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sasho [114]
3 years ago
15

A brass alloy is known to have a yield strength of 275 MPa (40,000 psi), a tensile strength of 380 MPa (55,000 psi), and an elas

tic modulus of 103 GPa (15.0×106 psi). A cylindrical specimen of this alloy 5.5 mm (0.22 in.) in diameter and 267 mm (10.52 in.) long is stressed in tension and found to elongate 7.0 mm (0.28 in.). On the basis of the information given, is it possible to compute the magnitude of the load that is necessary to produce this change in length? If not, explain why.
Engineering
2 answers:
Ugo [173]3 years ago
7 0

Answer:

Computation of the load is not possible because E(test) >E(yield)

Explanation:

We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if

E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:

Calculation of E(test is as follows)

E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm

=0.0262

Computation of E(yield) is given below:

E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027

Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).

frosja888 [35]3 years ago
5 0

Answer:

It's not possible to compute the magnitude of the load

Explanation:

Given data:

Yield strength = 275Mpa

Tensile strength = 380Mpa

Elastic modulus = 103Gpa = 103 * 1000MPa

Diameter = 5.5 mm

Length = 267 mm

Elongation = 7.0 mm

Calculating the strain test value using the formula'

<h3>ε(test) = ΔL/Lo</h3>

Substituting, we have

ε = 7.0/267

  =  0.0262

Calculating the yield strain using the formula

<h3>ε(yield) = бy/E </h3>

Substituting, we have

ε(yield) = 275/ 103 * 1000

             = 275/103000

            = 0.00267

From the calculation above, the strain test is greater than the yield. Therefore, it's not possible to compute the magnitude of the load.

<h3>ε(test) ⊃  ε(yield)</h3><h3></h3>
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Explanation:

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2 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

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W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

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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:

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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

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