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Alex777 [14]
3 years ago
6

A cone penetration test was carried out in normally consolidated sand, for which the results are summarized below: Depth (m) Con

e resistance, qc (MN/m2 ) 2.0 3.12 3.5 4.25 5.0 5.14 6.5 9.23 8.0 12.2 The average unit weight of the sand is 16.5 kN/m3 . Determine the friction angle at each depth using Eq. (3.52)
Engineering
1 answer:
Cerrena [4.2K]3 years ago
6 0

Answer:

hello your question is incomplete attached below is the missing equation related to the question  

answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°

Explanation:

<u>Determine the friction angle at each depth</u>

attached below is the detailed solution

To calculate the vertical stress = depth * unit weight of sand

also inverse of Tan = Tan^-1

also qc is in Mpa while σ0 is in kPa

Friction angle at each depth

2 meters = 40.389°

3.5 meters  = 38.987°

5 meters = 38.022°

6.5 meters = 39.869°

8 meters = 40.265°

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Three identical fatigue specimens (denoted A, B, and C) are fabricated from a nonferrous alloy. Each is subjected to one of the
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Answer:

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Explanation:

Given:

Specimen         σ_{max}                      σ_{min}

A                       +450                      -150

B                       +300                      -300

C                       +500                      -200

Solution:

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σ_{mA} =  (450 + (-150)) / 2

       =  (450 - 150) / 2  

       = 300/2

σ_{mA} = 150 MPa

σ_{mB}  = (300 + (-300))/2

        = (300 - 300) / 2

        = 0/2  

σ_{mB}  = 0 MPa

 

σ_{mC}  = (500 + (-200))/2

        = (500 - 200) / 2

        = 300/2

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Compute stress amplitude:

σ_{a} =  (σ_{max}  -  σ_{min})/2    

σ_{aA} =  (450 - (-150)) / 2

       =  (450 + 150) / 2

       = 600/2

σ_{aA} = 300 MPa

σ_{aB} =  (300- (-300)) / 2

       =  (300 + 300) / 2

       = 600/2

σ_{aB}  = 300 MPa

σ_{aC}  = (500 - (-200))/2

        = (500 + 200) / 2

        = 700 / 2

σ_{aC}   = 350 MPa

From the above results it is concluded that the longest  fatigue lifetime is of specimen B because it has the minimum mean stress.

Next, the specimen A has the fatigue lifetime which is shorter than B but longer than specimen C.

In the last comes specimen C which has the shortest fatigue lifetime because it has the higher mean stress and highest stress amplitude.

7 0
3 years ago
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