What are the unit vectors of slip for FCC and BCC materials?

AI Thread Summary
Materials with yield stresses that are highly strain-rate dependent are more prone to brittle fractures due to their inability to absorb energy effectively under rapid loading conditions. Toughness is defined differently across deformation types: for elastic deformation, it is resilience, or the ability to absorb and release energy; for plastic deformation, it is the total energy absorption until fracture; and for fast fracture, it is represented by the fracture toughness (K1c), indicating resistance to rapid failure. Materials can fail below their yield stress primarily through mechanisms like fatigue cracking, where repeated stress cycles lead to crack initiation and propagation, even when stresses are below yield levels. Understanding these concepts is crucial for material science engineering, especially in predicting failure modes and enhancing material performance. Additionally, assistance is sought for calculating the unit vector of slip in BCC systems, indicating ongoing inquiries into material behavior under stress.
foobag
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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.
 
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foobag said:
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.

foobag said:
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.

foobag said:
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.
 
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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