All about civil construction knowledge- PARAM VISIONS

How to calculate the correction for slope in surveying?/ How to make tape correction for slope in surveying?

 Eg:

The slope distance between two points measures 265m. The difference in the elevation between the two points is 8m. Determine the horizontal length by using the slope correction formula.


To find:

The horizontal distance = D =?


Calculation:




Correction for slope 

 Cs = - [ h²  ÷ 2L]

Where,

L = measured distance between two points = 265m.

h = Vertical difference in elevation between two points = 8m.

     Cs  = [8² ÷ (2 х 265) ]

            = [ 64  ÷ 530]

            = 0.1207m.


The horizontal distance between two points

D = [ Measured distance - slope correction]

    = [ 265m - 0.1207m.]

   = 264.879m.


Note:

1. The slope correction is always negative as the horizontal distance will be shorter than the measured slope distance.

2. The correction is applied when there is a difference in elevation between the two points.


To know:

The slope correction formula is derived by applying Pythagoras theorem in the equation.

Let us find the horizontal distance by the Pythagoras theorem.

L² = [D² + h²]

265² = [D² + 8²]

D² = [265² - 8²]

D² = 70,161

D = √ 70,161

    = 264.879m.


To understand A to Z of surveying, click here.


Thank you for going through these calculation steps. Have a good day 😄.



 


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How to make tape corrections for a pull at survey measurements?/ How to calculate true measurement by pull correction of a tape?

 Eg:

A 760m. distance is measured by applying a pulling force of  90N over the 30m. tape. The tape was made at 50N tensile force. Find the true length of the line if the cross-sectional area of the tape is 0.04cm². Modulus of elasticity = 21х 10⁶ N/cm².


To find:

The true distance between the points. = L =?


Calculation:




The true length 

 L = [ Measured length + pull correction.]

 L = [ L1 + Cp ]


Now,

The pull correction is calculated by the formula

Cp = [{( Pm - Po ) ÷ A х E } х L1]

Where,

Cp = Correction for pull = ?

Pm = Applied force over tape for pulling = 90N.

Po = Std. pulling force of tape = 50N.

A = Cross-sectional area of tape. = 0.04cm²

E = Modulus of elasticity of tape material =  21х 10⁶ N/cm².

L1 = Measured distance between the points = 760m.

Cp = [{( 90 - 50 ) ÷ 0.04 х  21х 10⁶ } х 760]

     = [{ 40 ÷ 0.84 х 10⁶ } х 760]

Cp = 0.0362m.


The true length 

 L = [ L1 + Cp ]

    = [ 760m. + 0.0362m.]

  L = 760.0362m.


Thank you for going through these calculation steps. Have a good day 😄.

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How to make temperature correction for a steel tape in surveying? / How to calculate true length for temperature correction?

 Eg:

A 30m. surveyor's steel tape is correct at 21°C. The distance measured by using this tape on a day when the temperature is 32°C is 230m. Calculate the true distance between the points by applying temperature correction. Coefficient of thermal expansion ∝ = 1.4 х 10⁻⁵ per ° C.


To find:

The true distance between the points.


Calculation:




The true length 

 L = [ Measured length + temp. correction.]

 L = [ L1 + Ct ]


Now,

The temperature correction is calculated by the formula

Ct = [ ∝ х ( Tm - To ) х L1 ]

   Where,

∝ = Coefficient of thermal expansion =  1.4 х 10⁻⁵

Tm = Field temp. = 32°C

To = Standard temp. = 21°C

L1 = Measured distance = 230m.

Ct = [  1.4 х 10⁻⁵ х ( 32°C - 21°C ) х 230m. ]

     = 0.0354m.


The true length between the points

   L  = [ L1 + Ct ]

       = [230m. + 0.0354m.]

    L = 230.0354m.


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What should be the maximum distance between two columns?

 Let us go through the answer for the FAQ "What should be the maximum distance between two columns?"

There is nothing as such a fixed measurement between two columns as it depends upon several factors.


Some of the factors that decide the distance between two columns are,

1. Total load ( LL + DL ) acting over the column.

2. Type of structural design.

3. Size of the column.

4. Size of the beams resting over the column.

5. Whether the column comes in a seismic zone or not.

6. Type of footing & SBC of soil.

7. No of storeys in the structure.

Considering all these points & safety factors, a structural engineer provides the most economical structural drawing for the construction.

If you are residing in an urban area, the best way is to hire a consultant for your construction works.

 In rural areas, where a local contractor or workers design the structures, the following are some of the points to be known by the house owners.

