Solid Mechanics Question - Help

[dislect]

hi all

i apologize in advance for the poor englsih, the question is translated from another language.

A round beam (radius R) is harnessed on the left side to a wall.
The beam is made up of two parts glued together by a thin layer of glue - the attachment plane is described in the picture as AB.
A 'glue fail' happens when the SIZE of the shear strain on plane AB is bigger the Sigma0 (some unkown value).

On the right side of the beam there's an equaly spred strain q=q0*e3 (note that q0 might be negetive as well).

What is the allowed value range of q0 so that there won't be a glue failure? (express your answer by using R and Sigma0)

this question was given as a challenge, and is a part of a course called solid mechanics 2 that deals more with shear strain and strain matrices. I really don't even have a clue of how to begin, i can't see how the radius R plays a part in the answer..

help, please?

 

[Mapes]

Hi dislect, welcome to PF. Do you know the relationship between force, area, and stress, and can you draw a free-body diagram and write the static equilibrium equations? These would go a long way in helping you solve the problem.

The translation looks pretty good, but "strain" should probably be "load" (or "force"), "distributed load," or "stress." Is is possible to find out which, since they imply different loading configurations?

 

[dislect]

Hi Mapes, thanks!

It's weird but I think they actually ment strain and not force. Well, that's the word they chose to describe it as. In the first course (Solid mechanics 1) we dealt a lot with forces and free-body diagrams. Now its more about strain tensors on different planes, and that's the part that bugs me because I know that the solution as something to do with it..

* I copied the parameter they signed for the maximum sheer strain that AB plane can tolerate (AKA Sigma0)

 

[Mapes]

OK, great. Now using the tangential force on AB, calculate the shear stress at the joint and compare this to $\sigma_0$ ($\sigma$ is almost universally used to denote stress, not strain).

 

[dislect]

Well that's the part I'm stuck in :)
How do I do that? how does the momentum take place if i don't know any distances on the beam?
Whem I'm summing all the forces do i count q0 as is or as q0*pi*R^2, and how do I get a range of values of q0 in such way?

 

[Mapes]

how does the momentum take place if i don't know any distances on the beam?

Moments? Don't worry about the moments; you just want to turn the tangential force v into a shear stress by dividing it by the cross-sectional area (which is not a circle!).

Whem I'm summing all the forces do i count q0 as is or as q0*pi*R^2, and how do I get a range of values of q0 in such way?

There's no way to tell without knowing what q0 is (e.g., a force, a stress...). I'm quite sure it's not a strain, despite the translation. I would assume it's a force.

 

[dislect]

If I translate it "as is" then q0 is said to be an equaly devided 'load' (that might be negetive as well)

Hmm.. about the sheer stress that's steping back to trigonometrics :)
How do I find that cross-sectional area? its sort of eliptic.

 

[Mapes]

How do I find that cross-sectional area? its sort of eliptic.

You got it!

 

[dislect]

So the cross-sectional area is 2*Pi*(R^2)
I still don't really get the question. It's like I'm solving it with you in baby steps and can't see the all picture :-(

 

[Mapes]

In the problem, you're pulling or pushing on a beam with circular cross section. There's an angled, glued joint within the beam that can only sustain a certain amount of shear stress. Your job is to calculate how hard you can pull or push on the beam without making the joint fail. So you need to connect the axial load $q_0$ to the rupture shear stress $\sigma_0$. Does this help?

 

[dislect]

So its something like V*2*pi*(R^2)*sin(30) = q0*pi*(R^2) ?

 

[Mapes]

So its something like V*2*pi*(R^2)*sin(30) = q0*pi*(R^2) ?

I'm not sure where this is coming from. Are you summing the forces in the $e_3$ direction? There's no 30° angle in this problem.

 

[dislect]

I'm not sure where this is coming from. Are you summing the forces in the $e_3$ direction? There's no 30° angle in this problem.

That what I tried.. casting V on e3

 

[dislect]

Mapes, could you help me with the static equilibrium part?
I fell like I'm getting lost in this question, and I have no other examples to rely on (its the first 'challenge question' that the lecturer gave). Plus, It's just the first part of the question :) in the second part they change q0 to pure moment.

 

[Mapes]

I think you need to find the relationship between V and q0. I'm not sure where sin(30) comes in. A couple hints I can give are that (1) V is smaller in magnitude than q0 (assuming q0 is the axial force applied to the beam end) and that (2) V equals q0 when the cut is horizontal but is zero when the cut is vertical. Does this help figure out the trigonometry part?

 

[dislect]

Good morning Mapes!
Me and a friend are going to sit and do our best with the question, and later on today i'll upload our solution trial for your confirmation :-)

thanks for all the patience, I really appreciate it!

btw, just today we realized that q0 is an already spread force, meaning we don't have to divide it by A and the same goes for the V(sheer)!

 

[dislect]

Ok, so as I said q0 is an already spread force and therefor -< the strain ( no need to divide by A).

This is what we did:

* Is our assumption correct about AC? if so, how can we justify it?
* Is our solution correct? :-)

Thanks Mapes!

 

[Mapes]

Right away, let me reiterate that a 3-4-5 right triangle does not contain any 30° angles.

It looks like you're getting closer.

EDIT: Try solving it for an arbitrary angle $\theta$. Think about how the answer must change as the angle approaches 0° or 90°.

 

[dislect]

Hmm I'm not sure I understand why its not a 30 degree angle..
and if its not, what does the 3-4-5 triangle give me and how do i express the n vector of plane AB?

