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Shear stress on a piece of Chalk 
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#1
Mar1514, 07:15 AM

P: 2

Hello Everyone
I'm trying to get a deeper understand about this, and I hope you guys can shed some light on the issue. I'm not an enginner, so this may sound a little dumb. Anyway, here it goes. Suposse I drop a piece of chalk. As it hits the floor, it will likely break into pieces, and I'd like to understand how some factors, such as, the chalk's length and the height from which it was dropped, can affect the number of pieces it'll break into. According to this, when the piece of chalk hits the floor, it will land slightly off center. Therefore, the piece will be under a shear stress, caused by gravity and the reaction force applied at the region of chalk that touches the floor first. As I understand (correct me if I'm wrong) the shear stress the piece of chalk will experience is: [itex]τ = \frac{F}{A}[/itex] Where F is the force (the gravity force) and A is the cross section area of the chalk. If τ is equal or bigger than the Shear strength of the chalk, then it will fail. But this doesn't depend directly on the chalk's length, and doesn't help me to determine into how many pieces it's likely to break. Another relation to consider, is the one presented on Giancoli's Physics for Science and Enginnering: ΔL = [itex]\frac{F Lo}{G A}[/itex] Where ΔL represents how much the chalk is "stretched" by the shear stress. The explanation contained on the link above states that: if ΔL is big engough, the chalk will fail. So, I'd like to know a few things: How can I relate the ΔL to the shear strength? I mean, how much must the chalk be stretched until it fails? Is there a way to predict where the chalk will break, and into how many pieces based on this? (I'd like something as simple as possible, there's no problem for me to consider the simplest possible scenario). Where can I find the shear strength for a piece of chalk? Any help is welcome. If you guys can point me something to help me, I'd be very glad. Thanks in advance. 


#2
Mar1514, 02:14 PM

P: 1,498

While the height of the release of the chalk has some bearing in this situation, it is the deceleration force produced as the chalk hits the surface that will break the chalk, and not the acceleration force that cause the chalk to fall. It a chalk is dropped onto a hard surface, such as a ceramic floor, the quick deceleration of the chalk will produce a large deceleration force. If dropped onto a soft surface such as a pillow, the deceleration is less as is the deceleration force. 


#3
Mar1514, 03:20 PM

P: 2

Nevertheless, I'd like to know if the length of the chalk will directly affect the outcome. I mean, if I apply the same shear stress on two pieces of chalk with different lengths, is it possible that the bigger one will break while the smaller one is intact? The deformartion is proportional to the length, as stated above. But the shear strength (the amount of shear stress the material can take before breaking) is the same for both cases (the big and the small pieces), isn't it? Which means that a bigger piece will deform more before breaking. The limit for failing is given by a stress or by the deformation? Or for both? That's what isn't clear for me. P.S.: I know that, if the piece of chalk is bigger, then it'll have more mass, and more kinect energy when gets to floor. So the force applied will by bigger, causing the stress to be bigger too, but it isn't my doubt. 


#4
Mar2714, 06:50 AM

P: 7

Shear stress on a piece of Chalk
You can use either stress or strain (more convenient than deformation) to decide if it breaks. The Tresca failure criterion tells you if there's enough stress to cause brittle failure. Stress isn't uniform over the cross section, but you can find it using some finite element analysis software, eg see fea compare for some free ones.
The longer piece can break more easily for two reasons I can think of: 1) It has more mass, therefore more kinetic energy to dissipate when it hits the floor. 2) It can experience larger stresses due to forces that tend to bend it. For a very long thin piece, you might be able to neglect your shear stress calculation and look at the tensile stress caused by the bending moment caused by the floor pushing one end. The number of pieces it breaks into is related to how much energy is absorbed by each breaking event. In an extreme case where the chalk dust is held together with almost no force (say, just gently pressed into shape), the first time it breaks won't absorb any energy so the pieces continue moving with almost all of their original kinetic energy and can break again and again back to dust. 


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