1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Serga [27]
3 years ago
14

A thin rim with a mean diameter of 1.2 m cross-section of 15 mm x 200 mm is subjected to an internal pressure of 10 MPa and rota

tes at 800 r/min. Calculate 1. the magnitude of the force and the stress in the rim if the density of the materials is 7800 kg/m3. 2. the total change in diameter. Young's modulus for the material is 200 GPa.
Engineering
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

The net centrifugal force over the rim is 30000N, the radial stress is 397887 Pa and the total change in diameter is 4.98 mm.

Explanation:

Lets first calculate the force in the rim due to the centrifugal force. For doing this we may assume that the centrifugal force is constant along with thick because of the thin thick.

Fc = m.ω^2/R

Where m is the mass, w the angular speed and R the mean radius.  The mass is computing by the rim density and its volume:

m=p.V

m=p*(A*R)

Where A is the cross-sectional area in meters:

m=((0.015m*0.200m)*0.6m)*(7800 kg/m^3)=28.08 kg

The angular speed in rad/s is:

ω = 800r/m . 1m/60s = 133.33 r/s

Thus the centrifugal force is:

Fc = (28.08 kg)*(133.33 rad/s)^2*(0.6m) = 299505N = 30000N

Note that the calculating value is the net contribution to the whole rim but the centrifugal force is distributed along the rim's external area:

fc = Fc / (2π .R .b)

Where b is rim's with equal to 200mm :

fc = 300000 N / (2π*0.6m*0.2m) = 397887 N/m^2

The centrifugal force can be taken as internal pressure:

Pfc = 397887 N/m^2 = 3978787 Pa

As both pressures act expanding the rim it can be summed:

Pt=Pi+Pfc

Pt = 10MPa+397887Pa= 10000000Pa+397887Pa= 10397887Pa

Then for a thinner thick the stress is calculated by:

Pt*d =2σr*t

Take into account that the stress σr is over the radial direction. Then solving for o and by replacing the total pressure:

σr = Pt.d/(2*t)

σr = 10397887 Pa / (2*0.015m*0.2m) = 415915480Pa = 415MPa

We know that the radial specific deformation ε is:

σr = E / εr

εr = σr / E

For a young modulus of 200GPa:

εr = 415MPa / 200GPa

εr = 415MPa / 200000MPa=0.002075

By definition the specific deformation is written in terms of the total change in the radius:

εr = Δr / R

Δr = R / εr =0.002075 * 1.2 m = 0.00249m

As we need the change in diameter:

Δd = 2Δr =0.00498m= 4.98mm [/tex]

You might be interested in
While there are many ways to solve this problem, one strategy is to calculate the volume of any metal's unit cell given its theo
IgorC [24]

Answer:

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

Explanation:

z = number of atoms

M = Molar mass of zirconium

N = Avogadro’s number

Vc = volume of zirconium unit cell

d = density

z=12x\frac{1}{6}+2x\frac{1}{2}+3=6

z = 6 atoms per unit cell

M = 91.224 g/mol

N = 6.023x10^{23}  atoms/mol

d = 6.51g/cm^{3}

V_{c}=\frac{zxM}{dxN}

V_{c}=\frac{6x(91.224g/mol)}{(6.51g/cm^{3}) x(6.023x10^{23}atoms/mol) }

V_{c}=1.396x10^{-22}  cm^{3} /unit.cell

V_{c}=1.396x10^{-28}  m^{3} /unit.cell

8 0
3 years ago
Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67
defon

Answer:

#include <iostream>

#include <string>

using namespace std;

bool isPalindrome(string str)

{

   int length = str.length();

   for (int i = 0; i < length / 2; i++)

   {

       if (tolower(str[i]) != tolower(str[length - 1 - i]))

           return false;

   }

   return true;

}

int main()

{

   string s[6] = {"madam", "abba", "22", "67876", "444244", "trymeuemyrt"};

   int i;

   for(i=0; i<6; i++)

   {

       //Testing function

       if(isPalindrome(s[i]))

       {

           cout << "\n " << s[i] << " is a palindrome... \n";

       }

       else

       {

           cout << "\n " << s[i] << " is not a palindrome... \n";

       }

   }    

       

   return 0;

}

5 0
2 years ago
What is the volicity of a rocket?
Marysya12 [62]

Answer:

7.9 kilometers per second

Explanation:

8 0
2 years ago
Read 2 more answers
An Otto cycle engine is analyzed using the cold air standard method. Given the conditions at state 1, compression ratio (r), and
lilavasa [31]

Answer:

Explanation:

The detailed and careful step by step calculation and analysis is as shown with appropriate formula in the attached files

4 0
2 years ago
A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

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}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

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}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =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}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0
3 years ago
Other questions:
  • How an AK 47 gun was works​
    14·1 answer
  • Determine the magnitude of the resultant force and the moment about the origin. Note: the symbol near the 140 N-m moment are not
    15·1 answer
  • An incompressible fluid flows along a 0.20-m-diameter pipe with a uniform velocity of 3 m/s. If the pressure drop between the up
    15·1 answer
  • A 0.39 percent Carbon hypoeutectoid plain-carbon steel is slowly cooled from 950 oC to a temperature just slightly below 723 oC.
    15·1 answer
  • Consider uniaxial extension of a test specimen. It has gauge length L = 22 cm (the distance between where it is clamped in the t
    6·1 answer
  • On July 23, 1983, Air Canada Flight 143 required 22,300 kg of jet fuel to fly from Montreal to Edmonton. The density of jet fuel
    8·1 answer
  • What structure was created to help prevent shipwrecks?
    9·1 answer
  • It is the tool used to measure the amount of electric current​
    6·2 answers
  • The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.
    11·1 answer
  • Why might many general contractors begin their careers as construction workers?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!