All categories

4 years ago

What does product integration refer to in an advanced manufacturing setting?

Engineering

2 answers:

creativ13 [48]4 years ago

6 0

Answer:

C!!

Explanation:

Combining all manufacturing processes to provide higher efficiency and fulfilling the requeriments.

inessss [21]4 years ago

3 0

Answer:

Product integration refer to in an advanced manufacturing setting as a (D) Bundling software with components to create a complete product

Explanation:

Product Integration means to integrate the product line. Let us go by basic terms. Integration means to join. So product integration in Advanced Manufacturing Setting means to join the manufacturing processes.

Now product have 2 vital stages apart from development design etc., which are supply and receipt of products. In any manufacturing stage, in ordrer to process a product, raw material comes in, product gets processed, and finally the product is fed further.

So basically Product Integration means integrating/connecting Receipts, which are inputs received from previous step, and Supply (Output of the current step).

You might be interested in

The shaft is made of A992 steel. It has a diameter of 1 in. and is supported by bearings at A and D, which allows free rotation.

zysi [14]

Answer:

the angle of twist of B with respect to D is -1.15°

the angle of twist of C with respect to D is 1.15°

Explanation:

The missing diagram that is supposed to be added to this image is attached in the file below.

From the given information:

The shaft is made of A992 steel. It has a diameter of 1 in and is supported by bearing at A and D.

For the Modulus of Rigidity G = 11 × 10³ Ksi = 11 × 10⁶ lb/in²

The objective are :

1) To determine the angle of twist of B with respect to D

Considering the Polar moment of Inertia at the shaft $J\tau$

shaft $J\tau$ = $\dfrac{\pi}{2}r^4$

where ;

r = 1 in /2

r = 0.5 in

shaft $J \tau$ = $\dfrac{\pi}{2} \times 0.5^4$

shaft $J\tau$ = 0.098218

Now; the angle of twist at B with respect to D is calculated by using the expression

$\phi_{B/D} = \sum \dfrac{TL}{JG}$

$\phi_{B/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

where;

$T_{CD} \ \ and \ \ L_{CD}$ are the torques at segments CD and length at segments CD

${T_{BC} \ \ and \ \ L_{BC}}$ are the torques at segments BC and length at segments BC

Also ; from the diagram; the following values where obtained:

$L_{BC}}$ = 2.5 in

$J\tau$ = 0.098218

G = 11 × 10⁶ lb/in²

$T_{BC$ = -60 lb.ft

$T_{CD$ = 0 lb.ft

$L_{CD$ = 5.5 in

$\phi_{B/D} = 0+ \dfrac{[(-60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{B/D} = \dfrac{[(-720 )] (30 )}{1079980}$

$\phi_{B/D} = \dfrac{-21600}{1079980}$

$\phi_{B/D} =$ − 0.02 rad

To degree; we have

$\phi_{B/D} = -0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{B/D} = -1.15^0}$

Since we have a negative sign; that typically illustrates that the angle of twist is in an anti- clockwise direction

Thus; the angle of twist of B with respect to D is 1.15°

(2) Determine the angle of twist of C with respect to D.Answer unit: degree or radians, two decimal places

For the angle of twist of C with respect to D; we have:

$\phi_{C/D} = \dfrac{T_{CD}L_{CD}}{JG}+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{C/D} = 0+\dfrac{T_{BC}L_{BC}}{JG}$

$\phi_{B/D} = 0+ \dfrac{[(60 \times 12 )] (2.5 \times 12 )}{ (0.9818)(11 \times 10^6)}$

$\phi_{C/D} = \dfrac{21600}{1079980}$

$\phi_{C/D} =$ 0.02 rad

To degree; we have

$\phi_{C/D} = 0.02 \times \dfrac{180}{\pi}$

$\mathbf{\phi_{C/D} = 1.15^0}$

3 0

3 years ago

The symmetrical load below is connected to a three-phase network. A line current of 25A has been measured. The load resistors ha

boyakko [2]

Answer:

