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Viktor [21]
3 years ago
7

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

tained in the field were as follows:
0-6 in. 4 blows; 6-12 in.: 6 blows; 12-18 in.: 6 blows. The tests were conducted using a USA-style donut hammer in a 6-in diameter boring using a standard sampler with the liner installed. The vertical effective stress at the test depth was 1500 lb/ft^2. Determine N1.60.
Engineering
1 answer:
scoray [572]3 years ago
8 0

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

                                              $= 71.82 \ kN/m^2$

                                             $ \sim 72 \ kN/m^2 $

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 = $ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

                      = 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                        = 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right) $

                         = 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

       = 13.14

       = 13

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Answer:

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P(F) + P(W) = 1

We are given P(F) = 0.5

Thus;

0.5 + P(W) = 1

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B) Now, from the problem statement, a phone call can either be made with a handheld(H) or mobile(M). Thus the sample space partition is {H, M} and we can express as;

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We are given P[F] = 0.5 and P[HF] = 0.2.

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0.2 + P(MF) = 0.5

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Explanation:

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0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

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and space mean speed is given by,

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1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

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here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

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Vt = 195+190+185+180/4

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Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

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Since it is a rigid container so the work done is zero.

Q = du    ( du - change in internal energy)

Q = m * C * dT      ( C = 4.186 KJ/KgK )

Q = 3 * 4.186 * (T₂ - T₁ )

900 = 12.558 * ( T₂ - 313 )

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