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SashulF [63]
3 years ago
15

Why is the experiment started with an initial load rather than starting at zero load?

Engineering
2 answers:
velikii [3]3 years ago
8 0

Answer:

Q1. Why is the percentage elongation in 2 inches greater than in 8 inches?

Q2. Describe the Type of Information which may obtained from a character of fracture

Q3. Explain the difference, on the basis of test result between the ultimate strength and The true stress at fracture, cite your specific value to enhance your explanation.

Q4. What is the purpose of using the spherically seated compression plate?

Q5. Why is the experiment started with an initial load rather than starting at zero load?

Following are the answers to each of the questions

Answer 1.

The larger the radius of an elastic material the lower the elongation because the radius is inversely proportional to the percentage of elongation. A wire with a lower radius has a higher elongation than a material with higher radius  

∈=PL/AE  

Where P is the force applied, L is the original length of the material, A is the cross-sectional area which is πr^2, E is the Elastic Modulus.  

Answer 2.  

A fracture is the separation of material into two or more pieces usually caused by ultimate load or stress. Fracture strength is the stress required for a material to fail.  

TYPES OF FRACTURE  

Brittle fracture: brittle materials are materials that are liable to break easily under load for example cast iron. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.  

In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated  

Ductile fracture: they are material that are capable of being molded in to any shape easily. In ductile fracture a plastic deformation occurs like cracks or rough surface sign will be visible before it attains it ultimate failure state.  

In ductile material, the cracks move slowly and in a gradual process together with plastic deformation. Cracks in ductile fracture will not grow unless there is increase in stress applied on the material  

Answer 3.  

The divergence in the values of true stress and engineering stress occurs only at large loads and displacements; or typically when the specimen is undergoing plastic deformation. That is because most materials have a elastic strain limit close to 0.2%. Note that the values of true stresses and strains are similar to their engineering counterparts below 0.2%.  

If a material that strain hardens is tested in tension, the true stress-true strain curve will keep increasing till the specimen fails as the cross section area keeps decreasing continuously (even during necking). However, the slope of engineering stress-strain curve becomes positive and negative before and after necking respectively . This is because necking reduces the material's ability to harden and hence take larger loads. Hence the load starts dropping but remember that one still measures the stress as the load divided by the original cross section area. If the instantaneous cross section area were taken into account, as was done for true stress, the load drop is compensated by the acute reduction in cross section area.  

In compression, the necking instability does not occur and this difference is not stark. the true stress-strain curve and engineering stress strain curve under compression more-or-less converge (unless specimen 'barrelling' is very prominent).  

Answer 4

They are designed to be placed at the center at a point where loading is exerted in an electromechanical or hydraulic universal test machine. It provides a hardened surface when performing complex compression test in which uniform stress distribution is critical. They can be used to test the variety of materials like concrete, metals, wood and composite.  

Answer 5

From ohms law, if the resistance is zero then the current that will flow through the experiment will be too high causing the wire to burn, since the level of current that the wire can tolerate has been produced by the manufacturer and if it exceed this limit, the wire will damage permanently. If there is no load at all this will be known as short circuit and might damage the source as well.

Explanation:

1. Elongation is the increase in length of a material after it has been stressed within the gauge length e.g. elastic materials like wires, spring and other materials that obeys Hooke's law.  

2. In brittle fracture no plastic deformation will occur before fracture i.e there are no signs of distress like cracks to notify that a certain material is about to fail.  

In brittle fracture cracks spread rapidly with little or no plastic deformation. These cracks continue to grow once it is initiated .

4. Spherically seated compression plate are mounted to the cross head electromechanical or hydraulic universal test machine.

5. The internal resistance of any typical wire is low compared to the resistance of the component that are likely to be used for the experiment.

horsena [70]3 years ago
7 0

Answer: The answer would depend on the experiment but in most cases zero load is not used to ensure that there is no overflow of current.

Explanation: The internal resistance of any typical wire is minute compared to the resistive components that you are likely to utilize in any given experiment. Using I=V/R (Ohms law) , if the resistance is zero then the current that will flow through the wire would be much too high, causing the wire to burn out. The level of currents that the wire can tolerate are provided by the manufacturers. If you exceed those limits, than the wire will surely get permanently damaged. If you have no load at all than this will be known as a short circuit and might damage the source as well. If you have other components in the circuit than the overflow of current might in effect damage them too. This could also damage the bread board if you are using one.

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Explanation:

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Answer:500,551.02

Explanation:

Given

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Final  enthaly of pump \left ( h_2\right )=550KJ/kg

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