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Solution video - H29.A
Solution video - H29.B
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HOMEWORK 29 - SP25
37 thoughts on “HOMEWORK 29 - SP25”
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When solving this problem, I first calculated the force and moment reactions at point B and then constructed the graphs based on that. The shear graph I found has two positively sloped lines both starting and ending at zero. I found the moment graph by looking at the moment at x=0 and then working my way to x=4.
I calculated a negative reaction moment at the left end of the beam. However, the area under my curve of the shear graph was not great enough to go from the negative moment start to the positive moment start. Should there be a positive moment at x=4, or should it be zero?
I believe it should be zero. Both my graphs began at zero and had positive values immediately afterwards.
Wouldn’t it be 80 though, because of the moment that is 80 CCW at the end of the board.
Nvm at the very end of the beam it would instantaneously jump from 80 to 0 due to the applied moment
I also calculated a negative reaction moment at the B end of the beam. However, I thought that the bending moment at the start would then be positive, because the moments would cancel each other out. This yielded a strange value for the ending moment however, so I wonder what the correct approach to solve this problem is.
The value should be zero and should include positive values
for this one can we solve it by calculating the areas instead of solving for each individual equation like we did in class? Just asking because it's a lot less work if we do it that way.
I believe it is wanted of us to do this problem using the new graphing method we learned in class in order to be able to solve problems like this one much more efficiently just as you said.
How did you solve for the y component of the force about B? When writing an equation for the Y Force components, I keep getting a result which says the Force equivalent of the load and the force applied P cancel, leading to Force Y at B being Zero
nvm
This is correct, there is no reaction force at B, only a reaction moment.
To solve for the y component of the force about B, I wrote out the sum of the forces in the y direction equation, for the entire beam, and set it equal to zero. Then, I did By + 40, for the distributed load force, -40, for the P force, therefore positive 40 and -40 cancel out to give the By force a value of zero.
I found Mondays Lecture super helpful. I was always slightly confused on this topic as a whole, but now it feels like I can look at a problem and instantly get an idea of what the shear force and bending moment diagrams will look like. If you are having trouble with this homework, you should definitely review the lecture/notes from 4/7.
Is there a systematic approach for handling different types of loads when constructing shear and moment diagrams? For example, is it better to split the beam into segments and apply force and moment balances like we did in Lecture 27, to write expressions for shear and moment like in Lecture 28, or to use integrals as shown in Lecture 29?
For me, I preferred to split the beams into segments since that's what we did in class and then each segment I label the lengths as "x" and then I go from there. As far as using integrals and all the other methods, I don't think it is necessary. Though, I don't think it'll hurt to try.
There are no reaction forces for this problem correct?
Never mind, just realized B would have a Y reaction
I made a FBD of the entire problem and found no reaction forces and a negative moment at B. Using this, I split the problem into two parts. The initial moment reaction in the beam cancels the moment at B. There is no initial shear stress since there is no reaction force at B either. For my shear diagram, I have two lines with positive slope that start at 0, peak at x=2, has a discontinuity at x=2, then ends at 0. For my moment diagram, I have two concave up lines that peak at x=2.
The first step to solving this problem it seems is to calculate the reaction forces/moment at B. From my calculations, it seemed that By and Bx was zero, and the only reaction was the moment reaction. However, this led to me getting strange values for starting and ending bending moments. Did anyone else have a similar issue?
Took me a while to realize that there was a reaction moment at B, had to adjust my answers to account for that. Luckily, it only ended up offsetting my final answers by the magnitude of the moment. It's pretty easy to adjust from there.
Although this problem could have been solved with the graphical method, I preferred using shear and moment equations then using the graphical method to go back and justify my answer.
I started by calculating the reaction moment at B, which is negative because of the cantilever setup. For the shear diagram, I noticed it begins at zero, increases linearly due to the distributed load, and drop at C because of the concentrated force P. The moment diagram starts at the reaction moment at B, curves upwards due to the distributed load, and increase at D because of the applied moment M.
Not sure if I am doing this correctly.
I found the moment around B, and found the Fy at 0, 2, and 4. Then I just did the same for moment and that gave me points, from which i just connected them. My moment diagram though was very similar to my shear diagram, just shifted down, so im not sure if it was done properly
Does the sign of the moment of B matters? Not sure if I need to make it positive or not?
The sign does matter since it opposes the couple at D and will be used to find bending moment. My approach for figuring out the sign is to assume a positive direction to better understand the signs if it turns out to be negative, but that's probably personal preference. You can find the moment at B by making a moment equilibrium at B that includes both the couple's moment at D and the reactionary moment at B.
I initially used the graphical method to sketch out my shear and moment diagrams, but I ended up re-finding my answer by writing out the shear and moment equations for each segment of the beam. Once I had those equations, it was super easy to plug them into Desmos to quickly double-check the shape and values of my graphs. Definitely helped confirm that my slopes and curvatures made sense.
When you guys were solving for the forces for the entire beam initially, did you also solve for the moment at B? Specifically, the actual moment value at point B, because I did and obtained a value of -80kN m. However, I'm not sure if I needed to. In past homework's, we didn't need to do it, but this homework also has a different reaction force compared to past reaction forces such as pins supports and roller supports.
I started by figuring out the force and moment reactions at point B, and then I built the graphs from there. The shear diagram ended up showing three segments with a positive slope that begin at zero. For the moment diagram, I checked the moment at x = 0 and then worked my way over to x = 4.
This question is fairly straight forward and similar to the last one. The only real thing which might trip you up is that the forces and the moments have to be zero not just one or the other. The you just need to go step by step and make your graphs curve the right way based on you slope for v(x).
Are we allowed to use the Graphical Method for this problem?
When solving this, I initially wasn’t sure about the effect of the CCW couple at the free end, but after working through it, I saw that it causes a positive jump in the moment diagram. Right before x is 4, the moment was -80 kNm, and the applied couple brings it up to zero, which makes sense since the beam ends w/ no moment.
I was wondering—if instead of having a concentrated moment M
M at the free end, we had a rotational spring with some stiffness, how would that affect the shear and moment diagrams? Would the problem still be statically determinate, or would we need additional methods to solve it?
One thing that almost tripped me up was treating the distributed load as a equivalent force even after solving for the reaction forces and moments at B. What caused me to rethink this was that since the equivalent load and the load at P would have canceled out, the shear graph would have stayed at zero.
I think that the example problems gone over in class on Monday were super helpful for solving this problem. It also helped me to read about the relationship between shear force and bending moments in the textbook to get a more intuitive understanding of what the graphs should look like in theory.
I know that the integral of the Shear force graph will produce the Moment graph. Since the Shear force graph comprises straight lines, I am inclined to think that the Moment graph will follow the path of a parabola. Am I correct in thinking this?
Think about it like this: let’s say you have a shear graph that is above the x axis (this should tell you that the graph of M is increasing wherever V is positive, Furthermore, when V INCREASING, M would concave up. So think about it, if V is positive and increasing, M would assume the shape of the right part of a parabola concave up.
I would recommend using your knowledge from calculus 1 and how first, second derivatives work.