Question

In a direct shear test on a sample of cohesionless sand, the vertical normal stress on the sample is 300 kPa and the horizont

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Answer #1

Let's Use Mohr's circle to solve for principal stresses and their orientation

Step 1:

Plot the normal and shear stress as shown below and draw the line which is passing exactly through the origin(As Cu=0)

the resultant line is Failure plane

And the angle of friction   \phi _f=34^o

Step 2:

Draw a line perpendicular to the failure plane which exactly intersects on the x-axis

The intersected point is Centre of Mohr's Circle

Step 3:

Draw the Mohr's Circle Which has a centre at C and passes through the point F as shown

Draw a horizontal line from P the intersected point on the circle is Pole(P)

Step 4:

Measure the Major and Minor Principal Stresses shown below

MINOR\,PRINCIPAL \,STRESS=20KPa

MAJOR\,PRINCIPAL \,STRESS=67.38KPa

Shear Stressi (30KPa,20KPa) POLE MINOR PRINCIPAL PLANE MAJOR PRINCIPAL PLAN 20KPa MINOR PRINCIPAL STRESS 67.38KPa MAJOR PRINC

Step 5:

Measure the Directions of Major and minor Principal plane

Shear Stressi (30KPa,20KPa)F/ Orientation of Major Principal Plan 34° 20KPa Orientation of minor Principal Plane 67.38KPa

b)

The major Drawback of the Direct shear test is it is not possible to attain the values of stresses and properties before the failure point.

So the soil should definitely fail by reaching failure point at failure shear stress to calculate Principal stresses.

It is not possible to Estimate Before Failure stress.


answered by: ANURANJAN SARSAM
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Answer #2

Зоо кра • 200 kPa T= ctotem o Griven, Т. 2 от 04 О. Qoy ka cao (Cohessionless sond). 2oo — Зоо n ф эфе erm (5) — ф-р+9 (фа 3وليه ما a . cosa= P I - R= a e => R= 200 coso C01133.694) » R=240.27 Sol r2 = 0C-R 7 = 422.32-240.27 10=192.65 kPa 1 m or oct

Feel free to ask for further clarification.

please up-vote if you liked this solution. As it will encourage me to solve more and more problems and help students.

Thank you,

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