ᐅ Calculated Settlement Behavior of a Single-Family House – Tolerance Range
Created on: 8 Jul 2019 17:12
A
aequitasHello everyone,
Due to challenging ground conditions (backfilling of a former pond with construction debris from the 1960s) and the expected settlement associated with this, we are currently deciding between using micropiles (approximately €23,000) or soil replacement (cost depends on depth and soil classification for disposal) for a new build.
Different settlement behaviors are expected (calculated) for each option:
1. Soil replacement with a 0.5 m (20 inches) gravel cushion: approximately 3.1 - 5.1 cm (1.2 - 2.0 inches)
2. Soil replacement with a 1.0 m (40 inches) gravel cushion: approximately 2.8 - 4.5 cm (1.1 - 1.8 inches)
3. Soil replacement with a 1.5 m (59 inches) gravel cushion: approximately 2.3 - 3.6 cm (0.9 - 1.4 inches)
4. Micropiles: approximately 1 cm (0.4 inches)
From experience, how much settlement is generally considered acceptable? We are planning a solid wood house with a 96 m² (1,033 ft²) footprint on a concrete slab.
If soil replacement would keep settlement within tolerable limits, we would need to refine costs through a thorough soil assessment.
I would appreciate any insights or shared experiences.
Best regards,
Stefan
Due to challenging ground conditions (backfilling of a former pond with construction debris from the 1960s) and the expected settlement associated with this, we are currently deciding between using micropiles (approximately €23,000) or soil replacement (cost depends on depth and soil classification for disposal) for a new build.
Different settlement behaviors are expected (calculated) for each option:
1. Soil replacement with a 0.5 m (20 inches) gravel cushion: approximately 3.1 - 5.1 cm (1.2 - 2.0 inches)
2. Soil replacement with a 1.0 m (40 inches) gravel cushion: approximately 2.8 - 4.5 cm (1.1 - 1.8 inches)
3. Soil replacement with a 1.5 m (59 inches) gravel cushion: approximately 2.3 - 3.6 cm (0.9 - 1.4 inches)
4. Micropiles: approximately 1 cm (0.4 inches)
From experience, how much settlement is generally considered acceptable? We are planning a solid wood house with a 96 m² (1,033 ft²) footprint on a concrete slab.
If soil replacement would keep settlement within tolerable limits, we would need to refine costs through a thorough soil assessment.
I would appreciate any insights or shared experiences.
Best regards,
Stefan
H
HilfeHilfe8 Jul 2019 18:12Option 4 seems to be the best choice as the costs are clear. How much do options 1-3 cost?
Option 1 would be nearly cost-neutral except for the gravel, as we need the same amount of excavation for Option 4 as well.
Option 2 would result in about 96 tons of excavated material after subtracting 0.5 m (20 inches) of excavation. With estimated disposal costs between 20-80€ per ton, the total would be roughly 1,920 - 7,680€.
Option 3 involves about 192 tons of excavated material, with disposal costs ranging from approximately 3,840 to 15,360€.
Gravel costs around 20€ per ton. With a density of 1.5 tons per cubic meter (60 pounds per cubic foot), this adds roughly 1,440€ for Option 2 (about 72 tons) and 2,880€ for Option 3 (about 144 tons).
If the calculation is correct, soil replacement would generally be the more economical choice—provided that the higher settlement is not an issue. What do you think?
Option 2 would result in about 96 tons of excavated material after subtracting 0.5 m (20 inches) of excavation. With estimated disposal costs between 20-80€ per ton, the total would be roughly 1,920 - 7,680€.
Option 3 involves about 192 tons of excavated material, with disposal costs ranging from approximately 3,840 to 15,360€.
Gravel costs around 20€ per ton. With a density of 1.5 tons per cubic meter (60 pounds per cubic foot), this adds roughly 1,440€ for Option 2 (about 72 tons) and 2,880€ for Option 3 (about 144 tons).
If the calculation is correct, soil replacement would generally be the more economical choice—provided that the higher settlement is not an issue. What do you think?
H
HilfeHilfe9 Jul 2019 06:11I would now calculate everything in centimeters.
For example, Option 2; worst case 7,680 for 4.5 cm (1.8 inches) = to reach about 1 cm (0.4 inches) we would be at €34,560.
Then, for the earthworks, there are certainly estimated costs, as you have quite a wide range with Options 2 and 3.
Who guarantees that it won’t be 8 cm (3.1 inches)? In the end, you could have a house that settles unevenly everywhere.
For example, Option 2; worst case 7,680 for 4.5 cm (1.8 inches) = to reach about 1 cm (0.4 inches) we would be at €34,560.
Then, for the earthworks, there are certainly estimated costs, as you have quite a wide range with Options 2 and 3.
Who guarantees that it won’t be 8 cm (3.1 inches)? In the end, you could have a house that settles unevenly everywhere.