Example & Verification Problem-5.32

CASTeR Example & Verification Problem Manual

1

CASTeR is a Geotechnical Engineering Software Suit developed and marketed by Technology Development Centre, 14/5 Gariahat Road, Kolkata 700019 India. Although every effort has been made to ensure the accuracy of this software, Technology Development Centre will not accept responsibility for any mistake, error or misrepresentation in or as a result of the usage of this software.

CASTeR Version 5.3

Prepared By - Technology Development Centre 14/5 Gariahat Road

Kolkata 700019 West Bengal

India www.tdcindia.com


2

Index

Section 1 Example Problems: 1. Square Root Time Fit 4 2. Time – Settlement 9 3. Mohr’s Circle 15 4. Pile Capacity 20 5. Grain Size 28 6. Bearing Capacity 32 7. Settlement Analysis 36

Section 2 Verification Problems: 1. Square Root Time Fit 44 2. Time Settlement 50 3. Mohr’s Circle 56 4. e – logP Curve 61 5. Pile Capacity 65 6. Grain Size 72 7. Bearing Capacity 76


3

Section 1

Example Problems


4

Square Root Time Fit

Example Problem from Prnciples of Geotechnical Engineering By Braja M. Das Fifth Edition

Publisher: Bill Stenquist First Reprint 2002 by Thomson Asia Pte. Ltd.

Page – 308 Problem : During a laboratory consolidation test , the time and dial gauge readings obtained from an increase in pressure on the specimen from 50 to 100 kN/m2 are given in the following table :

Time (min) Dial Gauge Reading (mm)

0 3.98

0.1 4.08

0.25 4.1

0.6 4.13

1 4.17

2 4.22

4 4.3

8 4.42

16 4.57

30 4.74

60 4.92

120 5.08

240 5.21

480 5.28

960 5.33

1440 5.39

Solution :

• Open CASTeR and from the Tools menu select clear screen. • Select Configure→Sq. Root Time Fit tab in the Plot Configuration Dialog box.

The dialog box with the Sq. Root Time Fit tab selected is shown below.


5

1. The No. of Plot in a Page is made to 1 as we need to draw only one plot in the page.

2. Least Count of Dial Gauge is taken to be 0.01 mm/div as we need to enter the dial gauge reading values and not the exact height of specimen as given in the book.

3. The Draw Graph Lines box is checked 4. The Write end Co-ord box is left unchecked. 5. Connect Points By Smooth Curve radio button is checked. 6. Connect Point By St. Line box is unchecked. 7. X axis Line Interval is made 1 Y axis Line Interval is made 0.1 8. X axis Label Interval is 2 Y axis Label Interval is made 0.1 [for proper clarity of

the plot] 9. The Dial Gauge Reading is checked for Increasing with time radio button. 10. The Input through Xcel Format option enables the user to put the input in the Xcel

table. Here this option is not enabled and the input dialog box in the ordinary format is shown below.

11. Press OK and the dialog box is closed.


6

• Select Tools→→→→Font and choose a proper Text style, colour of the text and its size. • Select Tools→→→→Lines and choose the line style and its colour. • Select Tools→→→→Graph Lines and choose the style and colour of the graph lines. • Click on the Dial Gauge vs. Time option in the Lab Test→→→→Consolidation→→→→Sq.

Root Time Fit menu and the respective dialog box opens. • The values given in the table of the given problem are inserted in the Sq. Root

Time Fit -Dial Gauge Reading vs. Time – Set 1 table. The values in the Dial Gauge Reading column in the table of the given problem are inserted under the Compression Dial Gauge Reading(R) boxes of the table and correspondingly the Elapsed time in the Time(t) in min. boxes as shown below. As the given problem has only 16 sets of values, the rest of the boxes have zero input values.

Dial Gauge Reading vs. Time Dialog box for Sq.Root Time Fit


7

• Press OK and the dialog box

closes. • Open the Consolidation→→→→Sq. Root Time Fit menu in the top menu bar and

select Other Data. The Other Related Data –1st Plot dialog box opens as shown below.

As nothing is mentioned about the BoreHole No. , Sample No. and Depth, the default values are accepted. But from the given problem the value for the Initial Pressure is taken to be 50 kN/m2 (converted through the use of F2 key)and the value for the Final Pressure is taken to be 100 kN/m2. For the Initial Height click the box and press F2 and insert the given value of 5.39 mm and in the same way insert the value of 3.98 mm in the box for Final Height. The given problem has Double Drainage system and hence the radio button for the Double Drainage is clicked. • Press OK and the dialog box closes. • Next Select Consolidation→→→→Sq. Root Time Fit→→→→Plot from the top menu bar and

the values are plotted on the screen as shown below.



8

• Press OK and the dialog box closes. • Open the Consolidation→→→→Sq. Root Time Fit menu from top menu bar and select

the Extrapolate→manual and extrapolate as per the instruction that comes up on the screen as shown. There is also the guided option to extrapolate but this has not been used here.

• The output value is given on the graph and the value of

Cv = 0.000738 cm2 /min = 1.23333310-5 cm2 / sec.


9

Time – Settlement

Example Problem from Prnciples of Geotechnical Engineering By Braja M. Das Fifth Edition Publisher: Bill Stenquist First Reprint 2002 by Thomson Asia Pte. Ltd. Page – 308

Problem: During a laboratory consolidation test, the time and dial gauge readings obtained from an increase in pressure on the specimen from 50 to 100 kN/m2 are given in the following table : Solution : • Open CASTeR and from the Tools menu select clear screen. • Select Configure→Time-Settlement tab in the Plot Configuration Dialog box.

The dialog box with the Tim • e-Settlement tab selected is shown below.

Time (min) Dial Gauge Reading (mm)

0 3.98

0.1 4.08

0.25 4.1

0.6 4.13

1 4.17

2 4.22

4 4.3

8 4.42

16 4.57

30 4.74

60 4.92

120 5.08

240 5.21

480 5.28

960 5.33

1440 5.39


10

1. The No. of Plot in a Page is made to 1 as we need to draw only one plot in the page.

2. Least Count of Dial Gauge is taken to be 0.01 mm/div as we need to enter the dial gauge reading values and not the exact height of specimen as given in the book.

3. The Draw Graph Lines box is checked 4. The Write end Co-ord box is left unchecked. 5. Connect Points By Smooth Curve radio button is checked. 6. Connect Point By St. Line box is unchecked. 7. Dial Gauge axis Line Interval is made 10 8. Dial Gauge axis Label Interval is made 10 [for proper clarity of the plot] 9. The Dial Gauge Reading is checked for Increasing with time radio button. 10. The Input through Xcel Format option enables the user to put the input in the

Xcel table. Here this option is not enabled and the input dialog box in the ordinary format is shown below.


• Select Tools→→→→Font and choose a proper Text style, colour of the text and its size. • Select Tools→→→→Lines and choose the line style and its colour.


11

• Select Tools→→→→Graph Lines and choose the style and colour of the graph lines • Click on the Dial Gauge Reading option in the Lab Test→→→→Consolidation→→→→

Time-Settlement menu and the respective dialog box opens. The values given in the table of the given problem are inserted in the Time Settlement –Dial Gauge Reading vs. Time – Set 1 table. The values in the Dial Gauge Reading column in the table of the given problem are inserted under the Compression Dial Gauge Reading(R) boxes of the table and correspondingly the Elapsed time in the Time(t) in min. boxes as shown below. As the given problem has only 16 sets of values, the rest of the boxes have zero input values.

