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3 years ago

The shear force diagram is always the slope of the bending moment diagram. a)True b)- False

Engineering

1 answer:

lara [203]3 years ago

8 0

Answer:

True

Explanation:

Shear force diagram is a diagram which is drawn by calculating the shear force either to the right of the section or to the left of the section .

shear force is also be define as the change of moment w.r.t to distance

V= $\frac{\mathrm{d}M }{\mathrm{d} x}$

where V is shear force and M is bending moment.

from the equation we can clearly say that shear force diagram is slope of bending moment diagram.

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2.4: Add a method called setValue(), and the description of setValue is: public int setValue(long searchKey) In this method, the

Yanka [14]

Answer:

Below is java code that must be used for the given question:

// highArray.java

// demonstrates array class with high-level interface

// to run this program: C>java HighArrayApp

////////////////////////////////////////////////////////////////

class HighArray

{

private long[] a; // ref to array a

private int nElems; // number of data items

//-----------------------------------------------------------

public HighArray(int max) // constructor

{

a = new long[max]; // create the array

nElems = 0; // no items yet

}

//-----------------------------------------------------------

public setValue find(long searchKey)

{ // find specified value

int j;

for(j=0; j<nElems; j++) // for each element,

if(a[j] == searchKey) // found item?

break; // exit loop before end

if(j == nElems) // gone to end?

return false; // yes, can't find it

else

return true; // no, found it

} // end find()

//-----------------------------------------------------------

public void insert(long value) // put element into array

{

a[nElems] = value; // insert it

nElems++; // increment size

}

//-----------------------------------------------------------

public void display() // displays array contents

{

for(int j=0; j<nElems; j++) // for each element,

System.out.print(a[j] + " "); // display it

System.out.println("");

}

//-----------------------------------------------------------

} // end class HighArray

////////////////////////////////////////////////////////////////

class HighArrayApp

{

public static void main(String[] args)

{

int maxSize = 100; // array size

HighArray arr; // reference to array

arr = new HighArray(maxSize); // create the array

arr.insert(77); // insert 10 items

arr.insert(99);

arr.insert(44);

arr.insert(55);

arr.insert(22);

arr.insert(88);

arr.insert(11);

arr.insert(00);

arr.insert(66);

arr.insert(33);

arr.display(); // display items

int searchKey = 35; // search for item

if( arr.find(searchKey) )

System.out.println("Found " + searchKey);

else

System.out.println("Can't find " + searchKey);

} // end main()

} // end class HighArrayApp

Explanation:

6 0

2 years ago

A large class with 1,000 students took a quiz consisting of ten questions. To get an A, students needed to get 9 or 10 questions

VMariaS [17]

Answer:

a. 0.11

b. 110 students

c. 50 students

d. 0.46

e. 460 students

f. 540 students

g. 0.96

Explanation:

(See attachment below)

a. Probability that a student got an A

To get an A, the student needs to get 9 or 10 questions right.

That means we want P(X≥9);

P(X>9) = P(9)+P(10)

= 0.06+0.05=0.11

b. How many students got an A on the quiz

Total students = 1000

Probability of getting A = 0.11 ---- Calculated from (a)

Number of students = 0.11 * 1000

Number of students = 110 students

So,the number of students that got A is 110

c. How many students did not miss a single question

For a student not to miss a single question, then that student scores a total of 10 out of possible 10

P(10) = 0.05

Total Students = 1000

Number of Students = 0.05 * 1000

Number of Students = 50 students

We see that 5

d. Probability that a student pass the quiz

To pass, a student needed to get at least 6 questions right.

So we want P(X>=6);

P(X>=) =P(6)+P(7)+P(8)+P(9)+P(10)

=0.08+0.12+0.15+0.06+0.05=0.46

So, the probability of a student passing the quiz is 0.46

e. Number of students that pass the quiz

Total students = 1000

Probability of passing the quiz = 0.46 ----- Calculated from (d)

Number of students = 0.46 * 1000

Number of students = 460 students

So,the number of students that passed the test is 460

f. Number of students that failed the quiz

Total students = 1000

Total students that passed = 460 ----- Calculated from (e)

Number of students that failed = 1000 - 460

Number of students that failed = 540

So,the number of students that failed is 540

g. Probability that a student got at least one question right

This means that we want to solve for P(X>=1)

Using the complement rule,

P(X>=1) = 1 - P(X<1)

P(X>=1) = 1 - P(X=0)

P(X>=1) = 1 - 0.04

P(X>=1) = 0.96

7 0

2 years ago

You are evaluating the lifetime of a turbine blade. The blade is 4 cm long and there is a gap of 0.16 cm between the tip of the

Tcecarenko [31]

Answer:

Explanation:

Given conditions

1)The stress on the blade is 100 MPa

2)The yield strength of the blade is 175 MPa

3)The Young’s modulus for the blade is 50 GPa

4)The strain contributed by the primary creep regime (not including the initial elastic strain) was 0.25 % or 0.0025 strain, and this strain was realized in the first 4 hours.

5)The temperature of the blade is 800°C.

6)The formula for the creep rate in the steady-state regime is dε /dt = 1 x 10-5 σ4 exp (-2 eV/kT)

where: dε /dt is in cm/cm-hr σ is in MPa T is in Kelvink = 8.62 x 10-5 eV/K

Young Modulus, E = Stress, $\sigma$ /Strain, ∈

initial Strain, $\epsilon_i = \frac{\sigma}{E}$

$\epsilon_i = \frac{100\times 10^{6} Pa}{50\times 10^{9} Pa}$

$\epsilon_i = 0.002$

creep rate in the steady state

$\frac{\delta \epsilon}{\delta t} = (1 \times {10}^{-5})\sigma^4 exp^(\frac{-2eV}{kT} )$

$\frac{\epsilon_{initial} - \epsilon _{primary}}{t_{initial}-t_{final}} = 1 \times 10^{-5}(100)^{4}exp(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )(800+273)K} )$

but Tinitial = 0

$\epsilon_{initial} - \epsilon _{primary}} = 0.002 - 0.003 = -0.001$

$\frac{-0.001}{-t_{final}} = 1 \times 10^{-5}(100)^{4}\times 10^{(\frac{-2eV}{8.62\times10^{-5}(\frac{eV}{K} )1073K} )}$

solving the above equation,

we get

Tfinal = 2459.82 hr

3 0

3 years ago

Am I alive I really need to know?

Nesterboy [21]

Hell no,cause i’m not

5 0

3 years ago

Six housing subdivisions within a city area are target for emergency service by a centralized fire station. Where should the new

Marina86 [1]

Answer:

Explanation:

Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F

To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.

$C_{x}=\frac{1}{6A} sum(({x_{i} +x_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$ where i=1 to N-1 and N=6

$C_{y}=\frac{1}{6A} sum(({y_{i} +y_{i+1})(x_{i}y_{i+1}-x_{i+1}y_{i}))$

A area of the polygon can be found by the following formula $A=\frac{1}{2} sum(x_{i} y_{i+1} -x_{i+1} y_{i})$ where i=1 to N-1

$A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]$

A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)

A=225.5 miles²

Now putting the value of area in Cx and Cy

$C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{x} =15.36$

For Cy

$C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]$

putting the values of x's and y's you will get

$C_{y} =22.55$

So coordinates for the fire station should be (15.36,22.55)

5 0

2 years ago