HGE 14

REVIEW INNOVATIONS
REFRESHER COURSE
1. A 450 kN load is transmitted by a column footing on the surface through a
rectangular footing 1.5 m × 2.5 m. Evaluate the pressure exerted by the
footing on a soil 2.7 m below the footing.
450 kN
450
𝑃=
1.5 + 2.7 2.5 + 2.7
𝑃 = 20.604 kN
𝐵

2.7 m 2
1

1.35 m 𝐵 1.35 m
2. An unconfined compression test was conducted to a soil sample having a diameter
of 50 mm. The failure load was 66 N. What is the value of the cohesion strength of the
clay in kPa.
66
𝜎3 = 0 𝜎1 = 𝜋 = 0.033614 MPa = 33.614 kPa
50 2
4
𝑐𝑢 𝑞𝑢 𝜎1 33.614
𝑐𝑢 = = = = 16.807 kPa
2 2 2
𝜎3 𝜎1
Sit 1. A two layer unconfined aquifer is shown in the figure.
3. Compute the equivalent coefficient of permeability.
∑ℎ𝑘 3.2 15 + 4.1 30
𝑘𝑒𝑞 = = = 3.8 m/day
𝐻 15 + 30

4. Find the hydraulic gradient.

ℎ 4
𝑖= = = 0.002
𝐿 2000

5. Determine the rate of flow from one stream to another per meter width.

𝑄 = 𝑘𝑖𝐴 = 3.8 0.002 45 1 = 0.342 m3 /day

Sit 2. The following table shows an approximate correlation between the standard
penetration test (SPT) and the unconfined compression strength of cohesive soils.
Consistency SPT Blows Unconfined compressive strength (kPa)
Very Soft 0–2 0 – 25
Soft 2–4 25 – 50
Medium 4–8 50 – 100
Stiff 8 – 16 100 – 200
Very Stiff 16 – 32 200 – 400

6. For an SPT of 5 blows, determine the approximate unconfined compression

strength of the clay
5−4 𝑞𝑢 − 50
=
8 − 4 100 − 50

𝑞𝑢 = 62.5 kPa
Sit 2. The following table shows an approximate correlation between the standard
penetration test (SPT) and the unconfined compression strength of cohesive soils.
Consistency SPT Blows Unconfined compressive strength (kPa)
Very Soft 0–2 0 – 25
Soft 2–4 25 – 50
Medium 4–8 50 – 100
Stiff 8 – 16 100 – 200
Very Stiff 16 – 32 200 – 400

7. For an SPT of 14 blows, determine the approximate cohesion of the clay.

14 − 8 𝑞𝑢 − 100
=
16 − 8 200 − 100
𝑞𝑢 = 175 kPa
𝑞𝑢 175
𝑐𝑢 = = = 87.5 kPa
2 2
Sit 2. The following table shows an approximate correlation between the standard
penetration test (SPT) and the unconfined compression strength of cohesive soils.
Consistency SPT Blows Unconfined compressive strength (kPa)
Very Soft 0–2 0 – 25
Soft 2–4 25 – 50
Medium 4–8 50 – 100
Stiff 8 – 16 100 – 200
Very Stiff 16 – 32 200 – 400

8. For a soil with cohesion 78 kPa, determine the consistency.

𝑐𝑢 = 78 kPa
𝑞𝑢 = 2𝑐𝑢 = 2 78 = 156 kPa
Sit 3. A footing 2 m x 3 m in plan and 0.50 m thick is designed to support a 0.60 m
square column. Due to architectural reasons, the column is located that its external
face is flush with the shorter edge of the footing. The column, however, is located
along the minor principal axis of the footing. The column load, including the weight of
the column itself is 50 kN. Assume concrete to weigh 24 kN/m3.
50 kN
9. Evaluate the total downward load on supporting ground. 𝑊𝑓
𝑃
𝑃 = 50 + 24 2 3 0.5 = 122 kN 𝑒
0.5 m
10. Evaluate the maximum pressure induced on the ground.
𝑃𝑒 = 50 1.5 − 0.3 3m
𝑒 = 0.4918 m < 𝐷/6
𝑃 6𝑒 122 6 ⋅ 0.4918
𝑞𝑚𝑎𝑥 = 1+ = 1+ = 40.333 kPa
𝐵𝐷 𝐷 2⋅3 3
𝑃 2m
11. Evaluate the minimum pressure induced on the ground.
𝑃 6𝑒 122 6 ⋅ 0.4918
𝑞𝑚𝑎𝑥 = 1− = 1− = 0.333 kPa
𝐵𝐷 𝐷 2⋅3 3
12. The relative density of loose soil, in percent, is: Relative Density
Description
B. 15 to 35 (%)
Very Loose 0 – 15
13. The relative density of dense soil, in percent, is:
Loose 15 – 35
C. 65 to 85
Medium Dense 35 – 65
14. The relative density of very dense soil, in percent, is: Dense 65 – 85
D. 85 to 100 Very Dense 85 – 100

15. A cohesionless soil is considered dense if the

SPT “N” value is between C. 30 to 50 Soil Packing SPT Blows
16. A cohesionless soil is considered loose if the Very Loose <4
SPT “N” value is between D. 4 to 10 Loose 4 – 10
Medium Dense 10 – 30
17. The relative density of very loose soil, in
percent, is between: A. 0 to 4 Dense 30 – 50
Very Dense > 50
18. Soil grains with particle size greater than 4.75 mm but less than 75 mm according
to USCS.
B. Gravel
19. Soil grains with particle size is greater than 75 mm according to USCS.
A. Cobbles
20. A practice of procedure used to asses the particle size distribution (also called
gradation) of a granular material. The size distribution is often of critical importance
to the way the material performs in use.
D. Sieve Analysis
21. The over consolidation ratio (OCR) of an over consolidated soil is:
C. Greater than 1
22. The over consolidation ratio (OCR) of a normally consolidated soil is:
A. Less than 1
Sit 2. From the following result of a sieve analysis, determine the following according
to USDA:
23. Percentage of sand.

24. Percentage of silt.

25. Classification of the soil.

 𝐹2 = 100%

𝐹0.05 = 60%

𝐹0.002 = 5%
Sit 2. From the following result of a sieve analysis, determine the following according
to USDA:
23. Percentage of sand.
𝐹0.002 = 5% 𝐹0.05 = 60% 𝐹2 = 100%
%𝑠𝑎𝑛𝑑 = 𝐹2 − 𝐹0.05 = 100% − 60% = 40%

24. Percentage of silt.

%𝑠𝑖𝑙𝑡 = 𝐹0.05 − 𝐹0.002 = 60% − 5% = 55%
= 𝐹0.002 = 5%

25. Classification of the soil.

∴ Silty Loam