Dial Gauge Reading vs. Time Dialog box for Time-Settlement



12

• Press OK and the dialog box closes. • Open the Consolidation→→→→Time – Settlement menu in the top menu bar and

select Other Data. The Other Related Data –1st Plot dialog box opens as shown below.

As nothing is mentioned about the BoreHole No. , Sample No. and Depth, the default values are accepted. But from the given problem the value for the Initial Pressure is taken to be 50 kN/m2 (converted through the use of F2 key)and the value for the Final Pressure is taken to be 100 kN/m2. For the Initial Height click the box and press F2 and insert the given value of 5.39 mm and in the same way insert the value of 3.98 mm in the box for Final Height. The given problem has Double Drainage system and hence the radio button for the Double Drainage is clicked. • Press OK and the dialog box closes. Next Select Consolidation→→→→Time–Settlement→→→→Plot from the top menu bar and the values are plotted on the screen as shown below.


13

• Open the Consolidation→Time–Settlement menu from the top menu bar and

select the Set Corrected Zero Reading and insert the value 1 min. in the dialog box.

• Press OK and the dialog box closes. • Open the Consolidation→Time–Settlement menu from the top menu bar and

select the Draw 1st Tangent→ΜΜΜΜanual and draw the first tangent as per the instruction that comes up on the screen as shown. There is also the guided option to draw the tangent.


14

• Similarly select Consolidation→Time–Settlement→Draw 2nd Tangent→ Manual and draw the tangent.

• The output values are given on the graph and the value of Cv = 0.0000092 cm2/sec

= 9.2333310-6 cm2/sec.


15

Mohr’s Circle Example Problem from Design Aids in SOIL MECHANICS AND

FOUNDATION ENGINEERING BY Shenbaga R Kaniraj

Fourth reprint Page No. – 137 , Q 9.2 Published by -Tata McGraw –Hill Publishing Company Limited

Data on CU test carried out on two saturated soil samples are given beow. Determine the total and effective stress shear strength parameters.

Sample1 Sample2 Consolidation pressure ( σσσσc ) 2.75 kg/cm2 4.25 kg/cm2 Back pressure (uB) 2 kg/cm2 2 kg/cm2

Confining pressure (σσσσ3 ) 4.75 kg/cm2 6.25 kg/cm2

Axial stress at failure (σσσσ1f ) 6.75 kg/cm2 9.25 kg/cm2 Pore water pressure at failure ( uf ) 3.75 kg/cm2 4.75 kg/cm2

Calculate the required stresses and plot them on a Mohr circle diagram to obtain the effective stress shear strength parameters for the clay. Solution: Effective stress parameters are given in the following table

Sample 1 Sample 2

(σ3 )= 4.75-3.75 1 kg/cm2 1.5 kg/cm2

σ1= 6.75-3.75 3 kg/cm2 4.5 kg/cm2

N.B - For entering the numerical values in the given units press F2 and there

will be automatic unit conversion. Here in this problem F2 is not required as all the values are in the default unit. • Open CASTeR and from the Tools menu and select clear screen. • Select Configure→Mohr Circle tab in the Plot Configuration Dialog box and

make the proper changes. 1. The No. of Plots in aPage is made to1 2. Write End Co-ord on Graph is left unchecked 3. Draw Graph Lines is checked


16

4. Graph Line Interval is 0.5 5. Graph Label Interval is 2 6. Input Through Xcel Format is left unchecked. This option enables the user to

make inputs in the Xcel table. 7. In Plot size, Small Plot radio button is checked. The completed Configure dialog box is shown below.

• From Lab Tests→→→→Shear Strength→→→→Mohr Circle menu from the top menu bar open the Mohr Circle Stress Values(Kg/cm2)-Set 1 dialog box and insert the σ1 and the σ3 values given in the problem (here these values are effective stress parameters) as shown in the following figure.

It is to be noted here that the stress parameters in CASTeR are not given any specific demarcation and hence can take both total and effective stress parameters as per the problem.


17

Here a table has been made as shown in the beginning of the solution to show the stress parameters at failure and these values are inserted in the dialog box as shown.

Stress Values (Kg/cm2) Dialog Box for Mohr Circle


• As two sets of values are provided in the problem the values for the other samples are given equal to zero.

• Click OK


18

• Open the Shear Strength→→→→Mohr Circle menu from top main menu bar and open the Other Data dialog box with heading Mohr Circle Other Values -Set 1

• As nothing is mentioned in the problem the default values are accepted.

• Click OK • Select the plot item in the Shear Strength→→→→Mohr Circle menu from the top

menu bar and the plot of the values is obtained as shown.

• Open the Shear Strength→→→→Mohr Circle menu from the top menu bar and select the Draw Tangent→→→→Manual and draw the tangent as per the instruction that comes up on the screen as shown above.


19

• The value of Ø and C from the plot are:

Ø = 30.29° and C = 0


20

Pile Capacity

Determine the safe load of a driven cast-in-situ pile, 40 cm in dia, 15 m long, installed in a cohesive deposit, as shown. The cut off level is 1.5 m below ground level.

Solution: N.B: Press F2 for insertion of numerical values in the given unit system • Open CASTeR. From the Tools menu and pick Clear Screen and refresh the

screen • Open a new .pil file • Select Configure→→→→Pile Capacity tab from the Plot Configuration Dialog box.

• Dimension Positioning Vertical Positioning is made 20 Horizontal Positioning is made 100

• Separate End Bearing Value is left unchecked as no separate end bearing value is given in the problem.

• Road Level is left unchecked as no information pertaining to the road level is mentioned in the problem.

• Location for Pdi Calculation Pdi at Layer Middle is checked since generally pile diameter value is taken from the value at the middle of each layer when no special instructions pertaining to this is mentioned.

The completed Dialog box is shown below.


21

• Press OK and the dialog box is closed.

• Open the Analysis→→→→Pile Capacity from the main menu and select Compression. Insert Pile Dia = 40 cm, Cut-off Level = -1.5 M, Termination Level = -15 M No. of Layers = 3, Max. Depth for Overburden Pressure = 20 x Pile Dia, GWT Level = 0, E.G.L Level = 0, F.O.S = 2.5 Reduction Factor (α) - As per me radio button is clicked Pile type The Driven Cast In-situ radio button is clicked.


22

Press OK and the dialog box is closed. The Layer Details dialog box comes up next and the input is given. Layer 1 Thickness = 2.5 , φ = 0 Average Cohesion = 50kN/m2 α = 0.9 Bulk Density = 1.80 T / m3, δ = 0 Earth Pressure Coeff (K) = 1 Layer 2 Thickness = 10.5, φ = 0 Average Cohesion = 25kN/m2 α = 1.0 Bulk Density = 1.80 T / m3, δ = 0 Earth Pressure Coeff (K) = 1 Layer 3 Thickness = 16, φ = 0 Average Cohesion = 100kN/m2 α = 0.45 Bulk Density = 1.80 T / m3,


23

δ = 0 Earth Pressure Coeff (K) = 1 The completed dialog box only for Layer 1 is shown below

• Open the Pile Capacity from the top menu and the next item Lateral is there. Since this problem has no lateral values the Lateral option box is not considered.

• Open the Pile Capacity from the top menu and select the next item Plot .The output is shown on the screen as given below


24

• Open the Pile Capacity from the top menu and select the next item View Output. The output is shown on the screen as given below.


