67°30' +180°]

                        = 247°30'

2. Line BC:

As you can observe in the drawing when you deduct the ㄥB from the BB of line AB, you will get FB of line BC.

Therefore 

FB of line BC = [BB of line AB - ㄥB]

                       = [247°30' - 94°45']

                       = 152°45' < 180°

BB of line BC  = [ FB + 180°]

                        = [ 152°45' +180°]

                        = 332°45'

3. Line CD:

FB of line CD = [BB of line BC - ㄥC]

                       = [332°45' - 54°30']

                       = 278°15' > 180°

BB of line CD  = [ FB - 180°]

                        = [ 278°15' -180°]

                        = 98°15'

4. Line DA:

FB of line DA = [BB of line CD - ㄥD]

                       = [98°15' - 76°30']

                       = 21°45' < 180°

BB of line DA  = [ FB + 180°]

                        = [ 21°45' +180°]

                        = 201°45'

Check:

FB of line AB  = [BB of line DA - ㄥA]

                       = [201°45' - 134°15']

                       = 67°30' ✔

Line	FB	BB
AB	67°30'	247°30'
BC	152°45'	332°45'
CD	278°15'	98°15'
DA	21°45'	201°45'

 

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How to calculate true bearing for magnetic declinations? / Magnetic declination in compass surveying.

 Let us review the examples to understand the calculation procedure of true bearings & magnetic bearings for a given magnetic declination.

Eg. 1. 

The magnetic bearing of a line is N46°15'E. If the magnetic declination is 6°30' East, calculate the true bearing.

Note:

TM    👉 True meridian

MM   👉 Magnetic meridian

RB     👉 Reduced bearing

WCB  👉 Whole circle bearing

Given data:

Reduced bearing = N46°15'E

Magnetic declination = 6°30'E

True bearing =?

As magnetic declination is towards the East, the value will be +ve. as shown above.

Here,

True bearing  

= [magnetic bearing + magnetic declination]

The given RB lies in the 1st quadrant. Therefore WCB = RB =46°15'

If you want to know, about the conversion of  RB to WCB, Go through the article 👇

👀. How do you convert reduced bearing to whole circle bearing?

True bearing 

= [46°15' + 6°30']

= 52°45'

As the true bearing is less than 90°, it lies in the 1st quadrant.

Therefore the position of bearing = N 52°45' E

Eg. 2.

The magnetic bearing of line AB is 147°30'. What will be the true bearing if the magnetic declination is 4°15' W.

Given data:

Magnetic bearing = 147°30'

Magnetic declination = 4°15' W

True bearing =?

As the magnetic declination is towards the West, the value will be -ve.

The given magnetic bearing is in the WCB system.

True bearing  

= [magnetic bearing - magnetic declination]

= [ 147°30' - 4°15']

= 143°15'

Eg. 3. 

The true bearing of a line AB is 217°30'. If the magnetic declination is 5°E, calculate the magnetic bearing.

Given data:

Magnetic declination = 5°E

True bearing = 217°30'

Magnetic bearing =?

As magnetic declination is towards the East, the value will be +ve. 

True bearing  = [magnetic bearing + magnetic declination]

Therefore,

Magnetic bearing

     = [ true bearing - magnetic declination]

     = [217°30' - 5°]

      = 212°30'

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Correction of bearings affected by local attraction./ How to correct bearings affected by local attractions?

 For the closed traverse ABCD, the following bearings were taken by compass. Find out which station is affected by local attraction & correct the bearings of the lines.

Sl. No.	Line	FB	BB
1.	AB	145°15'	325°45'
2.	BC	282°30'	102°0'
3.	CD	202°45'	22°45'
4.	DA	345°	165°0'

 

Let us draw the closed traverse ABCD from the above-given data.

As you can observe in the above drawing, the traverse is not closed properly at station A, due to the errors in the bearings affected by local attraction.

Calculation:

In the WCB system, the difference between the FB & BB should be 180°

So, let us find out the affected bearings by finding the differences between FB & BB for each line.

