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A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq’s theory is____Pa.
Correct answer is between '1300,1350'. Can you explain this answer?
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Verified Answer
A transmission tower carries a vertical load of 10000 kg, the vertical...
Concept:
Vertical stress increment at a any depth (z) at a radial distance (r) as per Boussines’q theory is:
Where,
KB = Boussinesq’s influence factor
Calculation:
At r = 0 m,
Q = 10000 × 9.81 = 98100 N
z = 6 m
Most Upvoted Answer
A transmission tower carries a vertical load of 10000 kg, the vertical...
Concept:
Vertical stress increment at a any depth (z) at a radial distance (r) as per Boussines’q theory is:
Where,
KB = Boussinesq’s influence factor
Calculation:
At r = 0 m,
Q = 10000 × 9.81 = 98100 N
z = 6 m
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Community Answer
A transmission tower carries a vertical load of 10000 kg, the vertical...
Calculation of Vertical Stress Increment Below Transmission Tower
Given data:
- Vertical load on transmission tower = 10000 kg
- Depth below the tower = 6 m
Using Boussinesq's theory, the vertical stress increment at a depth below a loaded area can be calculated using the formula:
\[ \Delta \sigma_v = \frac{qz}{\pi(1+v)} \left( \frac{r^2+z^2}{(r^2+z^2)^2} \right) \]
Calculations:
- Given load = 10000 kg = 10000*9.81 N = 98100 N
- Area of influence (r) = 0 (directly below tower)
- Poisson's ratio (v) for soil = 0.3 (assumed)
- Vertical stress increment at a depth (z) = 6 m
Substitute the values into the formula:
\[ \Delta \sigma_v = \frac{98100*6}{\pi(1+0.3)} \left( \frac{0^2+6^2}{(0^2+6^2)^2} \right) \]
\[ \Delta \sigma_v = \frac{588600}{\pi*1.3} \left( \frac{36}{36^2} \right) \]
\[ \Delta \sigma_v = \frac{588600}{4.084} \left( \frac{1}{36} \right) \]
\[ \Delta \sigma_v ≈ 144075.7 Pa ≈ 144.1 kPa \]
Therefore, the vertical stress increment at a depth of 6 m directly below the transmission tower as per Boussinesq's theory is approximately 144.1 kPa.
Given data:
- Vertical load on transmission tower = 10000 kg
- Depth below the tower = 6 m
Using Boussinesq's theory, the vertical stress increment at a depth below a loaded area can be calculated using the formula:
\[ \Delta \sigma_v = \frac{qz}{\pi(1+v)} \left( \frac{r^2+z^2}{(r^2+z^2)^2} \right) \]
Calculations:
- Given load = 10000 kg = 10000*9.81 N = 98100 N
- Area of influence (r) = 0 (directly below tower)
- Poisson's ratio (v) for soil = 0.3 (assumed)
- Vertical stress increment at a depth (z) = 6 m
Substitute the values into the formula:
\[ \Delta \sigma_v = \frac{98100*6}{\pi(1+0.3)} \left( \frac{0^2+6^2}{(0^2+6^2)^2} \right) \]
\[ \Delta \sigma_v = \frac{588600}{\pi*1.3} \left( \frac{36}{36^2} \right) \]
\[ \Delta \sigma_v = \frac{588600}{4.084} \left( \frac{1}{36} \right) \]
\[ \Delta \sigma_v ≈ 144075.7 Pa ≈ 144.1 kPa \]
Therefore, the vertical stress increment at a depth of 6 m directly below the transmission tower as per Boussinesq's theory is approximately 144.1 kPa.
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A transmission tower carries a vertical load of 10000 kg, the vertical stress increment at a depth 6 m directly below the tower as per Boussinesq’s theory is____Pa.Correct answer is between '1300,1350'. Can you explain this answer?
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