25

Output from Text File: **Sample Calculation for Reinforced Concrete Cir cular Pile as per IS:2911** Pile Dia : 0.400M Existing Ground level : 0.000M Existing GWT level : 0.000M Pile Cut-Off level : -1.500M Pile Termination level : -15.000M For maximum overburden pressure at Pile tip 20 x di a length of pile has been taken ****ULTIMATE END BEARING CAPACITY**** For Granular Soils Qeg = Ap(0.5 * D * W * Nr + Pd * Nq) where Ap = Cross sectional Area = 0.1256 Sq.M. D = Pile Stem Dia = 0.40 M W = Bulk Unit Wt. of soil at Pile tip = 0.80 T/Cu.M. Nr = Bearing Capacity factor = 0.00 Pd = Effective overburden pressure at pi le tip = 6.40 T/Sq.M. Nq = Bearing Capacity factor = 0.00 Ultimate End Bearing Capacity Qeg =0.126*(0.5*0.400*0.800*0.00+6.400*0.00 )=0.00 T For Cohesive Soils Qec = Ap * Nc * Cp where Ap = As defined above = 0.1256 Sq.M. Nc = Bearing Capacity factor = 9 Cp = Average Cohesion at pile tip = 10.20 T/Sq.M. Ultimate End Bearing Capacity Qec =0.126*10.196*9 = 11.53 T ****Total Ultimate End Bearing Capacity Qu = Qeg + Qec = 11.53 T **** ****Pd Level for this pile= -8.000 M **** Layer No.1 Effective overburden pressure due to this layer =0.000x1.800+2.500x(1.800-1) =2.000T/Sq.M. Layer No.2 Effective overburden pressure due to this layer =5.500x(1.800-1) =4.400T/Sq.M. Layer No.3 No contribution from this layer Total effective overburden pressure up to -8.000M l evel from EGL=6.400T/Sq.M.


26

****ULTIMATE SKIN FRICTION CAPACITY**** For Granular Soils Qsg = Sum[K * Pdi * tan(d) * Asi] for all layers where K = Earth Pressure Coeff. Pdi= Effective Overburden presure for ith layer d = Angle of wall friction for ith layer Asi = Surface area of pile stem for ith layer Negative Skin Friction Qsg(_ve) = Sum[0.5 * K * W * Ln * tan(d) * Asi] for all layers W = Bulk Unit Wt. of soil Ln = Thk. of Compressible layer K = As defined above d = As defined above Asi = As defined above but with Ln For Cohesive Soils Qsc = Sum[a * C * As] for all layers where a = Reduction factor C = Average Cohesion Asi = Surface area of pile stem for ith layer Negative Skin Friction Qsc(_ve) = Sum[S * Asi] f or all layers S = Shear strength Asi = As defined above but with Ln Layer no. 1: K : 1.00 Pdi : 1.00 T/Sq.M. tan(d) : 0.00 Asi : 1.26 Sq.M. Qsg = K*Pdi*tan(d)*Asi =1.00*1.00*0.00*1.26 = 0. 00 T Reduction factor(a) : 0.90 Average Cohesion(C): 5.10 T/Sq.M. Qsc = a*C*Asi =0.90*5.10*1.26 = 5.76 T Total net skin friction of this layer =[Qsg-Qsg(_v e)]+[Qsc-Qsc(_ve)] =[0.00-0.00]+[5.76-0.00]=5.76T/Sq.M. Layer no. 2: K : 1.00 Pdi : 4.20 T/Sq.M. tan(d) : 0.00 Asi : 13.19 Sq.M. Qsg = K*Pdi*tan(d)*Asi =1.00*4.20*0.00*13.19 = 0 .00 T Reduction factor(a) : 1.00 Average Cohesion(C): 2.55 T/Sq.M. Qsc = a*C*Asi =1.00*2.55*13.19 = 33.62 T Total net skin friction of this layer =[Qsg-Qsg(_ve )]+[Qsc-Qsc(_ve)] =[0.00-0.00]+[33.62-0.00]=33.62T/Sq.M. Layer no. 3: K : 1.00 Pdi : 6.40 T/Sq.M. tan(d) : 0.00 Asi : 2.51 Sq.M. Qsg = K*Pdi*tan(d)*Asi =1.00*6.40*0.00*2.51 = 0. 00 T Reduction factor(a) : 0.45 Average Cohesion(C): 10.20 T/Sq.M. Qsc = a*C*Asi =0.45*10.20*2.51 = 11.53 T


27

Total net skin friction of this layer =[Qsg-Qsg(_v e)]+[Qsc-Qsc(_ve)] =[0.00-0.00]+[11.53-0.00]=11.53T/Sq.M. Total Skin Friction Capacity Qus = Qsg + Qsc = 5 0.91 T Total Ultimate Pile Capacity Qu = Qus + Que = 62 .43 T ****Net Safe Pile Capacity Qu/FOS = 24.97 T****


28

Grain Size Example Problem from Principles of Soil Mechanics and Foundation

Engineering By V.N.S. Murthy Fifth Revised Edition Page – 790

Published by: UBS Publishers’ Distributors Ltd. Problem: Observations: The test results obtained from a sample of soil are given below. Mass of soil taken for analysis = 300 gm

Sl. No. IS sieves (mm) % finer 1 4.75 96.22 2 2.00 66.15 3 1.00 51.33 4 0.600 36.91 5 0.425 27.54 6 0.300 22.97 7 0.15 9.33 8 0.075 4.00 9 0.00

N.B: For entering the numerical values in the given units press F2 and there will be automatic unit conversion. Use of the unit conversion dialog box is discussed in the operational manual.

Here in this problem F2 isn’t required as all the values are in the default unit. Solution: • Open CASTeR and from the Tools menu select clear screen. • Select Configure→Grain Size tab in the Plot Configuration Dialog box and

make the proper changes.

1. No. of Curves in a Page is made to 1. 2. Connect Points by St. Line is left unchecked. 3. Connect Points by Smooth Curve is checked. 4. Draw Graph Lines is checked. 5. Write End Co-ord. On Graph is left unchecked. 6. Calculate Cc and Cu is checked. 8. Void Ratio Axis Line is made to 5 & Void Ratio Axis Label is made to 10.

The completed Configure dialog box is shown below.


29

• From the Lab Menu→→→→Grain Size menu from top menu bar click User Defined→→→→% Finer vs. Sieve Size and open the Finer-Size Value – Set 1 dialog box and insert the values in the table of the given problem in the %Finer and size (mm) boxes as shown in the following figure.


30

• As eight sets are available from the problem we have inserted the last set of values in the rest of the rows of the Xcel table.

• Click OK • Open the Grain Size menu from the top menu bar and select User

Defined→→→→Other Data when the dialog box with heading Finer-Size Other Values – Set 1 opens

As nothing is mentioned in the problem the default values are accepted.

• Click OK • Select the User Defined→→→→Plot item in the Grain Size menu from the top menu

bar and the plot of the values is obtained as shown.


31

• From the Plot the following values are obtained. Cu = 9.67 Cc = 0.929


32

Bearing Capacity Problem: A rectangular foundation 2.5 m wide and 3.5 long is to be placed at a depth of 1.7 m below ground level in a thick deposit of firm saturated clay. The water table is at 1.2 m below ground level. Determine the gross and net ultimate bearing capacities The soil parameters are Cu = 65 KN/ sq. m Øu = 0º. Saturated density =21.5 kN/M3 Problem as solved by CASTeR :

• CASTeR is assumed to be open. • Select Clear Screen from the Tools menu to clear the existing graph / plot on

the screen if any. • Select New option from the File menu. From File→→→→New menu create a .brn

file for calculation of bearing capacity in the desired folder. File name chosen will be reflected on the right side of the status bar at bottom.

• Select the Font, Line and Graph from the Tools menu and make proper selections. .

