the required documents is 3000
Answer:
class TriangleNumbers
{
public static void main (String[] args)
{
for (int number = 1; number <= 10; ++number) {
int sum = 1;
System.out.print("1");
for (int summed = 2; summed <= number; ++summed) {
sum += summed;
System.out.print(" + " + Integer.toString(summed));
}
System.out.print(" = " + Integer.toString(sum) + '\n');
}
}
}
Explanation:
We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.
In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.
With the given data we can proceed to calculate the compression stress:
![\sigma_c = \frac{P}{A}](https://tex.z-dn.net/?f=%5Csigma_c%20%3D%20%5Cfrac%7BP%7D%7BA%7D)
![\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}](https://tex.z-dn.net/?f=%5Csigma_c%20%3D%20%5Cfrac%7B160%2A10%5E3%7D%7B%5Cpi%2F4%2A0.05%5E2%7D)
![\sigma_c = 81.5MPa](https://tex.z-dn.net/?f=%5Csigma_c%20%3D%2081.5MPa)
Through Goodman's equations the combined effort by fatigue and compression is expressed as:
![\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma_a%7D%7BS_e%7D%2B%5Cfrac%7B%5Csigma_c%7D%7B%5Csigma_u%7D%3D%5Cfrac%7B1%7D%7BFs%7D)
Where,
Fatigue limit for comined alternating and mean stress
Fatigue Limit
Mean stress (due to static load)
Ultimate tensile stress
Security Factor
We can replace the values and assume a security factor of 1, then
![\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma_a%7D%7B320%7D%2B%5Cfrac%7B81.5%7D%7B400%7D%3D%5Cfrac%7B1%7D%7B1%7D)
Re-arrenge for ![\sigma_a](https://tex.z-dn.net/?f=%5Csigma_a)
![\sigma_a = 254.8Mpa](https://tex.z-dn.net/?f=%5Csigma_a%20%3D%20254.8Mpa)
We know that the stress is representing as,
![\sigma_a = \frac{M_c}{I}](https://tex.z-dn.net/?f=%5Csigma_a%20%3D%20%5Cfrac%7BM_c%7D%7BI%7D)
Then,
Where
=Max Moment
I= Intertia
The inertia for this object is
![I=\frac{\pi d^4}{64}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B%5Cpi%20d%5E4%7D%7B64%7D)
Then replacing and re-arrenge for ![M_c](https://tex.z-dn.net/?f=M_c)
![M_c = \frac{\sigma_a*\pi*d^3}{32}](https://tex.z-dn.net/?f=M_c%20%3D%20%5Cfrac%7B%5Csigma_a%2A%5Cpi%2Ad%5E3%7D%7B32%7D)
![M_c = \frac{260.9*10^6*\pi*0.05^3}{32}](https://tex.z-dn.net/?f=M_c%20%3D%20%5Cfrac%7B260.9%2A10%5E6%2A%5Cpi%2A0.05%5E3%7D%7B32%7D)
![M_c = 3201.7N.m](https://tex.z-dn.net/?f=M_c%20%3D%203201.7N.m)
Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm