Im not sure how to answer the three questions. Understanding Deformation Mechani
ID: 2074546 • Letter: I
Question
Im not sure how to answer the three questions.
Understanding Deformation Mechanisms of Materials Let us consider the simplest case of material deformation - that induced by a one dimensional uniaxial load, as shown for the cylindrical metal bar below As we know from our study of stress-strain relationships, under the influence of load P the bar will lengthen along the principle x-axis from its original length L, to final length L which is given by L,-(1+s,)L, where , is the strain in the x direction. Similarly. the bar will undergo contraction in all directions transversely perpendicular to the x axis, with original diameter d, contracting to final diameter d, given by: (1+5.)do-(1-ve,)do where vis Poisson's ratio. v:-g/Ey. We have learned that for most engineering materials in the elastic regime, Poisson's ratio is in the range of 0.25-0.33. However, in the plastic regime, Poisson's ratio changes to 0.50. Let us investigate these values of v and their implications Consider the volume of the bar as it undergoes deformation. The volumes , and v of the bar are given by the expressions: po = TO, Lo and Vr = Notice that because d, is a function of the Poisson's ratio vof the material, the volumes I, and I', are not necessarily identical. Different values of v will create different values of diameter d and hence different values of volume V nment 1- derive the value of Poisson's ratio necessary such that a material remains at constant volume as it deforms. Make use of the simplifying assumption: for small , 2- from (1) what can we then infer about the volume of a material as it undergoes elastic deformation? plastic deformation? 3- can you postulate mechanisms occurring at the microscopic (atomic) level of the material that would explain these observed behaviors? Hint: the key elements are: (a) volume behavior in the elastic regime. (b) deformation in the plastic regime is nonrecoverable (permanent)Explanation / Answer
1) Poissions Ratio =- Lateral strain /Linear strain =- y/x
Lateral strain =df/d0
df = Final diamter after plastic deformation
d0 = Initial diamter before deformation
df = (1+y) d0 = (1-x) d0
Lateral strain = (1-x) d0/ d0
Lateral strain =(1-x)
Linear Strain = change in length / Original Length
Lf =change in Length = (1+x) L0
L0 = Original Length
Linear Strain = (1+x) L0/L0
Linear Strain =(1+x)
Poissions Ratio = - (1-x) / (1+x)
2) Volume is a function of dimater and Lenght , the change in diamter and Lenght are functions of poission's ratio.
So depends upon the value of poission's ratio then we can define the deformation is Elastic or Plastic.
For low values of poission's ratio0.25-0.33 the deformation is elasatic and however for the values of 0.5 the deformation is Plastic.
Vf = Lf/L0(12) V0
3) longitudinal extension can be understood as a consequence of elastic atomic
bonds being stretched, it is harder to understand why materials should contract transversally.
The reason is, however, that in an isotropic material there are atomic bonds in all directions,
and when bonds that are not purely longitudinal are stretched, they create a transverse tension
which can only be relieved by transverse contraction of the mater.