# Find the minimum sum of products expression using Quine-McCluskey method of the function. F(A,B,C,D)= Î£ m(1,5,7,8,9,13,15)+ Î£ d(4,6,11)

Digital electronics deals with the discrete-valued digital signals. In general, any electronic system based on the digital logic uses binary notation (zeros and ones) to represent the states of the variables involved in it. Thus, Boolean algebraic simplification is an integral part of the design and analysis of a digital electronic system.

Explanation:

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8. 15 A manual arc welding cell uses a welder and a fitter. The cell operates 2,000 hriyr. The welder is paid $30/hr and the fitter is paid$25/hr. Both rates include applicable overheads. The cycle time to complete one welded assembly is 15. 4 min. Of this time, the arc-on time is 25%, and the fitter's participation in the cycle is 30% of the cycle time. A robotic arc welding cell is being considered to replace this manual cell. The new cell would have one robot, one fitter, and two workstations, so that while the robot is working at the first sta tion, the fitter is unloading the other station and loading it with new components. The fitter's rate would remain at $25/hr. For the new cell, the production rate would be eight welded assemblies per hour. The arc-on time would increase to almost 52%, and the fitter's participation in the cycle would be about 62%. The installed cost of the robot and worksta tions is$158,000. Power and other utilities to operate the robot and arc welding equipment will be $3. 80/hr, and annual maintenance costs are$3,500. Given a 3-year service life, 15% rate of return, and no salvage value, (a) determine the annual quantity of welded assem blies that would have to be produced to reach the break-even point for the two methods. (b) What is the annual quantity of welded assemblies produced by the two methods work. Ing 2,000 hryr?​

The annual quantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

To determine the break-even point between the manual arc welding cell and the robotic cell, we need to calculate the total costs for each method and then equate them.

For the manual arc welding cell:

Labor cost per hour = (welder's hourly rate x arc-on time) + (fitter's hourly rate x fitter's participation in the cycle) = ($30 x 0.25) + ($25 x 0.3) = $11.25 Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 =$11.25 x 15.4 / 60 = $2.89 Overhead cost per welded assembly = (labor cost per hour x (1 - arc-on time - fitter's participation in the cycle)) x cycle time per assembly / 60 = ($30 x 0.45) x 15.4 / 60 = $4.68 Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly =$2.89 + $4.68 =$7.57

Total cost per hour = total cost per welded assembly x production rate = $7.57 x 8 =$60.56

Total cost per year = total cost per hour x hours of operation per year = $60.56 x 2,000 =$121,120

For the robotic arc welding cell:

Labor cost per hour = fitter's hourly rate x fitter's participation in the cycle = $25 x 0.62 =$15.50

Labor cost per welded assembly = labor cost per hour x cycle time per assembly / 60 = $15.50 x 15.4 / 60 =$3.97

Overhead cost per welded assembly = power and utility cost per hour + annual maintenance cost / production rate = $3.80 +$3,500 / (8 x 2,000) = $3.80 +$0.22 = $4.02 Total cost per welded assembly = labor cost per welded assembly + overhead cost per welded assembly + (installed cost / (production rate x service life)) =$3.97 + $4.02 + ($158,000 / (8 x 3)) = $3.97 +$4.02 + $6,208.33 =$14.19

Total cost per hour = total cost per welded assembly x production rate = $14.19 x 8 =$113.52

Total cost per year = total cost per hour x hours of operation per year = $113.52 x 2,000 =$227,040

To find the break-even point, we set the total cost of the manual arc welding cell equal to the total cost of the robotic arc welding cell and solve for the annualquantity of welded assemblies:

$121,120 + x($7.57) = $227,040 + x($14.19)

$7.57x -$14.19x = $227,040 -$121,120

$-6.62x =$105,920

x = $105,920 /$6.62

x = 15,982.7

Therefore, the annualquantity of welded assemblies that would have to be produced to reach the break-even point for the two methods is approximately 15,983.

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Select the true statements regarding rigid bars. a. A rigid bar can bend but does not change length.
b. A rigid bar does not bend regardless of the loads acting upon it.
c. A rigid bar deforms when experiencing applied loads.
d. A rigid bar is unable to translate or rotate about a support.
e. A rigid bar represents an object that does not experience deformation of any kind.

option b and E are true

Explanation:

A lever is an example of a rigid bar that can rotate around a given point. In a rigid material, the existing distance does not change whenever any load is placed on it. In such a material, there can be no deformation whatsoever. Wit this explanation in mind:

option a is incorrect, given that we already learnt that no deformation of any kind happens in a rigid bar.

option b is true. A rigid bar remains unchanged regardless of the load that it carries.

option c is incorrect, a rigid bar does not deform with loads on it

option d is incorrect. A lever is a type of rigid bar, a rigid bar can rotate around a support.

option e is true. A rigid bar would not experience any deformation whatsoever.

Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.

Solution :

i. Slip plane (1 1 0)

Slip direction -- [1 1 1]

Applied stress direction = ( 1 0 0 ]

τ = 50 MPa    ( Here slip direction must be perpendicular to slip plane)

τ = σ cos Φ cos λ

τ = σ cos Φ cos λ

σ = 122.47 MPa

ii. Slip plane  --- (1 1 0)

Slip direction -- [1 1 1]

τ = σ cos Φ cos λ

σ = 122.47 MPa

iii. Slip plane  --- (1 0 1)

Slip direction --- [1 1 1]

τ = σ cos Φ cos λ

σ = 122.47 MPa

iv. Slip plane -- (1 0 1)

Slip direction  ---- [1 1 1]

τ = σ cos Φ cos λ

σ = 122.47 MPa

∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0

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

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Explanation:

Given data in question

bar yields load = 80 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 / /4 × D²

tensile stress at yield =  80 ×10³ / /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   / /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  / /4 × d²

average stress = 70 ×10³  / /4 × 0.010²

average stress = 892 MPa

As a means of preventing ice formation on the wings of a small, private aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine representative power requirements, consider nominal flight conditions for which the plane moves at 100 m/s in air that is at a temperature of -23 degree C. If the characteristic length of the airfoil is L = 2 m and wind tunnel measurements indicate an average friction coefficient of of C_f = 0.0025 for the nominal conditions, what is the average heat flux needed to maintain a surface temperature of T_s = 5 degree C?

Average heat flux=3729.82 W/

Explanation:

The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam material has a specific gravity, SG, of 3.1. You may assume that the dam is loosely attached to the ground at its base, though there is significant friction to keep it from sliding.Is the weight of the dam sufficient to prevent it from tipping around its lower right corner?

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = SGw * sin(30) * L * (h^2*y - h*y^2 + 1/3*y^3)(evaluated between 0 and h)

T1 = SGw * sin(30) * L * (h^2*h - h*h^2 + 1/3*h^3)

T1 = SGw * sin(30) * L * (h^3 - h^3 + 1/3*h^3)

T1 = 1/3 * SGw * sin(30) * L * h^3

To remain stable the equilibrant torque (Teq) must be of larger magnitude than the water pressure torque (T1)

1/4 * b^2 * h * L * SG > 1/3 * SGw * sin(30) * L * h^3

In an equilateral triangle h = b * cos(30)

1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

SG > SGw * 4/3* sin(30) * (cos(30))^2

SG > 1/2 * SGw

For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

This is avergae specific gravity, including holes.