τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] Clay 30 45 60 D) 75 CD test Find the cohesion, c, in pounds per square feet. [pause] In this problem, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] Clay 30 45 60 D) 75 CD test we are provided the major and minor principle stress values at failure, for two triaxial tests, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] Clay 30 45 60 D) 75 CD test And we know that the soil is a clay, in the consolidated, drained condition. σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] Clay 30 45 60 D) 75 CD test In problems like this one, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 the stress values are plotted, a semi circle is drawn, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 the cohesion is identified, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 And the rupture line is traced. The rupture line passes through the cohesion point and is tangent to the semicircle. σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 φ Also, the friction angle can be determined from the slope of the rupture line. σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 φ However, we don’t always know the cohesion, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 and may need data from a second test. σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 In this case, the rupture line is a line which is tangent to both semicircles, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 and the cohesion, c is determined last. c σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 Also, for a multiple choice problem, --- c σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] 30 45 60 D) 75 it can be helpful to consider the possible solutions while solving the problem. We’ll see how this works as we solve this problem. c σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] Back to our problem, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] For the first test, the confining stress is plotted, --- σ [lb/ft2] 400

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] and the axial stress is plotted, --- σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] and the two are connected with a semicircle. σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] For now, let’s assume the cohesion is zero and draw out our triangle using points from the origin, σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] center of the semicircle, σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] and point of tangency. σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] --- σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] The friction angle is identified, --- φ σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] and can be computed using --- c=0 φ σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 the arcsine of the radius of the semicircle, --- c=0 φ σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 divided by the distance to the center of the semicircle, --- c=0 φ σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 φ=sin-1 r x represented by the letters, --- c=0 φ σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] x σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 φ=sin-1 r x x [pause] x=circle center c=0 φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r x σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 φ=sin-1 r x and r. [pause] r x=circle center c=0 r=circle radius φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r x σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] φ=sin-1 σ1-σ3 σ1+σ3 only valid for c=0 φ=sin-1 r x However these equation for the friction angle only apply when the cohesion is zero. r x=circle center c=0 r=circle radius φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r c x φ only valid for c=0 r φ=sin-1 r x Considering the latter equation, we note that this equation is --- c>0 r x=circle center r=circle radius c φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r c x φ only valid for c=0 r Specific case of general equation only valid for c=0 φ=sin-1 r x a specific case of a larger, more general equation --- c>0 r x=circle center r=circle radius c φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r c x φ r c φ=sin-1 -tan-1 x general φ=sin-1 -tan-1 x equation x2+c2 τ [lb/ft2] Specific case of general equation only valid for c=0 φ=sin-1 r x shown here. [pause] If we shift our triangle, such that the cohesion is greater than zero, --- c>0 r x=circle center r=circle radius c φ x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r c x r c φ=sin-1 -tan-1 x general φ=sin-1 -tan-1 x equation x2+c2 τ [lb/ft2] We can calculate the friction angle as the difference between --- r c x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] r c x φ=α1-α2 r c φ=sin-1 general φ=sin-1 -tan-1 x equation x2+c2 τ [lb/ft2] φ=α1-α2 two other angles, --- r c x σ [lb/ft2] 400 2,000

τ [lb/ft2] α1 σ [lb/ft2] Find: c in [lb/ft2] α1 α2 r c x φ=α1-α2 r c general φ=sin-1 -tan-1 x equation x2+c2 τ [lb/ft2] α1 φ=α1-α2 alpha 1, α1 r α2 c x σ [lb/ft2] 400 2,000

τ [lb/ft2] α2 α1 σ [lb/ft2] Find: c in [lb/ft2] α1 α2 r c x φ=α1-α2 r general φ=sin-1 -tan-1 x equation x2+c2 α2 τ [lb/ft2] α1 φ=α1-α2 and alpha 2. α1 r α2 c x σ [lb/ft2] 400 2,000

τ [lb/ft2] α2 α1 σ [lb/ft2] Find: c in [lb/ft2] α1 α2 r c x φ=α1-α2 r general φ=sin-1 -tan-1 x equation x2+c2 α2 τ [lb/ft2] α1 φ=α1-α2 So we return to our given data. α1 r α2 c x σ [lb/ft2] 400 2,000

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] We plot the the data --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: c in [lb/ft2] σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] from the first test, plot the data --- σ [lb/ft2] 400 2,000

τ [lb/ft2] Find: c in [lb/ft2] σ σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] from the second test, --- σ 400 600 2,000 2,900

τ [lb/ft2] Find: c in [lb/ft2] σ σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] sketch out the rupture line,--- σ 400 600 2,000 2,900

τ [lb/ft2] ? Find: c in [lb/ft2] σ σ1 = 2,000 [lb/ft2] Test #1 Test #2 σ1 = 2,000 [lb/ft2] σ1 = 2,900 [lb/ft2] σ3 = 400 [lb/ft2] σ3 = 600 [lb/ft2] τ [lb/ft2] and identify the cohesion, even though we don’t know it’s value yet. ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 400 600 2,000 2,900 Since the slope of a line is constant, we can solve this problem by --- test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 test 2 test 1 c 400 600 2,000 defining and equating a friction angle 1 with a friction angle 2. test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 Substituting in our general equation, --- test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 we now have an equation entirely in terms of the radius of each semicircle, the distance to the center of each semicircle, and the cohesion. test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 Keeping in mind the units of r, x and c are in pounds per square feet, the value for r1 is 800, --- test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 the value for x1 is 1,200, --- test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 The value for r2 is 1,150, --- test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 and the value for x2 is 1,750. test 1 ? c σ 400 600 2,000 2,900

