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Hello everyone. I was just working on homework 1, problem 3 and I had a question. Assuming that this piece is in equilibrium, there should be a reaction force equal and opposite to the external load on the Bar AB. Can someone explain why we should only use the external load when we calculate the normal stress on the bar?

I suppose that the answer to your question depends on which FBD that you draw. Consider making a cut through bar AB, and keep the lower part of the bar. In that case you see that the axial load in AB is exactly equal to the 5000N applied force. However, if you make cuts at other locations on the bracket, you will expose other internal loads that would come into play in your equilibrium. The cut through AB is a good one, as it exposes the internal load in AB, which is what you want.

May I suggest that you pose homework-related questions like this on the discussion threads on the blog entry page? That will likely get you a quicker response as it will be more visible there.

I am starting Problem 1.1 and my first thought is to work from end E to cross section C, then to cross section A but the way the question is expressed feels like this is not the correct approach. Does anybody know if I am on the right track or if I need to rethink my approach.

You do not have to worry about which cross-section to start from - both (a) and (b) can be solved independently.

My advice would be to draw a detailed FBD for a particular cut and start resolving the force and moment equilibrium equations.

Is homework 02 due next wednesday night or friday night? On the document, it mentions friday but I thought the syllabus mentioned wednesdays as the norm.

It is due on Friday night i.e. 10th September 11:59 PM.

In problem 2.2, the question states ask to calculate the corresponding deformed lengths along the y and the x-axis. Is it safe to assume that we are just meant to find the change in length using the other two equations: upsilon x and z

Good morning, I had a conceptual question while doing HW 2.1: So in the given problem we are asked to find the stresses in all truss members and are given sigma (yield strength for axial loading). My question is what if we were asked to find shear loading on each truss member and were given tau (shear strength), would the axial loading free body diagrams have to be redrawn differently?

Hi David, just a note: Truss members are two force members i.e. they experience only axial stresses. So you don't have to worry about shear stresses in trusses.

I had a question regarding Lecture-book example 2.08 ( Prof. Krousgrill). In the second part of the question where it asks us to identify which material is stiffer, why do we consider the young's modulus as a parameter to compare the stiffness of materials?

when is the alphaTL term in the elongation equation positive like in the book and when is it negative like in the solution for HW 3.3

Hi Samuel,

The book follows the standard convention of taking the force on the body to be in tension. Hence if you follow the same convention in the solution, you should take the alpha TL term to be positive. However, if you assume the applied force to be in compression, you need to add a negative sign in front of the body to account for the change.

Where is Professor Kokini's zoom link for office hours?

Hi Zachary, I would suggest you email professor Kokini regarding the same.