Find: φ σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand 34˚ 36˚ 38˚

Find: φ σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand 34˚ 36˚ 38˚
Find the angle of internal friction. [pause] In this problem, ---

a sand sample was tested in a consolidated drained triaxial test, ---

and we’ve been given the confining stress and the deviator stress, at failure.

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 = 400 [lb/ft2] CD test
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] In problems like this one, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 = 400 [lb/ft2] CD test σ3 σ1
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] confining stress the confining stress, sigma 3, and the axial stress, sigma 1 --- axial stress σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] confining stress are plotted on a horizontal axis, and connected by a semi circle. axial stress σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] A rupture line is then traced tangent to that semicircle, --- σ [lb/ft2] σ3 σ1

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 = 400 [lb/ft2] CD test φ σ3 σ1
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] and the angle of internal friction is determined, by measuring the angle between that rupture line, and a horizontal line. This same angle is sometimes called the fiction angle. For this example, φ σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] the point where the rupture line and semicircle intersect was arbitrarily chosen, however, --- φ σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] any one of these points could have been chosen, --- φ σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] resulting in an arbitrary rupture line --- σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] φ5 φ4 φ3 and an arbitrary friction angle. φ2 φ1 σ [lb/ft2] σ3 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] φ5 φ4 φ3 However, if we had data from a second test, --- φ2 φ1 σ [lb/ft2] σ3 σ1

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 = 400 [lb/ft2] CD test σ3 σ3 σ1 σ1
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] φ5 φ4 φ3 we could plot it on the same graph, --- φ2 φ1 σ [lb/ft2] σ3 σ3 σ1 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] φ5 φ4 φ3 draw the rupture line, and be certain of it’s slope, and hence, it’s friction angle. φ2 φ1 σ [lb/ft2] σ3 σ3 σ1 σ1

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] τ [lb/ft2] φ5 φ4 φ3 Also, the cohesion of the soil would be known as the shear stress when there is no confining stress. φ2 φ1 c σ [lb/ft2] σ3 σ3 σ1 σ1

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 = 400 [lb/ft2] CD test
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] Returning to our problem, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 σ3 σ3 = 400 [lb/ft2] CD test
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] we have a confining stress at failure of 400 pounds per square feet, --- σ [lb/ft2]

τ [lb/ft2] σ [lb/ft2] Find: φ σ3 σ3 σ3 = 400 [lb/ft2] CD test σ3=400
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] so sigma 3 is added to the plot. σ [lb/ft2] σ3=400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ The axial stress at failure is the confining stress at failure plus the deviator stress at failure. σ [lb/ft2] σ3=400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ The values are plugged in, and --- σ [lb/ft2] σ3=400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] our axial stress at failure is 1,400 pounds per square feet, --- σ [lb/ft2] σ3=400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 σ3 = 400 [lb/ft2] CD test
Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 and added to the plot. σ [lb/ft2] σ3=400 σ1=1,400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 Next we draw our semicircle, however we don’t have a second set of test data to help determine the point of tangency, --- σ [lb/ft2] 400 1,400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 But we do know the soil is a sand in a consolidated drained condition, which means --- σ [lb/ft2] 400 1,400

Sand σ3 = 400 [lb/ft2] CD test Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 the cohesion is zero, and the rupture line passes through the origin of the graph. σ [lb/ft2] c=0 400 1,400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 The rupture line is plotted, --- σ [lb/ft2] c=0 400 1,400

Sand σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand τ [lb/ft2] σ1 = σ3 + Δσ = 1,400 [lb/ft2] σ1 φ and the angle of internal friction is identified. σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 φ 400 1,400 Sand c=0
For a cohesionless soil, --- σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ1 1+sin(φ) = σ3 1-sin(φ) σ3 σ3
Sand for c=0 only CD test Sand τ [lb/ft2] σ1 φ we can solve for the angle of internal friction using this equation σ [lb/ft2] c=0 400 1,400

Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 φ Where the friction angle is the arc sine of the quotient of sigma 1 minus sigma 3, divided by sigma 1 plus sigma 3. σ [lb/ft2] c=0 400 1,400

Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 φ 1,400 Plugging in 1,400 pounds per square feet in for the axial stress, --- σ [lb/ft2] c=0 400 1,400

Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 400 φ 1,400 and 400 pounds per square feet for the confining stress, --- σ [lb/ft2] c=0 400 1,400

Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 400 φ 1,400 φ=33.75˚ the friction angle computes to degrees. σ [lb/ft2] c=0 400 1,400

34˚ 36˚ 38˚ D) 40˚ Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 400 φ 1,400 φ=33.75˚ Looking back at our possible solutions, --- σ [lb/ft2] c=0 400 1,400

34˚ 36˚ 38˚ D) 40˚ Sand for c=0 only CD test Sand σ1-σ3 φ=sin-1 τ [lb/ft2] σ1+σ3 σ1 400 φ 1,400 φ=33.75˚ the answer is A. [pause] This equation for the friction angle can be derived --- Answer: A σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 Sand Sand
with the use of trigonometry. σ [lb/ft2]