τ ? Find: c in [lb/ft2] σ φ1 = φ2 φ2 φ1 r c φ=sin-1 -tan-1 x x2+c2 test 2 The value of c, the cohesion, is found four times in the equation. Rather than trying to solve for the cohesion directly, it would be easier --- test 1 ? c σ 400 600 2,000 2,900

τ φ1 = φ2 ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 400 600 2,000 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 test 2 to remember the left hand side and right hand side of the equation are equivalent, --- test 1 ? c σ 400 600 2,000 2,900

τ φ1 = φ2 ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 400 600 2,000 r r c c 1 -tan-1 2 sin-1 = sin-1 -tan-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 c φ1 φ2 test 2 And it may be easier to search for the value of c by guessing a value, --- test 1 ? c σ 400 600 2,000 2,900

τ φ1 = φ2 ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 400 600 2,000 r r c c 1 -tan-1 2 sin-1 = sin-1 -tan-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 c φ1 φ2 test 2 such as zero, and solving for--- test 1 ? c σ 400 600 2,000 2,900

τ φ1 = φ2 ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 41.81˚ 400 600 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 c φ1 φ2 test 2 friction angle 1, --- test 1 41.81˚ ? c σ 400 600 2,000 2,900

τ φ1 = φ2 ? Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 c 41.81˚ 41.08˚ r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 c φ1 φ2 test 2 and friction angle 2. test 1 41.81˚ 41.08˚ ? c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 test 2 test 1 > c 41.81˚ r r c c 1 -tan-1 2 sin-1 = sin-1 -tan-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ φ1 c φ1 φ2 test 2 In this case, we note friction angle 1 is larger. This mean our next value of cohesion we evaluate should be larger than 0. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 Now is where we turn to our possible solutions. And the question arises, what value should we next try for our cohesion? Should it be --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 30 pounds per square feet, --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 45 pounds per square feet, --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 60, --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 or 75? This is where test-taking strategy comes into play. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 30 pounds per square feet should not be the next value chosen for c, because if phi 1 or friction angle 1, does not equal phi 2, or friction angle 2, we know phi 1 will be greater than phi 2, as it was when we set c equal to zero, and the search direction will again be for a larger value of c. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 So by choosing c equal to 30, there is a 75% chance we’ll have to evaluate yet another value for the cohesion. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 Same is true for if we first try c equal to 75 pounds per square feet, --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 there will be a 75% chance we won’t be done with the problem. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 But if we choose an intermediate value such as c equal to 45 pounds per square feet, --- > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 or 60 pounds per square feet, there will be a 50% chance we won’t have to evaluate the expression an additional time. > test 1 41.81˚ 41.08˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 So if we evaluate the friction angles for a cohesion of 60 pounds per square feet, --- > test 1 41.81˚ 41.08˚ 60 c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 sin-1 -tan-1 2 = sin-1 -tan-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 phi 1 equals 38.89 degrees, > test 1 41.81˚ 41.08˚ 60 38.89˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 sin-1 -tan-1 2 = sin-1 -tan-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 and phi 2 equals 39.09 degrees. > test 1 41.81˚ 41.08˚ 60 38.89˚ 39.09˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 -tan-1 = sin-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 Since phi 2 is larger, we know to use a smaller value for cohesion. Unfortunately, being particular about which of our four possible solution to first evaluate did not pay off in this instance, but it is still a good habit to get into, especially if the equations are messy, and time is of the essence. > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 We’ll next try a cohesion value of 45 pounds per square feet --- > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c 45 σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 -tan-1 = sin-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 phi 1 equals 39.62 degrees, --- > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c 45 39.62˚ σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 -tan-1 = sin-1 x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 and phi 2 equals 39.60 degrees. > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c 45 39.62˚ 39.60˚ σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 Since these two values are virtually the same, > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c ~ 45 39.62˚ ~ 39.60˚ σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ φ2 φ1 30 test 2 45 60 test 1 D) 75 r r c c 1 -tan-1 2 sin-1 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 test 2 we know the cohesion is 45 pounds per square feet, --- > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c ~ 45 39.62˚ ~ 39.60˚ σ 400 600 2,000 2,900

τ φ1 = φ2 Find: c in [lb/ft2] σ AnswerB φ2 φ1 30 test 2 45 60 test 1 sin-1 -tan-1 2 sin-1 -tan-1 = x x x2+c2 x2+c2 1 2 1 2 r2=1,150 r1=800 x1=1,200 x2=1,750 φ2 τ 30 45 60 D) 75 φ1 c φ1 φ2 AnswerB test 2 and the answer is B. > test 1 41.81˚ 41.08˚ < 60 38.89˚ 39.09˚ c ~ 45 39.62˚ ~ 39.60˚ σ 400 600 2,000 2,900

( ) ? 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] 10 [ft] 20 ft φ=α1-α2 Clay = γsand dsand +γclay dclay A W S V [ft3] W [lb] 40 ft text wc = 37% ? Δh 20 [ft] (5 [cm])2 * π/4 ( ) H*C σfinal ρcn= 1+e σinitial log ‘ σ3 Sand τ [lb/ft2] σ1 φ c=0 400 1,400