FB & BB difference:

1. For line AB:

 FB - BB = [ 145°15' - 325°45']

               = - 180° 30'

Error = [ 180°30' - 180°] = +30'

The error exceeds 30' above the 180° 

So, there is a correction of  -30' 

2. For line BC:

FB - BB = [ 282°30' - 102°0']

               = 180° 30'

Error = [ 180°30' - 180°] = +30'

The error exceeds 30' above the 180° 

So, there is a correction of  -30' 

3. For line CD:

FB - BB = [ 202°45' - 22°45']

               = 180° 

Error = [ 180° - 180°] = 0°

There is no error for line CD

Hence there is no correction.

 4. For line  DA:

FB - BB = [ 345°0' - 165°0']

               = 180° 

Error = [ 180° - 180°] = 0°

There is no error for line DA

Hence there is no correction.

Line	Local attraction	Error	Correction
AB	Yes	+30°	-30°
BC	Yes	+30°	-30°
CD	No	0°	0°
DA	No	0°	0°

 

From the above calculation, you will find that there is no local attraction at stations D & C as there is a 0° error for line CD & DA.

So, The observed bearing of line CD = 202°45' is correct

       The observed bearing of line CB = 102°0' is correct

        The observed bearing of line DC = 22°45' is correct

        The observed bearing of line DA = 345°0' is correct

         

For station A

The observed bearing of line AD - observed bearing of line DA should be equal to 180°

i.e. [165°- 345°] = -180°

As there is no local attraction in station D & the difference in the bearings is 180°, which means there is no local attraction in station A.

Therefore FB of line AB = 145°15' is correct.

Now, 

For line AB

 BB - FB should be equal to 180°

BB = [180°+ FB]

      = 180° + 145°15'

       = 325°15'

    The corrected bearing of line BA = 325°15'

But the observed bearing of line BA = 325°45'

Correction for station B 

= [corrected bearing - observed bearing]

= [325°15' - 325°45'] = -30'

Observed FB of line BC = 282°30'

To correct the FB of line BC, we have to deduct -30' from the bearing as there is a correction of -30' for station B.

So corrected FB of line BC

    = [282°30'-30'] = 282°

Check:

As there is no local attraction in station C, the difference between the corrected FB & observed BB of line BC should be 180°

i.e. [282° - 102°] = 180° ✔

Now the traverse is closed correctly.

Line	Observed bearing	Correction	Corrected bearing
AB	145°15'	0°	145°15'
BA	325°45'	-30'	325°15'
BC	282°30'	-30'	282°0'
CB	102°0'	0°	102°0'
CD	202°45'	0°	202°45'
DC	22°45'	0°	22°45'
DA	345°0'	0°	345°0'
AD	165°0'	0°	165°0'

 

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How to design a wall footing by thumb rule method? / Calculating dimensions of wall footing by thumb rule.

 When we design a wall footing, the ratio of all the dimensions should be as shown in the below-given drawing.

Given:

Suppose, we take the width of the superstructure wall as 300mm. (i.e. T ), then all other dimensions should be as follows.

Calculation:

1. Concrete bed:

The width of the concrete bed

     = 3T = [ 3 × 300mm.]

              = 900mm.

The thickness of the concrete bed should not be less than 200mm.

2. Step-1 footing:

The width of the 1st footing

= 2.5T = [ 2.5 × 300mm.]

              = 750mm.

The depth of the 1st footing

= [(3T - 200mm.) ÷ 2]

= [( 3 × 300mm. - 200mm.) ÷ 2]

= [ 700mm. ÷ 2]

 = 350mm.

Note: The depth of the footing should be equal for the 1st & 2nd steps of footings.

3. Step-2 footing:

The width of the 2nd footing

= 2T = [ 2 × 300mm.]

              = 600mm.

The depth of the 2nd footing = The depth of the 1st footing = 350mm.

4. Plinth wall:

The width of the plinth wall

= 1.5T = [ 1.5 × 300mm.]

              = 450mm.

