Question

12.4 During a subsoil exploration program, undisturbed normally consolidated silty clay samples were collected in Shelby tubes. Following are the results of four drained, direct shear tests conducted on the clay samples with each having a diameter of 63.5 mm and height of 32 mm. Test no. 1 Normal force (N) 84 168 254 360 Shear force at failure (N) 28.9 59.6 89.1 125 a) Graph shear stress at failure vs. normal stress (10 points) b) Determine the drained angle of friction for the silty clay soil by using the line best-fit the shear stress at failure vs normal stress data. (10 points) Note: You may determine the drained angle of friction by using only one test results at a time and then report the average value. This method will lead to large errors sometimes. Best-fit method will effectively utilize all the data points and eliminate possible errors. c) If a square concrete foundation of 2 m size is designed with this clay as the foundation soil, how much horizontal force is allowed for this foundation for a factor of safety of 1.5? (10 points, hint: use Mohr-Coulomb failure criteria)

Answer 1

Normal Stress = Normal Force/ Cross sectional area

Shear Stress = Shear Force / Cross sectional area

From the tabular column, a graph of Normal stress on x-axis verses Shear stress on y-axis is plotted as shown in figure 1.

Horizontal Force resisted by square concrete foundation be "F_H".

F_H = Shear stress x Area of foundation

Shear stress depends on Normal stress and Normal stress depends on Normal load.

Since the normal load on the foundation is not given, let us say Normal load on foundation be N

From Mohr-coulomb failure criteria, we have

Shear stress = Normal stress x tan(phi) = N x 0.34810

F_H = (0.34810N) x (2m x 2m)

F_H = 1.3924N units

Since a factor of safety of 1.5 is given, then safe horizontal load allowed for the above foundation is (F_H)_Safe

(F_H)_Safe = 1.3924N/ 1.5

(F_H)_Safe = 0.92827N units.

les Normal Force (N) Normal Stress (N/mm²) Shear Force at failure 1 Shear Stress I at failure N/mm²) I-dmx 84N Ex 63.5² 84 0:02652 28.9 0.00913 2 168 0.05305 | 59.6 0.01882 3 254 0.08020 89.1 0.02813 4 360 | 0.11368 | 125 1 0.03947 Normal Stress (6)= Normal Force = cross sectional area 6= 0.02652 N/mm2 62 = 168 -> 6 = 0.05305 N/mm2 IX6 3.5" 633 254 + 65 = 0.0802 /mont 64 = 360 -7 62 = 0.11368 N/mm. Ix63:5² (Page 1 254 )-( 17 x 63.5²

4 y-axes Page No. Along x-axis, 1cm= 0.01 N/mm² Along y-axis, 1cm= 0.03 N/mm² + 1-0-0457 0.0427 0.039 0.0367 0.033 10-030 Ø = Drained angle of friction of silty day soil. Shear Stress (N/mm2) 0.027 0.024 0-021 6 g = 19 193°- 19° 11 35 0-018 -0-015+ 0.0121 0.009 0.0067 16.01 02 03 00 0.05 0.06 007 008 0.09 0.10 011 012 013 0.14 0.15 Normat Stress (N/mm2) ► figure-1 26- dous

a Shear Stress (Z) = Shear Force Cross sectional area 28.9. I x 63.5² TIE 0.00913 N/mm² Tz- 59.6 = 72 = 0.01882 N/mm? Ix635 Tz = 89.1 => 1T3 = 0·02813 N/mm2 1 x 63.52 T4 = 125 124 = 0.03947 N/mm² / x 63.5² Let us plot Normal stress on x-axis and Shear stress on y-axis as shown in figure (1 From the graph, we get tano = (0-03947 – 0-00913) – Dice 0-11368 – 0·02652) tan g = 0 3481020 - 49-193 Drained angle of friction Ø = 19.193° of silty day soil = Page 2

Hosigmtal force sesisted by square foundation concerte) f (2m x 2m) size be FH FH= Shear Stress) x Area of foundation Shear stress depends on normal stress and normal stress depends on normal load. Since normal load on the foundation is not given, let us say normal load on foundation = N from Mohr-Coulomb failure criteria, We have E = tan Ø Z= (tame)- N= 0.348100N 12=0.34810N units : FH = (0.34810N) x (2m x2m) [TH= 1.3924 N units Since a factor of safety of 1.5, is given Safe horisontal force allowed for above foundation = (FH) safe (Filsafe = 1.3924N units 1.5 T (FH) safe = 0.92827N writs / Page 3