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
Fantom [35]
3 years ago
7

Cng containers need to be inspected

Engineering
1 answer:
nevsk [136]3 years ago
3 0
Texas: The owner of the vehicle must prove that the container meets the inspection criteria of FMVSS 304. (Every 36 months or 36,000 miles, whichever occurs first.) 5. Utah: Inspection is required every 36 months or 36,000 miles, whichever occurs first.
You might be interested in
MCQ : What are the main the constituents of acid rains?
nexus9112 [7]

Answer:

a. Nitrogen Oxide

3 0
3 years ago
Read 2 more answers
To reduce the drag coefficient and thus improve the fuel efficiency of cars,the design of side rearview mirrors has changed dram
marissa [1.9K]

Answer:

Amount of fuel used per year is supposed to be 34150 KJ/kg

4 0
3 years ago
Use the map to answer the question.
djverab [1.8K]

Answer:

Captain falcon

Explanation:

7 0
2 years ago
Technician A says that the most commonly used combustion chamber types include hemispherical, and wedge. Technician B says that
Inessa05 [86]

Answer:

Technician A and Technician B both are correct.

Explanation:

Technician A accurately notes that perhaps the forms of combustion process most widely used are hemispherical and cross.

Technician B also correctly notes that in several cylinder heads, cooling system and greases gaps and pathways are found.

6 0
3 years ago
A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the s
Xelga [282]

Answer:

See explanation

Explanation:

Given The bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut: 600 MPa

Maximum temperature

Tmax: 500 °C

Bar side dimension

b: 150 mm

Alternating stress

σa: 100 MPa

Reliability

R: 0.999 Note 1.

Assumptions Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

Solution See Excel file Ex06-01.xls.

1 Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a.

S'e: 300 MPa = 0.5 * Sut

2 The loading is bending so the load factor from equation 6.7a is

Cload: 1

3 The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d

A95: 1125 mm2 = 0.05 * b * b Note 2.

dequiv: 121.2 mm = SQRT(A95val / 0.0766)

and the size factor is found for this equivalent diameter from equation 6.7b, to be

Csize: 0.747 = 1.189 * dequiv^-0.097

4 The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish.

Table 6-3 constants

A: 57.7

b: -0.718 Note 3.

Csurf: 0.584 = Acoeff * Sut^bCoeff

5 The temperature factor is found from equation 6.7f :

Ctemp: 0.710 = 1 - 0.0058 * (Tmax - 450)

6 The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab: 0.753

7 The corrected endurance limit Se can now be calculated from equation 6.6:

Se: 69.94 MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se: 70 MPa

8 To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading.

Sm: 540 MPa = 0.9 * Sut

9 The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles.

b: -0.2958 Note 4.

a: 4165.7

Plotting Sn as a function of N from equation 6.10a

N Sn (MPa)

1000 540 =aa*B73^bb

2000 440

4000 358

8000 292

16000 238

32000 194

64000 158

128000 129

256000 105

512000 85

1000000 70

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10 The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn.

At N = 103 cycles,

Sn3: 540 MPa = aa * 1000^bb

At N = 106 cycles,

Sn6: 70 MPa = aa * 1000000^bb

The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.

8 0
3 years ago
Other questions:
  • According to Manor, the example of the subway train in New York City is an example of which type of uniqueness?
    9·1 answer
  • a digital multimeter is set to read dc volts on the 4 volt scale the meter leads are connected to a 12 volt battery what will th
    14·2 answers
  • Describing Tasks for Stationary Engineers Click this link to view O*NET’s Tasks section for Stationary Engineers. Note that comm
    12·2 answers
  • Can someone please help me this is urgent!?
    12·2 answers
  • Technician A says that reinforcements may be made of plastic.
    6·1 answer
  • New ventures that are based on strategic value, such as valuable technology, are attractive while those with low or no strategic
    12·2 answers
  • AA
    10·1 answer
  • Name eight safety electrical devices including their functions and effects if not present.​
    15·1 answer
  • There are three homes being built, each with an identical deck on the back. Each deck is comprised of two separate areas. One ar
    7·1 answer
  • I NEED HELP!!!Situation: A client has hired Jose, a materials engineer, to develop a package for an item he has begun to market.
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!