From the UCS stress-strain plot, what is q_u?

The axial stress at zero strain

The final stress at large strain regardless of peak

How are c and phi obtained from direct shear tests at three or more normal stresses?

Plot peak tau_f against sigma, fit a straight line; intercept is c and slope angle is phi

Plot sigma against strain; slope gives c

Plot tau against displacement; the peak gives phi

Draw one Mohr circle and read c at the origin

For the same clay, how do the two tests differ in what they give?

UCS gives undrained c_u = q_u/2 quickly; direct shear (slow) gives drained c and phi

UCS gives phi and direct shear gives only c_u

Both give identical undrained parameters

UCS gives drained c and phi; direct shear gives c_u

2078 Baisakh · 2 marks

Draw UCS and direct shear results

Short answer

For UCS, plot axial stress vs strain to get qu, then draw Mohr circle through origin with diameter qu; the horizontal tangent gives cu = qu/2. For direct shear, plot shear stress vs displacement for each normal stress, then plot peak shear strength vs normal stress to obtain c and φ from the best-fit line.

Draw final test results of the Unconfined Compression Test and the Direct Shear Test for the same soil so that the strength parameters of the soil could be obtained.

Unconfined compression test (UCS)

Plot axial stress $\sigma_1$ against axial strain; the peak is $q_u$ . The Mohr circle drawn on the $\sigma$ - $\tau$ plane passes through the origin ( $\sigma_3 = 0$ ) with $\sigma_1 = q_u$ , and the envelope is a horizontal line tangent to the circle for $\phi_u = 0$ :

 UCS stress-strain          Mohr circle
 s1                         tau
  |   ___ qu                 |   ..--..
  |  /   \                 cu |--*'    '*--
  | /                         | /   C    \
  +--------- strain          -+-+--------+--- sigma
                              0 (s3=0)    qu

Direct shear test

Plot shear stress against horizontal displacement for each normal stress to get the peak $\tau_f$ . Then plot $\tau_f$ against normal stress $\sigma$ for three or more tests and draw the best-fit straight line:

 tau_f            /
   |            /  <- slope = phi
   |          /
 c |--------/
   |      .
   +--------------- sigma
   s1  s2  s3

Intercept $= c$ ; slope angle $= \phi$ . For the same clay, the UCS gives $c_u = q_u/2$ (total stress, quick), while the direct shear (slow) gives $c$ and $\phi$ under drained conditions.

✓ Answer checked by Claude Opus 5.5 (high effort)

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