Let us discuss the general requirement for the column size & its spacing in a single-storeyed residential building.

1. The minimum required size of the column should be 9"х 9".

2. When you provide a 9"х 9" column to the building, the distance between the two columns should be ≤ 4m or 13'.




3. The standard or most common size of the column for the single-storeyed residential building is 9"х 12"

4. When you provide a 9"х 12" column to the building, the distance between the two columns should be ≤ 5m or 16'.



5. In any condition the dia. of reinforcement should not be less than 12mm. - 4nos.

6. The dia. of lateral ties should be 8mm. as shown in the below drawing.



7. The minimum size of the beam should be  9"х 12" for the safe transfer of the loads.

8. For G +1 residential buildings, you should avoid 9"х 9" size columns in the structure. The general size adopted for the columns are 9"х 12" & 9"х 15".


Thank you for going through this article. Have a good day 😄.







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How to calculate the volume of earthwork from a contour map? / Finding the volume of earthwork in contour surveying.

 Eg:

Following are the different contours at 3m. intervals for a hill with their respective areas. 

Sl. No.

Contour

Area in

1

   450

   375

2

   447

   620

3

   443

   945

4

   441

   1365

5

   438

   1880

6

   435

   2570

  

 

 

 

Calculate the volume of the earth contained in the hill.




To find:

The volume of earth contained in the hill.


Calculation:

The volume of the earth is calculated by using the trapezoidal & prismoidal formulas for contour maps. The answer will be nearer to accuracy & they differ from each other.

Now, let us calculate the volume of the earth by using both of them to observe the difference.


1. Trapezoidal formula:

The volume of earth in the hill

V = D х [{(A0 + An) 2} + (A1 + A2 + A3 + ---- + An - 1)]

Here,

 D = Interval between the successive contours 

 D = {450 - 447} = 3


A0 = area of 1st contour line = 375m²

A1 = area of 2nd contour line = 620m²

Similarly,

    A2 = 945m²

     A3= 1365m²

       An  👉 The area of last contour = 2570m²

(An - 1) 👉 Prior to last contour = A4 = 1880m²


Now,

The volume of earth in the hill

V = D х  [{(A0 + An) 2} + {A1 + A2 + A3 + A4}]

    = 3 х [{(375 + 2570) 2} + {620 + 945 + 1365 + 1880}]

    = 3 х [{1472.5} + {4810}]

     = 3 х [6282.5]

 V  = 18,847.5m³


2. Prismoidal formula:

The volume of earth in the hill

V = D/3 [ 1st area + last area + 4 Σ even area + 2 Σ odd area ]

V =  (D ➗ 3)  х [{ (A0 + An) + 4  х ( A1 + A3 ) + 2  х ( A2 + A4)}]

   =  (3 ➗ 3)  х [{ (375 + 2570) + 4  х ( 620 + 1365 ) + 2  х ( 945 + 1880)}]

   = 1 х [{ ( 2945 ) + 4  х ( 1985 ) + 2 х ( 2825)}]

  = 1 х [{ 2945  + 7940 + 5650 }]

  = 1 х [16,535 ]

V = 16,535 m³


Note:

For the larger no. of contour data, having minimum intervals between the two contour lines, the prismoidal formula works better. 

For the lesser no. of contour data having maximum contour intervals, the trapezoidal formula works better.


Remember:

 Often there is confusion in selecting the areas in the prismoidal formulas. Some points to be known in selecting them are,

1. Each area is taken only once in the formula. No area is repeated twice.

2. The even area starts from the second area leaving the 1st one (A0). 👉 2nd, 4th, 6th, ...... so on. 

3. The odd area starts from the third area and leaves the last one (An). 👉 3rd, 5th, 7th,  ..... so on.

4. If you have only 3 contour data, the formula does not contain any odd areas.

      V = [ (D ➗ 3)  х { (A0 + An) + 4  х ( A1 ) + 2 х (0)}]

         = [ (D ➗ 3)  х { (A0 + An) + 4  х ( A1 )]


Note:

When we do earthwork filling, we can calculate the volume by using this method.

To understand A to Z of surveying, click here.


Thank you for going through these calculation steps. Have a good day 😄.

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How to calculate superelevation for a circular road? / Finding superelevation for a curved road.

 Eg:

Calculate superelevation for a road having a radius of a curve as 150m. The design speed is 60km/hr. The coefficient of lateral friction is 0.15.


To find :

Superelevation=e =?


