How to calculate the location of resultant earth pressure on a retaining wall?

Eg: Determine the lateral earth pressure at rest /unit length of the wall. Also, find the location of the resultant earth pressure on the retaining wall as shown below. Consider 𝛄w = 10KN/m³. 

Given data:

Depth of 1st layer = H1 = 2.5m.

The angle of friction = 𝜱1 = 32° 

Unit weight = 𝛄 = 16KN/m³

Depth of 2nd layer =H2 = 3m.

The angle of friction = 𝜱2 = 34° 

Saturated unit weight = 𝛄sat = 19KN/m³.

 Unit wt. of water 𝛄w = 10KN/m³. 

Calculation:

Coefficient of earth pressure at rest for the 1st layer

Ka01 = [1-sin𝜱1]

        = [1-sin32°]

        = [1-0.5299]

Ka01 = 0.47

Coefficient of earth pressure at rest  for the 2nd layer

Ka02 = [1-sin𝜱2]

        = [1-sin34°]

        = [1-0.5591]

Ka02 = 0.4408

The intensity of pressure at layer B

p01 = [ Ka01 x 𝛄 xH1]

       = [ 0.47 x 16 x 2.5]

p01 = 18.80KN/m².

The intensity of pressure at layer C due to 1st layer

p02 =  [ Ka02 x 𝛄 xH1]

       = [ 0.4408 x 16 x 2.5]

p02 = 17.632KN/m².

The intensity of pressure at layer C due to 2nd layer

p03 =  [ Ka02 x 𝛄¹ xH2]

Where 𝛄¹ = submerged unit wt.

                = [𝛄sat -𝛄w]

                = [19 - 10] 

                = 9KN/m³.

   p03  = [ 0.4408 x 9 x 3]

   p03  = 11.016KN/m². 

Pressure due to water

 u = [𝛄w x H2]

    = [10 x 3]

 u = 30KN/m².

Note: The pr. due to water in the 1st layer will be 0, as the water table is only in the 2nd layer.

Now, let us draw the pressure distribution diagram from the above-calculated values.

Total pressure of triangle ABD

P1= [1/2 x base x height]

    = [1/2 x 18.8 x 2.5]

P1= 23.5KN.

The total pressure of rectangle BCFE

P2= [Base x height]

     = [17.632 x 3]

P2= 52.896KN.

The total pressure of triangle EFG

P3= [1/2 x 11.016 x 3]

P3= 16.524KN.

The total pressure of the triangle EGH

P4= [1/2 x 30 x 3]

P4= 45KN.

The total active earth pressure 

P = [P1 + P2 + P3 + P4]

   = [23.5 + 52.896 + 16.524 + 45]

P = 137.92KN.

Resultant active earth pressure:

The centroid of the total active earth pressure acts horizontally at a distance of Ȳ from the base of the pressure diagram.

For a triangular pressure diagram, the pressure acts at 1/3rd distance from the base of the diagram. For the rectangular diagram, the acting pressure will be at 1/2 the distance from the base.

( According to Rankine theory)

Therefore,

PȲ = [P1Ȳ1 + P2Ȳ2 + P3Ȳ3 + P4Ȳ4]

 = [{23.5 x (3m. + 1/3 x 2.5m.)} +{52.896 x (1/2 x3m )} +{16.524 x (1/3 x 3m.)} +{30 x (1/3 x 3m.)}]

 = [{23.5 x 3.833} + {52.896 x 1.5} + {16.524 x 1} + {30 x 1}]

 = [90.076 + 79.344 + 16.524 + 30]

PȲ = 215.944

Ȳ = [215.944 ➗ P]

    = [215.944 ➗ 137.92]

  Ȳ = 1.565m.

The total earth pressure of 137.92KN  acts horizontally at a distance of 1.565m. from the base of the wall.

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How to calculate the distance between 2 points having coordinates?/ Finding distance by using the distance formula.

The formula for distance (d )between the two points is 

d = [ √ ( x2 -x1)² + (y2 - y1)²]

Where, (x1, y1) & (x2, y2) are the coordinates of those two points.

Note: This formula is to calculate the straight line distance in a 2D plane.

Now, let us understand this with derivation & an example.

Let us assume the distance between the two points A & B as d.

