Detemination of Angle of Internal Friction

Problem 1. A sample of dry sand was tested in direct shear apparatus under a normal load of 36 kg. The sample failed under a shearing load of 58 lb. The sample size was 2" * 2".
What is the angle of internal friction?
Exam Name: Sylhet Gas Field

Solution:
We know that,
            τ = C + δ tanφ
  Here, τ = (36 * 2.2 ) / (2*2)
              = 19.8 lb/in²
          C = 0, because sample failed
           δ = 58 / (2*2)
              = 14.5 lb/in²
           φ = ?

  ∴ 19.8 = 0 + 14.5 tanφ
          φ = 53.78⁰
             = 53⁰ 47'

Answer:    φ = 53⁰ 47'

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Problem 2. Find the value of angle of internal friction of a soil sample which is subjected to direct shear test, 2" * 2" size, normal force was 36 kg with a shear force at failure of 58 lb?
Exam Name: Gas Transmission Company 2016

Solution:
We know,
              τ = C + δ tanφ
    Here, τ = (36 * 2.2 ) / (2*2)
                = 19.8 lb/in²
            C = 0, because sample failed
             δ = 58 / (2*2)
                = 14.5 lb/in²
             φ = ?
∴ 19.8 = 0 + 14.5 tanφ
          φ = 53.78⁰
             = 53⁰ 47'

Answer:    φ = 53⁰ 47'

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