Normally, we keep the depth of the plinth wall between 450mm. to 650mm., depending upon the site conditions.

Note: The offset of all the steps should be equally distributed on both sides.

For eg:

The extra length of the concrete bed

= [900mm. - 750mm.]

 = 150mm.

Offset or projection on both sides 

= [150mm.  ÷ 2] = 75mm.

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How to calculate the size of the footing?/ How to design a footing for the given building load?

Let us calculate the size of the footing for the following given data.

Eg:

Calculate the length & width of the footing for the building load of  5000KN. The building has 12nos. of footings as shown below. The bearing capacity of soil is 11,297 kg/ m.² 

If you want to know, how to calculate the bearing capacity of any construction site, Go through the article 👇

👀.  How to calculate the bearing capacity of soil at the site?

Given data:

Building load = 5000KN.

Bearing capacity of soil = 11,297kg./ m.²

No. of footings = 12 nos.

Calculation:

Area of all the footings 

= [ building load ÷ bearing capacity of soil]

First, we have to convert the building load into kg. to have the same unit.

As you know, 1KN = 101.971kg.

Therefore building loads in kg.

     = [5000 × 101.971]

     = 509,855 kg.

Area of all the footings

= [ 509,855kg. ÷ 11,297kg/./ m.²]

= 45.132 m.² 

So, the area of one footing

= [area of all footings  ÷ no. of footing]

= [ 45.132 m.²   ÷ 12]

= 3.76 m.² 

Size of the footing:

1. The size of the square footing:

Area = [length × breadth] 

          = [ L × B]

In square footings, L = B.

So, 

3.76 m.² = L²

L = √ 3.76 = 1.94m.

The size of the square footing will be 1.94m. × 1.94m.

2. The size of the rectangular footing:

Suppose, we need the breadth of the footing as 1.5m.

        Area =  [ L × B] 

  3.76 m.² = [ L  × 1.5m.]

              L =  [3.76 m.² ÷ 1.5m.]

                = 2.51m.

The size of the rectangular footing will be 2.51m. × 1.5m.

Go through the article 👇

👀. How to calculate the size of the columns?

Note: This method holds good for the smaller residential buildings having equally distributed loads.

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How to calculate the bearing capacity of soil at the site?/ Finding SBC of soil at the construction site.

 Let us go through the procedure to find out the SBC of soil at any construction area to design a footing.

Procedure:

1. Excavate a footing pit on the construction site of the required depth.

2. Take an iron or concrete cube & note down the dimensions of the cube.

3. Drop the cube from a known height over the bottom soil layer of the excavated pit. This process should be repeated several times at different locations of the pit. The measurement of free fall height should be kept equal for all the pits.

4. The free fall of the cube creates an impression over the soil surface. You have to note down all the readings of impression depth and the average should be calculated.

5. This process should be repeated for at least 3 different footing pits excavated at different locations of the site. Please remember that the excavated depth of all the footings should be the same.

Calculation:

Let us take a concrete cube of size 15cm. × 15cm. × 15cm.

The concrete cube is dropped from a freefall height of 1.5m. at different footing pits & the readings of impressions were noted down as shown below.

Sl. No.	Footing pit	Depth of impression                       in m.				Average value of impression
		Read-1	Read-2	Read-3	Read-4
1.	Pit-1	0.024	0.018	0.025	0.027	0.0235
2.	Pit-2	0.019	0.028	0.023	0.024	0.0235
3.	Pit-3	0.026	0.026	0.022	0.025	0.0248
	Total average depth of impression					0.0239

 

1. Resistance of soil: 

Resistance of soil

R = [( W × h ) ÷ d]

Where,

R = Resistance of soil

W = Weight of concrete cube.

h = Height of freefall of cube.

d = average depth of impression.

Here,

 h = 1.5m.

 d = 0.0239m. (As calculated in the above table.)

W = 8.1kg. ( Picked from below linked article.)

If you want to know how to calculate the weight of any concrete cube, Go through the article 👇

👀.  How to calculate the volume & weight of a concrete cube?

