ELASTIC DEFORMATION.

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Hookes Law The following topics will be discussed in this presentation: 1. Hookes law 2. Elastic behaviour of materials by stretching a spring and producing.

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Presentation transcript:

ELASTIC DEFORMATION

Elasticity Materials bodies which get stretched on the application of a force and regain its original length and shapes on the removal of the deforming force are called elastic bodies. This property of the material bodies is known as elasticity. Length of the body before applying the stretching force is called unstretched length.

Length of the body after applying the stretched force is called the stretched length.

Difference between the stretched and the unstretched length of the body is known as extension.

Hooke’s law A material obeys Hooke’s Law if, the extension is directly proportional to the stretching force within the limit of proportionality. Extension – load graph The table below shows how the extension of s spring varies with force applied (load). Load /N 0 1 2 3 4 5 Extension /mm 0 10 20 30 40 58

The extension – load graph shown below is plotted the information given on the table

OX is the region where Hooke’s law is obeyed, that is extension is directly proportional to the stretching force. X is the point of limit of proportionality. (The point beyond which the extension of an object is no longer proportional the load producing it is called limit of proportionality.) E is its elastic limit. Up to E, the spring behaves elastically and returns to its original length when the load is removed. Beyond E the spring will not regain its original length when the stretching force is removed and the spring is left permanently stretched hen the load is removed.

Spring constant of spring can be calculate by using formula: Force = spring constant × extension (F = Ke) Spring constant of this spring is: K = F/e = 2 / 20 = 0.1 N/mm

Example question: A spring of original length 3 Example question: A spring of original length 3.0 cm is extended to a total length of 5.0 cm by a force of 8.0 N. (a) Calculate the spring constant of the spring. Ans: extension = 5 – 3 = 2 cm F = Ke K = F/e = 8/2 = 4N/cm (b) Assuming the limit of proportionality of the spring has not been reached; calculate the force needed extend it to a total length of 6.0 cm. Ans: extension = 6 – 3 = 3cm F = Ke = 4 × 3 = 12N