Statics: Lesson 61 - Shear Moment Diagram, The Equation Method
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I've learned more about Shear Moment Diagrams in less than an hour by this man than 2 days by my professor. I am very grateful for this, thank you Sir!
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A little confused. How did you determine the 8' and 4' from the concentrated load of 54? I figured the midpoint of the 12' for the concentrated load of 54 would be 6'. Thank you
because its a triangle
For triangles, the centroid is 2/3 of the length (2/3*12=8)from the short lines and 1/3 of the length from the tall lines(1/3*12=4)
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DR. Hanson, thank for an incredible lecture on the Shear Moment Diagram using The Equation Method. These lecture makes Shear and Moments Diagrams seems very easy.
I don't know if shear moment diagram is easy or you're just better at teaching them. Neither way, THANK YOU SO MUCH!
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you can actually use the graphical method for this and find the area of a parabola. For the part above the x axis All you is height*length*2/3 = 18*6.93*2/3 = 83.14
real structural analysis teacher! I am satisfied! I REGRET that I lost much of my time in irrelevant Africa, Ethiopia politics.
Whilst doing the global equilibrium... why did you put the centre of the mass 8m from A. If its at the centre should it not be just located at 6m from A?
forget statics, i’m coming back to this while taking strengths of materials. great video.
0:55 Possible exception? On problem 7-53, the shear graph crosses the axis with a slope, not a parabolic curve, yet I didn't see how to continue with the graphing method because I needed to know where x=0 (which I think requires having the shear equation) in order to calculate the area of the triangles (it's not neatly on a line of discontinuity where we are given the x-value). This is due to an additional moment on the beam making the support forces unequal. If any of you see how I'm wrong, please point it out because I'd much rather use the graphing method as much as possible.
Please tell me how u solved y=mx+b ... Specifically, why u divided both 9 and 12 by '3' .. where did that 3 come from
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Can you do a course on Machine Design?, I think it is basically the course right after Solids.
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sir what if the given triangle load is from zero
increasing up unto the center but then from center to the right end becomes decreasing to zero, how to make an equation?
What if in this example the triangle distributed force would be flipped the other way around? This method doesn't seem to work in that case, if the slope is -3/4 and that the starting point is (0, 9) v(x)=0 has no solution.
Thank you thank you thank you thank you thank you thank you 😭
What do you do when you have a distributive force on the top and bottom of the beam at the same place?
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When doing moment on A. Why is (54)×(8) negative ? Won't it be moving clockwise direction and so positive ?
Thank you so much!!! You make learning statics so much fun:)
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Mr. Hanson can we imaging distributed load like rain and graphic like umbrella?
I wish I have this clip and internet as well as TH-cam when I was talking Statics Course 30 years ago.
lets say there was also a point load acting on the beam, then you have to take the cut áfter the point load right?? or can it be at any point? because what if you took the cut too early on? this is where you have to use Macaulay's method right?
Great video as always, just to let some of you know though, it would probably be in your best interest to use calculus and integrate to get the equations instead of cutting the beam. It would save you alot more time on exams, and its not hard integration either, its just integrating a linear function. Could easily find videos on how to do it, def saves lots of time tho instead of drawing a FBD and cutting just to find the max points and what not.
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excuse me professor, where to get the 3 as the denominator in finding the equation y?
The denominator is 4.
You can use ratio and proportion method
i wish I could have watched this during my undergraduate Mechanics course! :(
Thank you
This was enlightening. Thank you so much.
What if the triangle was flipped? Would the force be “(1/2)*(9 - (9/12)*x)*x?
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Good day sir, is this the Shear and Moment Diagrams using Shear and Moment Functions? Thank you.
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i have a question. so you know how he says the graph should always go back to zero, iv come across a question where the moment bending diagram did not go to zero however it went to -25 as there was starting ACW moment on the left which was 25 and a CW moment on the right which was +25.
How did you get 8 and 4 from the total base
I didn't know that the centroid of a triangle is 2/3 of the base. So I figured out an x so that the areas on either side would be equal. In other word determine an x so that (XxY)/2= 27 Which is half the size of the total triangle. X by my calculations would be 8.485, Which is pretty close, but still not the same. Why isn't this correct?
stunning!
you are heisenberg of statics
do you have videos on moment are theorem for deflection and virtual work?
Why is the distance on the distributed load not half which is 6?
Hello sir.. while you calculated the external reactions Ay and By, you didn't consider the distributed load. Can you please explain why because I'm confused!
He converted the distributed load into a point load acting at the centroid of the triangle. So .5x12x9=54kn and it acts at 2/3rds along the length or 8m.