• Select Analysis from the main menu. Select Configure→Bearing Capacity from the top menu and choose the desired configurations as shown. 1. Analysis Method As Per General Shear Failure is checked as the problem considers

General Shear Failure As Per Local Shear Failure is left unchecked, as Local Shear Failure

is not considered here Based Upon Relative Density is left unchecked Based Upon Void Ratio is left unchecked Based Upon Condition is left unchecked 2. User Supplied Nc, Nq , Ng is left unchecked .

The completed Configure dialog box is shown below


33

• Select Analysis→→→→Bearing Capacity→→→→Shear Failure from the top menu and select the box for General Shear Failure. The dialog box for the Shear failure Dialog opens.

1. The dialog box opens with Ton-M as the default unit. 2. The inputs are as follows. : 3. Type of footing = Rectangular 4. From the Size of Footing insert

Depth (Df) = 1.7 m, Length (L) = 3.5 m, Width (B) = 2.5, 5. Cohesion = 65 kN/m2, 6. α = 0, 7. Soil type = clay 8. Above Base of Footing Bulk Density = 21.5 kN/M3 9. Below Base of Footing Bulk Density = 21.5 kN/M3 10. F.O.S = 2.5 11. Under Levels insert


34

E.G.L = 0, G.W.T = -1.2m

The completed General Shear dialog box is shown below.

• Click OK. • Open Bearing Capacity Menu and choose View Output. The entire data of

the problem with the safe bearing capacity is shown in the output. **Sample Calculation for Bearing capacity** Failure Mode - General Shear Footing Type - ISOLATED RECTANGULAR SIZE : 3.50M x 2.50M Depth of Foundation : 1.70M Existing Ground level : 0.00M Ground Water Table level: -1.20M Bulk Density (W)of Soil Above Footing Base : 2.19 T/Cu.M. Submerged Density (W')of Soil Above Footing Base : 1.19 T/Cu.M. Bulk Density (W)of Soil Below Footing Base : 2.19 T/Cu.M. Cohesion C : 6.63 T/Sq.M factor of Safety : 2.50


35

Shape factor Depth Factor Inclination Factor Sc = 1.143 Dc = 1.136 Ic = 1.000 ****For General Shear Failure**** Ultimate Net B.C. = Qult_n = C*Nc*Sc*Dc*Ic Bearing Capacity Factors Nc = 5.14 Thus Qult_n = 44.23 T/Sq.M. Allowable Bearing Capacity (Qall) = 17.69 T/Sq.M.


36

SETTLEMENT ANALYSIS Example Problem from Theory and Practice of Foundation Design By - N.N SOM and S.C. DAS Publisher : Prentice Hall of India Private Ltd. New Delhi - 110001 Page – 159 Problem: A 4m x 4m foundation is placed 1m below G.L in the stratified sandy deposit as depicted in the following figure . Calculate the settlement of the foundation .

N.B In the book the size of the foundation is given as 5m x 5m which is a printing mistake . The problem as solved in the book is with 4m x 4m and hence we consider 4m x4m in solving the problem through CASTeR. Also note that we have considered 5m as the total thickness for the first layer as against 6m as shown in the figure given in the book. Problem as solved by CASTeR Open CASTeR and from the Tools menu select Clear Screen. Select the Font, Line and Graph from the Tools menu and make proper selections.


37

Select Analysis from the main menu. Select Configure→→→→Settlement tab from the Plot Configuration Dialog box . The Dialog box with the Settlement tab put on is shown below .

User’s Influence Factor - Generally CASTeR takes up the Influence Factor from the I.S Code . But if the user has to specify a design value of the Influence Factor , then this option can be used . How to Calculate Settlement- Open a new Settlement Analysis Text File ( * .stl) from File→New in CASTeR . The creation of new file is very important as otherwise the calculations cannot be written in the output. Click on the Analysis→→→→Settlement tab from the top menu bar and from the pull-down menu select Footing Parameters . A dialog box as shown opens and the parameters related to Footing are inserted.


38

Click on the Settlement tab on the top menu bar and from the pull-down menu select Soil Parameters. The Soil Parameter dialog box opens as shown and the soil parameters pertaining to the 3 layers of soil as shown in the problem are entered. The soil parameters as inserted for the first layer of soil is shown.

Layer 1 : Layer Thk.(M) = 5 , Bulk Density of Soil in this Layer ( T/M3)=18 kN/m3 , Immediate Settlement : SCPT Method , Static Cone Resistance(Kg./cm2)= 8000kN/m2 Stress Distribution Method : 2:1 Method P.W.P Corr. Factor = 1 Layer 2 : Layer Thk.(M) = 5 , Bulk Density of Soil in this Layer ( T/M3)= 19 kN/m3 , Immediate Settlement : SCPT Method


39

Static Cone Resistance(Kg./cm2)= 10000kN/m2 , Stress Distribution Method : 2:1 Method P.W.P Corr. Factor = 1 Layer 3 : Layer Thk.(M) = 5 , Bulk Density of Soil in this Layer ( T/M3)= 20 kN/m3 , Immediate Settlement : SCPT Method , Static Cone Resistance(Kg./cm2)= 12000kN/m2 , Stress Distribution Method : 2:1 Method P.W.P Corr. Factor = 1 N.B.-In CASTeR the F2 button is used for unit conversion. If the given values are not in the default unit system, press F2 and the unit conversion dialog box opens which enables the user to insert the values in the given unit system. Click on the Settlement tab on the top menu bar and from the pull-down menu select Calculate . CASTeR will perform the calculations based on the data supplied and it will store it in the (.stl) text file . Click on the Settlement tab on the main menu bar and from the pull-down menu select View Output . The output of the calculations are shown.

** CALCULATION FOR SETTLEMENT ** Footing Size:

Breadth = 4.000 M Length = 4.000 M Depth of footing = 1.000

Existing Ground Level = 0.000 M Ground Water Table Level = -1.000 M


40

Applied Pressure at Footing Base = 10.190 T/Sq.M ****Parameters for Layer no. 1**** Immediate Settlement As per SCPT method. Ref.Cl.#9 .1.2 IS:8009(Pt-I) Stress Increment Method - 2:1 Slope Theory Layer thickness = 5.000 M Saturated Density = 1.835 T/Cu.M Static Cone Penetration Value(Ckd) = 81.572 Kg/S q.Cm ****Parameters for Layer no. 2**** Immediate Settlement As per SCPT method. Ref.Cl.#9 .1.2 IS:8009(Pt-I) Stress Increment Method - 2:1 Slope Theory Layer thickness = 5.000 M Saturated Density = 1.937 T/Cu.M Static Cone Penetration Value(Ckd) = 101.964 Kg/ Sq.Cm ****Parameters for Layer no. 3**** Immediate Settlement As per SCPT method. Ref.Cl.#9 .1.2 IS:8009(Pt-I) Stress Increment Method - 2:1 Slope Theory Layer thickness = 5.000 M Saturated Density = 2.039 T/Cu.M Static Cone Penetration Value(Ckd) = 122.357 Kg/ Sq.Cm ****End of Input Parameters **** ****Increment of Stresses for all layers**** Increment in Stress(2V:1H Slope) at centre of lay er 1= 4.529 T/Sq/M. Increment in Stress(2V:1H Slope) at centre of lay er 2= 1.479 T/Sq/M. Increment in Stress(2V:1H Slope) at centre of lay er 3= 0.679 T/Sq/M. ****Effective Stresses for all all layers**** Effective Stress at centre of layer 1= 3.506 T/Sq/M . Effective Stress at centre of layer 2= 7.520 T/Sq/M . Effective Stress at centre of layer 3= 12.462 T/Sq/ M. ****Layer 1 Settlement**** Immediate Settlement of this layer 1 = 9.507 mm. Pore Water Pressure Correction Factor for layer no. 1 = 1.000 ****Layer 2 Settlement**** Immediate Settlement of this layer 2 = 4.414 mm. Pore Water Pressure Correction Factor for layer no. 2 = 1.000 ****Layer 3 Settlement**** Immediate Settlement of this layer 3 = 1.800 mm.


