Question

04:- A sand sample is tested in a direct shear device. The vertical normal stress on the sample is 300 kN/m². The horizontal shear stress at failure is equal to 210 kN/m² a. Compute the angle of internal friction for the sand and the magnitudes of the principal stresses at failure. b. During a consolidated drained triaxial test, a sample of the same sand failed at a principal stress difference of 130 kN/m². Determine the magnitudes of the principal stresses at failure.
subject / soil mechanics

i want the solution as soon as possible

Answer 1

Solution Given :- verticle normal stress - 300 kN/m² Horizontal shear stress = 210 kn/m2 In direct shear test, As we know that, THE ct In tand equo on where, = Horizontal shear stress normal stress angle of internal friction cohesion of soll (for sand, c=0) Here, test is conducted on sand, socc=o) substitute the value en above equation C = 210 0 + 300,tand tand 210 300 34.382 So, angle of internal friction is 34.9920

6 principle stress at failure, a at failure 45't $ 2 Where ø ir angle of internal frictim 62 a = 45°+ 34,992 2 to, 62.4969 As we know that, 5t83 on Cosaa where, 6, ) principle stress - normal stress, on 6, +63 300 + 2 (-3) cos (2x 62,436) ( 6-3)* (-0,5725) 300 6+3 + 2 Ans, T where, てこ (5-63) sinza e= horizontal shear stress, Coming sin (2x62.496" (5-3) = 256,338 210 3

substitute the value of eqr ③ in ear ② (6,+)+(-_^)*(-0,5725) 300 300 = (8+62) - 0.5735 X 256.338 6+ 3 = 894.0 equ 6 As we know that, 6 = 62 tan² (45+ 2) +20 tan (45+ 2) put the value in above equation 61 - 83 ton? (45+ 34,992) 6,5 3,6889 62 substitute the value of eqh ③ en eqh @ 6, + = 894,0 3,68 & 9 3 + 3 = 894.0 894,0 4.688962 63 = 190.663 kN/m2 Ans and By ear @ Ans = 703.337KN/m²

(6) liven, principle stress difference = 130 KN 1m2 6- 63 = 130 kN/m2 eq" As we know that from ean o 6 = 3.6889 2 6 substitute the value in ean 8-83=130 3,6889 83-63=130 2.688 9 63 2 2=70,661 KNim' - ANS and from ean 6 ans 6 = 200.661 kN/m² please give positive rating !