In a right-angle triangle ABC,

AB² = [BC² + AC²]

(By Pythagoras theorem)

Here,

Side AB = d

Side AC = (y2 -y1)

Side BC = ( x2 -x1)

Therefore,

d² = [(x2 -x1)² + (y2 - y1)²]

d = [√ ( x2 -x1)² + (y2 - y1)²]

Eg:

Find the distance between the two points A & B, having coordinates (12, 10) & (-7, -5) respectively.

Distance between AB

d = [√ {x2 -x1}² + {y2 - y1}²]

   =  [√ {12- ( -7)}² + {10 - ( -5)}²]

   =  [√ {19}² + {15}²]

  =   [√ {19}² + {15}²]

  =  [√ {361 + 225}]

  =  [√ 586]

 d = 24.207 units.

 Note: 

1. If the values of x1 & x2 or y1 & y2 are interchanged, the answer will be the same.

2. Whatever the unit of coordinates, the derived distance shall be denoted in the same unit.

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How to calculate the intensity of active earth pressure?

 Eg:

Calculate the intensity of active earth pressure at a depth of 12m. in dry cohesionless soil. The coefficient of earth pressure is 1/3 & unit weight = 16KN/m³.

Given data:

Depth = H = 12m.

Earth pressure coefficient = Ka = 1/3

Unit wt. = Ɣ = 16KN/m³.

To find:

The intensity of active earth pressure = Pa =?

Calculation:

The intensity of  active earth pressure is calculated by the formula

 Pa = [KaƔ H]

       = [ 1/3 Х 16 Х 12]

  Pa = 64 KN/m².

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How to calculate the active earth pressure on a retaining wall?

 Eg:

Determine active earth pressure on the retaining wall as shown in the below drawing. 

Consider 𝛄w = 10KN/m³. 

Given data:

Depth of 1st layer = H1 = 3m.

The angle of friction = 𝜱1 = 34° 

Unit weight = 𝛄 = 16KN/m³

Depth of 2nd layer =H2 = 3m.

The angle of friction = 𝜱2 = 39° 

Saturated unit weight = 𝛄sat = 18KN/m³.

Calculation:

Coefficient of earth pressure for 1st layer

Ka1 = [1-sin𝜱1 / 1+ sin𝜱1]

       =  [(1-sin34°) / (1+ sin34°)]

       = [(1-0.5592) / (1+ 0.5592)]

       = [0.4408/1.5592]

 Ka1 = 0.283

Coefficient of earth pressure for 2nd layer

Ka2 = [1-sin𝜱2 / 1+ sin𝜱2]

       =  [(1-sin39°) / 1+ (sin39°)1]

       = [(1-0.6293) / (1+ 0.6293)]

       = [0.3706/1.6293]

 Ka2 = 0.228

Intensity of pressure at layer B

p1 = [ Ka1 x 𝛄 xH1]

     = [ 0.283 x 16 x 3]

p1 = 13.584KN/m².

Intensity of pressure at layer C due to 1st layer

p2 =  [ Ka2 x 𝛄 xH1]

     = [ 0.228 x 16 x 3]

p2 = 10.944KN/m².

The intensity of pressure at layer C due to 2nd layer

p3 =  [ Ka2 x 𝛄¹ xH2]

Where 𝛄¹ = submerged unit wt.

                = [𝛄sat -𝛄w]

                = [18 - 10] 

                = 8KN/m³.

   p3   = [ 0.228 x 8 x 3]

          = 5.472KN/m².

Pressure due to water

 u = [𝛄w x H2]

    = [10 x 3]

 u = 30KN/m².

Note: The pr. due to water in the 1st layer will be 0, as the water table is only in the 2nd layer.

Now, let us draw the pressure distribution diagram from the above-calculated values.

Total pressure of triangle ABD

P1= [1/2 x base x height]

    = [1/2 x 13.584 x 3]

P1= 20.376KN.

Total pressure of rectangle BCFE

P2= [Base x height]

     = [10.944 x 3]

P2= 32.83KN.

The total pressure of triangle EFG

P3= [1/2 x 5.472 x 3]

P3= 8.208KN.

The total pressure of the triangle EGH

P4= [1/2 x 30 x 3]

P4= 45KN.