41

Pore Water Pressure Correction Factor for layer no. 3 = 1.000 Total Settlement of all layers below Footing bas e with P.W.P correction factors= 15.722 mm. Depth Correction factor= 0.970 Rigidity Correction factor= 0.800 **** Total Settlement of all layers below Footing base with depth correction factor= 15.250 mm.**** **** Total Settlement of all layers below Footing base with rigidity factor= 12.200 mm.****


42

NOTE _____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________


43

Section 2 (Verification Problem)


44

Square Root Time Fit

Verification Problem from Soil Mechanics and Foundations By Dr. B.C. Punmia Publisher: Standard Book House 1705 – A, Nai Sarak, Delhi – 6 Fifth Edition: Feb 1979 Page – 360

Problem: TO DETERMINE THE COEFFICIENT OF

CONSOLIDATION(C v) (Method suggested by Taylor) Data obtained from test is as follows. Pressure Range: 1 kg/cm2 to 2 kg/cm2 Initial Height of specimen (under pressure of 1 kg/cm2) = 17.38 mm Final Height of specimen (under pressure of 2 kg /cm2) = 16.11 mm

Elapsed Time t (min) Dial Reading R (10-2 mm)

0 340 0.25 360

1 370 2.25 378

4 386 6.25 394

9 402 12.25 410

16 416 20.25 422

25 426 36 434 49 440 60 445 120 454 180 456 300 459 480 462 1440 467


45

Solution: • Open CASTeR and from the Tools menu select clear screen. • Select Configure menu→Sq. Root Time Fit tab in the Plot

Configuration dialog box. The dialog box with the Sq. Root Time Fit tab put on is shown below.

1. The No. of Plots in a Page is made to 1 as we need to draw only one plot in the page.

2. Least Count of dial gauge is kept to 0.002 mm/div (this value has no implication and hence the default value is accepted.)

3. The Draw Graph Lines box is checked 4. The Write end Co-ord box is left unchecked. 5. The Connect Points by Smooth Curves box is checked. 6. The Connect Points by St. Line box is left unchecked. 7. X axis Line Interval is made 1 Y axis Line Interval is made 0.1 8. X axis Label Interval is 2 Y axis Label Interval is made 0.1 [for proper clarity of

the plot] 9. The Dial Gauge Reading is checked for Increasing with time....


46

10. The Input through Xcel Format option enables the user to put the input in the Xcel table. Here this option is not enabled and the input dialog box in the ordinary format is shown below.

11. Press OK and the dialog box is closed. • Select Tools→→→→Font and choose a proper Text style, colour of the text and its size. • Select Tools→→→→Lines and choose the line style and its colour. • Select Tools→→→→Graph Lines and choose the style and colour of the graph lines. • Click on the Dial Gauge vs. Time option in the Lab tests→→→→Consolidation→→→→.Sq.

Root Time Fit menu and the respective dialog box opens. • The values given in the table of the given problem are inserted in the Sq. Root

Time Fit-Dial Gauge Reading vs. Time – Set 1 table: The values in the Dial Reading R column in the table of the given problem are inserted under the Compression Dial Gauge Reading(R) boxes of the table and correspondingly the Elapsed time in the Time(t) in min boxes as shown below. As only 19 sets of data are provided in the given problem, the rest of the boxes in the dialog box are given input of zero values.



47

• Press OK and the dialog box closes. • Open the Consolidation→→→→Sq. Root Time Fit menu from the top menu bar and

select Other Data. The Other Related Data – 1st Plot dialog box opens as shown below.

As nothing is mentioned about the BoreHole No., Sample No. and the Depth the default values are accepted. But from the given problem the value for the Initial Pressure is taken to be 1 Kg/cm2

and the value for the Final Pressure is taken to be 2 Kg/cm2. For the Initial Height click the box and press F2 and insert the given value of 17.38 mm and in the same way insert the value of 16.11mm in the box for Final Height. The given problem has Double Drainage system and hence the radio button for the Double Drainage is clicked.



48

• Press OK and the dialog box closes. • Next Select Plot in the Consolidation→→→→Sq. Root Time Fit from the top menu

bar and the values are plotted on the screen as shown below.

• Open the Consolidation→→→→Sq. Root Time Fit menu from the top menu bar and select the Extrapolate→ΜΜΜΜanual and extrapolate as per the instruction that comes up on the screen as shown. There is also the guided option to extrapolate but this has not been used here.

• Please note that to compare with the curve given in the book and as shown above you will have to enter only 14 sets of values from the starting i.e. data set for 15th to 19th are to be put as zero.

• The output values are given on the graph and the value of Cv = 0.021310sq. cm2/min

= 3.56 x 10-4 cm2/sec.


49

Verification:

Remarks From Book

3.94 x 10-4 cm2 / sec

From CASTeR

3.56 x 10-4 cm2 / sec

Difference 0.38 10-4 cm2 / sec

This difference is due to lack of expertise in doing the extrapolation, higher the expertise, lower will be the difference.


50

Time Settlement

Verification Problem from Soil Mechanics and Foundations by Dr. B.C. Punmia Publisher: Standard Book House

1705 – A, Nai Sarak, Delhi – 6 Fifth Edition: Feb 1979 Page – 360

Problem: TO DETERMINE THE COEFFICIENT OF

CONSOLIDATION(C v) (Method suggested by Casagrande) Data obtained from test is as follows. Pressure Range: 1 kg/cm2 to 2 kg/cm2 Initial Height of specimen (under pressure of 1 kg/cm2) = 17.38 mm Final Height of specimen (under pressure of 2 kg /cm2) = 16.11 mm

Elapsed Time t (min) Dial Reading R (10-2 mm)

0 340 0.25 360

1 370 2.25 378

4 386 6.25 394

9 402 12.25 410

16 416 20.25 422

25 426 36 434 49 440 60 445 120 454 180 456 300 459 480 462 1440 467

Solution: • Open CASTeR and from the Tools menu select clear screen. • Select Configure→Time–Settlement tab from the Plot Configuration Dialog

box The Dialog box with the Time–Settlement tab put on is shown below.


51

1. The No. of Plots in a Page is made to 1 as we need to draw only one plot in the page.

2. Least Count of Dial Gauge is kept to 0.002 mm/div ( as shown in the table, but this has no implication and hence the default value is accepted. )

3. The Draw Graph Lines box is checked 4. The Write end Co-ord box is left unchecked. 5. The Connect Points By Smooth Curve box is checked. 6. The Connect Points By St. Line box is left unchecked. 7. Dial Gauge axis Line Interval is made 50 8. Dial Gauge axis Label Interval is made 50 [ for proper clarity of the plot ] 9. The Dial Gauge Reading is checked for Increasing with time. 10. The Input through Xcel Format option enables the user to put the input in the Xcel

table but the use of this option is not shown in this problem. Hence this option is not enabled and the input dialog box in the ordinary format is shown below.



