Question 5 [a] A tapered vane shown in the Figure has dimensions d=65mm, h = 130

Question

Question 5 [a] A tapered vane shown in the Figure has dimensions d=65mm, h = 130mm and = 45°. It was inserted to a depth of 3m below ground level in a homogeneous soft clay, and 50 Nm torque was required to fail the clay. Torque components required to fail the soil along a cylindrical surface is d2hCu/2 and along a conical surface is dou/(12 cos(a)). Determine the undrained cohesion C of the clay. b A sand was consolidated to an effective confining stress of 300 kPa and then sheared drained ina direct shear device. The shear stress at peak state was 200 kPa and at residual state was 160 kPa. [1) Determine the peak, critical state friction angles. 2) What is the dilation angle at the peak state, if this soil can be modelled using (a) Taylor, and (b) Coulomb models 3) If an undrained triaxial test was carried out on this sand would you expect a positive or negative excess pore water pressure at failure? Explain why. Answers: Cu (kPa) Peak (Deg) Pc (Deg) "peak (Deg)

Explanation / Answer

Answer 5 a

For obtaining Cu value we will equate torque applied for shear failure to the equation given in question

T= d^2 h Cu/2 + d^3Cu/(12 Cos (alpha))

50 = 8.627 x 10^-4 Cu + 1.016 x 10^-4 Cu

Thus Cu = 51851.08 N/ m2

Answer 5 b

Shear stress at peak = 200 kpa

Horizontal normal stress assumed to be zero

Thus from direct shear stress effective maximum normal stress at failure = 0+ 2*200= 400 kpa

For sand using columb equation for shear stress T= c + × tan ¶ , where T is shear stress , c is cohesion , × is effective normal stress and ¶ is angle of internal friction

For sand c=0

Thus 200= 400 tan ¶

Thus peak friction angle = 26.56 degree

Shear stress at residual state = 160 kpa

Thus critical angle of friction tan € = 160/400 thus € = 21.8 degree

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