A series of three direct shear tests has been conducted on a certain saturated soil. Each test was performed on a 2.375-inch dia
meter, 1.00-inch tall sample. The results of these tests are as follows:
Test Number Normal Stress Shear Stress
1 2435 1656
2 4870 3571
3 7305 4578
Determine the friction angle and cohesion intercept.
Test Number Normal Stress Shear Stress
1 2435 1656
2 4870 3571
3 7305 4578
Determine the friction angle and cohesion intercept.
1 answer:
Answer:
Go to explaination for the details of the answer.
Explanation:
Angle of friction (friction angle) is a measure of the ability of a unit of rock or soil to withstand a shear stress. It is the angle (φ), measured between the normal force (N) and resultant force (R), that is attained when failure just occurs in response to a shearing stress (S).
The cohesion intercept is a used when we want to describe the shear strength soils. The definition is mainly derived from the Mohr-Coulomb failure criterion and it is used to describe the non-frictional part of the shear resistance which is independent of the normal stress.
Please check attachment to Determine the friction angle and cohesion intercept.
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Answer:
Matlab code with step by step explanation and output results are given below
Explanation:
We have to construct a Matlab function that creates a row vector "countValues" with elements 1 to endValue. That means it starts from 1 and ends at the value provided by the user (endValue).
function countValues = CreateArray(endValue)
% Here we construct a row vector countValues from 1:endValue
countValues = 1:endValue;
% then we transpose this row vector into column vector
countValues = countValues';
end
Output:
Calling this function with the endValue=11 returns following output
CreateArray(11)
ans =
1
2
3
4
5
6
7
8
9
10
11
Hence the function works correctly. It creates a row vector then transposes it and makes it a column vector.
Answer:
a)Cycle time = 2.37 min
b)Numbers of workers =21
c)Stations on the line =24
Explanation:
Given that
Total work content time(TWC) = 50 min
Production rate Rp= 24 units/hr
manning level will be close =1.5
Line balancing efficiency =0.94
a)
Cycle time
Cycle time = 2.37 min
b)
Numbers of workers ,W
W= 21
Numbers of workers =21
c)
Stations on the line(n)
Lets find service time Ts
Ts = Cycle time - Time for repositioning
Ts = Tc- Tr
Ts= 2.37 - 9/ 60 min
Ts= 2.22 min
We know that efficiency
n=23.94 ⇒n=24
n=24
Stations on the line =24