ENGINEERING COLLEGE

What are the general principles of DFA? What are the steps to minimize the number of parts for an assembly?

Answers

Answer 1

Answer:

Answer Explanation : The general principles for design for assembly (DFA) are,

MINIMIZE NUMBER OF COMPONENT
USE STANDARD COMMERCIALLY AVAILABLE COMPONENTS
USE COMMON PARTS ACROSS PRODUCT LINES
DESIGN FOR EASE OF PART FABRICATION
DESIGN PARTS WITH TOLERANCE THAT ARE WITHIN PROCESS CAPABILITY
MINIMIZE USE OF FLEXIBLE COMPONENT
DESIGN FOR EASE OF ASSEMBLY
USE MODULAR DESIGN
REDUCE ADJUSTMENT REQUIRED

STEPS TO MINIMIZE THE NUMBER OF PARTS

USE OF INCORPORATE HINGS
USE OF INTEGRAL SPRINGS
USE OF SNAP FITS
USE OF GUIDES BEARINGS
USE OF COVERS

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Answer:

The solution is given in the attachments.

A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

Answers

Answer:

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Explanation:

Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

solution

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ / $\pi$ /4 × D²

tensile stress at yield = 80 ×10³ / $\pi$ /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ / $\pi$ /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ / $\pi$ /4 × d²

average stress = 70 ×10³ / $\pi$ /4 × 0.010²

average stress = 892 MPa

If you answer the whole question and show your work/coding I will rate 5 stars/brainliest!!! Walnut Orchard has two farms that grow wheat and corn. Because of different soil conditions, there are differences in the yields and costs of growing crops on the two farms. The yields and costs are shown in the following table. Each farm has 100 acres available for cultivation. 11,000 bushels of wheat and 7,000 bushels of corn must be grown. Please have an LP model to minimize the total cost while meeting the demand and solve it with Lindo or Excel. You need to have all parts of a model: notation, objective function, constraints, and sign restrictions.

Answers

Answer: 18,100

Explanation: two farms that grow wheat and corn.

The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.

Answers

Answer:

Specific weight of the pop, $w_(s) = 8619.45 N/m^(3)$

Density of the pop, $\rho_(p) = 8790.76 kg/m^(3)$

$g_(s) = 8.79076$

$w_(w) = 9782.36 N/m^(3)$

Given:

Volume of pop, V = 360 mL = 0.36 L = $0.36* 10^(-3) m^(3)$

Mass of a can of pop , m = 0.369 kg

Weight of an empty can, W = 0.153 N

Solution:

Now, weight of a full can pop, W

W' = mg = $0.369* 9.8 = 3.616 N$

Now weight of the pop in can is given by:

w = W' - W = 3.616 - 0.513 = 3.103 N

Now,

The specific weight of the pop, $w_(s) = (weight of pop)/(volume of pop)$

$w_(s) = (3.103)/(0.36* 10^(- 3)) = 8619.45 N/m^(3)$

Now, density of the pop:

$\rho_(p) = (w_(s))/(g)$

$\rho_(p) = (86149.45)/(9.8) = 8790.76 kg/m^(3)$

Now,

Specific gravity, $g_(s) = (\rho_(p))/(density of water, \rho_(w))$

where

$g_(s) = (8790.76)/(1000) = 8.79076$

Now, for water at $20^(\circ)c$ :

Specific density of water = $998.2 kg/m^(3)$

Specific gravity of water = $0.998 kg/m^(3)$

Specific weight of water at $20^(\circ)c$ :

$w_(w) = \rho_(20^(\circ))* g = 998.2* 9.8 = 9782.36 N/m^(3)$

In same Fig. A-3, a force F, having a slope of 2 vertical 3 horizontal, produces a clockwise moment of 330 ft-lb about A and a counterclockwise moment of 420 ft-lb about B. Compute the moment of F about C.

Answers

The moment of force at the given slope about point C is 210 ft-lb.

The given parameters:

Vertical slope, = 2
Horizontal slope = 3
Clockwise moment = 330 ft-lb
Counterclockwise moment = 420 ft-lb

The magnitude of the two moments are in the following simple ratio;

330:420 = 11:14 (divide through by 30)

the A coordinate = (0, 5)
the B-coordinate = (5,0)

The line of action of the force passes line AB at the final following coordinates;

total ratio of 11:14 = 11 + 14 = 25

$= (11 )/(25) * 5, \ \ (14)/(25) * 5\n\n= (2.2, \ 2.8)$

The position of C = (3, 1)

The resultant position of point C = (3 - 2.2, 2.8-1) = (0.8, 1.8)

The moment of force at the given slope about point C is calculated as;

3(1.8) + 2(0.8) = 7

Recall that this is the simplest form of the moment produced by the force.

Moment about C = 7 x 30 = 210 ft-lb

Thus, the moment of force at the given slope about point C is 210 ft-lb.

Learn more about moment of force here: brainly.com/question/6278006

Which statement best describes how power and work are related?O A. Power is the ability to do more work with less force.
O B. Power is a measure of how quickly work is done.
O C. Power and work have the same unit of measurement
O D. Power is the amount of work needed to overcome friction.
Pls answer quick

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