 Resistance of soil 

R = [( W × h ) ÷ d]

    = [( 8.1kg. × 1.5m. ) ÷ 0.0239m.]

    = 508.37 kg.

The soil has a resistance of 508.37kg for the surface area of the given concrete cube.

Therefore, 

 Resistance of soil per unit area

= [R ÷ A]

Where,

A = surface area of the cube

    = [ 0.15m. × 0.15m.]

    = 0.0225 sqm.

Resistance of soil per unit area

= [508.37kg ÷ 0.0225 sqm.]

= 22,594 .20 kg/sqm.

2. SBC of soil :

  SBC of soil

= [R ÷ (A × FOS ]

      OR

= [ resistance of soil per unit area ÷ FOS]

Where,

SBC 👉 Safe bearing capacity.

FOS 👉 Factor of safety.

The factor of safety is taken between 2 to 3 depending upon the structural design.

Let us take FOS = 2 for the calculation purpose.

  SBC of soil 

= [ 22,594.2 kg/sqm. ÷ 2]

= 11,297 kg/sqm.

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Estimating by center line method./ How to make building estimation by center line method?

 Let us estimate the below-given building drawing by the centerline method.

Calculation:

Procedure:

First, we have to calculate, the center-to-center distance of all the walls of the building above the plinth level.

c/c length of the long wall 

     = [ room-1 breadth + partition wall thickness + room-2 breadth + (2nos. × 1/2 × wall thickness)]

     = [3.5m. + 0.3m. + 4.5m. +  (2nos. × 1/2 × 0.3m.)]

     = 8.6m. ( as shown in the above drawing).

c/c length of the short wall

 = [ room length  + (2nos. × 1/2 × wall thickness)]

     = [5m. +  (2nos. × 1/2 × 0.3m.)]

     = 5.3m.

Length of the partition wall

= room length = 5m.

Points to remember 👇

We have not added the width of the wall to the partition wall length. This is because we have already taken that part of the wall in the c/c length of the long wall. As you can observe in the above drawing, the black shaded portion will get repeated, if we add them to the partition wall.

You have to remember this point while doing the estimation by the centerline method.

 Now, the total centerline length of the wall

= [(2nos. × c/c length of longwall) + (2nos. × c/c length of the short wall) + length of partition wall]

=  [(2nos. × 8.6m.) + (2nos. × 5.3m.) + 5m.]

= [17.2m. + 10.6m. +5m.]

= 32.80 m.

1. Excavation:

 The volume of excavation for the foundation

= [ centerline length × breadth × depth]

= [32.80m. × 1.2m. × (0.15m. + 0.55m.)]

= 27.552 cum.

2. PCC :

The volume of PCC for the foundation

= [ centerline length × breadth × thickness]

= [32.80m. × 1.2m. × 0.15m.]

= 5.904 cum.

3. Foundation:

The volume of the building foundation

= [ centerline length × breadth × depth]

= [32.80m. × 0.9m. × 0.55m.]

= 16.236 cum.

4. Plinth:

The volume of the plinth wall

= [ centerline length × breadth × depth]

= [32.80m. × 0.6m. × 0.45m.]

= 8.856 cum.

5. Superstructure:

The volume of the wall above the plinth level

= [ centerline length × breadth × height]

= [32.80m. × 0.3m. × 3m.]

= 29.52 cum.

Note: The estimation is done in an elaborate way so that you will understand the concept clearly. Actually, the estimation is done in a table format as shown below.

Sl. No.	Description	Nos.	Length in m.	Breadth in m.	Depth in m.	Qty.  in cum
1.	Excavation for foundation.	1	32.80	1.2	0.70	27.552
2.	PCC for foundation.	1	32.80	1.2	0.15	5.904
3.	Foundation	1	32.80	0.9	0.55	16.236
4.	Plinth wall	1	32.80	0.6	0.45	8.856
5.	Superstructure wall.	1	32.80	0.3	3.00	29.52

 

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