The total active earth pressure 

P = [P1 + P2 + P3 + P4]

   = [20.376 + 32.83 + 8.208 + 45]

P = 106.414KN.

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How to calculate the shear strength of a soil?

 Eg:

The unit cohesion of soil is 12KN/m² normal stress is 22KN/m². & the angle of internal friction is 32°. Calculate the shear strength of the soil.

Given data:

The cohesion of soil = C = 12KN/m².

Normal stress = 𝝈 = 22KN/m².

The angle of internal friction = 𝜱 = 32°

To find:

Shear strength of soil = 𝞽 =?

Calculation:

Shear strength of soil is calculated by the formula

𝞽 = [C + (𝝈 tan𝜱)]

   = [12 + (22✖ tan32°)]

   = [ 12 + (22✖ 0.6248)]

   = [12 + 13.747]

𝞽 = 25.747 KN/m².

 

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How the orientation of the column is decided? / How to position & orient the column in buildings?

 Following are the 3 points to be considered while deciding the orientation of the column.

1. The directional ratio of the column should be nearer to 60:40 for the x & y-axis.

This helps to resist the lateral loads on the building from all directions. 

What does that mean?

As you can observe in the above drawing,

Columns having their major axis in the x-direction  👉  C1, C2, C3, C4, & C7

             Total no. of columns in x-direction = 5 nos.                                                  

Columns having their major axis in the y-direction  👉  C5, C6, C8, & C9

            Total no. of columns in y-direction = 4 nos.

The ratio by percentage = 56: 44 ✔ OK

2. The structural & architectural requirements.

As you can observe in the above drawing,

Columns C2, C4, C5, & C8 intruded into the room space of the building. From a structural point of view, designing those columns might be efficient & economical. When you observe those columns from an architectural point of view, they are not aesthetic & appealing.  

So, to satisfy the structural as well as architectural requirements, columns are designed to have the width of the wall. 

Usually, columns are oriented in such a way that it falls in line with the masonry walls having no offsets in any direction. 

3. The sectional length of the column should be in the major plane of bending. This helps to enhance the moment-resisting capacity of the column.

Again, your mind raises the question, what does that mean? 😄

1. Column C1:

As you can observe in the above drawing,

Column C1 supports the beams B1 & B4

The span of B1 = 3.5m. > span of B4 = 2.5m.

Therefore, the moment created by B1 will be greater than that of B4.

So, the major axis of the column should be oriented in alignment with beam B1.

2. Column C2:

Column C2 supports the beams B1, B2, & B4.

Here, beams B1 & B2 are on the same axis, but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B1 & B2 

   = [5m -3m.] = 2m. < span of B4 = 2.5m.

 The moment created by B4 will be greater than  (B1 + B2).

So, the major axis of the column should be oriented in alignment with beam B4.

3. Column C3:

Column C3 supports the beams B2 & B4

The span of B2 = 5m. > span of B4 = 2.5m.

Therefore, the moment created by B2 will be greater than B4.

So, the major axis of the column should be oriented in alignment with beam B2.

4. Column C4:

Column C4 supports the beams B2, B3, & B4.

Here, the beams B3 & B4 are on the same axis, but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B3 & B4 

   = [4m -2.5m.] = 1.5m. < span of B2 = 5m.

 The moment generated by B2 > net moment of (B3 + B4).

So, the major axis of the column should be oriented in alignment with beam B2.

5. Column C5:

Column C5 supports the beams B1, B2, B3 & B4.

Here, the beams B1 & B2 are on the same X-axis, but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B1 & B2 

   = [5m -3.5m.] = 1.5m. 

The beams B3 & B4 are on the same Y-axis but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B3 & B4 

   = [4m -2.5m.] = 1.5m. 

 The moment created by [B1 + B2] =  [B3 + B4].

Now, what should be the orientation? 😲

You can orient the column in any direction as an equal moment will be created.

But to satisfy the directional ratio of 60:40 (condition-1), the major axis of the column should be kept in the y-direction.

6. Column C6:

Column C6 supports the beams B1, B3, & B4.

Here, the beams B3 & B4 are on the same axis, but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B3 & B4 

   = [4m -2.5m.] = 1.5m. < span of B1 = 3.5m.