52

• Select Tools→→→→Font and choose a proper Text style, colour of the text and its size. • Select Tools→→→→Lines and choose the line style and its colour. • Select Tools→→→→Graph Lines and choose the style and colour of the graph lines. • Click on the Dial Gauge Reading option on Lab Tests→→→→Consolidation→→→→

Time–Settlement from top menu and the respective dialog box opens. • The values given in the table of the given problem are inserted in the Time-

Settlement –Dial Gauge Reading vs. Time – Set 1 dialog box table. The values in the Dial Reading R column in the table of the given problem are inserted under the Compression Dial Gauge Reading (R) boxes of the table and correspondingly the Elapsed time in the Time(t) in min boxes as shown below. As only 19 sets of data are provided in the given problem, the rest of the boxes in the dialog box are given input of zero values.



53


• Press OK and the dialog box closes. • Open the Consolidation→→→→Time–Settlement from the top menu bar and select

Other Data. The Other Related Data – 1st Plot dialog box opens as shown below.

As nothing is mentioned about the BoreHole No., Sample No. and the Depth, default values are accepted. But from the given problem the value for the Initial Pressure is taken to be 1 Kg/cm2

and the value for the Final Pressure is taken to be 2 kg/cm2. For the Initial Height click the box and press F2 and insert the given value of 17.38 mm and click on ACCEPT button and in the same way insert the value of 16.11mm in the box for Final Height.


54

The given problem has Double Drainage system and hence the radio button for the Double Drainage is clicked. • Press OK and the dialog box closes. • Next Select Consolidation→→→→Time–Settlement→→→→ Plot and the values are plotted

on the screen as shown below.

• Open the Consolidation →→→→Time–Settlement from the top menu bar and select the Set Corrected Zero Reading and insert the value 1 min. in the Parabola Starting Pt. Selection-1st Plot dialog box.

• Press OK and the dialog box closes. • Open the Consolidation→→→→Time–Settlement menu from the top menu bar and

select the Draw 1st Tangent→Μ→Μ→Μ→Μanual and draw the first tangent as per the instruction that comes up on the screen as shown. There is also the guided option to draw the tangent but this has not been used here. Here the top tangent is drawn as the first tangent.


55

• Similarly select Consolidation→→→→Time–Settlement→→→→Draw 2nd Tangent→ Manual and draw the tangent at the bottom part of the plotted curve.

• The output values are given on the graph and the value of Cv = 0.0002723 cm2/sec.

Verification:

Remarks From Book 2.71 x 10-4 cm2 / sec

From CASTeR

2.723 x 10-4 cm2 / sec

Difference 0.01 x 10-4 cm2 / sec

This difference is due to lack of expertise in drawing the tangents. Higher the expertise, lower will be the difference.


56

Mohr’s Circle Verification Problem from “ Soil Mechanics: Principles And Practice “

By Graham Barnes Second Edition, Page No. - 214 Worked example - 7.4 Published by -

Problem: TO FIND THE SHEAR PARAMETERS FROM TEST OF TRIAXIAL TEST. The results of a consolidated Undrained triaxial compression test on a sample of fully saturated clay are given below. Each specimen has been consolidated to a back pressure of 200 kN/m2.

Parameter (kN/m2) Specimen 1 Specimen2 Specimen 3 Cell pressure 300 400 600 Pore pressure at failure 146 206 280

Calculate the required stresses and plot them on a Mohr circle diagram to obtain the effective stress shear strength parameters for the clay. Solution: At failure: The parameters are given in the following table

Specimen 1 Specimen 2 Specimen 3 σσσσ3 300 400 600 Uf 146 206 280 σσσσ1- σσσσ3 326 416 635 σσσσ1 626 816 1235 σσσσ`1 480 610 955 σσσσ`3 154 194 320

: N.B - For entering the numerical values in the given units press F2 and there will be automatic unit conversion. Here in this problem F2 is not required as all the values are in the default unit. • Open CASTeR and from the Tools menu select clear screen. • Select Configure→→→→Mohr Circle tab in the Plot Configuration dialog box and

make the proper changes. 1. The No. of Plots in a Page is made to 1 2. Write End Co-ord on Graph is left unchecked 3. Draw Graph Lines is checked 4. Graph Line Interval is 1 5. Graph Label Interval is 1


57

6. Input through Xcel Format is left unchecked. This option enables the user to make inputs in the Xcel table.

7. In Plot Size, Small Plot is checked.


• From the Lab Tests→→→→Shear Strength→→→→Mohr Circle from the top menu bar open the Stress dialog box and insert the σ1 and the σ3 values as shown in the following figure. It is to be noted here that the stress parameters in CASTeR are not given any specific demarcation and hence can be both total or effective stress parameters as per the problem. Here a table has been made as shown in the beginning of the solution to show the stress parameters at failure and these values are inserted in the dialog box as shown.


58



• As three sets of values are provided in the problem the values for the other

samples are given zero. • Click OK • Open the Shear Strength→→→→Mohr Circle from the top menu bar and open the

Other Data dialog box. As nothing is mentioned in the problem the default values are accepted.


59

• Click OK • Select the Plot item in the Shear Strength→→→→Mohr Circle from the top menu bar

and the plot of the values is obtained as shown.

• Open the Shear Strength→→→→Mohr Circle from the top menu bar and select the

Draw Tangent→→→→Manual and draw the tangent as per the instruction that comes up on the screen as shown.


60

• The value of Ø ´ and c´ from the plot are: Ø ´ = 28.76° and c´ = 0.146 kg/cm2 = 14.6 kN/m2

Verification:

Remarks

From Book 28.5° 15 kN/m2

From CASTeR

28.76° 14.6 kN/m2

Difference 0.26 0.4

This difference is due to lack of expertise in drawing the tangent, higher the expertise, the lower the difference.


61

e – logP Curve Verification Problem from “Problems in Soil Mechanics”

By B. P Verma Seventh Edition Page – 82, Problem No. 4.13 Published By: Khanna Publishers, 2-B Nath Market, Nai Sarak, Delhi – 110006 Page – 82

Problem: TO FIND COMPRESSION INDEX Cc In a consolidation test on undisturbed clay specimen, the following results were tabulated.

Pr. Kg / cm2 Void Ratio 0.2 0.953 0.4 0.948 0.8 0.938 1.6 0.92

3.2 0.878 6.4 0.789 12.8 0.691 3.2 0.719 0.8 0.754 0.2 0.791 0.0 0.890

Plot these results and draw the compression curve. Solution: N.B - For entering the numerical values in the given units press F2 and there will

be automatic unit conversion. Here in this problem F2 is not required as all the values are in the default unit.

• Open CASTeR and from the Tools menu select clear screen. • Select Configure→→→→eLogP tab in the Plot Configuration Dialog box and make

the proper changes. 1. The No. of Curves in a Page is made to 1. 2. Connect Points is checked 3. Draw Graph Lines is checked 4. Void Ratio Axis Line Interval is made 0.025 5. Void Ratio Axis Label Interval is made 0.025 6. Dial Gauge Least Count is made 0.001 cm/div 7. Dial Gauge Reading is made Increasing with Pressure. 8. Dry Weight Method of Mv Calculation is checked. The completed Configure dialog box is shown below.


62

• Select Consolidation→→→→e-LogP→→→→e & p Values menu and open the corresponding

Void Ratio & Pressure Value – Set 1 dialog box and insert the e (void ratio) and the p (pressure) values as shown in the following figure.


63

• The initial void ratio e0 is = 0.975 (Since nothing is mentioned about the initial

void ratio we have assumed e0 =0.975 • The final e1 – p1 and the final e2 – p2 values are two sets of void ratio and pressure

values during release of pressure. Here in the problem 4 sets of values during release of pressure are given. But as CASTeR has the provision of accepting only two sets of values during release of pressure, we have selected the following value e1 = 0.719 p1 = 0.2 kg /cm2 & e2 = 0.791 p2= 0.2kg/cm2

• Click OK • The next dialog box Other Data comes either from the Consolidation→→→→e-LogP

from the top main bar or from the menu that comes up by right clicking the mouse. As nothing is mentioned in the problem the default values are accepted.

