# Material Science Questions

by foobag
Tags: material, science
 P: 76 Hey guys I have a few questions to ask: 1.) Why are materials whose yield stresses are highly strain-rate dependent more susceptible to brittle fractures than those materials whose yield stresses do not exhibit marked strain-rate dependence? 2.) How would you define 'toughness' in the case of elastic deformation; plastic deformation; and fast fracture? 3.) Under what conditions can materials that generally exhibit a yeilding stress and final fracture stress can still fail below the yeild stress. I appreciate the help in this matter since I am not too familiar with material science engineering.
P: 21
 Quote by foobag 1) Why are materials whose yield stresses are highly strain-rate dependent more susceptible to brittle fractures than those materials whose yield stresses do not exhibit marked strain-rate dependence?
- Unlike the next two questions, this one is actually more difficult for me to answer because although the very fact itself is obvious, I can think of several explanations, but none of them seem to really explain it as well as I would like. So I'll leave this question for someone else or maybe you'll or I'll eventually figure it out.

 Quote by foobag 2) How would you define 'toughness' in the case of elastic deformation; plastic deformation; and fast fracture?
- Toughness in general is defined as a material's ability to absorb energy until fracture. Now since you need three definitions, I'll give you what I believe to be the answer:
1) For elastic deformation, toughness must be defined as resilience. That is to say it must be the ability of a material to absorb energy and to release it upon unloading. In a graphical sense, it can be defined as the area under the stress-strain curve until the yield strength.
2) For plastic deformation, toughness must be defined as the definition I first gave (the ability to absorb energy until fracture). In a graphical sense, it can be defined as the area under the entire stress-strain curve.
3) For fast fracture, I would define toughness as the plain strain fracture toughness (the K1c value), or just fracture toughness in general. The higher a material's K1c, the less likely it is to exhibit fast fracture. Fracture toughness would thus define a material's ability to resist fast fracture.

 Quote by foobag 3) Under what conditions can materials that generally exhibit a yeilding stress and final fracture stress can still fail below the yield stress.
- There are quite a few ways this can happen, but the one that stands out most is fatigue cracking or just plain crack propagation.
1) Fatigue cracking will occur when the material experiences a number of stress cycles (the stresses can be less than the yield stress) that impose cleaving, sliding, or tearing of stress concentrations (which can be external features or internal pre-existing cracks/voids). In the case of external features (such as sharp corners), the stress cycles will eventually nucleate a crack at the feature which will then propagate.
2) For crack propagation, once the material's fracture toughness is exceeded, fast fracture will occur. This applies to all cracks and voids in general, but also in the case of the nucleated fatigue crack.
 P: 3 I have gone through a paper where the authors have given the unit vector of slip of a FCC material which has slip system as (1-1 1)[1-1 2], the unit vector of slip is given as (sqrt(-1/3),sqrt(-2/3),0) now I have to find out the similar thing for BCC system (110)[111], but am unable to find that out....please can anyone help me understand how do i get these unit vector of slip.

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