 

[dislect]

Ok I'll organize my questions:

*Why does this triangle don't have a 30 degree angle? i thought that any 3-4-5 triangle has 30-60-90 degrees in it.
*The overall solution for the question is right? I mean besides the angle part
*In the second part of the question that change the strain q0 into pure moment T. Out of curiosity, where would the glue fail first on plane AB (while applying moment T)? My intuition says that it should fail all over the edges of plane AB simultaneously but I can't really explain it. Is it true? and if so, how can I justify my saying?

Once again, I'ts important for me to emphasize how much I appreciate your help Mapes :-)

 

[dislect]

Sorry for spamming but I just got it.. its 53.1 degrees!
now given that i corrected the degree, is the solution ok?
(my other questions are still relevant :) )

 

[Mapes]

What is your revised answer? (Try solving for an arbitrary angle $\theta$, then you can plug in the correct angle. This makes your solution more general and also let's you check the extreme cases of $\theta =0^\circ$ and $\theta =90^\circ$.)

For the case of the moment, try drawing a free-body diagram and looking at what the distribution of forces will be at the joint. (It won't just be a single force any more.)

 

[dislect]

fixed version:

http://up189.siz.co.il/up1/mdhzyjini2zj.jpg

and after looking at the free-body diagram for the moment i think that the first place to fail is the exact middle of the AB plane.

 

[Mapes]

That looks a lot better.

and after looking at the free-body diagram for the moment i think that the first place to fail is the exact middle of the AB plane.

Really? It seems like this point lies on the neutral axis, where bending stresses are zero.

 

[dislect]

Ok i'll take one step back and see first if I got the moment instead of q0 part right :shy:
This is what I did the find the max pure moment T that the beam can tolerate on the glued part AB:

http://up185.siz.co.il/up1/ynzt4hdyjoxn.jpg

(notice the * I got a bit confused there..)

Now according the the moment formula it depends on R so the first to break point on plane AB should probably be A or B. The thing is, I can't find any formula that connects the distance from the applyied moment to a certain point and that's why I first thought that A and B would fail together but it contradicts my intuition :uhh:

 

[Mapes]

You have got to distinguish between stress and strain. Stress is load per unit area and has units of pressure. Strain is normalized deformation and is unitless. Mixing these up is such a big deal that I tend to stop reading at this point. It signals major confusion.

It would be easier to follow your solution if you solved for an arbitrary joint angle $\theta$.

 

[dislect]

You got to understand that its just translation problems. I just tryed to stay unified in my choice of word (strain=stress).
I can't really see a reason to choose an arbitrary joint angle for the transformation, its not such a long solution.

 

[dislect]

Mapes, I need to submit this question by tommorow.
Could you please help me with one last push?
Given that the solution is for Stress and not strain, is it correct?
And more importat, where on plane AB would the glue fail first? is it on point B because its closer to where the moment is applyied?

 

[Mapes]

OK, I just took a look at your solution, substituting "stress" for "strain." Your answer looks like it's in the right ballpark. Did you check whether the shear would be higher at the midpoint of the joint at the surface?

 

[dislect]

I'm having some difficulties understanding which point you mean, so here's a diagram of the mid section (transformed):
http://up186.siz.co.il/up2/yzxizmmgyj4t.jpg

by midpoint did you mean c,d or e?

 

[Mapes]

I'd check the shear stress at points e and d too, in case that's where the maximum shear occurs (I don't know).

 

[dislect]

hmm I wish I could but me and my friends don't know how.. how do we check the sheer stress in a certain point?

 

[dislect]

Could you do the last part with me?
I can't figure out which point on plane AB would fail first. Its past the due date but I really want to know.. I guessed its B but I have no explanation

 

[Mapes]

Here's a possible approach:

1. Define angle $\phi$ as rotating from axis 1 towards axis 2 around axis 3, and angle $\theta$ as rotating from axis 1 towards axis 3 around axis 2. In other words, axis 2 corresponds to $\phi=\pi/2$, $\theta=0$, and axis 3 corresponds to $\phi=0$, $\theta=\pi/2$.

2. Identify the stress state at the surface from torsion to be

$$\bold{\sigma}=\tau_0\left[\begin{array}{ccc} 0& 0& \sin\phi\\ 0 & 0 & -\cos\phi\\ \sin\phi & -\cos\phi & 0\\ \end{array}\right]$$

where $\tau_0=TR/J$.

3. To find the shear stress on a certain plane, define a direction cosine matrix $\bold{a}$ that gives the new axes when multiplied by the old axes. Double check that this is

$$\bold{a}=\left[\begin{array}{ccc} \cos\theta& 0& \sin\theta\\ 0 & 1 & 0\\ -\sin\theta& 0 & \cos\theta\\ \end{array}\right]$$

where $\theta$ is the angle of the joint, and we know that $\sin\theta=\frac{4}{5}$ and $\cos\theta=\frac{3}{5}$. Then the stress transformed to the joint plane is

$$\bold{\sigma '}=\bold{a\sigma ' a^T}$$

The shear components of $\bold{\sigma '}$ (hint: two of them are $-\cos\phi\sin\theta$ and $\sin\phi(\cos^2\theta-\sin^2\theta)$) represent the shear stresses on the joint; by checking different values of $\phi$, you can check where the stress is highest and what its value is to solve the problem. This is a generalized technique that can be used to solve any joint problem for any loading configuration.

 

[dislect]

Wow, thanks!

So in this question there's a little dilema.
As I understand, all I really care about is a(1,3) and a(3,1) because they are the sheer stress on the direction that I am looking for, right?
$\sin\phi(\cos^2\theta-\sin^2\theta)$
with the given data leaves me with -0.28sin(phi) so the max values is when the angle phi is 90 or 270, and that means A or B on my AB plane.
Meaning, it depends on the direction of the moment T? :)