Question

(12 marks) (a) In a triaxial test for sand (c) the stress increment (s)-2.5o. What is the angle of internal frictica (o (b) A consolidated undrained (CU) triaxial test was performed on two samples of a clay and the results are: Test No. Cell pressure (kNm) Devilo stress ( km) Pore water pressure (kN) G; 1 150 500 220 (i) The Mohe circles and the failure line in terms effective stresses are drwn as shown in the figure below. List the steps followed and the calculations done in der to draw the circle (No need to draw the circle) From the figure below, find the show strength parameters in terms of effective stresse. Gui) From the figure below and for Test2: What is the neemal stress and shear stress at failure on failure plane lost? (iv) If a third test was performed on an identical sample with o'y-300 kPa, what would be the normal stress and shear stress at failure on failure planet. By the aid of sketch show how can you answer this is the steps to be followed. No need to draw a circle 400 350 300 250 200 (kPa) 150 100 0 50 100 150 200 250 300 350 400 450 500 550 600 700 750 800 ol' (kPa)

Answer 1

The solution for this problem is attached below:

problem cas, Sand C=o. A6= 2.5. ! o Oy +2583= 3.503 Ag Sing= & sing = 07 ooz 6, 603 e out (351) 2:45 (3.5+1) 8 पह = Sin ( (3) 33.94 Yol 16) 1) GL. stress 4.ou & from 100 consolidateds undrained urtarral test: construction of mour's circle: Draw na y co-ordinates of the graph with taking noimal strength on & aris & Shear strength on y axis. * ptol the 4 pore pressue ù measured, Subhract principle plou the wolves of 0;&e and taking 04.05 as centre and 5, 103 as the radres draw the circle joing 5, 168 points. t. Like wire draw morir's circles for all test samples f. Draw a line loethe ordinakee akiocos tangent to all moli's circles. The intersection Line with the ordinate akú is wred to pland gle with slespect to horicarwal angle of obliquity. 2 OP 03.031 this & angle the cohesion shot on motorola Vamsi Krishna K i known

og 150-54 from rest No.1 03 - 150 karan Oto -150+190 lockey 0023 Oz thoning ieb eo artom por qoray SP WOO * wowote lower of A = 190 kN/m U 280 kayman 03 h = 0 .U in OL = 150-80 03 = 10 kN/mr. to ni 340 KUAM tot beesten i seren ser se qui 10 260 test No.2. deallov stresses, efteweise we can draw a cucle for Fli) shear strength parameters (col) in terms of effective to pay zostani(us+ ) a ac'tan(45) 179 for test. I Oz = 0 . U = 150.80 = 70 KN or mote M wb wil Avec riso = 340-60 st my of box story hoilt of to More -80 = 26 O KAYA 0,= 0;' tanus +1) +20'tan (ust ) Vio ot on motorola one macro msi Krishna Kotni

260 = 40.tan(45 + + 2c!tan (454 from test. 2 uso ko 40 - 500 KM 450+500 = 950 kalon U = 220 Kayombo MUP. o to O- 730 kNam YOU 220 950 COMP8) not soroso 18 E': OU 450 - 220 > 130 kerton 01'50, tan" (457) +2e" tar(uspero use ') 750 - 250 tan(use ') + 2e', tan(useg) - . => 470 = 160.tan" ſus e) Hon 45+ B) = 1.7139 45*6' 59.13 = 24.47 Substituting in a 260 = 70.to/1.7139)* + 20'.(1.7459) 3.427 - 54.37 Based on inspection, & c'= 15,861 KN/m, (iii) from the figure it is clearly evident that 0 = 870 KYME 1 a = 225 KNO 225 shot on motor Vamsi Krishna Kot 570

znot, of . ox (iv). og' = 300 kPa (70 ...09: Os'tan' (us t') + 20' tam usa 0,' = 300. tan(45+ # 28.15.867 .tam (45+ 29,147 0 = 935.402 Kolm, Oss o 29.47 2 2 04 C002 030 " Normal Shress 935.402 Kaymo shear stress - cto tang - 15.867 + 935.402. tan (29.40) 8 0.02 ( pow) = 544.44 Kolme py