1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
rusak2 [61]
3 years ago
11

A large part in a turbine-generator unit operates near room temperature and is made of ASTM A470-8 steel ( ). A surface crack ha

s been found that is roughly a semi-ellipse, with surface length 2c= 50 mm and a depth a= 15 mm. the stress normal to the plane of the crack is 250 MPa, and the member width and thickness are large compared to the crack size. What is the safety factor against brittle fracture? Should the power plant
Engineering
1 answer:
mr Goodwill [35]3 years ago
3 0

Answer: safety factor= 1.26

safety factor does not exceed 2. Therefore, it is not safe to operate.

Explanation:

Express the stress infinity factor (k) that relate to the crack length (Q) applied stress for the centre crack of the plate.

K= Fsq √ πa/Q

Here,

the dimension less function quality is F

Crack length is a

Q= 1 + 1.48(a/c)^1.65

The surface length is c

c= 25mm

a= 15mm

Substitute c= 25mm and a=15mm into the equation

Q= 1 + 1.48(15mm/25mm)^1.65

Q= 1 + 1.48(0.6)^1.65

Q= 1 + 0.630

Q= 1.630

Find the stress intensity factor of the corresponding ratio of 0.4 and 0.12

Where

F= 1.12

S= 250mpa

Q= 1.63

a= 15mm

Substitute into the qequation

K= (1.12) (250mpa) √π(15mm× (1/1000mm))/1.63

K= (1.12) (250mpa)√47.1mm× (1/1000mm))/1.63

K= 2.80mpa √0.0471× (1m)/1.63

K= 280mpa × 0.170√m

K= 476mpa√m

Calculate the safety

Xk= Kk/K

Where fraction toughness is Kk

From the table of fraction toughness, corresponding tensile property for metal at room temperature, select this fraction toughness for ASTM A470-8 steel.

Substitute 47.6mpa√m for k

60mpa√m for Kk

Xk= 60mpa√m/47.6mpa√m

Xk= 1.26

In conclusion, from the above result, safety factor does not exceed 2. Therefore, it is not safe to operate.

You might be interested in
A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq
Sever21 [200]

Answer:

a) Solid solution strengthening and alloying,  Precipitation hardening, work hardening

b) Absence of enough  crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

<u>A) strengthening mechanism</u>

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as:  Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as :  Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0
3 years ago
3. If nothing can ever be at absolute zero, why does the concept exist?
JulijaS [17]
The reason has to do with the amount of work necessary to remove heat from a substance, which increases substantially the colder you try to go. To reach zero kelvins, you would require an infinite amount of work.
6 0
2 years ago
0 - 1"<br> -20<br> -15<br> -10<br> 5<br> 0 1 2 3<br> 0
faust18 [17]

Answer:

#WeirdestQuestionOfAllTime

Explanation:

8 0
3 years ago
Read the following statement.
alekssr [168]
The correct answer is A
5 0
4 years ago
Read 2 more answers
How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an
iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

or

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0
3 years ago
Other questions:
  • 6.27 LAB: Convert to binary - functions Write a program that takes in a positive integer as input, and outputs a string of 1's a
    5·1 answer
  • Write the definition of a function printLarger, which has two int parameters and returns nothing. The function prints the larger
    14·1 answer
  • (I'm bored and have points to blow so here is a riddle. Don't try to look it up, I made it up just now.)
    14·2 answers
  • What material has the highest strength to weight ratio?
    9·2 answers
  • In your new role at Wayne Industries, you have been given the freedom to propose and develop your own project ideas. You have an
    14·1 answer
  • A piece of corroded steel plate was found in submerged ocean vessel. It was estimated that the area of plate to be ~38 inch 2, a
    10·1 answer
  • 2. An AC current of
    9·1 answer
  • On-site oil storage containers must be marked “Used Oil.”<br> True<br> False
    12·1 answer
  • Which of the following is an example of a hardwood? A maple B spruce C pine D fir
    15·1 answer
  • What do one drop, rockers, and steppers all have in common?.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!