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This issue has been bothering me for a few days now. It's actually part of my homework, but I've already finished them and in a sense they're not part of the question here. What I would like to know is more of a theoretical question about this formulation.

Now my professor's book and many online resources (sample) maintain the same mathematical relation, hence my dilemma.


Consider the following:


Under axial loading, the stresses can now be computed for any section. The question posed is what if we make an imaginary section at an inclined, its normal defined by an angle $\theta$. I will not write every detail here, but its clear that we can use $A$ as a placeholder for the area and we can define $\sigma_x$ to be $\sigma_x = \frac{F}{A}$ and therefore in the diagram below we will use only stress.


Now, I will refrain working with vector quantities, I will only use scalar relations. If you take a look at this triangle, its easy to see that the area is projected to the section and we can get the relationship:

$$ A' = \frac{A}{\cos(\theta)} $$

Now the rest of the equations go by the books:

$$ T = \frac{F}{A'} = \frac{F}{A}\,\cos(\theta) = \sigma_x\,\cos(\theta) $$ $$ T_n = T\,\cos(\theta) $$ $$ T_s = T\,\sin(\theta) $$

My dilemma starts here. First of all notice how in each of the above expressions the value for the length of the object is not taken into consideration. Also, we're working with areas and forces hence the length itself is not part of the equation at all even as a contained information. Now that would be all fine, if it were not for the following section(to simplify, I've taken the origin to be at the center of the object).

Section C

Now as you can see this section of the object gets very interesting. Clearly the formula for $A'$ does not hold anymore because the area itself is shrinking on the left side. To make things even worse, notice how the area on the right appears. The two areas are clearly linked, in this case, the sum of the two should be equal to the original area - from the fact that the height appearing on the right is equal to the height subtracting from the left. Calculating the angle at which the breaking point is made, that is when the section plane cuts the corner is easily calculated using:

$$ \tan(\theta_k) = \frac{l}{h} $$

Where $ \theta_k $ stands for the critical angle, just so we can refer to it. Now the question that is on my mind is what is the distribution of the stresses on this section. Clearly there is stress on the left, there's stress on the right. Now I am confused as to the distribution of stress at this angle.

My Work

Now, I've attempted a solution here. First, I am in a tight spot between assumptions. Clearly the original force $F$ must still be present here. But the force distribution is confusing me at this point. The best idea I could come up with, is the following. I would be grateful for your review.


Problem 3

I've only written here $F_1$ and $F_2$ for visualization purposes, because personally I don't think that we will need the $F_2$. Knowing that $F_1 = F$ it's obvious that they will cancel each other out, and they will also cancel the moments they create against the newly formed centroid on the trapezoid. Hence I think the only remaining task is to compute the areas of each individual piece and compute stresses.


Computing the individual areas for the cuts, since we won't need the area at where supposedly $F_2$ was located, we will ignore that for the time being. Though it will come naturally as we try to find the left area.

The inclined plane area is the easiest to find, clearly, if we project the length $l$ to the top of the smaller rectangle we can see that using trigonometric relations we can find:


Multiply that by the base and you've got the area $A'=b l\,'$. To find the areas of the left and right sections one must notice that the amount we deviate from the original height we add to the right side. So if we're to project the length like we did last time we get this:


From the figure above it's clear that the following relation holds true:

$$ h = h' + 2\,h_2 $$

From trig we know $h'=l\cot(\theta)$. From this, $h_2$ is defined by:

$$ h_2 = \frac{1}{2}(h - l\cot(\theta)) $$

Now $ h_1 = h - h_2 $ and after some mathmagic we should get the following:

$$ h_{1/2} = \frac{1}{2}(h\pm l\,\cot(\theta)) $$

Now let's multiply this by the base and we've got ourselves $A_1 = h_1\,b$.


Now we have two stresses, namely $\sigma_{x_1}$ and $T$. Let's analyze the relations. The forces clearly have an eccentric deviation from the centerline. Now, is this going to be a problem? I seem to think that it won't be, and I base my logic on the fact that they will cancel the bending moments as we said above. The formulas are:

$$\sigma_{x_1} = \frac{F}{A_1}$$ $$T = \frac{F}{A'} $$

Now obviously these are general equations, one could always dig deeper and I believe we could even express these stresses as a function of elementary stresses, namely $\sigma_x$ and some form of this other tangential stress $F/(l\,b)$.


To be completely accurate, the formula for $\theta_k$ seems to be a bit more complex, since there are four corners on the cut (in our case). Knowing that paired corners are symmetric to the x-axis we can use this equation:

$$\theta_k(n) = |\arctan(\frac{l}{h}) + \pi\,n|$$

where $n\in(-2,-1,0,1)$. Is this ugly or is it just me?


I hope to get some review on the ideas here, I'd really love to know how would this get solved correctly. And please someone, help me fix the appendix formula.

share|cite|improve this question
Usually the force stress relationships only apply to infinitesimally small volumes of materials. In any case, the balance of forces is what drives these relationships and here your question is more about geometry than about mechanics. – ja72 Jul 14 '14 at 15:10

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