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A direct shear test was performed on a specimen of dry sand. The specimen was 50 mm by 50 mm square and 25 mm thick (height). A normal stress of 192 kPa was applied to the specimen and the shear stress at failure () was 120 kPa.
a. What is the effective friction angle of the sand?
b. For a normal stress of 200 kPa, what shear force (F) will be required to cause failure of the specimen? (Report answer in N).

Respuesta :

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To solve this problem it is necessary to apply the concepts related to the normal effort and the shear effort due to failure. The shear stress can be defined based on normal stress and effective friction angle. Mathematically it can be defined as

[tex]\tau_f = \sigma_n tan\phi[/tex]

Where

[tex]\tau_f =[/tex] Shear stress at failure

[tex]\sigma_n =[/tex] Normal stress

[tex]\phi =[/tex] Effective friction angle

PART A) Our values are given as ,

[tex]\sigma_n = 192kPa[/tex]

[tex]\tau_f  = 120kPa[/tex]

Replacing at the previous equation we have,

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]120 = 192tan\phi[/tex]

[tex]\phi = tan^{-1}(\frac{120}{192})[/tex]

[tex]\phi = 32.27°[/tex]

Therefore the effective friction angle of the sand is 32.27°

B) Using the maximum possible effort and the angle previously given we can calculate the maximum shear force, from which from its definition it is possible to find the force.

[tex]\tau_f = \sigma_n tan\phi[/tex]

[tex]\tau_f = 200tan(32\°)[/tex]

[tex]\tau_f = 124.97kPa[/tex]

With the value of the shear stress from its basic definition we can find the force. By definition the shear stress is given by

[tex]\tau_f = \frac{F}{A}[/tex]

Re-arrante to find the Force

[tex]F = A\tau_f[/tex]

The shear stress is a function of the Force and the Area, therefore, the area would be the square of the sides (50mm)

Replacing in the equation we have to,

[tex]F = 124.97*10^{-3}*(50*50)[/tex]

[tex]F = 312.425N[/tex]

Therefore the shear force required to cause failure of the specimen is 312.425N

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