HELP ASSAPPP

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago

with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d

Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

To know more about kVp visit:-

brainly.com/question/17095191

#SPJ4

5 0

1 year ago

Someone has suggested that the air-standard Otto cycle is more accurate if the two polytropic processes are replaced with isentr

omeli [17]

Answer:

q_net,in = 585.8 KJ/kg

q_net,out = 304 KJ/kg

n = 0.481

Explanation:

Given:

- The compression ratio r = 8

- The pressure at state 1, P_1 = 95 KPa

- The minimum temperature at state 1, T_L = 15 C

- The maximum temperature T_H = 900 C

- Poly tropic index n = 1.3

Find:

a) Determine the heat transferred to and rejected from this cycle

b) cycle’s thermal efficiency

Solution:

- For process 1-2, heat is rejected to sink throughout. The Amount of heat rejected q_1,2, can be computed by performing a Energy balance as follows:

W_out - Q_out = Δ u_1,2

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_2 - T_L) = R*(T_2 - T_L)/n-1 - q_1,2

- Using polytropic relation we will convert T_2 = T_L*r^(n-1):

c_v*(T_L*r^(n-1) - T_L) = R*(T_1*r^(n-1) - T_L)/n-1 - q_1,2

- Hence, we have:

q_1,2 = T_L *(r^(n-1) - 1)* ( (R/n-1) - c_v)

- Plug in the values:

q_1,2 = 288 *(8^(1.3-1) - 1)* ( (0.287/1.3-1) - 0.718)

q_1,2= 60 KJ/kg

- For process 2-3, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

Q_in = Δ u_2,3

q_2,3 = u_3 - u_2

q_2,3 = c_v*(T_H - T_2)

- Again, using polytropic relation we will convert T_2 = T_L*r^(n-1):

q_2,3 = c_v*(T_H - T_L*r^(n-1) )

q_2,3 = 0.718*(1173-288*8(1.3-1) )

q_2,3 = 456 KJ/kg

- For process 3-4, heat is transferred into the system. The Amount of heat added q_2,3, can be computed by performing a Energy balance as follows:

q_3,4 - w_in = Δ u_3,4

- Assuming air to be an ideal gas, and the poly-tropic compression process is isentropic:

c_v*(T_4 - T_H) = - R*(T_4 - T_H)/1-n + q_3,4

- Using polytropic relation we will convert T_4 = T_H*r^(1-n):

c_v*(T_H*r^(1-n) - T_H) = -R*(T_H*r^(1-n) - T_H)/n-1 + q_3,4

- Hence, we have:

q_3,4 = T_H *(r^(1-n) - 1)* ( (R/1-n) + c_v)

- Plug in the values:

q_3,4 = 1173 *(8^(1-1.3) - 1)* ( (0.287/1-1.3) - 0.718)

q_3,4= 129.8 KJ/kg

- For process 4-1, heat is lost from the system. The Amount of heat rejected q_4,1, can be computed by performing a Energy balance as follows:

Q_out = Δ u_4,1

q_4,1 = u_4 - u_1

q_4,1 = c_v*(T_4 - T_L)

- Again, using polytropic relation we will convert T_4 = T_H*r^(1-n):

q_4,1 = c_v*(T_H*r^(1-n) - T_L )

q_4,1 = 0.718*(1173*8^(1-1.3) - 288 )

q_4,1 = 244 KJ/kg

- The net gain in heat can be determined from process q_3,4 & q_2,3:

q_net,in = q_3,4+q_2,3

q_net,in = 129.8+456

q_net,in = 585.8 KJ/kg

- The net loss of heat can be determined from process q_1,2 & q_4,1:

q_net,out = q_4,1+q_1,2

q_net,out = 244+60

q_net,out = 304 KJ/kg

- The thermal Efficiency of a Otto Cycle can be calculated:

n = 1 - q_net,out / q_net,in

n = 1 - 304/585.8

n = 0.481

6 0

3 years ago

What is 4 principles of experimental design

Mrrafil [7]

Answer:

manipulation, control , random assignment, and random selection

Explanation:

7 0

1 year ago

Read 2 more answers

Complete function PrintPopcornTime(), with int parameter bagOunces, and void return type. If bagOunces is less than 3, print "To

weqwewe [10]

Answer:

#include <iostream>

using namespace std;

void PrintPopcornTime(int bagOunces) {

if(bagOunces < 3){

cout << "Too small";

cout << endl;

}

else if(bagOunces > 10){

cout << "Too large";

cout << endl;

}

else{

cout << (6 * bagOunces) << " seconds" << endl;

}

int main() {

PrintPopcornTime(7);

return 0;

}

Explanation:

Using C++ to write the program. In line 1 we define the header "#include <iostream>" that defines the standard input/output stream objects. In line 2 "using namespace std" gives me the ability to use classes or functions, From lines 5 to 17 we define the function "PrintPopcornTime(), with int parameter bagOunces" Line 19 we can then call the function using 7 as the argument "PrintPopcornTime(7);" to get the expected output.

8 0

3 years ago