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scoundrel [369]
3 years ago
9

For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta

nts: the shear modulus G, the bulk modulus K, and the Lames’ constants, μ and λ.
Suppose that G = E/[2(1+v)].
Prove that μ and λ, and K, satisfy the following relations:

μ = G = 3K(1-2v)/[2(1+v)], λ = Ev/ [(1+v)(1-2v)] = 2vG/ (1-2v),
K = E/ 3(1-2v) = λ+(2/3)μ

To prove it, apply pure shear stress τ, and connect it to pure shear strain γ by G, ie. τ = Gγ. Then in the 45° orientation, consider its normal stresses σ, and σ: are related to its normal strains ε, and e, through E and v in Hooke's law. We can easily establish this relation.
Engineering
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

khgy

Explanation:

nbv

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Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

 

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

5 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
Write a simple phonebook program that reads in a series of name-number pairs from the user (that is, name and number on one line
Vlad1618 [11]

Answer:

import java.util.HashMap;

import java.util.Map;

import java.util.Scanner;

public class PhoneBook {

   public static void main(String[] args) {

       Scanner in = new Scanner(System.in);

       Map<String, String> map = new HashMap<>();

       String name, number, choice;

       do {

           System.out.print("Enter name: ");

           name = in.next();

           System.out.print("Enter number: ");

           number = in.next();

           map.put(name, number);

           System.out.print("Do you want to try again(y or n): ");

           choice = in.next();

       } while (!choice.equalsIgnoreCase("n"));

       System.out.print("Enter name to search for: ");

       name = in.next();

       if (map.containsKey(name)) {

           System.out.println(map.get(name));

       } else {

           System.out.println(name + " is not in the phone book");

       }

   }

}

6 0
3 years ago
A negative normal strain can be considered to increase or decrease volume depending on the coordinate system used. a)True b)- Fa
geniusboy [140]

Answer:

The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is

b) False

Explanation:

Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.

Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.

6 0
3 years ago
A 15,000 kg satellite is orbiting at 800 km above the Earth’s surface. What is the potential energy of the satellite?
vredina [299]

Answer:

-8.34\times 10^{11}\ \text{J}

Explanation:

m = Mass of satellite = 15000 kg

h = Distance above Earth = 800 km

R = Radius of Earth = 6371 km

M = Mass of Earth = 5.972\times 10^{24}\ \text{kg}

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Potential energy is given by

U=-\dfrac{GMm}{R+h}\\\Rightarrow U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 15000}{(6371+800)\times 10^3}\\\Rightarrow U=-8.34\times 10^{11}\ \text{J}

The potential energy of the satellite is -8.34\times 10^{11}\ \text{J}.

6 0
3 years ago
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