Calculation:




The formula to calculate the superelevation of the road is

[e + f ] = [V² / 127R]

Where,

e = rate of superelevation =?

f = lateral friction factor =  0.15

V = Velocity of vehicle in km / hr. = 60km/hr

R = Radius of a circular curve in meters.= 150m.

[e + f ]=  [V² / 127R]

[e + 0.15] = [60² ÷ 127 х 150 ]

[e + 0.15] = [3600 ÷ 19,050]

[e + 0.15] = 0.189

e = [0.189 - 0.15]

   = 0.0389

 Required slope = 1/0.0389

                            = 1 in 25.70

  Superelevation in %

   = [0.0389  х 100]

    = 3.89%


According to IRC, the maximum allowable superelevations for mixed traffic over hills & terrains is 7%.

e = 3.89% < 7% hence safe.


Thank you for going through these calculation steps. Have a good day 😄.


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How to calculate the volume of water in a pond by a contour map?/ Finding volume of a contour lines.

 Eg:

Following are the different contours at 2m. intervals in a pond up to a water level with their respective areas. 

Sl. No.

Contour

Area in

1

   550

   775

2

   552

   920

3

   554

   1145

4

   556

   1565

5

   558

   1980

6

   560

   2650

7

   562

   3425

 

 

Calculate the volume of water in the pond.




To find:

The volume of water in the pond.


Calculation:

The volume of water is calculated by using the trapezoidal & prismoidal formulas for contour maps. The answer will be nearer to accuracy & slightly differ from each other.

Now, let us calculate the volume of water by using both of them to observe the difference.


1. Trapezoidal formula:

The volume of water in the pond

V = D х [{(A0 + An) 2} + (A1 + A2 + A3 + ---- + An - 1)]

Here,

 D = Interval between the successive contours 

 D = {562 - 560} = 2


A0 = area of 1st contour line = 775m²

A1 = area of 2nd contour line = 920m²

Similarly,

A2 = 1145m², A3= 1565m², A4 = 1980m², 

       An  👉 The area of last contour = A6 =3425m²

(An - 1) 👉 Prior to last contour = A5 = 2650m²


Now,

The volume of water in a pond

V = D х  [{(A0 + A6) 2} + {A1 + A2 + A3 + A4 + A5}]

    = 2 х [{(775 + 3425) 2} + {920 + 1145 + 1565 + 1980 + 2650}]

    = 2 х [{2100} + {8260}]

     = 2 х [10,360]

 V  = 20,720m³


2. Prismoidal formula:

The volume of water in the pond

V = D/3 [ 1st area + last area + 4 Σ even area + 2 Σ odd area ]

V = [ (D 3)  х { (A0 + An) + 4  х ( A1 + A3 + A5 ) + 2  х ( A2 + A4)}]

   = [ (2 ➗ 3)  х { (775 + 3425) + 4  х ( 920 + 1565 + 2650 ) + 2  х ( 1145 + 1980)}]

   = [0.6667 х { ( 4200 ) +  4  х ( 5135 ) + 2 х ( 3125)}]

  = [0.6667 х { 4200  + 20540 + 6250 }]

  = [0.6667  х 30,990 ]

V = 20, 660 m³


Thank you for going through these calculation steps. Have a good day 😄.

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How to calculate the volume of earthwork for embankment?/ Earthwork calculation of embankment by mean area method.

 Eg:

Calculate the volume of earthwork for the embankment 75m. long having 4m. width at the top. Depth of embankment at every 15m. interval is 1.45m., 1.70m., 1.95m., 2.25m., 2.55m., &2.80m. The side slope on both sides is 1:2.


To find:

We have to calculate the total volume of earthwork in embankment by using the mean area method.





Procedure:

Consider an embankment section having 15m. length = L1 as shown in the above drawing. 

1. First, you have to calculate the sectional area A1 for the initial depth d1.

The formula for calculating A1 = [ bd1 + d1²S]


Before proceeding further, Go through the article 👇

👀.  How to calculate the sectional area of an embankment?

To understand the formula derivation &  clear concept of calculation.


2. Similarly, the sectional area A2 should be calculated for the given depth d2 after a 15m. interval.

The volume of earthwork for an embankment for the 1st. 15m. length

        = V15 = [ Am ✖ L1]

Where the mean area 

            Am  = [(A1 +A2) ➗ 2]

3. In a similar way, the volume of earthwork for every 15m. embankment section should be calculated.

4. The total volume of earthwork for embankment 

    V = [V15 + V30 +V45 + V60 + V75]


Calculation:





1. Volume V15:

     A1 = [ bd1 + d1²S]

          =  [ (4m.  1.45m.) + (1.45²  2)]

           = [5.8m² + 4.205m²]

     A1 = 10.005m².


