Geotechnical Engineering II: Footing Design and Settlement Analysis

This assignment focuses on the design of isolated footings for a commercial building in Dubai, considering the soil properties of stiff clay. The solution begins with determining appropriate footing sizes for three columns (C1, C2, and C3) based on given column loads and soil strength at 3m depth, using the formula A=F/P. It then calculates both short-term and long-term settlement for the largest footing, considering the saturated unit weight and void ratio, and evaluates whether the settlement exceeds allowable limits. Furthermore, the assignment addresses footing design with a higher soil strength, recalculating footing sizes using a safety factor of 2 and calculating settlement for both the largest and smallest footings. The solution includes professional engineering drawings for the largest footing, providing plan and cross-section views, and a plan showing the overall footing layout. Finally, the assignment checks the adequacy of the smallest footing (C1) to withstand a horizontal load, analyzing the moments generated and the soil's ability to resist the load. References from Bo (2019) and geotechnicaldirectory.com are included.

a) Determine the appropriate footing sizes for the given building.
The slab area = 17 x 30 = 510m2
Soil type - stiff clay
Compression strength = 350kN/ m2
The recommended footing sizes for the columns are as calculated below, (Bo, 2019)
Formula for calculating pressure is applied force( F )divide by area( A)∨P=F / A
That is where theunit of pressure kN
sq m∨ pounds per square foot – sometimes express
¿ squareinch as psi
( pounds per square inch)∨ksi(kilo pounds per square inch) .

Since we arelooking for the area of the footing , solving the equation P=F / A for area yields A=F /P
soil strength at 3m provided as 350 kN/m2
C1
A=F/P= (690)/(350x2kN/m2)= 1 sq m
C2
A=F/P= (1275)/(350x2kN/m2)= 1.8 sq m
C3
A=F/P= (1965)/(350x2kN/m2)= 2.8 sq m
b. Calculate the short term and long-term settlement for the largest footing.
∆h e f −ei
= h
1+ei
where e i∧e f arethe initial∧final voids ratio .e
f −ei
Thus ∆ =h h
1+ei
(Geotechnicaldirectory.com, 2020)
Saturated unit weight =
Ww  Ws
Vv  Vs
= 17+20
1+1 = 18.5 Kn/ M3
Total stress σzz=2 ×18+3 × 22+ 2×19.44=140.89 kN /m²
Pore water pressure uw=5 ×10 kN / m ²=50 kN /m ²
Effective stress σ ' zz=σzz−uw=140.89−50=90.89 kN /m
= −0.09 m
b) Does settlement exceed maximum allowable settlement for isolated footings? If so, what
are your recommendations to keep settlement within required limits.

The settlement does not exceed the allowed value for isolated footings

Part B
a) Determine the appropriate footing sizes for the given building.
soil strength at 3m dept= 3.0mpa, 3000 kN/m2
C1
A=F/P= (690)/(3000kN/m2)= 0.23 sq m
C2
A=F/P= (1275)/(3000kN/m2)= 0.425 sq m
C3
A=F/P= (1965)/(3000kN/m2)= 0.655 sq m
Using safety factor of 2
C1= 0.46 sq m
C2= 0.95 sq m
C3= 1.31 sq m
b. Calculate the settlement for the largest and smallest footings.
∆h e f −ei
= h
1+ei
where e i∧e f arethe initial∧final voids ratio .e
f −ei
Thus ∆ =h h
1+ei
Saturated unit weight=17.8 kN /m3
Upper states
Total stress σzz = 2 ×18+3 ×22+6 × 19.44=218.67 kN /m ²
Pore water pressure uw = 9 ×10 kN /m²=90 kN /m²
Effective stress
Below states
σ′zz = σ zz −u w=218.67−90=128.67 kN /m²
Total stress σ zz=60+2× 22+3 ×22+6 ×19.44=286.67 kN / m²
Pore water pressure u w= 11 ×10 kN /m ²=110 kN /m ²
Effective stress σ ' zz=σ zz−u w=286.67−110=176.67 kN /m²

Settlement values
The soil∈the second is normally consolidated∧thus:
∆ e 2=−C c × log10 (σ ' F /σ ' I)
∑h ek ∆ k
∆ =S
1+e i
¿ 0.061 m
c. Provide professional engineering drawings for i) plan and cross section of
largest footing. You can assume appropriate thickness and reinforcement. ii)
plan showing footings layout.
Largest footing size = 1.31 sq m, assumed width and thickness as 1.14 mby 1.14 m

Figure 1 FOOTING DESIGN, check AutoCAD file for clear view
Plan showing footings layout

Figure 2footings layout
d. Check the adequacy of the size of the footing carrying column C1 if it is subjected to
a horizontal load of 100 kN at the ground level.
Footing area = 0.46 sq m
Horizontal force = 100KN

Moments at footing = 100x3 = 300kNM
The footing is not designed to withstand horizontal weights, the weight will be transferred to the
soil profile walling of the column. Since the soil has a strength of 3000kN/m2 along its profile,
hence the exposed height is adequate to outdo the horizontal force.

References
Bo, M. (2019). Editorial: in situ tests in geotechnical engineering. Geotechnical Research, 6(1),
1-2. doi: 10.1680/jgere.2019.6.1.1
Fhwa.dot.gov. (2020). Retrieved 29 February 2020, from
https://www.fhwa.dot.gov/engineering/geotech/pubs/reviewguide/checklist.pdf
Geotechnicaldirectory.com. (2020). A handy reference for use in geotechnical analysis and
design. Retrieved 29 February 2020, from
http://geotechnicaldirectory.com/publications/courses/geotechformulas.pdf