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3 years ago

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A sign structure on the NJ Turnpike is to be designed to resist a wind force that produces a moment of 25 k-ft in one direction.

The axial load is 30 kips. Soil conditions consist of a normally consolidated clay layer with following properties; su=800 psf, andγsat= 110 pcf. Design for a FOS of 3. Assume frost depth to be 3ft below grade

1 answer:

Jlenok [28]3 years ago

7 0

Solution :

Finding the cohesion of the soil(c) using the relation:

$c = \frac{q_u}{2}$

Here, $q_u$ is the unconfined compression strength of the soil;

$c = \frac{800}{2}$

= 400 psf

∴ The cohesion value is greater than 0

So the use of the angle of internal friction is 0

Referring to the table relation between bearing capacity factors and angle of internal friction.

For the angle of inter friction $0^\circ$

$N_c = 5.14$

$N_q = 1.0$

$N_r = 0$

Therefore,

$q_{ult} = (400 \times 5.14 )+(110 \times 3 \times 1.0)+(0.5 \times 100 \times 13 \times 0)$

= 2386 psf

∴ Allowable bearing capacity $q_{a} = \frac{Q_{allow}}{A}$

$=\frac{30}{B^2}$

∴ $q_a = \frac{q_{ult}}{F.O.S}$

$\frac{30}{B^2} = \frac{2386}{3}$

∴ B = 0.2 ft

Therefore, the dimension of the square footing is 0.2 ft x 0.2 ft

$=0.04 \ ft^2$

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If an object is near the surface of the earth the variation of its weight with distance from the center of the earth can ofteb b

zalisa [80]

Answer:

h=32.1 km

Explanation:

<em>solution:</em>

using newton law of gravitational attraction and newton second law:

$W=\frac{Gmm_{E} }{r^{2} } \\a=\frac{Gm_{E}}{r^{2}} \\W=ma\\$

$m_{E}= mass of earth$

r= distance between two masses

at sea level

a=g

$r=R_{E}$

$a=\frac{Gm_{E}}{r^{2}}$ .............................(1)

$Gm_{E} =gR_{E}^2$ .........................(2)

by substituting (2) and (1) $a=g\frac{R_{E}^2 }{r^{2} }$ acceleration due to gravity at a distance r from the centre of the earth in terms of g (sea level)

so the weight of the object at a distance r from the centre of the earth (W=ma)

W=mg(Re^2/r^2)..........(3)

h the height above the surface of the earth: r=Re+h

putting the value of r in eq (3)

W=mg(Re/Re+h)^2

W=0.99 mg

solving for height h:

h=Re(1/√0.99)-(1))

h=32.1 km

4 0

3 years ago

1 // Lab 2 tryIt2A 2 #include 3 using namespace std; 4 5 int main() 6 { int x = 1, y = 3; 7 int X = 2, Y = 4; 8 9 cout <<

padilas [110]

Answer:

Here is the complete program:

#include <iostream>

using namespace std;

int main()

{ int x = 1, y = 3;

int X = 2, Y = 4;

cout << "tryIt 2A" <<endl;

cout << x << y << endl;

cout << "x" << "y" << endl;

cout << X << " " << Y << endl;

cout << 2 * x + y << endl;

cout << 2 * X + Y << endl;

//cout << x + 2*y << endl;

cout << "x = ";

cout << x;

cout << " y = ";

cout << y;

return 0;

}

Explanation:

I will explain the code line by line in the comment with each line of code and the output of each cout statement.

int x = 1, y = 3;

This statement assigns value 1 to integer variable x and 3 to int variable y

int X = 2, Y = 4;

This statement assigns value 2 to integer variable X and 4 to int variable Y As C++ is a case sensitive language so variable x and y are different from variables X and Y.

cout << "tryIt 2A" <<endl;

This statement has cout which is used to display output on the screen. So the output displayed by this cout statement is:

tryIt2A

cout << x << y << endl;

This statement will print the values stored in x and y variables. So output displayed by cout statement here is 1 and 3. As there is not space or next line specified in the statement so output displayed will look like this:

13

cout << "x" << "y" << endl;

This statement will display x and y but these are not the variable x and y. They are enclosed in double quotation marks so they are treated as strings not variables so the output displayed is:

xy

cout << X << " " << Y << endl;

This statement will print the values stored in X and Y variables. So output displayed by cout statement here is 2 and 4. As there is space " " specified in the statement so 2 and 4 are displayed with a space between them so the output displayed will look like this:

2 4

cout << 2 * x + y << endl;

This statement has an arithmetic operation in which 2 is multiplied by the values stored in variable x and then the result is added by value of y. So 2*1 = 2 and 2 + 3 = 5. So the result produced by this cout statement is:

5

cout << 2 * X + Y << endl;

This will work same as above cout statement but the only difference is that the values of capital X and Y variables are calculated here. So 2 * 2 = 4 and then 4 + 4 = 8. The result produced by this cout statement is:

8

//cout << x + 2*y << endl;

This is a comment because before this statement // is written which is used for single line comment. So compiler ignores comments and will not compile this statement.

cout << "x = ";

This will display "x = " as it is not variable but it is treated as a line to be displayed on the screen. So cout statement displays:

x =

cout << x;

This will print the value stored in x variable as there are no double quotes around x so it is a variable which contains value 1. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1

cout << " y = ";

This will display "y = " as it is not variable but it is treated as a line to be displayed on the screen. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen

x = 1 y =

cout << y;

This will print the value stored in y variable as there are no double quotes around y so it is a variable which contains value 3. In the above statement there is no endl so the output of this cout statement is displayed with the output of previous cout statement. So the following line is displayed on screen:

x = 1 y = 3

So the output of the entire program along with the program is attached as screenshot.

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4 years ago

What information does the api symbol or donut provide?.

kondor19780726 [428]

Answer:

The API "Donut" identifies oils that meet current API engine oil standards.

Explanation:

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3 years ago

Technician A says that the distributor cap provides a connection point between the rotor and each individual cylinder plug wire.

Vera_Pavlovna [14]

Answer:

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Explanation:

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3 years ago

Explain how collecting mean values (X-bar) from samples drawn from a dataset, regardless of the statistical distribution of the

Pani-rosa [81]

Answer:

answer is given below

Explanation:

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The given phenomenon is described in the central limit theory. In other words, if we repeatedly take independent random samples of size n from any population, when n is large, the sample distribution is the normal distribution pattern.

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$\mu _\bar x = \mu$ .............1

and here standard deviation of the sample means is

$\mu _\bar x = \frac{\sigma }{\sqrt{n}}$ .........2

and This theory is found elsewhere in the field of statistics. Although the central limit theory may seem abstract and devoid of any application, this theory is actually important for statistical practice.

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3 years ago