    A2 = [ bd2 + d2²S]

          =  [ (4m.  1.70m.) + (1.70²  2)]

          = [6.8m² + 5.78m²]

    A2 = 12.58m².


     V15 = [{(A1 +A2) 2} L1]   

             = [{(10.005 + 12.58)  2}  15]   

     V15 = 169.39m³.


2. Volume V30:

     A3 = [ bd3 + d3²S]

          =  [ (4m.  1.95m.) + (1.95²  2)]

           = [7.8m² + 7.605m²]

     A3 = 15.405m².


   V30 = [{(A2 +A3)  2}  L2]   

           = [{(12.58 + 15.405)  2}  15]   

   V30 = 209.88m³.

  NOTE: The area A2 becomes the surface sectional area for the next 15m. length of the embankment.


3. Volume V45:

     A4 = [ bd4 + d4²S]

          =  [ (4m.  2.25m.) + (2.25²  2)]

           = [9.0m² + 10.125m²]

     A4 = 19.125m².


   V45 = [{(A3 +A4)  2}  L3]   

           = [{(15.405 + 19.125)  2}  15]   

   V45 = 258.975m³.


4. Volume V60:

     A5 = [ bd5 + d5²S]

          =  [ (4m.  2.55m.) + (2.55²  2)]

           = [10.20m² + 13.005m²]

     A5 = 23.205m².


   V60 = [{(A4 +A5)  2}  L4]   

           = [{(19.125 + 23.205)  2}  15]   

   V60 = 317.475m³.


5. Volume V75:

     A6 = [ bd6 + d6²S]

          =  [ (4m.  2.80m.) + (2.80²  2)]

           = [11.20m² + 15.68m²]

     A6 = 26.88m².


   V75 = [{(A5 +A6)  2}  L4]   

           = [{(23.205 + 26.88)  2}  15]   

   V75 = 375.637m³.


 The total volume of earthwork for embankment 

    V = [V15 + V30 +V45 + V60 + V75]

       = [169.39m³ + 209.88m³ + 258.975m³ + 317.475m³ + 375.637m³]

       = 1331.357


Note

The calculation is done in an elaborate way to understand the concept clearly. Once known, the calculation should be done in a table format as shown below.


Thank you for going through these calculation steps. Have a good day 😄.


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How to calculate the sectional area of an embankment?/Calculating the area of earthwork in an embankment.

Let us calculate the sectional area of the embankment as shown below.




Given data:

Breadth of embankment = b = 8m.

Depth of embankment at the given section = d = 3.5m.

Embankment slope = S = 1:2


Calculation:

The sectional area of embankment is calculated by the formula

 A = [bd + d²S]

    = [ (8m. x 3.5m.) + ( 3.5² x 2)]

    = [28m². + 24.5m²]

    = 52.50 m²


Now, the question that comes to your mind is,

   How is the formula [bd +  d²S] derived?

Let us observe the drawing as shown below.




To calculate the sectional area of the embankment, we have divided it into 2 triangles ABC & DEF on either side & a rectangle BCED at the center.

As you know,

The slope 1: S means for every 1 unit rise, the value of the run will be S units.

Therefore, if the height of the triangle is d, then the length of the base of the triangle will be d x S.


If you have any doubt regarding the run, rise, gradient, or slope calculations, Go through the article 👇

👀.  How to calculate gradient, run & rise in civil construction?


Now, 

The sectional area 

A = [ (area of rectangle) + (2nos. x area of triangles)]

    = [(breadth x height) + ( 2nos. x 1/2 x base x height )]

    = [( b x d) + ( 2 x 1/2 x d x S x d )]

     = [ bd +  d²S] 

Usually, the slope of 1:2 is provided for the embankments. In some cases, slopes of 1:1.5, 1:1 are taken in earthworks.

Whatever might be the value of slope S, you will get the correct sectional area by using the formula.


Thank you for going through these calculation steps. Have a good day 😄.






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How to calculate the volume of a brick chimney?/ Calculating the quantity of material in a brick chimney.

 Let us calculate the material volume of the brick chimney as shown below.




Given data:

The outer diameter of the chimney at the bottom section = D1 = 3m.

The outer diameter of the chimney at the top section = d1 = 1.2m.

The brick wall thickness of the chimney = 0.3m.