If we draw out our semicircle and rupture line as before, --- σ [lb/ft2] c=0 400 1,400

And we add a point at the center of the circle, --- σ [lb/ft2] c=0 400 1,400

And a third point and the tangent, --- σ [lb/ft2] c=0 400 1,400

we can draw out the a right triangle. σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 φ 400 1,400 Sand φ
The friction angle is this interior angle of the triangle. φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 φ 400 1,400 Sand φ
We make a copy of this triangle and --- φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ x σ1 σ3 σ3 CD test σ1 φ 400 1,400 Sand φ
we define x as the --- φ x φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ x σ1 σ3 σ3 CD test σ1 φ 400 1,400
x=circle center σ3 σ3 Sand CD test Sand τ [lb/ft2] σ1 φ distance from the origin to the center of the semicircle, --- φ x φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ x σ1 σ1+σ3 σ3 σ3 CD test σ1 φ 400 1,400
x=circle center σ1+σ3 σ3 σ3 Sand = 2 CD test Sand τ [lb/ft2] σ1 φ which is average of the confining and axial stresses. φ x φ σ [lb/ft2] c=0 400 1,400

x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 CD test Sand τ [lb/ft2] σ1 φ Plugging in 400 pounds per square feet for the confining stress, --- φ x φ σ [lb/ft2] c=0 400 1,400

x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand τ [lb/ft2] σ1 φ and 1,400 pounds per square feet for the axial stress, --- φ x φ σ [lb/ft2] c=0 400 1,400

x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] σ1 φ our value of x is 900 pounds per square feet. φ x φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ1+σ3 σ3 σ3 CD test σ1 φ 400 1,400
x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] σ1 φ --- φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ r σ1 σ1+σ3 σ3 σ3 CD test σ1 φ 400 1,400
x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] r=circle radius σ1 φ r Next we define the radius of the semi circle as r. φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ r σ1 σ1+σ3 σ3 σ3 CD test σ1-σ3 σ1 400
x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] r=circle radius σ1-σ3 σ1 = 2 r Which equals the axial stress minus the confining stress, all divided by 2. φ 900 φ σ [lb/ft2] c=0 400 1,400

x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] r=circle radius σ1-σ3 σ1 1,400[lb/ft2] = 2 400[lb/ft2] r After plugging in the values like before, we learn the --- φ 900 φ σ [lb/ft2] c=0 400 1,400

x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] r=circle radius σ1-σ3 σ1 1,400[lb/ft2] = 2 400[lb/ft2] r radius is 500 pounds per square feet. =500[lb/ft2] φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ1+σ3 σ3 σ3 CD test σ1-σ3 σ1 400
x=circle center σ1+σ3 σ3 σ3 Sand = 400[lb/ft2] 2 1,400[lb/ft2] CD test Sand = 900 [lb/ft2] τ [lb/ft2] r=circle radius σ1-σ3 σ1 1,400[lb/ft2] = 2 400[lb/ft2] 500 --- =500[lb/ft2] φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 φ 400 1,400 Sand 500
Keeping in mind this is a right triangle, --- φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ σ1 σ3 σ3 CD test σ1 φ 400 1,400 r
φ=sin-1 r x σ3 σ3 Sand CD test Sand τ [lb/ft2] σ1 φ 500 the friction angle is simply the arc sine of the opposite leg of the triangle divided by the hypotenuse of the triangle. φ 900 φ σ [lb/ft2] c=0 400 1,400

φ=sin-1 r x σ3 σ3 Sand 500[lb/ft2] CD test Sand τ [lb/ft2] σ1 φ 500 Our values for are added for the radius, --- φ 900 φ σ [lb/ft2] c=0 400 1,400

φ=sin-1 r x σ3 σ3 Sand 500[lb/ft2] CD test Sand 900[lb/ft2] τ [lb/ft2] σ1 φ 500 and for the distance to center, --- φ 900 φ σ [lb/ft2] c=0 400 1,400

τ [lb/ft2] σ [lb/ft2] Find: φ φ=33.75˚ σ1 σ3 σ3 CD test σ1 φ 400 1,400
φ=sin-1 r x σ3 σ3 Sand 500[lb/ft2] CD test Sand 900[lb/ft2] τ [lb/ft2] φ=33.75˚ σ1 φ 500 and we compute the same friction angle as before. φ 900 φ σ [lb/ft2] c=0 400 1,400

( ) ? γclay=53.1[lb/ft3] Index σ’v = Σ φ γ Δ d H*C σfinal ρcn= log
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 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 ‘

Find: φ σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand 34˚ 36˚ 38˚

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Find: φ σ3 = 400 [lb/ft2] CD test Δσ = 1,000 [lb/ft2] Sand 34˚ 36˚ 38˚

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