Soils and Materials

Question 1

Moisture content calculation

Answer 1

((wet - dry weight)/wet) x 100

Question 2

Calculating liquid limit

Answer 2

The liquid limit is defined as the
water content at which the groove
cut into the soil will close over a
distance of 13 mm following 25
blows.

Question 3

parameters gained from shear test

Answer 3

Cohesion
angle of internal friction

Question 4

Soil description order

Answer 4

Strength
Colour
Structure
Soil type
Additional info

Question 5

Determination of Liquid Limit tests (atterberg limits)

Answer 5

One-point Casagrande method (line in soil closing 13mm after 25 blows)

Cone penetrometer method (moisture content at which a standard cone will
penetrate a distance of 20 mm in 5 seconds)

Question 6

Atterberg Limits

Answer 6

basic measure of the critical water contents of a fine-grained soil

L– for low plasticity soil (liquid limit < 50)
H– for high plasticity soil (liquid limit > 50)

Liquid limit
Plastic limit PI=0.73(LL-20)
Shrinkage limit

Question 7

What is Cohesion

Answer 7

property of like molecules (of the same substance) to stick
to each other due to mutual attraction (e.g. soil particles).

Question 8

what is Adhesion

Answer 8

the property of different molecules or surfaces to cling to each
other (e.g. soil & water).

Question 9

What is the proctor test and how is it performed

Answer 9

used to obtain the maximum dry density of compaction and the optimum moisture content

The soil is mixed with varying amounts of water, then compacted in 3 equal layers in a standard mould using a 2.5kg rammer

Parameters gained:
The maximum dry density and optimum moisture content is
determined from the results of the test.

Question 10

what is Compaction

Answer 10

process of increasing the bulk density of a soil or aggregate by removing the air voids trapped within. The density achieved is very much dependent on the moisture content

Question 11

Importance of the proctor test

Answer 11

• It gives density that must be achieved in the field.
• Provides the moisture range that allows for minimum compaction
  effort to achieve density.
• Provides data on the behaviour of the material in relation to various
  moisture contents.
• It is not possible the determine if a soil density test passes or fails
  without it.

Question 12

Determination of coefficient of permeability, (k) tests

Answer 12

The constant-head test (For permeable soils (k>10-3 mm/sec)) FOR MEDIUM TO COURSE SOILS

The falling-head test (variable-head), (For less permeable soils (k<10-3
mm/sec)) FOR SAND

Question 13

Unit weight of water

Answer 13

1000 kg/m3

Question 14

Specific gravity of soil

Answer 14

Gs = Density soil / Density of water

typically varies between 2.6 and 2.8

enables the volume of solid particles to be
calculated from mass or weight.

Question 15

coefficient of permeability calculation

Answer 15

From Darcy Law

𝑘 =𝑞/𝐴i

q = flow rate = mm3 of liquid / time in seconds
A = cross sectional area of sample
i = h/L = change in head height mm/length of sample mm

Question 16

Loads on masonary

Answer 16

compressive and shar loads

Parallel flexure and normal flexure

Question 17

Compaction proactor test calcs

Answer 17

moisiture content (w)= wet - dry / dry
bulk density (p) = sample mass in grams (m) / volume of mould in m3 (v)
dry density = 100p/100+w

plot on graph to find max dry density and optimum water content

Question 18

Liquid limit calculation

Answer 18

Calculate moisure content = Ww/Ws x 100
Ww = mass of wet
Ws = Mass of solid

plot

LL = blows at 25

Question 19

Sieve test Cu (uniformity) coefficient

Answer 19

Typical values for Cu:
* Cu >5 indicates a well-graded soil.
* 1< Cu <4 indicates a good soil.
* Cu =1 indicates that all the particles are in the same size.

Question 20

Sieve test Cc (curvature) coefficient

Answer 20

Typical values for Cc:
* 0.5< Cc <2 indicates a well-graded soil
* Cc <0.1 indicates a possible gap-graded soil

Question 21

Geological strata on whitecliff bay in chronological order

Answer 21

White chalk
Reading beds
London clay
Bagshot sands
Bracklesham sands

Question 22

Hard engineering examples

Answer 22

Gabions
Concrete sea wall
rock armour
tetrapods
Gryones

Question 23

Soft engineering examples

Answer 23

Creating offshore sand dunes
Dredging
Beach replenishing

Question 24

Shear box test pocess

Answer 24

1) Internal box measured
2) loading weight determined
3) Box filled with sample in 10mm layers
4) topper and weight added
5) motor and timer started
6) shear stress values taken every 15 secs for several mins until guage fluctuates
7) failure occurs when guage drops

Question 25

importance of shear box test

Answer 25

As normal load is directly proportional to the shear loa, the shear stress can be determined from the results

Angle of internal friction and cohesion are also determined from results collected in the test. These are used in the design process for engineering applications such as foundations and retaining walls.

Question 26

Weaknesses of the shear box test

Answer 26

Does not demonstrate the weakest plane
Failure plane may not be the weakest plane
Not a uniform stress
Uncontrollable variables

Question 27

specific gravity of soil grain

Answer 27

between 2.6 and 2.8

Question 28

Determination of Liquid Limit
Cone penetrometer method

Answer 28

hollow stem filled with lead shot. 80g mass of assembly height, 30 degree cone, 5mm high

Question 29

Determination of Liquid Limit
One-point Casagrande method