 The moment created by B1 will be greater than  (B3 + B4).

So, the major axis of the column should be oriented in alignment with beam B1.

7. Column C7:

Column C7 supports the beams B1 & B3

The span of B3 = 4m. > span of B1 = 3.5m.

Therefore, the moment created by B3 will be greater than B1.

So, the major axis of the column should be oriented in alignment with beam B3.

8. Column C8:

Column C8 supports the beams B1, B2, & B3.

Here, beams B1 & B2 are on the same axis, but in opposite directions.

Therefore, the net moment along this axis is generated by the difference in the span of  B1 & B2 

   = [5m -3.5m.] = 1.5m. < span of B3 = 4m.

 The moment generated by B3 > net moment of (B1 + B2).

So, the major axis of the column should be oriented in alignment with beam B3.

9. Column C9:

Column C9 supports the beams B2 & B3

The span of B2= 5m. > span of B3 = 4m.

Therefore, the moment created by B2 will be greater than B3.

So, the major axis of the column should be oriented in alignment with beam B2.

I think you understood the concept.👍

Note: In multi-storeyed buildings having complicated designs, several other factors are considered to decide the column orientation.

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How to calculate the load transfer from a slab to the beam?/Calculating the load over the beam from a one-way slab.

  Let us calculate the total load over beams B1 & B2 as shown below.

Given data:

A span of beam B1= 5500mm. = 5.5m.

A span of beam B2 = 2500mm. = 2.5m.

Sectional dimension of all the beams = 230mm. x 450mm. = 0.23m. x 0.45m.

Calculation:

Lx /Ly = 5500mm. / 2500mm. 

           = 2.2 > 2.

Therefore it is a one-way slab.

The load distribution of the one-way slab over the beams is as shown below.

1. Beam - B1:

Total load over the beam B1

= [Self wt. of the beam + superimposed load from the slab]

Here, 

a.) Self-wt. of the beam /m.

= [(area of cross-section) × density of RCC]

= [ (0.23m. x 0.45m.) × 25 KN/m³]

= 2.588 KN/m.

Total self-wt. of the beam

= [ beam wt./m × span of the beam]

= [2.588KN/m × 5.5m.]

= 14.234 KN.

Factored self-wt. of the beam

= [1.5 × 14.234]

= 21.351KN.

b.) Load transferred from slab to the beam B1

  = [1/2 x (area of slab) x W]

Before proceeding further, Go through the article 👇

👀.  Understanding the concept of the load distribution from slab to beam.

Where all the load distribution formula is derived. 

= [ 1/2 x (5.5  x 2.5) x 12.94]

The value of W is taken from the article

How to calculate the total load over the RCC slab?

= 88.96KN.

Total factored load over the beam B1

= [Factored self wt. of the beam + factored load from the slab]

= [21.351 + 88.96]

= 110.311KN.

2. Beam - B2:

Total load over the beam B2

= [Self wt. of the beam]

Here, 

a.) Self wt. of the beam

= [(area of cross-section) × density of RCC]

= [ (0.23m. x 0.45m.) × 25 KN/m³]

= 2.588 KN/m.

Factored self wt. of the beam 

= [1.5 × 2.588KN/m.]

= 3.88 KN/m.

Total factored self-wt. of the beam

= [ factored wt./m × span of the beam]

= [3.88KN/m × 2.5m.]

= 9.70 KN.

Total factored load over the beam B2

= 9.70KN.

Note: Beam-B2 does not carry any load from the slab.

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How to calculate the total load over the RCC beam?/ Calculating the superimposed & dead load over the beam.

 Let us calculate the total load over beams B1 & B2 as shown below.

Given data:

A span of beam B1= 4500mm. = 4.5m.

Span of beam B2 = 3000mm. = 3.0m.

Sectional dimension of all the beams = 230mm. x 450mm. = 0.23m. x 0.45m.

Calculation:

Lx /Ly = 4500mm. / 3000mm. 

           = 1.5 < 2.

Therefore it is a two-way slab.

The load distribution of the 2-way slab over the beams is as shown below.

1. Beam - B1:

Total load over the beam B1

= [Self wt. of the beam + superimposed load from the slab]

Here, 

a.) Self wt. of the beam

= [(area of cross-section) × density of RCC]

= [ (0.23m. x 0.45m.) × 25 KN/m³]

= 2.588 KN/m.

