Cie6003 Geotechnical Engineering And Ground Assessment Answers
Carry out bearing capacity calculations for shallow foundations for the scheme, alongside calculations for deep and piled foundations, so that a comparison can be made of the relative dimensions of the shallow and piled foundation solutions.
Outline the technical and practical problems likely to be experienced when implementing the solution to the ground problems at the given site.
Answer:
Introduction:
The soil bearing capacity for shallow foundation can be calculated using the Terzaghi’s ultimate bearing capacity equation. When the unit weight and the cohesion of the soil in addition to the surcharge are considered, the equation below can be applied:
qu= cNcSc + γ DNqSq + 0.5BγNγS γ (Das, 2017)
Where:
- quis the ultimate bearing capacity
- D is the depth of the foundation
- B is the width of the foundation, and
- Nc, Nq, Nγ, and, Sc, Sq and Sγ are bearing capacity and shape factors respectively (Das, 2014).
In Development Area 1, a pad footing is used to support loads of up to 800 KN. In addition to that, a strip foundation supports a perimeter wall load of up to 70 KN/m.
For square footing:
Qu = 1.3cNc + γ DNq + 0.4 γ BγNγ
Taking D= 2.10m; B= 3m and the angle of friction φ =20° Nc =17.69, Nq= 7.44 and Nγ= 3.64.
Also, assume a unit weight of 20KN/m3 for clay soil.
Therefore, qu = (1.3×35×17.69) + (20×2.1×7.44) + (0.4×20×3.64)
= 804.895 + 312.480 + 29.120
=1146.495 KN/m2
Pile foundation
According to Skempton, for a deep foundation, Nc = 9. Therefore;
Qu= αcAf + 9CAb.
For a pile of diameter B and embedded depth D
Ab= B2 and Af =
Therefore, Qu =αBDc +2.25 B2c
With Cu=35KN/m2 and using a concrete pile of B= 400mm and D= 10 meters. Taking the adhesive factor α =0.60
Qu =(0.60×0.4×10×35) + (2.25× 0.42 × 35)
=84+ 39.58
=123.58 KN/m2
Task 2.2
The engineering properties of a site should be improved in cases where the ground quality is not up to standard. This can be done by reducing the volume of voids in the soil, reducing pore water pressure and using stronger materials. The main methods of achieving this are consolidation and compaction. Consolidation is the improvement of the soil bearing capacity of a soil mass by the application of load over a given period.
Alluvial soil deposits have a large of amount of water that heavily reduces its soil bearing capacity. Therefore, to increase its bearing capacity, consolidation is used. If the ground quality is not improved, settlement tend to happen as the soil mass is consolidated as the result of the load of the structure.
Stress increment because of the new structure will cause a reduction in the void ratios. As a result, the height of the soil layer reduces. For a laterally confined soil
?H= mvH0?P
Where:
- mv is the volume of compressibility
- H0is the initial thickness of the soil
- ?P is the change in pressure.
By definition, the compression index Cc is:
Cc= (Hirkane et al., 2014)
Therefore;
?e =Cc log10
But = = =
Therefore, consolidation settlement Sc is:
Sc = H0 log10
Task 2.3
Geotechnical engineers are faced with numerous challenges when attempting to implement the various ground improvement solution. Some of these challenges include:
- Unpredictable weather. When trying to improve the soil quality through the expulsion of water, raining conditions may reverse the process hence nullifying all the work (Hirkane et al., 2014). Additionally, swelling of clay due to moisture may interfere with the works.
- Improvement, which entails the importation of soils with better engineering properties, incurs high costs that increases the cost of construction.
- In addition, incidences of uneven settlement after consolidation and compaction are common.
References
Das, B. M. (2017). Shallow foundations: bearing capacity and settlement. CRC press.
Hirkane, S. P., Gore, N. G., & Salunke, P. J. (2014). Ground Improvement Techniques. International Journal of Inventive Engineering and Sciences, 2, 11-13.
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