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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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4 0
3 years ago
. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th
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Answer:

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step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

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Given data

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step three:

substituting into the expression we have

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