The line voltage of the three phase network is 346.41 V

Explanation:

Star Connected Load

Resistance, R₁ = R₂ = R₃ = 18 Ω

For a star connected load, the line current = the phase current, that is we have

$I_L = 25 \, A = I_{Ph}$

Whereby the the voltage across each resistance = $V_R$ is given by the relation;

$V_R$ = $I_{Ph}$ × R

Hence;

$V_{Ph}$ = $V_R$ = $I_{Ph}$ × R = 25 × 8 = 200 V

Therefore we have;

The line voltage, $V_{L}$ = √3 × $V_{Ph}$ = √3 × 200 = 346.41 V.

Hence, the line voltage of the three phase network = 346.41 V.

3 0

4 years ago

Aerospace engineers who work for certain government agencies are often required to have security clearance. Explain two reasons

yuradex [85]

Answer:

Two reasons that justify the requirement for security clearance for aerospace engineers working for government agencies are;

1) Such engineers have access to data regarding the blueprint, components, method of construction, functionality status, new systems design, future systems design, inventory of systems and aeronautical systems database which are sensitive information that are of high importance to the federal government

2) Such engineers take part in the testing of aeronautic equipment, and will require security clearance to be able to input data results into the data base of the aeronautic equipment

Explanation:

3 0

3 years ago

A horizontal curve is being designed for a new two-lane highway (12-ft lanes). The PI is at station 250 + 50, the design speed i

kogti [31]

Answer

given,

Speed of vehicle = 65 mi/hr

= 65 x 1.4667 = 95.33 ft/s

e = 0.07 ft/ft

f is the lateral friction, f = 0.11

central angle,Δ = 38°

The PI station is

PI = 250 + 50

= 25050 ft

using super elevation formula

$e + f = \dfrac{v^2}{rg}$

$0.07 + 0.11 =\dfrac{95.33^2}{r\times 32.2}$

$r = \dfrac{95.33^2}{32.2\times 0.18}$

r = 1568 ft

As the road is two lane with width 12 ft

R = 1568 + 12/2

R = 1574 ft

Length of the curve

$L = \dfrac{\piR\Delta}{180}$

$L = \dfrac{\pi\times 1574\times 38}{180}$

L = 1044 ft

Tangent of the curve calculation

$T = R tan(\dfrac{\Delta}{2})$

$T = 1574 tan(\dfrac{38}{2})$

T = 542 ft

The station PC and PT are

PC = PI - T

PC = 25050 - 542

= 24508 ft

= 245 + 8 ft

PT = PC + L

= 24508 + 1044

=25552

= 255 + 52 ft

the middle ordinate calculation

$MO = R(1-cos\dfrac{\Delta}{2})$

$MO = 1574\times (1-cos\dfrac{38}{2})$

MO = 85.75 ft

degree of the curvature

$D = \dfrac{5729.578}{R}$

$D = \dfrac{5729.578}{1574}$

D = 3.64°

8 0

4 years ago

The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 4930

Verizon [17]

Answer:

64640.92 psi

Explanation:

True stress ( psi ) True strain

49300 0.11

61300 0.21

Determine the true stress necessary to produce a true plastic strain of 0.25

бT1 = 49300

бT2 = 61300

бT3 = ?

∈T1 = 0.11

∈T2 = 0.21

∈T3 = 0.25

note : бTi = k ∈Ti^h

∴ 49300 = k ( 0.11 )^h ----- ( 1 )

61300 = k ( 0.21)^h ------ ( 2 )

solving equations 1 and 2 simultaneously

49300/61300 = ( 0.11 / 0.21 )^h

0.804 = (0.52 )^h

next step : apply logarithm

log ( 0.804 ) = log(0.52)^h

h = log 0.804 / log (0.52)

= 0.33

back to equation 1

49300 = k ( 0.11 )^0.33

k = 49300 / (0.11)^0.33

= 102138

therefore бT3 = K (0.25)^h

= 102138 ( 0.25 )^ 0.33

= 64640.92

5 0

3 years ago