• Click OK • Select the Plot item in the Consolidation→→→→e-LogP menu and the plot of the

values is obtained • We have a provision in CASTeR of intelligent scaling by which the scale is fixed

according to the values provided by the user. Since the plot obtained by the


64

default scale is not very clear we select the Intelligent Scale from Tools menu(refer to the manual) and a clear picture of the plot is obtained as shown below.

• Select the Consolidation→→→→e-LogP→→→→Draw Tangent for Cc

and draw the tangent at the lower straight part of the curve by reading the CASTeR INSTRUCTION NOTICE as shown below. The final tangent gives the value of Cc = 0.327676 Verification:

Cc Remarks

From Book 0.325 From CASTeR 0.328

Difference 0.00267

This difference is due to lack of expertise in drawing the tangent. Higher the expertise, the

lower will be the difference.


65

Pile Capacity Verification Problem From ”Design Aids in SOIL MECH ANICS AND

FOUNDATION ENGINEERING “ By Shenbaga R Kaniraj Fourth reprint, Page No. – 443, Q 15.2 Published By -Tata McGraw –Hill Publishing Company Limited

Problem: Determine the ultimate load capacity of a driven circular pipe pile. The following are the pile and soil data. D= 45cm, L= 22m, Cu = 0.4 kg/cm2, φu = 0 , γsat= 1.85T/m2. The soil is deep homogeneous clay and the groundwater is at the level of the ground surface. Solution: N.B: Press F2 for insertion of numerical values in the given unit system • Open CASTeR. From the Tools menu and pick Clear Screen and refresh the

screen • Open a new file from File→New with File type *.PIL • Select Configure→Pile Capacity tab in the Plot Configuration Dialog box.

• Dimension Positioning Vertical Positioning is made 20 Horizontal Positioning is made 100

• Separate End Bearing Value is left unchecked as no separate end bearing value is given in the problem.

• Road Level is left unchecked as no information pertaining to the road level is mentioned in the problem, though it has got no implication in the calculation..

• Location for Pdi Calculation Pdi at Layer Middle is checked no special instructions pertaining to this is mentioned.

The completed Dialog box is shown below.


66

Press OK and the dialog box is closed • Open the Analysis→→→→Pile Capacity from the main menu and select Compression.

Insert Pile Dia = 45 cm, Cut-off Level = -1.5 M, Termination Level = -22 M, No. of Layers = 1 , Max. Depth for Overburden Pressure = 25 * Pile Dia, GWT Level = 0, E.G.L Level = 0, F.O.S = 2.5, Reduction Factor α - As per me radio button is clicked since the factors used in the book are to be entered. Pile Type The Driven Cast In-Situ radio button is clicked.


67

Press OK and the dialog box is closed. The Layer Details for Pile dialog box comes up next and the input is given.

Thickness = 25, φ = 0, Average Cohesion = 0.4 kg / cm2, α = 1, Bulk Density = 1.85 T / m3, δ = 0, Earth Pressure Coeff (K) = 1

The completed dialog box is shown below


68

• Open the Pile Capacity from the top menu and select the next item Plot. The output is shown on the screen as given below

Open the Pile Capacity from the top menu and select the next item View Output. The output is shown on the screen as given below Output from Text File:- **Sample Calculation for Reinforced Concrete Cir cular Pile as per IS:2911** Pile Dia : 0.450M Existing Ground level : 0.000M Existing GWT level : 0.000M Pile Cut-Off level : -1.500M Pile Termination level : -22.000M


69

For maximum overburden pressure at Pile tip 14 x di a length of pile has been taken ****ULTIMATE END BEARING CAPACITY**** For Granular Soils Qeg = Ap(0.5 * D * W * Nr + Pd * Nq) where Ap = Cross sectional Area = 0.1590 Sq.M. D = Pile Stem Dia = 0.45 M W = Bulk Unit Wt. of soil at Pile tip = 0.85 T/Cu.M. Nr = Bearing Capacity factor = 0.00 Pd = Effective overburden pressure at pi le tip = 5.36 T/Sq.M. Nq = Bearing Capacity factor = 0.00 Ultimate End Bearing Capacity Qeg =0.159*(0.5*0.450*0.850*0.00+5.355*0.00 )=0.00 T For Cohesive Soils Qec = Ap * Nc * Cp where Ap = As defined above = 0.1590 Sq.M. Nc = Bearing Capacity factor = 9 Cp = Average Cohesion at pile tip = 4.00 T/Sq.M. Ultimate End Bearing Capacity Qec =0.159*4.000*9 =5 .72 T ****Total Ultimate End Bearing Capacity Qu = Qeg + Qec = 5.72 T **** ****Pd Level for this pile= -6.300 M **** Layer No.1 Effective overburden pressure due to this layer =0.000x1.850+6.300x(1.850-1) =5.355T/Sq.M. Total effective overburden pressure up to -6.300M l evel from EGL=5.355T/Sq.M. ****ULTIMATE SKIN FRICTION CAPACITY**** For Granular Soils Qsg = Sum[K * Pdi * tan(d) * Asi] for all layers where K = Earth Pressure Coeff. Pdi= Effective Overburden presure for ith layer d = Angle of wall friction for ith layer Asi = Surface area of pile stem for ith layer Negative Skin Friction Qsg(_ve) = Sum[0.5 * K * W * Ln * tan(d) * Asi] for all layers W = Bulk Unit Wt. of soil Ln = Thk. of Compressible layer K = As defined above d = As defined above Asi = As defined above but with Ln


70

For Cohesive Soils Qsc = Sum[a * C * As] for all layers where a = Reduction factor C = Average Cohesion Asi = Surface area of pile stem for ith layer Negative Skin Friction Qsc(_ve) = Sum[S * Asi] f or all layers S = Shear strength Asi = As defined above but with Ln Layer no. 1: K : 1.00 Pdi : 2.68 T/Sq.M. tan(d) : 0.00 Asi : 31.09 Sq.M. Qsg = K*Pdi*tan(d)*Asi =1.00*2.68*0.00*31.09 = 0.00 T Reduction factor(a) : 1.00 Average Cohesion(C): 4.00 T/Sq.M. Qsc = a*C*Asi =1.00*4.00*31.09 = 124.34 T Total net skin friction of this layer =[Qsg-Qsg(_ve)]+[Qsc-Qsc(_ve)] =[0.00-0.00]+[124.34-0.00]=124.34T/Sq.M. Total Skin Friction Capacity Qus = Qsg + Qsc = 1 24.34 T Total Ultimate Pile Capacity Qu = Qus + Que = 13 0.07 T ****Net Ultimate Pile Capacity Qu/FOS = 52.03 T* ***

Verification: Here L / D = 22 / 0.45 = 48.89 Using alpha method: Cu = 0.4 kg / cm2 ( = 4 T/m2) , is medium clay. The Adhesion factor for driven piles in clay (After Tomlinson, 1970) For piles driven through clay without any overlying strata = 1.0 for stiff clay Therefore for medium clay also alpha = 1.0 is more appropriate. The total skin friction load on the pile Qsu = summation through 0 to L of (perimeter of pile x skin friction resistance) = π x 0.45 x 22 x 1 x 4 = 124 kg/cm2 The ultimate point load of piles into clays is given by Qpu = Undrained cohesion of soil at the pile base x bearing capacity factor x area base

= 4 x 9 x π x 0.45 x 0.45 /4 = 6 T Therefore Qu = 124T + 6 T = 130 T


71

Qu Remarks

From Book 130. T

From CASTeR 130.07 T

Difference 0.07 T

This difference is due to the high precision used in the computation within CASTeR.


