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kap26 [50]
3 years ago
13

A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

mm will experience only elastic deformation when a tensile load of 1810 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.46 mm.
Engineering
1 answer:
IrinaVladis [17]3 years ago
5 0

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

l = 3161.025289 / 7240

l = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

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Show that for a linearly separable dataset, the maximum likelihood solution for the logisitic regression model is obtained by fi
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Answer:

Answer for the question:

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is explained in the attachment.

Explanation:

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3 years ago
What impact does modulus elasticity have on the structural behavior of a mechanical design?
devlian [24]

Answer with Explanation:

The modulus of elasticity has an profound effect on the mechanical design of any machine part as explained below:

1) Effect on the stiffness of the member: The ability of any member of a machine to resist any force depends on the stiffness of the member. For a member with large modulus of elasticity the stiffness is more and hence in cases when the member has to resist a direct load the member with more modulus of elasticity resists the force better.

2)Effect on the deflection of the member: The deflection caused by a force in a member is inversely proportional to the modulus of elasticity of the member thus in machine parts in which we need to resist the deflections caused by the load we can use materials with greater modulus of elasticity.

3) Effect to resistance of shear and torque: Modulus of rigidity of a material is found to be larger if the modulus of elasticity of the material is more hence for a material with larger modulus of elasticity  the resistance it offer's to shear forces and the torques is more.

While designing a machine element since the above factors are important to consider thus we conclude that modulus of elasticity has a profound impact on machine design.

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3 years ago
This assignment is designed to test your understanding of graphical user interface components and Event Driven programming in Ja
lys-0071 [83]

Answer:

Java program is given below

Explanation:

JavaPadGUI.java

package javapad.gui;

import java.awt.BorderLayout;

import java.awt.Container;

import java.awt.Dimension;

import java.awt.event.ActionEvent;

import java.awt.event.ActionListener;

import java.awt.event.WindowAdapter;

import java.awt.event.WindowEvent;

import java.io.File;

import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

import javax.swing.JButton;

import javax.swing.JFrame;

import javax.swing.JLabel;

import javax.swing.JOptionPane;

import javax.swing.JPanel;

import javax.swing.JScrollPane;

import javax.swing.JTextArea;

public class JavaPadGUI extends JFrame implements ActionListener{

  //constants

  private static final String TITLE = "Macrosoft JavaPad XP";

  private static final String SLOGAN = "Macrosoft: Resistance is futile";

 

  //button names

  private static final String NEW = "New";

  private static final String LOAD = "Load";

  private static final String SAVE = "Save";

  private static final String QUIT = "Quit";

 

  private static final String FILENAME = "hardcode.txt";

  //messages

 

  private static final String QUIT_MSG = "Quitting, Save?";

  private static final String LOAD_ERR = "Could not access file " + FILENAME;

 

 

  //fields

  private JTextArea txtContent;

  private JButton butNew, butLoad, butSave, butQuit;

 

  public JavaPadGUI()

  {

      super(TITLE);

      createUI();

      setSize(new Dimension(400, 300));

      setDefaultCloseOperation(EXIT_ON_CLOSE);

         

  }

 

  private void createUI()

  {

      Container c = getContentPane();

      c.setLayout(new BorderLayout(30,30));

     

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      JPanel butPanel = new JPanel();

      butPanel.add(butNew = new JButton(NEW));

      butPanel.add(butLoad = new JButton(LOAD));

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      butPanel.add(butQuit = new JButton(QUIT));

     

      //set command names for the buttons, these will be when button is clicked

      butNew.setActionCommand(NEW);

      butLoad.setActionCommand(LOAD);

      butSave.setActionCommand(SAVE);

      butQuit.setActionCommand(QUIT);

     

      JPanel sloganPanel = new JPanel();

      sloganPanel.add(new JLabel(SLOGAN));

     

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      txtContent.setLineWrap(true);

         

      JScrollPane scroll = new JScrollPane(txtContent);

     

      //now add all components

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          Scanner s = new Scanner(new File(FILENAME));

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RunJavaPad.java

package javapad;

import javapad.gui.JavaPadGUI;

public class RunJavaPad {

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      JavaPadGUI gui = new JavaPadGUI();

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7 0
3 years ago
Steam at a pressure of 100 bar and temperature of 600 °C enters an adiabatic nozzle with a velocity of 35 m/s. It leaves the noz
butalik [34]

Answer:

Exit velocity V_2=1472.2 m/s.

Explanation:

Given:

At inlet:

P_1=100 bar,T_=600°C,V_1=35m/s

Properties of steam at 100 bar and 600°C

        h_1=3624.7\frac{KJ}{Kg}

At exit:Lets take exit velocity V_2

We know that if we know only one property inside the dome  then we will find the other property by using steam property table.

Given that dryness or quality of steam at the exit of nozzle  is 0.85 and pressure P=80 bar.So from steam table we can find the other properties.

Properties of saturated steam at 80 bar

   h_f= 1316.61\frac{KJ}{Kg} ,h_g= 2757.8\frac{KJ}{Kg}

So the enthalpy of steam at the exit of turbine  

h_2=h_f+x(h_g-h_f)\frac{KJ}{Kg}

h_2=1316.61+0.85(2757.8-1316.61)\frac{KJ}{Kg}

 h_2=2541.62\frac{KJ}{Kg}

Now from first law for open system

h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w

In the case of adiabatic nozzle Q=0,W=0

3624.7+\dfrac{35^2}{2000}+0=2541.62+\dfrac{(V_2)^2}{2000}+0

V_2=1472.2 m/s

So Exit velocity V_2=1472.2 m/s.

4 0
4 years ago
What is a semiconductor whose electrical properties are based on the electronic structure inherent in the pure materials? MATSE
OleMash [197]

Answer:

intrinsic semiconductors

Explanation:

An intrinsic semiconductor is also known as a pure conductor. In such a semiconductor there are no impurities, that is why it is said to be pure.

It has some of these properties:

1. Electrical conductivity is only based on temperature

2. The quantity of electrons is the same as the number of holes in the valence bond

3. Electrical conductivity is not on the high side

4. These materials exist in their pure forms.

6 0
3 years ago
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