Factored self wt. of the beam 

= [1.5 × 2.588KN/m.]

= 3.88 KN/m.

Total factored self-wt. of the beam

= [ factored wt./m × span of the beam]

= [3.88KN/m × 4.5m.]

= 17.46 KN.

b.) Load transferred from slab to the beam B1

=  [1/2 × ( Lx Ly - Lx²/2) × W] 

Before proceeding further, Go through the article 👇

👀.  Understanding the concept of the load distribution from slab to beam.

Where all the load distribution formula is derived 

= [1/2 × ( 3 × 4.5 - 3²/2) × 12.94] 

The value of W is taken from the article

How to calculate the total load over the RCC slab?

 = [ 1/2 × ( 9 ) × 12.94]

= 58.23KN.

Total factored load over the beam B1

= [Self wt. of the beam + load from the slab]

= [ 17.46 + 58.23]

= 75.69 KN.

2. Beam - B2:

Total load over the beam B2

= [Self wt. of the beam + superimposed load from the slab]

Here, 

a.) Self wt. of the beam

= [(area of cross-section) × density of RCC]

= [ (0.23m. x 0.45m.) × 25 KN/m³]

= 2.588 KN/m.

Factored self wt. of the beam 

= [1.5 × 2.588KN/m.]

= 3.88 KN/m.

Total factored self-wt. of the beam

= [ factored wt./m × span of the beam]

= [3.88KN/m × 3m.]

= 11.64 KN.

b.) Load transferred from slab to the beam B2

=  [ (Lx²/4) x W]

= [ (3²/4) x 12.94]

= [2.25 x 12.94]

= 29.11KN.

Total factored load over the beam B2

= [Self wt. of the beam + load from the slab]

= [ 11.64 + 29.11]

= 40.755 KN.

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How the load is transferred from the slab to the beams?/Understanding the concept of the load distribution from slab to beam.

 The load transfer from the slab to the beam can be determined using yield line theory.

But, the commonly adopted practice is

1. For a one-way slab 👉  The load is split equally between the beams that support the longer span of the slab.

When Ly/Lx ≥ 2, the slab is considered a one-way slab.

Suppose if the total load of the slab is Wu, the load over the individual beam B1 = Wu/2

If the total load per unit area of the slab is W, 

The load transferred to the individual beam B1

= [1/2 x area of slab  x W]

= [1/2 x Ly x Lx x W]

2. For a two-way slab 👉  The load is transferred to all 4 sides of the slab. Usually, the load of a triangular area is taken by the beam at the shorter side (Lx) & a load of the trapezoidal area is distributed to the longer side (Ly) of the slab. 

When Ly/Lx < 2, the slab is considered a two-way slab.

Now, let us understand the load distribution concept in a two-way slab from the following drawing.

1. Load transferred to the beam B1

= [ trapezoidal area x total load per unit area of the slab (W)]

Here,

Area of trapezoid AEFD

= [average length x perpendicular width]

= [{(side AD + side EF) / 2 } x (length of DG)]

= [{(Ly + (Ly - Lx) /2}  x (Lx / 2)]

= [{ 2Ly/2 - Lx/2} x (Lx / 2)]

= [Lx Ly /2 - Lx²/4]

= [1/2 ( Lx Ly - Lx²/2)]

Therefore,

The load transferred to the individual beam B1

= [1/2 ( Lx Ly - Lx²/2) x W]

2. Load transferred to the beam B2

= [ triangular area x total load per unit area of the slab (W)]

Here,

Area of triangle DFC

= [1/2 x base x height]

= [1/2 ✖ Lx ✖ Lx/2]

= Lx²/4

Therefore,

The load transferred to the individual beam B2

= [ Lx²/4 x W]

Note: This method will be accurate only when the load over the slab will be uniformly distributed.

Now, my question to you is, "what will be the load distribution pattern in a square-shaped slab?"

In a square-shaped slab Lx = Ly 

Ly /Lx = 1 < 2, the slab is considered a two-way slab.

Here, the load will be transferred equally to all 4 sides. 