72

Grain Size

Verification Problem from Principles of Geotechnical Engineering

By Braja M. Das Fifth Edition Publisher: Bill Stenquist First Reprint 2002 by Thomson Asia Pte. Ltd. Page – 38

Following are the results of a sieve analysis. Make the necessary calculations and draw a particle – size distribution curve.

U.S. sieve Opening ( mm) Percent finer

4 4.75 100 10 2.00 94.5 20 0.850 86.3 40 0.425 74.1 60 0.25 54.9 80 0.18 38.1 100 0.15 9.3 200 0.075 1.7 Pan - 0

solution: .B - For entering the numerical values in the given units press F2 and there will be automatic unit conversion. Use of the unit conversion dialog box is discussed in the operational manual. Here in this problem F2 isn’t required as all the values are in the default unit.

• From the Tools menu select clear screen. • Select Configure→Grain Size tab in the Plot Configuration Dialog box and

make the proper changes.

1. No. of Curves in a Page is made to1 2. Connect Points by Smooth Curve is checked 3. Connect Points by St. Line is left unchecked 4. Draw Graph Lines is checked 5. Write End Co-ord. On Graph is left unchecked 6. Calculate Cc and Cu is checked 7. % Finer axis Line is made to 5. 8. % Finer axis Label is made to 5.


73


• From the Lab Tests→→→→Grain Size→→→→User Defined from the top menu bar select

% finer vs. Sieve Size option and insert the values in the table of the given problem in the %Fine and size (mm) boxes as shown in the following figure.


74

• As eight sets are available from the problem we have inserted the last set of values

in the rest of the rows of the Xcel table. • Click OK • Open the Grain Size→→→→User Defined menu in the main menu bar and open the

other data dialog box. As nothing is mentioned in the problem the default values are accepted.

• Click OK • Select the Plot item in the Grain Size→→→→User Defined from the top menu bar and

the plot of the values is obtained as shown.


75

• From the Plot the following values are obtained Cu = 1.910 Cc = 0.67

Verification:

Cc Cu Remarks

From Book 0.71 1.8

From CASTeR 0.67 1.910

Difference 0.04 0.11

This difference is due to high precision used in the calculations and plotting within CASTeR where as in the book there is a high percentage of approximation in decimal digits in plotting and calculations.


76

Bearing Capacity Verification problem From TheTheory and Practice of Foundation

Design By N.N Som and S.C Das Publisher: Prentice Hall of India Private

Limited ISBN – 81 – 203 – 2190-1 Page- 135

Problem: A rectangular footing with a plan area of 1.4 m x 2 m is to be placed at

a depth of 2 m below the ground surface. The footing would be subjected to a load inclined at 10° To the vertical. The subsoil is clayey, sandy silt with saturated unit weight of 18 kN/M3, and c = 10 kN/M2 and φ= 30º. Assuming the rate of loading is such that drained condition prevails, compute the magnitude of load the footing can carry if the water table is at the base of the footing. Fs = 3

Solution:

• CASTeR is assumed to be open. • Select Clear Screen from the Tools menu to clear the existing graph / plot on

the screen if any. • Select New option from the File menu. In the File→New box create a .brn

file for calculation of bearing capacity in the desired folder. File name chosen will be reflected on the right side of the status bar at bottom of your screen.

• Select Configure→Bearing Capacity from the top menu and choose the desired configurations as shown. 1. Analysis Method

As Per General Shear Failure is checked as the problem considers General Shear Failure

As Per Local Shear Failure is left unchecked, as Local Shear Failure is not considered here

Based Upon Relative Density is left unchecked Based Upon Void Ratio is left unchecked Based Upon Condition is left unchecked

2. User Supplied NC, Nq, Ng is left unchecked. The completed Configure dialog box is shown below


77

2. Select Analysis→→→→Bearing Capacity from the top menu and select Shear Failure. The dialog box for the general shears failure analysis opens. The dialog box opens with Ton_M as the default unit. The inputs are as follows. :

• Type of footing = Rectangular • From the size of footing insert

• Depth (Df) = 2 m, • Length (L) = 2 m, • Width (B) = 1.4,

• Cohesion = 10 kN/m2 • α = 10 • Ø = 30° • Soil type = other (because it is neither clay nor sand but c- Ø soil) • Above Footing Base Bulk Density = 18 kN/M3 • Below Footing Base Bulk Density = 18 kN/M3 • F.O.S = 3 • Under Levels insert

• E.G.L = 0 ,


78

• G.W.T = -2m • Under Layer choose the radio button Single Layer.

The completed Shear Failure Dialog box is shown below.

3. Click OK 4. Open Bearing Capacity Menu and choose View Output. The entire data of the

problem with the safe bearing capacity is shown in the output. **Sample Calculation for Bearing capacity** Failure Mode - General Shear Footing Type - ISOLATED RECTANGULAR SIZE : 2.00M x 1.40M Depth of Foundation : 2.00M Existing Ground level : 0.00M Ground Water Table level: -2.00M Bulk Density (W)of Soil Above Footing Base : 1.84 T/Cu.M. Bulk Density (W)of Soil Below Footing Base : 1.84 T/Cu.M.


79

Cohesion C : 1.02 T/Sq.M Inclination Angle of Load with vertical : 10.00 factor of Safety : 3.00 Shape factor Depth Factor Inclination Factor Sc = 1.140 Dc = 1.495 Ic = 0.790 Sq = 1.140 Dq = 1.247 Iq = 0.790 Sg = 0.720 Dg = 1.247 Ig = 0.444 Water Table Correction Factor w' = 0.50 Effective surcharge at base level q = 3.67 T/Sq.M. ****For General Shear Failure**** Ultimate Net B.C. = Qult_n = C*Nc*Sc*Dc*Ic + q*(Nq- 1)*Sq*Dq*Iq + 0.5*B*W*Ng*Sg*Dg*Ig*W' Angle of Internal Friction(phai) : 30.00 Bearing Capacity Factors Nc = 30.16 Nq = 18.41 Ng = 22.42 Thus Qult_n = 118.99 T/Sq.M. Allowable Bearing Capacity (Qall) = 39.66 T/Sq.M.

Verification:

5. Open Bearing Capacity Menu and choose Range. The respective dialog box

opens. The following inputs are made.

Min Width is made 0.75 Max Width is made 2.5 Step is made 0.5 Min Length is made 0.75 Max Length is made 0.75 Step is made 0.5

The completed Range dialog box is shown below.

(qnet)safe Qult Remarks From Book

380 kN/M2 1166.8 kN/M2

From CASTeR

39.66T/M2

118.99kT/M2

Difference 2.38% 0.042%

This difference is due to high precision used in the calculations within CASTeR where as in the book there is a high percentage of approximation in decimal digits in plotting and calculations.


80

The final output with the bearing capacity calculations as well as the table showing the stresses for the given Range of values is shown. From this table we can adjust the size of the footing to bear safe stress. P.S It is to be noted that the safe load obtained from the Bearing Capacity calculations as done by CASTeR should be checked against safe settlement calculations and only then the final safe load could be decided upon. N.B: The variation in the values with the calculated values is due to the high precision used in CASTeR


81

NOTE _____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________


82

NOTE _____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

Documents

Example & Verification Problem-5.32