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3 years ago

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A tensile test was operated to test some important mechanical properties. The specimen has a gage length = 1.8 in and diameter =

0.8 in. Yielding occurs at a load of 30,000 lb. The corresponding gage length = 1.8075 in, which is the 0.2 percent yield point. The maximum load of 56,050 lb is reached at a gage length = 2.35 in. Determine (a) yield strength, (b) modulus of elasticity, and (c) tensile strength. (d) If fracture occurs at a gage length of 2.5 in, determine the percent elongation. (e) If the specimen necked to an area=0.35 in^2, determine the percent reduction in area.

1 answer:

oee [108]3 years ago

5 0

Answer:

a) 60000 psi

b) 1.11*10^6 psi

c) 112000 psi

d) 30.5%

e) 30%

Explanation:

The yield strength is the load applied when yielding behind divided by the section.

yield strength = Fyield / A

A = π/4 * D^2

A = 0.5 in^2

ys = Fy * A

y2 = 30000 * 0.5 = 60000 psi

The modulus of elasticity (E) is a material property that is related to the object property of stiffness (k).

k = E * L0 / A

And the stiffness is related to change of length:

Δx = F / k

Then:

Δx = F * A / (E * L0)

E = F * A / (Δx * L0)

When yielding began (approximately the end of the proportional peroid) the force was of 30000 lb and the change of length was

Δx = L - L0 = 1.8075 - 1.8 = 0.0075

Then:

E = 30000 * 0.5 / (0.0075 * 1.8) = 1.11*10^6 psi

Tensile strength is the strees at which the material breaks.

The maximum load was 56050 lb, so:

ts = 56050 / 0.5 = 112000 psi

The percent elongation is calculated as:

e = 100 * (L / L0)

e = 100 * (2.35 / 1.8 - 1) = 30.5 %

If it necked with and area of 0.35 in^2 the precent reduction in area was:

100 * (1 - A / A0)

100 * (1 - 0.35 / 0.5) = 30%

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Answer:

(a) attached below

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Explanation:

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$S_{2} =10 kVA$

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$pf_{2}=0.8$ $leading$

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$\alpha_{1}=cos^{-1} (0.6)=53.13$ °

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The negative sign means that the load 2 is providing reactive power rather than consuming

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or in the polar form

$S_{c} =26.92$ °

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The relationship between Apparent power S and Current I is

$S=VI^{*}$

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

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Current of the combined load can be found by

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$X_{C} =3*(480)^2/14*10^3$ Ω

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Current flowing in the capacitor is

$I_{cap} =V/X_{C} =480/49.37=9.72 A$

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$I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A$

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3 years ago

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import java.io.FileNotFoundException;

import java.io.PrintWriter;

import java.util.Scanner;

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public static void main(String[] args) throws FileNotFoundException

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lastname = sc.next();

strSalary = sc.next();

status = sc.next();

cityState = sc.next();

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for(i=0; i<strSalary.length() && c != '.'; i++)

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c = strSalary.charAt(i);

if(!Character.isDigit(c))

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System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

}

}

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{

if(!Character.isDigit(strSalary.charAt(i)))

{

System.out.println("Char at position " + (i+1) + " in salary is not a digit");

incorrectRecords++;

continue;

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System.out.println("Char at position " + (i+1+1) + " in salary is not a digit");

incorrectRecords++;

continue;

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System.out.println("Period is in the wrong position. Expected at " + (strSalary.length()-3) + " but found at " + (i+1));

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6 0

3 years ago

A small pad subjected to a shearing force is deformed at the top of the pad 0.08 in. The height of the pad is 1.38 in. What is t

Aleksandr-060686 [28]

Answer:

The shear strain is 0.05797 rad.

Explanation:

Shear strain is the ratio of change in dimension along the shearing load direction to the height of the plate under application of shear load. Width of the plate remains same. Length of the plate slides under shear load.

Step1

Given:

Height of the pad is 1.38 in.

Deformation at the top of the pad is 0.08 in.

Calculation:

Step2

Shear strain is calculated as follows:

$tan\phi=\frac{\bigtriangleup l}{h}$

$tan\phi=\frac{0.08}{1.38}$

$tan\phi= 0.05797$

For small angle of $\phi$ , $tan\phi$ can take as $\phi$ .

$\phi = 0.05797 rad.$

Thus, the shear strain is 0.05797 rad.

7 0

3 years ago