The load distributed over the beam-B1 = Load over the beam-B2

The load over the individual beam ( B1 or B2)

 = [ triangular area x total load per unit area of the slab (W)]

=  [ Lx²/4 x W] or [ Ly²/4 x W]

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How to calculate the total load over the RCC slab?/Calculating the factored load of a RCC slab.

 Let us calculate the total load over the RCC slab as shown below.

Given data:

Length of slab = 4500mm.=4.5m.

Width of slab = 3000mm. = 3.0m.

Slab thickness = 125mm. = 0.125m.

Calculation:

Total load over the slab

= [dead load + live load]

Where,

1. Dead load 

= [self wt. of slab + floor finish]

a. Self wt. of slab

   = [(volume of the slab) x density]

   = [(length x breadth x thickness) x 25KN/m³]

   = [( 4.5m. x 3.0m. x 0.125m.) x 25KN/m³]

   = [1.6875m³  x 25KN/m³]

  = 42.19KN.

b. Total load of floor finish

= [area of slab x wt. of floor finish /m².]

= [ (length x breadth)  x 1.5KN/m²]

= [(4.5m. x 3.0m.) x 1.5KN/m²]

= [13.5m² x 1.5KN/m²]

= 20.25KN.

Total dead load 

= [self wt. of slab + floor finish]

= [42.19KN + 20.25KN]

= 62.44KN.

2. Total live load over the slab

   = [area of slab x live load /m².]

   =  [ (length x breadth)  x 4KN/m²]

  = [(4.5m. x 3.0m.) x 4KN/m²]

  = [13.5m² x 4KN/m²]

 = 54KN.

Total load over the slab

= [dead load + live load]

= [62.44 + 54]

= 116.44 KN 

Total load over the slab in kg.

= [116.44 KN x 101.97]

= 11,873.39 kg.

Factored load of RCC slab

Wu = [1.5 x total load]

= [1.5 x 116.44]

Wu = 174.66 KN.

Factored load of RCC slab per unit area

W = [ total factored load ÷ area of slab]

      = [ 174.66  ÷ (4.5 x 3)]

 W = 12.94KN/m²

   To go through all types of structural design articles, click here.

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How to calculate the volume & weight of material in a Hyva truck?/ Calculating the quantity of sand in a hyva truck.

 Let us calculate the quantity of sand in a hyva truck as shown below.

As you can observe in the above image, the truck body is tapered at one end & the sand is filled above the container level of the truck.

So, the length of the container should be measured at the top ( L1) & bottom (L2) of the body (inner length) & average length should be calculated.

Now the question is, how to find the depth of the sand as it is above the container level?

You have to insert 16mm. steel rod at different locations from the top of the sand, until it touches the bottom level. At least 3 readings should be taken to find the average depth (D). If you take more readings at different points, it will increase the accuracy of the calculated average depth.

Now let us consider a hyva truck with measurements as shown below.

Given data:

Top length = L1 = 5m.

Bottom length L2 = 4.3m.

Width = W =2.25m.

Depth h1 = 1.95m.

Depth h2 = 2.15m.

Depth h3 = 1.9m.

Calculation:

The volume of sand 

V = [average length х breadth х average depth]

Here, 

Average length

L = [L1 +L2 ➗ 2]

   = [5m. + 4.3m. ➗ 2]

  = [9.3m ➗ 2]

 L= 4.65m.

Average depth

D = [h1 +h2 +h3➗ 2]

    = [1.95m. + 2.15m. +1.9m.➗ 3]

   = [6 ➗ 3] 

  D = 2.0m.

The volume of sand 

V = [ L х W х D]

     = [ 4.65m. х 2.25m. х 2m.]

   V = 20.925 m³.

The vol. of sand in cu. ft

  V = [20.925 х 35.315]

  V = 738.97 ft³

The vol. of sand in brass

= [738.97 ft³ ➗ 100]

= 7.39 brass.

The wt. of sand in a ton

= [volume in m³ х density in metric ton/ m³]

= [20.925m³  х 1.6 ton/ m³]

 = 33.48 tons.

Note: The density of sand varies from 1.5ton/m³. to 1.9ton/m³. depending upon the type & particle size of sand. However, for calculation purposes, 1.6 tons/ m³ is considered.

To go through all types of quantity estimate articles, click here.

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

 

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