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Anna007 [38]
3 years ago
6

Consider uniaxial extension of a test specimen. It has gauge length L = 22 cm (the distance between where it is clamped in the t

esting fixture) with a cross section of diameter D = 1.8 mm. You expect the material to have a Young’s modulus E = 205 GPa and ultimate tensile strength of σu = 420 MPa. You would like to conduct a tensile test such that the test takes 2 minutes to go from initial loading to fracture. In completing the calculations below, assume the material’s stress-strain curve is perfectly linear until fracture.
Required:
a. At what force do you expect the specimen to fracture?
b. At what strain rate should you pull the specimen, in mm/minute? Assume the material undergoes brittle fracture and its stress-strain curve is perfectly linear to fracture.
c. What is the elongation (in inches) of the specimen in the moment just prior to fracture?
Engineering
1 answer:
Katena32 [7]3 years ago
3 0

Answer:

A) 4800.6 N

B) 0.2255 mm/min

C) 8.6792 inches

Explanation:

Given data :

gauge Length = 22 cm = 220 mm

cross section diameter/thickness = 1.8 mm

young's modulus ( E ) = 205 GPa

Ultimate tensile strength ( σu ) = 420MPa

Time = 2 mins

A) what force would fracture the specimen

σU = Ff /A

where A (area) = thickness * width ( unknown )  hence we assume a width of 6.35 mm

Ff = σU × A =  420 * 1.8 * 6.35 = 4800.6 N

B) calculate the strain rate

Ultimate tensile strength ( σU ) = εE

ε = σU / E = 420/205000 = 0.00205

also  ε ​​​​​​ =  ΔL / Li = ( Lf - Li) / Li = (Lf / Li) - 1

 therefore  ε ​​​​​​ + 1 = (Lf / Li) = 1 + 0.00205 = 1.00205

 Li = gauge length = 220 mm

Hence :  Lf = 1.00205 × gauge length =  1.00205 * 220 =  220.451 mm

  strain rate

(Lf - Li) / Time = (0.451) / 2 min = 0.2255 mm/min

C) Elongation in inches prior to fracture

ΔL = ε × Li = 0.451 mm

Lf = 220 mm + 0.451 mm = 220.451 mm =  8.6792 inches

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Answer:

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Explanation:

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Circumference of rear wheel =  \pi d=26\piπd=26π

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Converting to mph

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Answer:

Java program is given below. You can get .class after you execute java programs, You can attach those files along with .java classes given , Those .class files are generated ones.

Explanation:

//Rectangle.java class

public class Rectangle {

private int x;

private int y;

private int width;

private int height;

// constructs a new Rectangle with the given x,y, width, and height

public Rectangle(int x, int y, int w, int h)

{

this.x=x;

this.y=y;

this.width=w;

this.height=h;

}

// returns the fields' values

public int getX()

{

return x;

}

public int getY()

{

return y;

}

public int getWidth()

{

return width;

}

public int getHeight()

{

return height;

}

// returns a string such as “Coordinate is (5,12) and dimension is 4x8” where 4 is width and 8 is height. public String toString()

public String toString()

{

String str="";

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str+="Coordinate is ("+x+","+y+")";

str+=" and dimension is : "+width+"x"+height;

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return str;

}

public void changeSize(int w,int h)

{

width=w;

height=h;

}

}

======================

//main.java

class Main {

public static void main(String[] args) {

//System.out.println("Hello world!");

//create an object of class Rectangle

Rectangle rect=new Rectangle(5,12,4,8);

//print info of rect using toString method

System.out.println(rect.toString());

//chamge width and height

rect.changeSize(3,10);

//print info of rect using toString method

System.out.println(rect.toString());

}

}

==========================================================================================

//Output

Coordinate is (5,12) and dimension is : 4x8 Area is 32

Coordinate is (5,12) and dimension is : 3x10 Area is 30

========================================================================================

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