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mina [271]
2 years ago
7

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de

termine: I. the engineering stress and strain, and [2] II. the true stress and strain
Engineering
1 answer:
Margaret [11]2 years ago
4 0

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

  1. L = 20 cm d x 1 = 0.21 cm
  2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
  3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
  4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
  5. = 1561.84 MPa.

Read more about the diameter :

brainly.com/question/358744

#SPJ1

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Which phrase best describes a safety-critical system? A. a system that faces a very high risk of failure B. a system isolated fr
KIM [24]

Answer:

B.

Explanation:

A safety-critical system (SCS) or life-critical system is a system whose failure or malfunction may result in one (or more) of the following outcomes: death or serious injury to people. loss or severe damage to equipment/property.

7 0
3 years ago
Write a C program that will update a bank balance. A user cannot withdraw an amount ofmoney that is more than the current balanc
GarryVolchara [31]

Answer:

Explanation:

Sample output:

BANK ACCOUT PROGRAM!

----------------------------------

Enter the old balance: 1234.50

Enter the transactions now.

Enter an F for the transaction type when you are finished.

Transaction Type (D=deposit, W=withdrawal, F=finished): D

Amount: 568.34

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 25.68

Transaction Type (D=deposit, W=withdrawal, F=finished): W

Amount: 167.40

Transaction Type (D=deposit, W=withdrawal, F=finished): F

Your ending balance is $1609.76

Program is ending

Code to copy:

// include the necessary header files.

#include<stdio.h>

// Definition of the function

float withdraw(float account_balance, float withdraw_amount)

{

// Calculate the balace amount.

float balance_amount = account_balance - withdraw_amount;

// Check whether the withdraw amount

// is greater than 0 or not.

if (withdraw_amount > 0 && balance_amount >= 0)

{

// Assign value.

account_balance = balance_amount;

}

// return account_balance

return account_balance;

}

// Definition of the function deposit.

float deposit(float account_balance, float deposit_amount)

{

// Check whether the deposit amount is greater than zero

if (deposit_amount > 0)

{

// Update account balance.

account_balance = account_balance + deposit_amount;

}

// return account balance.

return account_balance;

}

int main()

{

// Declare the variables.

float account_balance;

float deposit_amount;

float withdrawl_amount;

char input;

// display the statement on console.

printf("BANK ACCOUT PROGRAM!\n");

printf("----------------------------------\n");

// prompt the user to enter the old balance.

printf("Enter the old balance: ");

// Input balance

scanf("%f", &account_balance);

// Display the statement on console.

printf("Enter the transactions now.\n");

printf("Enter an F for the transaction type when you are finished.\n");

// Start the do while loop

do

{

// prompt the user to enter transaction type.

printf("Transaction Type (D=deposit, W=withdrawal, F=finished): ");

// Input type.

scanf(" %c", &input);

// Check if the input is D

if (input == 'D')

{

// Prompt the user to input amount.

printf("Amount: ");

// input amount.

scanf("%f", &deposit_amount);

// Call to the function.

account_balance=deposit(account_balance,deposit_amount);

}

// Check if the input is W

if (input == 'W')

{

printf("Amount: ");

scanf("%f", &withdrawl_amount);

// Call to the function.

account_balance = withdraw(account_balance,withdrawl_amount);

}

// Check if the input is F

if (input == 'F')

{

// Dispplay the amount.

printf("Your ending balance is $%.2f\n", account_balance);

printf("Program is ending\n");

}

// End the while loop

} while(input != 'F');

return 0;

}

the picture uploaded below shows the program screenshot.

cheers, i hope this helps.

5 0
3 years ago
Example – a 100 kW, 60 Hz, 1175 rpm motor is coupled to a flywheel through a gearbox • the kinetic energy of the revolving compo
rjkz [21]

Answer:

1200KJ

Explanation:

The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.

P (rotor-loss) = 3 x K.E

P = 3 x 300 = 900 KJ

After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;

KE = 300 KJ

Since it is in opposite direction, it will also add up to rotor loss

P ( rotor loss ) = 900 + 300 = 1200 KJ

7 0
3 years ago
A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro
Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= \frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt

= \int\limits^a_b {(5-3t)} \, dt

5t - \frac{3t^2}{2} +c

v2 = 5x2 -  3x2 + c

= 10-6+c

= 4+c

s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c

S2 - S1

=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0
2 years ago
A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can
Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a)   ln(T2/T1)=μβ         β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π  radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π  

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb     T1=80 lb   μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0
3 years ago
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