The product of two factors is 4,500. If one of the factors is 90, which is the other factor?

Compare the use of a low-strength, ductile material (1018 CD) in which the stress-concentration factor can be ignored to a high-

kicyunya [14]

Answer:

Step 1 of 3

Case A:

AISI 1018 CD steel,

Fillet radius at wall=0.1 in,

Diameter of bar

From table deterministic ASTM minimum tensile and yield strengths for some hot rolled and cold drawn steels for 1018 CD steel

Tensile strength

Yield strength

The cross section at A experiences maximum bending moment at wall and constant torsion throughout the length. Due to reasonably high length to diameter ratio transverse shear will be very small compared to bending and torsion.

At the critical stress elements on the top and bottom surfaces transverse shear is zero

Explanation:

See the next steps in the attached image

5 0

4 years ago

What is the foundation

vova2212 [387]

Answer:

foundation is ........see more

7 0

3 years ago

The files provided in the code editor to the right contain syntax and/or logic errors. In each case, determine and fix the probl

Yanka [14]

Question Continuation

public class DebugOne3{

public static void main(String args){

System.out.print1n("Over the river");

System.out.pr1ntln("and through the woods");

system.out.println("to Grandmother's house we go");

}

Answer:

Line 2: Invalid Syntax for the main method. The main method takes the form

public static void main (String [] args) { }

or

public static void main (String args []) { }

Line 3: The syntax to print is wrong.

To print on a new line, make use of System.out.println(".."); not System.out.print1n();

Line 4:

To print on a new line, make use of System.out.println(".."); not System.out.pr1ntln();

Line 5:

The case of "system" is wrong.

The correct case is sentence case, "System.out.println" without the quotes

The correct program goes, this:

public class DebugOne3{

public static void main(String [] args){

System.out.println("Over the river");

System.out.println("and through the woods");

System.out.println("to Grandmother's house we go");

}

Explanation:

3 0

3 years ago

What are materials engineers trying to discover when they study different materials?

coldgirl [10]

What do materials science and engineering graduates do?
Almost two-thirds of materials science graduates are in employment six months after graduation.

The skills developed during a materials science degree allow graduates enter a range of sectors, including working as engineering professionals and in design and marketing roles.

Destination Percentage
Employed 60.4
Further study 24.5
Working and studying 5.1
Unemployed 4
Other 6.1
Graduate destinations for materials science and engineering
Type of work Percentage
Engineering and building 23.9
Marketing, PR and sales 11.9
Business, HR and financial 11.7
Technicians and other professionals 10
Other 42.5

5 0

3 years ago

Read 2 more answers

Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and

Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒ P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0

4 years ago