ALLOWABLE FORMULA, CHART, FIGURES

For SOIL MECHANIC EXAMS

Academic year 2014-2015

Third year Student

Lecturer: M. Chener S. Qadr

Soil Classification

Table 1: U.S Standard Sieve Sizes

Sieve No Opening (mm) Sieve No Opening (mm)

4 4.75 35 0.500

5 4 40 0.425

6 3.35 50 0.355

7 2.8 60 0.250

8 2.36 70 0.212

10 2.00 80 0.180

12 1.70 100 0.150

14 1.40 120 0.125

16 1.18 140 0.106

18 1.00 170 0.090

20 0.850 200 0.075

25 0.710 270 0.053

30 0.600

Figure 1

AASHTO Soil Classification

Table 2

General

Classification

Granular materials

(35 or less of total sample passing No.200)

Group classification

A-1 A3

A-2

A-1-a A-1-b A-2-4 A-2-5 A-2-6 A-2-7

Sieve analysis

(percentage passing)

No.10 50 max.

No.40 30 max. 50 max. 51 min.

No.200 15 max. 25 max. 10 max. 35 max. 35 max. 35 max. 35 max. Characteristics of fraction

passing No.40

Liquid Limit 40 max. 41 min. 40 max. 41 min.

Plastic Index 6 max. NP 10 max. 10 max. 11 min. 11 min.

Usual types of significant

constituent materials

Stone fragments,

gravel, and sand

Fine sand Silty or Clayey gravel and sand

General subgrade rating Excellent to good

Table 3

General classification Silt-Clay materials

(More than 35% of total sample sassing No.200)

Group classification A-7

A-7-5a

A-4 A-5 A-6 A-7-6b

Sieve analysis (percentage passing

No.10

No.40

No.200 36 min. 36 min. 36 min. 36 min. Characteristics of fraction passing No.40 Liquid Limit 40 max. 41 min. 40 max 41 min. Plastic Index 10 max. 10 max. 11 min. 11 min. Usual types of significant constituent materials Silty Soils Clayey Soils General subgrade rating Fair too Poor

a For A-7-5, PI ≤ LL-30

b For A-7-6, PI > LL-30

200 200( 35) 0.2 0.005(LL 40) 0.01( 15)(PI 10)GI F F

USCS Soil Classification

Figure 2

Figure 3

Figure 4

IN Situ Stress

Vertical Stress Due to POINT LOAD

5/222

12

3 1

2 ( ) 1

(1)

Pz

z rz

Por z I Equation

z

Table 4 Variation of I1 for various values of r/z ( I1 for Equation 1)

r/Z I1 r/Z I1 r/Z I1 r/Z I1

0 0.4775 0.28 0.3954 0.80 0.1386 3.00 0.0015

0.02 0.4770 0.30 0.3849 0.85 0.1226 3.20 0.0011

0.04 0.4765 0.32 0.3742 0.9 0.1083 3.40 0.00085

0.06 0.4723 0.34 0.3632 0.95 0.0956 3.60 0.00066

0.08 0.4699 0.36 0.3521 1.00 0.0844 3.80 0.00051

0.1 0.4657 0.38 0.3408 1.20 0.0513 4.00 0.00040

0.12 0.4607 0.40 0.3294 1.40 0.0317 4.20 0.00032

0.14 0.4548 0.45 0.3011 1.60 0.0200 4.40 0.00026 0.16 0.4482 0.50 0.2733 1.80 0.0129 4.60 0.00021

0.18 0.4409 0.55 0.2466 2.00 0.0085 4.80 0.00017

0.20 0.4329 0.6 0.2214 2.20 0.0058 5.00 0.00014

0.22 0.4242 0.65 0.1978 2.40 0.0040

0.24 0.4151 0.7 0.1762 2.60 0.0029

0.26 0.4045 0.75 0.1565 2.80 0.0021

Vertical Stress due to Vertical Line-Load

32

2 2 2( )

qzz

x z

(2)

2

zor Iv equation

qz

where Iv from table

Table 5 Variation of /( / )z q z with x/z ( Iv for Equation 2)

x/z /( / )z q z x/z /( / )z q z x/z /( / )z q z

0 0.637 0.8 0.237 1.7 0.042

0.1 0.624 0.9 0.194 1.8 0.035

0.2 0.589 1.0 0.159 1.9 0.030

0.3 0.536 1.1 0.130 2.0 0.025

0.4 0.473 1.2 0.107 2.2 0.019

0.5 0.407 1.3 0.088 2.4 0.014

0.6 0.344 1.4 0.073 2.6 0.011

0.7 0.287 1.5 0.060 2.8 0.008

1.6 0.050 3.0 0.006

Vertical Stress Caused by a Horizontal Line-Load

2

2 2 2

2

( )

q xzz

x z Or (3)

HqZ

zI equation

Table 6:Variation of /( / )z q z with x/z ( IH for Equation 3)

x/z /( / )z q z x/z /( / )z q z x/z /( / )z q z

0 0 0.4 0.189 0.9 0.175

0.1 0.062 0.5 0.204 1.0 0.159

0.2 0.118 0.6 0.207 1.5 0.090

0.3 0.161 0.7 0.201 2.0 0.051

0.8 0.189 3.0 0.019

Vertical Stress Due to Strip Load (Finite Width and Infinite Length)

22 2

222 2 2 2

4tan tan

/2 /24

BBz x zz zq

zx B x B Bx z B z

10.4

,s

s can be found from table

zI

q

I

Vertical Stress Caused by a Rectangular Loaded Area

3z qI

Where I3 Can be found either using Table 10.8 or Using figure 1

FIGURE 1

Stress increase caused by uniformly loaded area

z netq Ic

Where Ic can be calculated from X/R and Z/R

Vertical Stress Due to Embankment Loading

Or

2

where I2 is function of B1/z and B2/z.

2 10.15

,

can be found from Figure

zI

q

I

Chapter 10 Shear Strength of Soil

1 3 1 3

1 3

cos(2 )2 2

Sin (2 )2

n

f

Chapter 11: Consolidation

1 2 1 2

2 2 2

1

log loglog

c

e e e eC

0.009 (LL 10)CC 0.007 (LL 10)CC 3 4 3 4

4 3 4

3

log loglog

S

e e e eC

Calculation of Consolidation Settlement.

For normally Consolidated Clay NCC

0 C

0

0

0

0

1

log

log( )1

eS Hc e

o

e Cc

HCcS

c eo

For over consolidate Clay ( O.C.C )

0 C

0

0

log( )1

o

H CsS

c e

0 0c

0

0

log( ) log( )1 1

c

co o

H C H Cs cS

c e e

For Under Consolidated Clay

0  C

0log( )1 co

H CcS

c e

2

1

log1

s

p

C H tS

e t

( )Sc tU

Sc

1 1

1

vv

o

am

e

in KPa or MPa

. .c vS m H z v

w v

KC

m

2

1

log

eC

t

t

2

vv

dr

CT t

H 4

6t m b

av