Answered

Get the answers you need at Westonci.ca, where our expert community is always ready to help with accurate information. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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

Sagot :

batolisis

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:

Determine the friction angle at each depth

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°

View image batolisis
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.