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3 years ago

The minimum requirements for engineering documents are enumerated in

Engineering

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23' from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

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The yield strength of mild steel is 150 MPa for an average grain diameter of 0.038 mm ; yield strength is 250 MPa for average gr

djyliett [7]

Answer:

Explanation:

Hall-Petch equation provides direct relations between the strength of the material and the grain size:

σ=σ0+k/√d , where d- grain size, σ- strength for the given gran size, σ0 and k are the equation constants.

As in this problem, we don't know the constants of the equation, but we know two properties of the material, we are able to find them from the system of equations:

σ1=σ0+k/√d1

σ2=σ0+k/√d2 , where 1 and 2 represent 150MPa and 250MPa strength of the steel.

Note, that for the given problem, there is no need to convert units to SI, as constants can have any units, which are convenient for us.

From the system of equations calculations, we can find constant: σ0=55.196 MPa, k=18.48 MPa*mm^(0.5)

Now we are able to calculate strength for the grain diameter of 0.004 mm:

σ=55.196+18.48/(√0.004)=347.39 MPa

The strength of the steel with the grais size of 0.004 mm is 347.39 MPa.

6 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2

IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length $l$ for the deformation, we use the following relation;

$l$ = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

$l$ = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

$l$ = 3161.025289 / 7240

$l$ = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0

3 years ago

Which branch of engineering studies the physical behavior of metallic elements?

olga55 [171]

Material engineering studies the physical behavior of metallic elements.

Answer: Option C

<u>Explanation: </u>

Material Engineering is the creation and learning about the materials at an atomic level. An engineers from this branch focus on material and model its characteristics using the computer.

Also, they combine the knowledge of solid-states, metallurgy, chemistry and ceramics to the application level. It also has a great role in building the future with the advancing study in nanotechnology, biotechnology, etc. Simply, these are meant to have vivid applications in future life.

6 0

3 years ago

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The term _______________refers to the science of using fluids to perform work.

Alla [95]

The term is hydraulics.

5 0

3 years ago

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Products that are in the process of being manufactured but are not yet complete are called:

vazorg [7]

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

6 0

3 years ago