Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] 600 750 900 D) 1,200 Find the axial stress at failure when the confining stress is 150 kilopascals, for a sand sample tested in the consolidated drained condition. [pause] In this problem, ---

Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] 600 750 900 D) 1,200 data has been provided from a triaxial shear test for the sand sample in the consolidated un-drained condition.

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] We begin by plotting the confining stress at failure from the undrained test, σ [kPa]

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] 70 kilopascals. [pause] Next we compute the --- σ [kPa] 70

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ axial stress at failure to be --- σ [kPa] 70

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ the confining stress at failure, plus, --- σ [kPa] 70

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ the deviator stress at failure, to get --- σ [kPa] 70

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] 170 kilopascals, --- σ [kPa] 70

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] and the point is plotted. [pause] A semi circle is drawn between --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] these two stress values. Keep in mind, --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] this is an un-drained test, and the problem asks to find, --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] the major principle stress during the drained condition. So the given stress values for the --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ1 = σ3+Δσ = 170 [kPa] un-drained test include the pore water pressure. The effective stress values can be computed --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] σ3’ = σ3-u σ1 = σ3+Δσ σ1’ = σ1-u = 170 [kPa] by subtracting the pore water pressure from the total stress, for both sigma 3, the confining stress, and sigma 1, the axial stress. σ [kPa] 70 170

τ [kPa] ? σ [kPa] Find: σ1 [kPa] for CD test at failure σ1 = σ3+Δσ if σ3 = 150 [kPa] CU test data: σ3 = 70 [kPa] Sand d= 5.1 [m] Δσ= 100 [kPa] τ [kPa] ? σ3’ = σ3-u σ1 = σ3+Δσ σ1’ = σ1-u = 170 [kPa] However we still need to compute the pore water pressure for the un-drained soil sample. σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: u = ρw * g * d Sand d= 5.1 [m] τ [kPa] σ3’ = σ3-u σ1’ = σ1-u We do so by multiplying --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] τ [kPa] σ3’ = σ3-u σ1’ = σ1-u the density of water --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u σ1’ = σ1-u times the gravitational acceleration, --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u σ1’ = σ1-u times the depth beneath the groundwater table. [pause] The pore water pressure at this depth is --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u 50 kilopascals. [pause] σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u Plugging in our pore water pressure, --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u and our total stress values, the effective confining and --- σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u and axial stress values are 20 kilopascals and 120 kilopascals, respectively. 20[kPa] 120[kPa] σ [kPa] 70 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u These points are plotted on the graph, and connected --- 20[kPa] 120[kPa] σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u by a semicircle. [pause] The pore water pressure --- 20[kPa] 120[kPa] σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u of 50 kilopascals shifted the semicircle --- 20[kPa] 120[kPa] σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] 1,000 [kg/m3] CU test data: u = ρw * g * d Sand d= 5.1 [m] 9.81 [m/s2] τ [kPa] σ3’ = σ3-u = 50 [kPa] σ1’ = σ1-u to the left, on our plot. 20[kPa] 120[kPa] σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] Now we have a semicircle for the ---- σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] un-drained condition, --- undrained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] and for the drained condition. The problem asks about the drained condition when --- undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] the confining stress is 150 kilopascals. undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] Since sand is a coarse grain, cohesionless soil, a point is plotted at the origin, --- undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] and the rupture line is traced, tangent to the drained semicircle, --- undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] φ and the angle of internal friction is identified. [pause] undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] σ1-σ3 CU test data: φ=sin-1 σ1+σ3 Sand d= 5.1 [m] τ [kPa] φ Since the cohesion is zero, the angle of internal friction can be computed --- undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] φ=sin-1 σ1-σ3 σ1+σ3 CU test data: Sand d= 5.1 [m] τ [kPa] φ by plugging in the appropriate stress values. [pause] The angle of internal friction is --- undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] φ=sin-1 σ1-σ3 σ1+σ3 CU test data: Sand = 45.58˚ d= 5.1 [m] τ [kPa] φ 45.58 degrees. undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] φ=sin-1 σ1-σ3 σ1+σ3 CU test data: Sand = 45.58˚ d= 5.1 [m] τ [kPa] 45.58˚ The problem asks for the axial stress, when the confining stress is 150 kilopascals. undrained drained σ [kPa] 20 70 120 170

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] 45.58˚ So 150 kilopascals is plotted on the graph, and the rupture line is --- σ [kPa] 20 120 150

τ [kPa] σ [kPa] Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] 45.58˚ extended outward. Next, a semicircle is drawn tangent to rupture line and passing through 150 kilopascals. σ [kPa] 20 120 150

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: Sand d= 5.1 [m] τ [kPa] 45.58˚ This graph is too small to see the value of the axial stress, --- σ [kPa] 20 120 150 σ1

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 45.58˚ but it can be solved mathematically by knowing the friction angle, --- σ [kPa] 20 120 150 σ1

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 1+sin(φ) 45.58˚ σ1=σ3 1-sin(φ) and confining stress, and by known the cohesion is zero. [pause] σ [kPa] 20 120 150 σ1

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 1+sin(φ) 45.58˚ σ1=σ3 1-sin(φ) The values are plugged in, and our axial stress computes to --- σ [kPa] 20 120 150 σ1

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 1+sin(φ) 45.58˚ σ1=σ3 1-sin(φ) 900 kilopascals. [pause] = 900 [kPa] σ [kPa] 20 120 150 σ1

τ [kPa] σ [kPa] σ1 Find: σ1 [kPa] for CD test at failure if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 1+sin(φ) 600 750 900 D) 1,200 σ1=σ3 1-sin(φ) Upon reviewing or possible solutions, --- = 900 [kPa] σ [kPa] 20 120 150 σ1

τ [kPa] σ1 Find: σ1 [kPa] for CD test at failure AnswerC if σ3 = 150 [kPa] CU test data: 1+sin(φ) σ1 = Sand 1-sin(φ) σ3 d= 5.1 [m] τ [kPa] 1+sin(φ) 600 750 900 D) 1,200 σ1=σ3 1-sin(φ) the answer is C. = 900 [kPa] AnswerC 20 120 150 σ1

( ) ? 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