Height of chimney = h = 20m.


Calculation:

The geometrical shape of the chimney is a hollow frustum.

Therefore,

The volume of the hollow frustum is calculated by the formula

V =  [h / 3 { (A1 + A2 + √ A1A2 ) -  (a1 + a2 + √ a1a2)}]


Where,

A1 = Outer sectional area of the chimney at the bottom.

A2 = Inner hollow sectional area of the chimney at the bottom.

a1 = Outer sectional area of the chimney at the top.

a2 = Inner hollow sectional area of the chimney at the top.

h = Height of the chimney.


Now,

A1 = [𝜋D1² ÷ 4]

      = [ 3.142 x 3² ÷ 4]

     = 7.069m²


A2  = [𝜋D2² ÷ 4]

 Here, 

 Inner diameter D2 

     = [Outer dia. D1 - (2 x wall thickness)]

      = [ 3m. - (2 x 0.3m.)]

 D2 = 2.4m.

A2 = [ 3.142 x 2.4² ÷ 4]

      = 4.524m²


Similarly,

a1 = [𝜋d1² ÷ 4]

      = [ 3.142 x 1.2² ÷ 4]

     = 1.131m²


a2 = [𝜋d2² ÷ 4]

      = [ 3.142 x 0.6² ÷ 4]

      = 0.283m²


The material volume of chimney

V =  h / 3  x [ {A1 + A2 + (√ A1A2 )} - {a1 + a2 + (√ a1a2)}]

    =  20 / 3  x [{ 7.069 + 4.524 + (√ 7.069 x 4.524 )} - {1.131 + 0.283 + (√ 1.131  x 0.283)}]

    =  20 / 3 x [{ 17.248 } - { 1.979 }]

    = 20 / 3 x [15.269]

  V  = 101.79 m³.


Once, the volume of the chimney is calculated, you can calculate the required no. of bricks & mortar for the construction.

To know how to estimate the materials, Go through the article 👇

👀. Brick, cement, & sand calculations for modular brick masonry.


Thank you for going through these calculation steps. Have a good day 😄.

    

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Basic rules for the design of beams./ Basic rules to calculate beam dimension as per IS-456.

 Let us go through the basic rules for the design of beams as per IS456-2000.


1. Beam dimension:

a) Effective depth: The minimum effective depth of the beam can be taken as 1/10th. of the clear span of the beam.

b) Breadth: The minimum required breadth should be between 1/2 to 2/3 of the total depth of the beam.




Eg:

If the span of the beam is 5000mm. ( as shown above), 

The effective depth  d = [1/10 × 5000mm.]

                                   = 500mm.

If the effective cover of the beam is 40mm.,

Beam depth D = [effective depth + effective cover]

                        = [ 500mm. + 40mm.]

                     D   = 540mm.

The minimum breadth b = [1/2 × 540mm.]

                                     b = 270mm.


2. Beam reinforcement:

a) Minimum reinforcement:

The minimum area of tensile reinforcement should not be less than as calculated by the formula

As = 0.85bd / fy

Where, 

As = minimum area of tensile reinforcement.

 b = breadth of beam.

d = effective depth of the beam.

fy = characteristic yield strength of steel.


b) Maximum reinforcement:

The maximum area of  reinforcement should not be more than 0.04bD.


Eg:

If fy of steel = 415, 

Minimum reinforcement

As = [0.85bd / fy]

      = [( 0.85 × 270mm. × 500mm.) ÷ 415]

      = 276.50 mm²

     (Beam dimension is taken from Eg-1.)

Maximum reinforcement

= 0.04bD

 = [0.04 × 270mm. × 540mm.]

= 5832 mm².


 3. Side face reinforcement:




If the depth of the beam exceeds 750mm., side face reinforcement should be provided along with the vertical faces on either side of the beam.

The total area of reinforcement should not be less than 0.1% of the web area of the beam. Spacing between the bars should not exceed 300mm.


4. Spacing of reinforcement:

The vertical & horizontal spacing of reinforcement is covered in detail in a separate article as linked below.

👀.  What should be the minimum distance between reinforcement bars?


5. Deflection control:




The vertical deflection limit of the beam should not be more than the values given below.


a) For span up to 10m:

Cantilever beam: The ratio between span/effective depth should be ≤ 7

Simply supported beam: The ratio between span/effective depth should be ≤ 20

Continuous beam: The ratio between span/effective depth should be ≤ 26


b) For spans above 10m.:

The ratio of span/effective depth obtained above should be multiplied by 10/span in m.


