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3 years ago

14

Primary productivity is not limited by _______. a. the availability of sunlight b. time c. the availability of nutrients d. acce

ss to water

1 answer:

enot [183]3 years ago

3 0

Primary productivity is not limited by a. Time

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Write the 5 important of profession education in a point

tensa zangetsu [6.8K]

did it help ? I was going to put the same things so

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3 years ago

Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

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3 years ago

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kirill115 [55]

Answer:

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3 years ago

Modify the Rainfall Statistics program you wrote for Programming Challenge 2 of Chapter 7 . The program should display a list of

rjkz [21]

Answer:

#include<iostream>

#include <iomanip>

using namespace std;

const int NUM_MONTHS = 12;

double getTotal(double [], int);

double getAverage(double [], int);

double getLargest(double [], int, int &);

double getSmallest(double [], int, int &);

double getTotal(int rainFall,double NUM_MONTHS[])

{

double total = 0;

for (int count = 0; count < NUM_MONTH; count++)

total += NUM_MONTH[count];

return total;

}

double getAverage(int rainFall,double NUM_MONTH[])

{getTotal(rainFall,NUM_MONTH)

average= total/NUM_MONTHS;

return average;

}

double getHighest(int rainFall, double NUM_MONTHS[]) //I left out the subScript peice as I was not sure how to procede with that;

{

double largest;

largest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month++ ){

if ( values[month] > largest ){

largest = values[month];

return largest;

}

double getSmallest(int rainFall, double NUM_MONTHS[])

{

double smallest;

smallest = NUM_MONTHS[0];

for ( int month = 1; month <= NUM_MONTHS; month){

if ( values[month] < smallest ){

smallest = values[month];

return smallest;

}

int main()

{

double rainFall[NUM_MONTHS];

for (int month = 0; month < NUM_MONTHS; month++)

{

cout << "Enter the rainfall (in inches) for month #";

cout << (month + 1) << ": ";

cin >> rainFall[month];

while (rainFall[month] < 0)

{

cout << "Rainfall must be 0 or more.\n"

<< "Please re-enter: ";

cin >> rainFall[month];

}

}

cout << fixed << showpoint << setprecision(2) << endl;

cout << "The total rainfall for the year is ";

cout << getTotal(rainFall, NUM_MONTHS)

<< " inches." << endl;

cout << "The average rainfall for the year is ";

cout << getAverage(rainFall, NUM_MONTHS)

<< " inches." << endl;

int subScript;

cout << "The largest amount of rainfall was ";

cout << getLargest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl;

cout << "The smallest amount of rainfall was ";

cout << getSmallest(rainFall, NUM_MONTHS, subScript)

<< " inches in month ";

cout << (subScript + 1) << "." << endl << endl;

return 0;

}

8 0

4 years ago