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2 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.

1 answer:

scoray [572]2 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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A house is losing heat at a rate of 1700 kJ/h per °C temperature difference between the indoor and the outdoor temperatures. Exp

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2 years ago

Tensile Strength (MPa) Number-Average Molecular Weight (g/mol)

IceJOKER [234]

Answer:

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

Explanation:

The question can be well structured in a table format as illustrated below:

Tensile Strength (MPa) Number- Average Molecular Weight (g/mol)

82 12,700

156 28,500

The tensile strength and number-average molecular weight for two polyethylene materials given above.

Estimate the number-average molecular weight that is required to give a tensile strength required above. Using the data given find TS (infinity) in MPa.

<u>SOLUTION:</u>

We know that :

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

where;

$T_S$ = Tensile Strength

$T_{S \infty}$ = Tensile Strength (Infinity)

$M_n$ = Number- Average Molecular Weight (g/mol)

SO;

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

From equation (1) ; collecting the like terms; we have :

$T_{S \infty} =82+ \dfrac{A}{12700}$

From equation (2) ; we have:

$T_{S \infty} =156+ \dfrac{A}{28500}$

So; $T_{S \infty} = T_{S \infty}$

Then;

$T_{S \infty} =82+ \dfrac{A}{12700} =156+ \dfrac{A}{28500}$

Solving by L.C.M

$\dfrac{82(12700) + A}{12700} =\dfrac{156(28500) + A}{28500}$

$\dfrac{1041400 + A}{12700} =\dfrac{4446000 + A}{28500}$

By cross multiplying ; we have:

$({4446000 + A})* {12700} ={28500} *({1041400 + A})$

$(5.64642*10^{10} + 12700A) =(2.96799*10^{10}+ 28500A)$

Collecting like terms ; we have

$(5.64642*10^{10} - 2.96799*10^{10} ) =( 28500A- 12700A)$

$2.67843*10^{10} = 15800 \ A$

Dividing both sides by 15800:

$\dfrac{ 2.67843*10^{10} }{15800} =\dfrac{15800 \ A}{15800}$

A = 1695208.861

From equation (1);

$82= T_{S \infty} - \dfrac{A}{12700} ---- (1)$

Replacing A = 1695208.861 in the above equation; we have:

$82= T_{S \infty} - \dfrac{1695208.861}{12700}$

$T_{S \infty}= 82 + \dfrac{1695208.861}{12700}$

$T_{S \infty}= \dfrac{82(12700) +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{1041400 +1695208.861 }{12700}$

$T_{S \infty}= \dfrac{2736608.861 }{12700}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

From equation(2);

$156= T_{S \infty} - \dfrac{A}{28500} ---- (2)$

Replacing A = 1695208.861 in the above equation; we have:

$156= T_{S \infty} - \dfrac{1695208.861}{28500}$

$T_{S \infty}= 156 + \dfrac{1695208.861}{28500}$

$T_{S \infty}= \dfrac{156(28500) +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{4446000 +1695208.861 }{28500}$

$T_{S \infty}= \dfrac{6141208.861}{28500}$

$\mathbf{T_{S \infty } \ \approx 215.481 \ MPa}$

We are to also estimate the number- average molecular weight that is required to give a tensile strength required above.

If the Tensile Strength (MPa) is 82 MPa

Definitely the average molecular weight will be = 12,700 g/mol

If the Tensile Strength (MPa) is 156 MPa

Definitely the average molecular weight will be = 28,500 g/mol

But;

Let us assume that the Tensile Strength (MPa) = 181 MPa for example.

Using the same formula:

$T_S = T_{S \infty} - \dfrac{A}{M_n}$

Then:

$181 = 215.481- \dfrac{1695208.861 }{M_n}$

Collecting like terms ; we have:

$\dfrac{1695208.861 }{M_n} = 215.481- 181$

$\dfrac{1695208.861 }{M_n} =34.481$

1695208.861= 34.481 $M_n$

Dividing both sides by 34.481; we have:

$M_n$ = $\dfrac{1695208.861}{34.481}$

$\mathbf{M_n = 49163.56431 \ g/mol }$

5 0

3 years ago