Thank you for going through these calculation steps. Have a good day 😄.




        





 




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Calculating forces in wires by Lami's theorem./ Finding tension in wires using Lami's theorem.

 Eg:

Calculate the tensional force developed in the wires  AC & BC in the following drawing using Lami's theorem.




To find:

The tensional forces  developed in the wires AC & BC

Calculation:

Let us draw a line QD parallel to the support PB as shown below. The line QD should pass through intersection point C.


As ∠ B = 65°, the ∠BCD will be equal to 65°

Note: 

The alternate angles of a line cutting the two parallel lines are always equal to one another.


 ∠Q =90° as line QD is perpendicular to the support QP.

In triangle AQC, 

∠C = [180°- ( ∠Q + ∠A )]

       = [180°- ( 90° + 40° )]

       = 50°

As an applied force is in the vertical axis, ∠QCE = ∠DCE = 90°

( As shown in the above drawing)


Now, let us draw a free-body diagram of forces as shown below.




∠ACE = [∠ACQ + ∠QCE]

            = [ 50° + 90°]

             =  140° 

(As shown in the above drawing)


∠BCE = [∠BCD + ∠DCE]

            = [ 65° +  90°]

             =  155° 


∠ACB =  [360° - (∠ACE + ∠BCE)]

 (As total value of a angle = 360°)

∠ACB =  [360° - ( 140° +  155° )]

             = 65° 


By applying Lami's theorem,

T1 / sin155° = T2 / sin140° = 430N / sin65°

Where, T1 & T2 are the tensional forces in wires.

Therefore,

T1 = [(430N ➗  sin65°) x sin155°]

     = [ (430N ➗ 0.906 ) x 0.422]

     = [ 474.613N x 0.422 ]

  T1 = 200.287N


T2 = [(430N ➗  sin65°) x sin140°]

     = [ (430N ➗ 0.906 ) x 0.643 ]

     = [ 474.613N x 0.643 ]

  T2 = 305.075N


Tensional force in wire AC = T1 = 200.287N

Tensional force in wire BC = T2 = 305.075N


     Thank you for going through these calculation steps. Have a good day 😄.





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How to calculate the tension developed in ropes using Lami's theorem?/ Lami's theorem application for calculating forces.

 Eg:

Calculate the tension developed in the ropes AC & BC in the following drawing using Lami's theorem.




To find:

Tension developed in the ropes AC & BC

Calculation:

Let us draw a line DE, parallel to the support AB as shown below. The line DE should pass through intersection point C.



As ∠ A = 42°, the ∠DCA will be equal to 42°

Note: 

The alternate angles of a line cutting the two parallel lines are always equal to one another.


Similarly,  ∠ECB = ∠B = 35°

As an applied load is in the vertical axis, ∠DCF = ∠ECF = 90°

( As shown in the above drawing)



Now, 

∠ACF = [∠ACD + ∠DCF]

            = [ 42° + 90°]

             =  132° 

(As shown in the above drawing)


∠BCF = [∠BCE + ∠ECF]

            = [ 35° +  90°]

             =  125° 


∠ACB =  [360° - (∠ACF + ∠BCF)]

 (As total value of a angle = 360°)

∠ACB =  [360° - ( 132° +  125° )]

             =  103° 


By applying Lami's theorem,

T1 / sin125° = T2 / sin132° = 300N / sin103°

Where, T1 & T2 are the tensions in ropes.

Therefore,

T1 = [(300N ➗  sin103°) x sin125°]

     = [ (300N ➗ 0.9743 ) x 0.819 ]

     = [ 307.91N x 0.819 ]

  T1 = 252.18N


T2 = [(300N ➗  sin103°) x sin132°]

     = [ (300N ➗ 0.9743 ) x 0.743 ]

     = [ 307.91N x 0.743 ]

  T2 = 228.77N


Tension in rope AC = T1 = 252.18N

Tension in rope BC = T2 = 228.77N


     Thank you for going through these calculation steps. Have a good day 😄.

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How to calculate the relative density of soil? / Calculating density index of soil.

 Eg:

Calculate the relative density of a soil sample for the following data.

The actual void ratio of soil is 0.58, emax =1 & emin = 0.43.


Given data:

Void ratio = e = 0.58

Maximum void ratio = emax =1

Minimum void ratio = emin = 0.42


Note: 

Void ratio of cohesionless soil in its loosest state =  emax

 Void ratio of cohesionless soil in its densest state =  emin

Void ratio of cohesionless soil in its natural state = e


To find:

Relative density = Dr = Id?

