Find: angle of failure, α Clay CD Test σ3 = 150 [kPa] σ1 = 450 [kPa] c = 50[kPa] Clay 40˚ 45˚ 50˚ D) 55˚ Find the angle of failure, alpha. [pause] In this problem, ---

Find: angle of failure, α Clay CD Test σ3 = 150 [kPa] σ1 = 450 [kPa] c = 50[kPa] Clay 40˚ 45˚ 50˚ D) 55˚ soil data from a tri-axial shear test has been provided, including ---

Find: angle of failure, α Clay CD Test σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] Clay 40˚ 45˚ 50˚ D) 55˚ the confining stress and ---

Find: angle of failure, α Clay CD Test σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 40˚ 45˚ 50˚ D) 55˚ the axial stress.

Find: angle of failure, α Clay CD Test σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 40˚ 45˚ 50˚ D) 55˚ The failure angle, alpha, ---

Find: angle of failure, α Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 40˚ 45˚ 50˚ D) 55˚ is the angle made between the failure plane and the plane normal the the cylinder’s axis.

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] Next we’ll sketch out the axial stress and confining stress values --- σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] and connect them with a mohr’s circle. 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] [pause] Next, the cohesion value, of 50 kilopascals, is plotted--- 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] on the the vertical axis --- 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] and the rupture line, is traced. 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] Now we identify the point of tangency and --- 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] the center of the circle, --- 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] draw a cord between these two points, and 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] define 2 times alpha as as the angle between this cord and the horizontal axis. 2*α 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] The remainder of this problem is simply an exercise in trigonometry. 2*α 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] 2 times alpha is equal to the supplement angle of --- β1 2*α 50 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] Beta 1 plus Beta 2. β1 2*α 50 β2 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 σ3 = 150 [kPa] Clay CD Test α σ3 = 150 [kPa] σ1 = 450 [kPa] σ3 c = 50[kPa] σ1 τ [kPa] 2*α=180˚-β1-β2 [pause] Both Beta 1 and Beta 2 are part of right triangles, --- β1 2*α 50 β2 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 σ3 σ1 50 150 450 2*α=180˚-β1-β2 which will be helpful later. A variable x, --- β1 2*α 50 β2 150 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 x σ3 σ1 50 150 450 2*α=180˚-β1-β2 is defined as the distance from the origin and the center of the semi-circle, --- β1 2*α 50 β2 150 x 450 σ [kPa]

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 r x σ3 σ1 50 150 450 2*α=180˚-β1-β2 And a variable r is defined as the radius of the circle. β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ [kPa] α β1 β2 r x σ3 σ1 x= 2*α=180˚-β1-β2 Where x is half the sum of the major and minor principle stress values, --- β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ1-σ3 σ [kPa] α β1 β2 r x σ3 2*α=180˚-β1-β2 and r is half the difference of the major and minor principle stress values. β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ1-σ3 σ [kPa] α β1 β2 r x σ3 2*α=180˚-β1-β2 Substituting in our stress values, we find that x --- β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ1-σ3 σ [kPa] α β1 β2 r x σ3 2*α=180˚-β1-β2 equals 300 kilopascals, and r --- β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ1-σ3 σ [kPa] α β1 β2 r x σ3 2*α=180˚-β1-β2 equals 150 kilopascals. β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ1+σ3 σ1-σ3 σ [kPa] α β1 β2 r x σ3 c = 50[kPa] σ1 τ [kPa] 2*α=180˚-β1-β2 As was given, we know the cohesion is 50 kilopascals. β1 2*α r 50 β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 r c x σ3 σ1 x=300 [kPa] r=150 [kPa] α c = 50[kPa] σ3 σ1 τ [kPa] 2*α=180˚-β1-β2 Since the smaller right triangle has side length of c and x, --- β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] α β1 β2 r c x σ3 σ1 x=300 [kPa] r=150 [kPa] α c = 50[kPa] σ3 σ1 τ [kPa] 2*α=180˚-β1-β2 we know the length of the hypotenuse is the square root of x squared plus c squared. x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β1 β2 r c x σ3 σ1 x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] σ3 σ1 τ [kPa] 2*α=180˚-β1-β2 That way, Beta 1 can be defined as the arccos of the quantity r divided by the square root of x squared plus c squared. x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β1 β2 r c x σ3 σ1 x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] σ3 σ1 τ [kPa] 2*α=180˚-β1-β2 The values are plugged in and we find Beta 1 --- x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β1 β2 r c x σ3 σ1 x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 σ1 τ [kPa] 2*α=180˚-β1-β2 equals 60.45 degrees. x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 β1 β2 r x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x 2*α=180˚-β1-β2 From the diagram, we notice Beta 2 is equal as the arctan of c divided by x. x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 β1 β2 r x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x 2*α=180˚-β1-β2 c and x are plugged in, and Beta 2 --- x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 β1 β2 r x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x =9.46˚ 2*α=180˚-β1-β2 equals 9.46 degrees. To solve for our angle of failure, alpha, --- x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 β1 β2 r x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x =9.46˚ 2*α=180˚-β1-β2 we plug in Beta 1, --- x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 β1 β2 r x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x =9.46˚ 2*α=180˚-β1-β2 and Beta 2, and alpha --- x2+c2 β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 α=55.05˚ x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x =9.46˚ 2*α=180˚-β1-β2 equals 55.05 degrees. [pause] x2+c2 α=55.05˚ β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 α=55.05˚ x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x 40˚ 45˚ 50˚ D) 55˚ =9.46˚ 2*α=180˚-β1-β2 Looking back at our possible solutions, --- x2+c2 α=55.05˚ β1 2*α r 50 c β2 150 450 σ [kPa] x

τ [kPa] Find: angle of failure, α σ [kPa] β1=cos-1 α β2=tan-1 α=55.05˚ x=300 [kPa] r β1=cos-1 r=150 [kPa] α x2+c2 c = 50[kPa] =60.45˚ σ3 c β2=tan-1 σ1 τ [kPa] x 40˚ 45˚ 50˚ D) 55˚ =9.46˚ 2*α=180˚-β1-β2 the answer is D. x2+c2 α=55.05˚ β1 2*α r AnswerD 50 c β2 150 450 σ [kPa] x

( ) ? τ [lb/ft2] γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d ˚ H*C σfinal Find: σ’v ρc d = 30 feet (1+wc)*γw wc+(1/SG) σ’v = Σ φ γ Δ d ˚ d Sand 10 ft γT=100 [lb/ft3] 100 [lb/ft3] φ=α1-α2 10 [ft] 20 ft Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] τ [lb/ft2] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘ φ c=0 400 1,400 σ3 Sand σ1