Elasticity
This presentation is the property of its rightful owner.
Sponsored Links
1 / 11

elasticity PowerPoint PPT Presentation


  • 82 Views
  • Uploaded on
  • Presentation posted in: General

elasticity. Elasticity. Elasticity is a branch of Solid mechanics that deals with the elastic behavior of solids. It is the property of material of a body which regains its original shape and size.

Download Presentation

elasticity

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Elasticity

elasticity


Elasticity1

Elasticity

  • Elasticity is a branch of Solid mechanics that deals with the elastic behavior of solids. It is the property of material of a body which regains its original shape and size.


Elasticity

Suppose that we pull on the ends of a bar with a force F , as shown in the figure. We say the bar is in the tension. The internal forces in the bar resist the tension forces and holds the bar together.

One of these internal forces is called Tensile Stress.


Tensile stress

Tensile Stress

Tensile Stress – the ratio of the magnitude of the applied force F to the cross-sectional area A.

Stress = Force/Area

= F/A

Another internal force on the bar is called Tensible Strain.


Tensile strain

Tensile Strain

Tensile Strain – defined as the ration of the change in length, to the initial length before the force was applied. This strain refers to an increase in length.

Strain = change in length/ initial length

= Δℓ

Aℓ

The strain is the fractional change in length. As such dimensions. The bar vould deform also if it were under compressive force, but this time it would be compressed rather than stretched, the bar is under compressive strain.


Elasticity

Compressive Strain – opposite of tensile strength, causes the bar to be compressed, rather than stretched.

The amount of strain an object undergoes depends on the amount of stress applied to it. If the stress is not too great, the strain is observed to be proportional to the stress. The ratio of a stress to the corresponding strain is called an elastic modulus. Young’s modulus.

Y = stress/strain

= (F/A)

= Δℓ

Lo


Young s modulus has the units as stress n m

Young’s Modulus has the units as stress, N/m ²


Elasticity

Elastic Limit – refers to the maximum stress that can be applied to an object without its breaking up, or being permanently deformed.

When its elastic limit is exceeded, the object may or may not be far from breaking. Brittle substances like glass break at or near their elastic limits. Bone is another material that remain elastic until its breaking point is reached, and so it cannot be permanently changed in size or shaped by applying a force. Most metals can be deformed a great deal beyond their elastic limits, a property known as Ductility.


Elasticity

A graph of the elongation of an iron rod versus the tension applied to it is shown. At first the graph is a straight line, which corresponds to elastic behavior. Past the elastic limit the graph flattens out, which means that each increase in tension by a given amount produces a greater increase in length than it did below the elastic limit. The rod stretches more readily. If the tension is removed after having exceeded the elastic limit, the rod remains longer than it was originally. It has undergone plastic deformation. The ultimate strength of the rod is the greatest tension it can withstand, and it corresponds to the highest point on the curve.


Example

Example

A weightlifter raises a weight of 600 N over her head. Assuming that each of her legs supports the same weight and that the legs are parallel and vertical, determin the amount by which each femur (thigh bone) compresses. Each femur has an effective cross-sectional area of 7.5 x 10-4 m2 and a leght of 0.52m.


A nswer

Answer

Given:

total force exerted on the two femur bones F 600N

cross-sectional area of femur A 7.5 x 10-4 m2

length of femur L 0.52m

Y = stress/strain = (F/A) / (ΔL/Lo)

ΔL = (Flo) / (Y A)

= (300 N) (0.52) / [7.5 x 10-4 m2N/m2) 7.5 x 10-4m2)]

= 3 x 10-5 m


  • Login