Relative density is also referred to as density index = Id.




Calculation:

Relative density is calculated by the formula

Dr = [ ( emax - e  ) ➗ ( emax - emin )] x 100

     = [ ( 1 - 0.58  ) ➗ ( 1 - 0.43 )] x 100

     = [ 0.42 ➗ 0.57 ] x 100

Dr = 0.73 x 100 = 73


The relative density= Dr = 73


    Thank you for going through these calculation steps. Have a good day 😄.

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Derivation of formula to set out curves by the method of ordinates from a long chord./ Deriving long chord method for ordinates to draw a curve.

 When two roads intersect each other, we set out a curve to connect the roads by measuring the offsets or ordinates as shown below.



As you can observe in the above drawing, AP & BP are the two roads intersecting at point P. By surveying, the two roads are joined by setting out a curve T1CT2 for smooth transportation.

 Now, let us derive a formula to calculate the ordinates (or offsets) from a long chord to set out a curve.




 As you can observe in the above drawing,

Long chord = TIT2 = L

The radius of the curve = OT1, OT2, or OE = R

Mid-ordinate = CD =Oo

Ordinates at a distance x from the mid-ordinate = EF = Ox


Now,

Mid-ordinate = Oo = CD = [OC - OD]

In triangle T1DO,

T1O² = [T1D² + OD²]

( By Pythagoras theorem. )

 or

OD² = [T1O² - T1D²]

Here, T1D = Half of long chord = L/2, TIO = Radius = R

Therefore

OD² = {R² - (L/2)²}

OD = {√ R² - (L/2)²}


1. Mid-ordinate 

  = Oo = CD = [OC - OD]

    Here, OC = Radius of the curve = R

Substituting the values of OC & OD,

Oo = [ R - √ R² - (L/2)²]   ----------- ① 

 

From the above drawing,    

EF=GD=Ox

GD = [OG - OD]

or

OG = [GD + OD]

      Substituting the values of GD & OD

OG = [Ox + {√ R² - (L/2)²}]

or

Ox = [OG - {√ R² - (L/2)²}] ---------- ©


In triangle OEG

OE² = EG² + OG²

( By Pythagoras theorem. )

 Here,

EG = Distance of ordinates from mid-ordinate over long chord = x

OE = Radius of curve = R

Therefore,

R² = x² + OG²

OG² = R² -x²

OG = √ (R² -x²)

Substituting the value of OG in the above derived equation, ©

 Ox = [√ (R² -x²)  - {√ R² - (L/2)²}]  ----------- ② 


The equations ① & ② are the formulas required to find out the mid-ordinate, ordinates, radius, etc. to set out a curve.


Let us rewrite the formulas for further reference as follows

Mid-ordinate = Oo = [ R - √ R² - (L/2)²] 

Ordinates = Ox = [√ (R² -x²)  - {√ R² - (L/2)²}]


Note: Whatever might be the distance of x over the long chord, you will get the measurement of ordinates Ox from that point.

To understand the concept correctly, go through the solved problems as linked below.

👀.  How to calculate the offsets for a curve from a long chord?

👀.  How to calculate the ordinates for a curve by the long chord method?


Thank you for going through these calculation steps. Have a good day 😄.

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How to calculate the offsets for a curve from a long chord? /Finding ordinates from a long chord to set out a curve.

  Eg:

Calculate the offsets from a long chord at 8m. interval to set out a curve. The length of the long chord is 80m. & the mid-ordinate measures 5m.


Given data:

Mid-ordinate = Oo = 5m.

Length of long chord = L = 80m.

Ordinate interval over long chord = x = 8m.


To find:

Length of ordinates at 8m. intervals to set out a curve.


Calculation:





 Ordinate at given intervals is calculated by the formula

  Ox = [√ (R² - x²) - ( R - Oo) ]

 Where x 👉 Distance of ordinates over the long chord. = 8m.

            R 👉 Radius of the curve =?

           Oo 👉 Length of mid-ordinate = 5m.


Here, the value of R is unknown. So, we have to find the value of R by using the following formula.

      Oo = [ R - √ {R² - ( L/2)²} ]

     5m. = [ R - √ { R² - ( 80/2)²} ]

     5 =  R - √ { R² - 40²}

    √ { R² - 40²} = (R -5)


By squaring the above equation

[√ { R² - 40²}]² = (R -5)²

 R² - 40²  = [R² - (2 x R  x 5) + 5²]

Note: Solved as per algebra equation  (a -b)² = [a² - (2 x a  x b) + b²]

R² - 1600  = [R² - 10R  + 25]

Cancelling R² on both sides,

- 1600 = -10R + 25

- 1600 -25 = - 10R

R = 1625 / 10

R = 162.50 m.


