Lec-07 by dr irshad ahmad departement .pptx

...

1. 1 CE-313 (2 Credit Hours) Geotechnical Engineering-II Shear Strength of Soils Instructor: Dr Irshad Ahmad Lecture -07 Department of Civil Engineering University of Engineering and Technology, Peshawar

2. 2 Why Shear Strength? Most geotechnical failures involve shear type failure of the soil. This is due to the nature of soil, which is composed of individual soil particles that slide. (i.e. shear past each other) when the soil is loaded. The shear strength of soil is required in many different types of engineering analyses, such as bearing capacity of shallow and deep foundations, slope stability analyses, and design of retaining walls.

3. 3 Bearing Capacity failure in foundations Slope failure Retaining wall failure Sample uses of shear strength

4. 4 Shear failure? Shear strength > (Applied) Shear stress (Safe) Shear Strength = Applied Shear stress (failure occurs) FOS = (Available) shear strength / (Applied) Shear stress

5. 5    =  tan     T = N tan [ T =  N ] [  = tan] Dividing each side by shear Area =  tan   Coulomb’s Theory of Soil Shear Strength (1776) Basic Concept

6. 6 Shear Strength

7. Coulomb’s Theory of Soil Shear Strength (1776) f =c +  f tan  

8. f =c +  f tan   Coulomb’s Theory of Soil Shear Strength (1776)  f  f failure plane c      c       θ  1  f  3  3  1   Failure Plane

9. 9 Modified Coulomb’s Theory f =c ꞌ +  f ꞌ tanꞌ cꞌ & ꞌ effective shear strength parameter   c ꞌ ꞌ     c ꞌ ꞌ    

10. 10 = c ꞌ + ꞌtanꞌ = c ꞌ +( - u)tanꞌ = 113 kPa   30  120 kPa 295 kPa 12 kPa

11. 11 Mohr-Coulomb’s Theory

12. 12 12 ꞌ     C A B Failure Envelope c ꞌ < f   = f   Does not Exist Mohr-Coulomb’s Failure Theory/Failure Envelop Mohr (1990) presented a theory for rupture in materials that contended that a material fails because of a critical combination of normal stress and shearing stress, and not from either maximum normal or shear stress alone. Point A: The state of stress (combination of , ) in the soil mass represented by point A is safe. Point B: The soil is at the verge of failure for combination of ,) represented by point B. Just on the failure envelop. Point C: While state of stress represented by point C is not possible. The soil will fail before the soil attain the state of stress represented by point C. Safe zone f =cꞌ +  f ꞌ tanꞌ  

13. 13 20      A B For stress state A shear strength Available =20 + 100tan(30)=77.7 kPa, shear stress applied =77.7 kPa, FOS = 77.7/77.7 = 1 For stress state B =20 + 200 tan(30) = 135.5 kPa Shear stress applied=100 kPa FOS = 135.5/100=1.35 This implies that it is not the max Shear stress which causes failure rather it is a unique combination of  and  which cause failure.   77.7 kPa 100 kPa Find factor of safety against shear failure for two stress states: A( =100 kPa, =77.7kPa), B(200 kPa, 100 kPa). The soil shear strength parameters are c=20 kPa, =30. 135.5 kPa 200 kPa 100 kPa Example

14. 14 Mohr-Coulomb’s Failure Theory A soil sample is subjected to lateral stress  3 and vertical stress of  1 . The horizontal and vertical planes on which  1 and  3 act are called the Principal Planes because no shear stresses are applied on these planes. The  1 and  3 are the Principal stresses.  1 and  3 are respectively the major principal stress and minor principal stress. The failure occurs on the plane shown in the figure at an angle  with horizontal. The stresses acting on the failure plane are  f , and  f . Let us draw the Mohr circle on the next slide. failure plane  ꞌ 1 = Major Principle stress  ꞌ 3 = Minor Principle stress  ꞌ 1  ꞌ 3 Major Principle Plane Minor Principle Plane

15. 15 Mohr-Coulomb’s Failure Theory Mohr Circle representation has the advantage that the circle represents stress states at any orientation of the plane. You can locate the failure plane orientated at  (2  on Mohr circle). The stressed there are represented by coordinate defined by  f ,  f. The shear strength Parameters for the soil sample are c and . Now let us draw the failure envelope. The line defined by  f = c+ f tan should must touch the Mohr circle tangentially at coordinate defined by  f ,  f Not Possible

16. 16 Mohr-Coulomb’s Failure Theory All stress states corresponding to different orientations of planes on the Mohr circle are safe except for the stress state defined by  f ,  f on a plane defined by  where failure occurs.

17. 17 Mohr-Coulomb’s Failure Theory- In terms of Mohr Circles             Failure envelop Safe Safe Fail Dose not exist  

18. 18 Mohr-Coulomb’s Failure Theory- In terms of Mohr Circles Note that failure envelop may be straight or slightly curved as in the case shown in figure.

19. 19 “Top half” of Mohr Circle

20. 20 Relation between Shear strength Parameters (c  ,  ) and Effective Principal stresses (1 , 3) Note : The effective stresses at failure can only be determined when Mohr circle touches tangentially the Failure envelope.

21. 21 F C B A 2θ <BCF =180 - 2 θ  ꞌ 3  ꞌ 1  ꞌ f f   ½(  ꞌ 1 +  ꞌ 3 ) ½(  ꞌ 1 -  ꞌ 3 ) ½(  ꞌ 1 -  ꞌ 3 )cos(180-2 θ ) Considering triangle ∆ FCB f = ½(  ꞌ 1 -  ꞌ 3 ) sin(180 - 2 θ ) f = ½(  ꞌ 1 -  ꞌ 3 ) sin2 θ  ꞌ f = AC – BC  ꞌ f = ½(  ꞌ 1 +  ꞌ 3 ) - ½(  ꞌ 1 -  ꞌ 3 )cos(180-2 θ )  ꞌ f = ½(  ꞌ 1 +  ꞌ 3 )+ ½(  ꞌ 1 -  ꞌ 3 )cos2 θ    ꞌ c ꞌ  ꞌ   sin2 θ -cos2 θ  Considering triangle ∆FCB +90+(180-2 θ ) = 180 θ=45+/2

22. 22 ½(  ꞌ 1 +  ꞌ 3 ) c ꞌ cot ꞌ  ꞌ F C B A 2θ <C =180 - 2 θ  ꞌ 3  ꞌ 1  ꞌ f f   ½(  ꞌ 1 -  ꞌ 3 )  ꞌ c ꞌ   O  ꞌ Considering tringle ∆FOC sin  ꞌ = = sin  ꞌ + 2cꞌcosꞌ = ( = tan 2 (45+ + 2c ꞌ tan(45+   From trigonometry ∆FOC  ꞌ + (180 - 2 θ ) + 90 =180 θ = 45 + ꞌ /2 = tan 2 (45+  

23. 23 θ  1  f  3  3  1   Failure Plane Minor Principle Stress Major Principle Stress Major Principle Plane Minor Principle Plane