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3 years ago

Select the correct answer.

Engineering

2 answers:

Lilit [14]3 years ago

8 0

Answer:

Explanation:

Confidential data is not supposed to be shared amongst others.

geniusboy [140]3 years ago

5 0

Answer:

Explanation:

Revealing confidential data is illegal first of all. And second business isn't allowed to share info due to the confidentiality law. All the others aren't unethical, they are just precautions for the safety of other employees.

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Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320

lesya [120]

Answer:

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

Explanation:

The complete statement of this question is "Five bolts are used in the connection between the axial member and the support. The ultimate shear strength of the bolts is 320 MPa, and a factor of safety of 4.2 is required with respect to fracture. Determine the minimum allowable bolt diameter required to support an applied load of P = 450 kN"

Each bolt is subjected to shear forces. In this case, safety factor is the ratio of the ultimate shear strength to maximum allowable shear stress. That is to say:

$n = \frac{S_{uts}}{\tau_{max}}$

Where:

$n$ - Safety factor, dimensionless.

$S_{uts}$ - Ultimate shear strength, measured in pascals.

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

The maximum allowable shear stress is consequently cleared and computed: ( $n = 4.2$ , $S_{uts} = 320\times 10^{6}\,Pa$ )

$\tau_{max} = \frac{S_{uts}}{n}$

$\tau_{max} = \frac{320\times 10^{6}\,Pa}{4.2}$

$\tau_{max} = 76.190\times 10^{6}\,Pa$

Since each bolt has a circular cross section area and assuming the shear stress is not distributed uniformly, shear stress is calculated by:

$\tau_{max} = \frac{4}{3} \cdot \frac{V}{A}$

Where:

$\tau_{max}$ - Maximum allowable shear stress, measured in pascals.

$V$ - Shear force, measured in kilonewtons.

$A$ - Cross section area, measured in square meters.

As connection consist on five bolts, shear force is equal to a fifth of the applied load. That is:

$V = \frac{P}{5}$

$V = \frac{450\,kN}{5}$

$V = 90\,kN$

The minimum allowable cross section area is cleared in the shearing stress equation:

$A = \frac{4}{3}\cdot \frac{V}{\tau_{max}}$

If $V = 90\,kN$ and $\tau_{max} = 76.190\times 10^{3}\,kPa$ , the minimum allowable cross section area is:

$A = \frac{4}{3} \cdot \frac{90\,kN}{76.190\times 10^{3}\,kPa}$

$A = 1.640\times 10^{-3}\,m^{2}$

The minimum allowable cross section area can be determined in terms of minimum allowable bolt diameter by means of this expression:

$A = \frac{\pi}{4}\cdot D^{2}$

The diameter is now cleared and computed:

$D = \sqrt{\frac{4}{\pi}\cdot A}$

$D =\sqrt{\frac{4}{\pi}\cdot (1.640\times 10^{-3}\,m^{2})$

$D = 0.0457\,m$

$D = 45.7\,mm$

The minimum allowable bolt diameter required to support an applied load of P = 450 kN is 45.7 milimeters.

5 0

3 years ago

The Hoover Dam is 221 m tall and 379 m wide. Approximating it as a flat plate, determine the effective resultant force on the da

IrinaVladis [17]

Answer:

2165800 Pa

Explanation:

See it in the pic.

3 0

2 years ago

Outline the structure of an input-output model (including assumptions about supply and demand). What is an inverse matrix? Why i

pishuonlain [190]

Answer:

Explanation:

C.1 Input-Output Model

It is a formal model that divides the economy into 2 sectors and traces the flow of inter-industry purchases and sales. This model was developed by Wassily Leontief in 1951. In simpler terms, the inter-industry model is a quantitative economic model that defines how the output of one industry becomes the input of another industrial sector. It is an interdependent economic model where the output of one becomes the input of another. For Eg: The Agriculture sector produces output using the inputs from the manufacturing sector.

The 3 main elements are:

Concentrates on an economy which is in equilibrium

Deals with technical aspects of production

Based on empirical investigations and assumptions

Assumptions

2 sectors - " Inter industry sector" and "final sector"

Output of one industry is the input for another

No 2 goods are produced jointly. i.e each industry produces homogenous goods

Prices, factor suppliers and consumer demands are given

No external economies or diseconomies of production

Constant returns to scale

The combinations of inputs are employed in rigidly fixed proportions.

Structure of IO model

See image 1

Quadrant 1: Flow of products which are both produced and consumed in the process of production

Quadrant 2: Final demand for products of each producing industry.

Quadrant 3: Primary inputs to industries (raw materials)

Quadrant 4: Primary inputs to direct consumption (Eg: electricity)

The model can be used in the analysis of the labor market, forecast economic development of a nation and analyze economic developments of various regions.

Leontief inverse matrix shows the output rises in each sector due to a unit increase in final demand. Inverting the matrix is significant since it is a linear system of equations with unique solutions. Thus, the final demand vector for the required output can be found.

C.2 Linear programming problems

Linear programming problems are optimization problems in which objective function and the constraints are all linear. It is most useful in making the best use of scarce resources during complex decision makings.

Primal LP, Dual LP, and Interpretations

Primal linear programming: They can be viewed as a resource allocation model that seeks to maximize revenue under limited resources. Every linear program has associated with it a related linear program called dual program. The original problem in relation to its dual is termed as a primal problem. The objective function is a linear combination of n variables. There are m constraints that place an upper bound on a linear combination of the n variables The goal is to maximize the value of objective functions that are subject to the constraints. If the primal linear programming has finite optimal value, then the dual has finite optimal value, and the primal and dual have the same optimal value. If the optimal solution to the primal problem makes a constraint into a strict inequality, it implies that the corresponding dual variable must be 0. The revenue-maximizing problem is an example of a primal problem.

Dual Linear Programming: They represent the worth per unit of resource. The objective function is a linear combination of m values that are the limits in the m constraints from the primal problem. There are n dual constraints that place a lower bound on a linear combination of m dual variables. The optimal dual solution implies fair prices for associated resources. Stri=ong duality implies the Company’s maximum revenue from selling furniture = Entrepreneur’s minimum cost of purchasing resources, i.e company makes no profit. Cost minimizing problem is an example of dual problems

See image 2

n - economic activities

m - resources

cj - revenue per unit of activity j