Now, putting the value of R in the above formula,

1. 1st ordinate at 8m. interval

  O8 =  [√ ( R² - x²) - ( R - Oo) ]

          =  [√ (162.5² - 8²) - ( 162.5 - 5)]

          = [√ 26,342.25 - (157.5)]

          = [162.30 - 157.50]

  O8   = 4.80m.


Similarly,

2. 2nd ordinate at 16m. interval

  O16  =  [√ (162.5² - 16²) - ( 162.5 - 5)]

          = [√ 26,150.25 - (157.5)]

          = [161.71 - 157.50]

  O16 = 4.21m.


3. 3rd ordinate at 24m. interval

  O24  =  [√ (162.5² - 24²) - ( 162.5 - 5)]

          = [√ 25,830.25 - (157.5)]

          = [160.717 - 157.50]

  O24 = 3.218m.


4. 4th ordinate at 32m. interval

  O32  =  [√ (162.5² - 32²) - ( 162.5 - 5)]

          = [√ 25,382.25 - (157.5)]

          = [159.318 - 157.50]

  O32 = 1.818m.


Note:

As the length of the long chord L = 80m., half of the long chord L/2 = 40m. So the length of the 5th ordinate at 40m. distance should be equal to zero.


Check:

6. 5th ordinate at 40m. interval

  O40 = [√ (162.5² - 40²) - ( 162.5 - 5)]

          = [√ 24,806.25 - (157.5)]

          = [157.50 - 157.50]

  O40 = 0m




Note:

By symmetry, the length of the ordinates on another half of the curve will be the same at a given interval over the long chord.


Thank you for going through these calculation steps. Have a good day 😄.

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How to calculate the ordinates for a curve by the long chord method? /Calculating offsets from a long chord to set out a curve.

 Eg:

Calculate the ordinates on a long chord at 10m. interval for a circular curve of radius 450m. & long chord of 100m.


Given data:

Radius of curve = R = 450m.

Length of long chord = L = 100m.

Ordinate interval over long chord = x = 10m.


To find:

Length of ordinates at 10m. intervals to draw a curve.


Calculation:





1. Length of Mid-ordinate = Oo 

                                       = [ R - √ {R² - ( L/2)²} ]

                                       = [ 450m. - √ {450² - ( 100/2)² }]

                                       = [ 450m. - √ 200000 ]

                                       = [450m. - 447.214m.]

                              Oo = 2.786m.


 Ordinate at various intervals is calculated by the formula

  Ox = [√ R² - x² - ( R - Oo) ]

Where x is the distance of ordinates over the long chord.


2. 1st ordinate at 10m. interval

  O10 =  [√ ( R² - x²) - ( R - Oo) ]

          =  [√ (450² - 10²) - ( 450 - 2.786)]

          = [√ 202,400 - (447.214)]

          = [449.888 - 447.214]

  O10   = 2.674m.


Similarly,

3. 2nd ordinate at 20m. interval

  O20 = [√ (450² - 20²) - ( 450 - 2.786)]

          = [ √ 202,100 - 447.214]

          = [449.555 - 447.214]

     O20 = 2.341m.


4. 3rd ordinate at 30m. interval

  O30 = [√ (450² - 30²) - ( 450 - 2.786)]

          = [ √ 201,600 - 447.214]

          = [448.998 - 447.214]

  O30 = 1.784m.

         

   5. 4th ordinate at 40m. interval

  O40 = [√ (450² - 40²) - ( 450 - 2.786)]

          = [ √ 200,900 - 447.214]

          = [448.21 - 447.214]

     O40 = 1.005m.


Note:

As the length of the long chord is 100m., the length of half of the long chord i.e. L/2 = 50m. So the length of the 5th ordinate at 50m. distance should be equal to zero.


Check:

6. 5th ordinate at 50m. interval

  O50 = [√ (450² - 50²) - ( 450 - 2.786)]

          = [ √ 200,000 - 447.214]

          = [447.214 - 447.214]

       O50 = 0m.


Now, let us draw the curve by calculated measurement of ordinates as shown below.




Note:

By symmetry, the length of the ordinates on another half of the curve will be the same at a given interval over the long chord.


Thank you for going through these calculation steps. Have a good day 😄.

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