Srinidhi's Blog

Weekly Learning Log #1 (2/22-2/27)

Hello everyone. I wanted to try out something new, so this post will be the first (hopefully) weekly learning log where I’ll write about what I learned that week in both my classes and other means (news articles, youtube videos, etc). This post is for the Week of 2/22-2/28.

Chemistry (CHEM 20B)

This week, I had my 2nd chem midterm on Friday (2/27). It covered concepts from entropy to acid-base calculations. Below is a summary from my Midterm 2 concepts quizlet.

Entropy

Entropy is the possibilities of position. The entropy of gas is higher than liquid than solid since there’s more positions possible. If \(\Delta\)S > 0: more possibilites (spontaneous). If \(\Delta\)S < 0: less possibilities.

Gibbs Free Energy

This the measure of reversible work done at constant temperature and pressure.

Reaction Quotient Q

Acid-Base

Reactions with Acid-Base

Statics (MAE 101)

I also had a statics quiz today (3/2), so I did a lot of statics studying throughout the week. The professor taught 3 chapters over the course of 3 lectures, so I didn’t really get the material and spent my whole weekend trying to learn it. I think I got an A on the quiz today so I am very glad the studying paid off.

Chp. 14 (Internal Forces & Moments in Beams)

If we have a beam, we could cut the beam at a section to determine the internal forces & moments. These are V (shear force), P (axial force), and M (moment). To determine these internal forces, we must first draw a FBD of the whole beam with all the reactions and solve for them using equilibrium. Afterwards, we cut the beam to determine V, P and M through force balance/equilibrium again.

Another (easier) way to find V, P and M is through the differential method, where \(V = -\frac{dW}{dx}\) and \(M = \frac{dV}{dx}\). In this method, we begin by drawing a FBD and using equilibrium/force balance to solve for the reactions. Then, we can section off the beam. For example, let’s say we have a beam 4 meters long, and in the first half of the bar, there is a distributed force W. We could separate the bar into 2 sections, the half section (\(0 \leq x < 2\) ) with the distributed force, and the other half section (\(2 \leq x < 4\)).

For V1, we would have \(V_{1} = V_{A} - \int_{0}^{x} W(x) \,dx\) where VA is any external (vertical) reaction present at the beginning of the section. \(M_{1} = M_{A} + \int_{0}^{x} V(x) \,dx\), where MA is any external couple present at the beginning of the section. The results will give us V1(x) and M1(x).

For V2, we would have \(V_{2} = V_{A} + V_{x=2} - \int_{0}^{x} W(x) \,dx\), where VA is any external (vertical) reaction present at the beginning of the section and we add that to \(V1_{x=2}\). Similarly, \(M_{2} = M_{A} + M_{x=2} + \int_{2}^{x} V(x) \,dx\) where MA is any external moment present at the beginning of the section.

Chp. 9 (Stress and Strain)

These chapter took me a while to understand the concepts and problems. In short, stress is the amount of force over a specific area. Normal stress is that normal to the plane, while shear stress is that tangential to the plane. We can get normal stress using \(\sigma = \frac{P}{A}\) where P is the axial force and A is the cross-sectional area. The problem solving strategy is drawing an FBD and solving for P. After we get P, we want to draw another FBD right at the plane cut P (different P than axial force), and drawing the stress reactions. We do another force balance here to solve for the normal and shear stresses.

Strain would be how much it’s stretched or shortened when a force is applied. Usually, in these problems, we look for the length by which it changes. The prroblem solving strategy is pretty similar to before, in regards to drawing an FBD and getting P. Here, we have the change in length to be \(\delta = \varepsilon \cdot L\) where \(\varepsilon\) is the strain and L be the original length. Another way it’s commonly written is \(\varepsilon = \frac{dL' - dL}{dL}\) where dL’ is the final length and dL is the original length.

Chp. 10 (Axial loaded bars)

Here, we combine the concepts from chapter 9. The problem solving strategy is pretty much writing down all your knowns and trying to connect the equations in a way to get what you’re solving for. One thing I struggled with here, was the comptability equation when having statically indeterminate problems. This type of problem means that there are more unknowns than number of equations. Thus, we have to use the comptability equation involving \(\delta\). To make it easier to see this, let me write down all the equations we now have. \(\delta = \varepsilon \cdot L\), \(\sigma = E \cdot \varepsilon = \frac{P}{A}\), \(\delta = \frac{PL}{EA}\)

In particular, we use the last definition of \(\delta\) for the compatability equation. If we have it so the bars are side-by-side, it is \(\delta_{1} = \delta_{2}\). If it’s end-to-end, it is \(\delta_{1} + \delta_{2}\). For side-by-side, the bars are sandwiched by top and bottom plates, so they must be squished or stretched by the same amount. For end-to-end, their total change has to be 0 to fit in the space. If one bar is squished, the other bar must stretch.

Conclusion

That’s pretty much it for this week’s learning writeup. It’s already pretty long, so I’m gonna keep it to just classes. I didn’t really do any Chem 20L or ECE 102 studying since I had exams for these classes coming up, but this week I’m gonna do a lot more studying for them.