SA07U: Shear & Moment Equations

[@hollytang8769]

This is the most straight forward clearest explanation I have gotten 😭❤

[@issswore]

The clearest explanation I ever seen . Thank you so much !!

[@musa8ballpoolyt644]

Really 100%awesome lecture

[@nakamura7346]

8:41 It should be Cy not By in second equation. And you should solve the third equation before you can solve the second

[@joshpord]

At 00:18 shouldn't the M(x) = -5000+1000x-50x^2 ? As shown, the 5000 has an x multiplying it also, making the moment M(x) = -4000x-50x^2

[@DrStructure]

Yes, you are right. Thanks for noticing and mentioning the error.

[@mesaplayer9636]

I had noticed that too, glad I am not going crazy yet.

[@nathanbanda8877]

Sweet lecture.Makes me understand easily.

[@traincrazy7361]

8:39 in your sum of forces, your By should be Cy but other than that, great video

[@harisabdullah2980]

Awesome keep going 😍🙌🙌

[@shutnawi6635]

thanks this was incredibly helpful

[@whereitreallycounts36]

at around 7:20 doesn't make sense. what happened to the 4kn force of B? why isn't it from 6 to 10kn and then from 1-kn to 4kn?

[@DrStructure]

Please elaborate. How would you draw the free-body diagrams for the beams Dan write its shear and moment equations?

[@ShogunleEsther]

Thanks for this

[@clintcallender9502]

I have a question. I saw other persons when they cut the beam after mid span, being given the distance from origin to mid span, they will use the distance plus x2. Example, in the second example, in segment 2 after cuttting the beam after mid span, the 375 support force distance from the cut whenndoing moments would be 5m+x. Can this be done? If you do this, you get a different equation for moments

[@DrStructure]

X is the distance from the origin of the coordinate system to the cut point. If we select the left end of the beam as the origin of the coordinate system, the expression given in the video will be obtained. But we get a different equation if we change the position of the origin. For example, if we place the origin of the coordinate system 5 meters to the right of the left support, and cut the beam in the second segment, instead of (5 + x) for the moment arm for the support reaction, we should use x. Although the equations will be different for different origin location, they will give the same moment values when evaluated.

[@clintcallender9502]

​​@@DrStructureif you can, what would the equation be for the second segment if you do it that way, you get 625-125x? Or i am wrong. Also if so, the range of x would change right, as in it would be from 1 to 5, as oppose to yours which would be from 5 to 10?

[@DrStructure]

You are correct. If we place the origin of the coordinate system 5 meters to the right of the left support, the moment equation for the right segment of the beam becomes: M(x) = 625 - 125x But the range of x would be between zero and five, not 1 and 5.

[@OmarHadani]

Can you explain how the reaction forces were obtained please? It just goes from having Ay and By to the solution for Ay, so you somehow got By but you didn’t show that

[@DrStructure]

The support reactions are calculated using the static equilibrium equations. You can find explanations and examples in the previous (prerequisite) lectures. Use the online course (link in the video description field) to explore the course content further.

[@kinjazz5050]

Nice video. Thank!

[@INAMmathematicivilengineering]

Which app you are using for creating the leacture please tell

[@DrStructure]

We mainly use Adobe Illustrator and Camtasia Studio.

[@INAMmathematicivilengineering]

@@DrStructure suggest any app for creating leacture in android mobile, I am very grateful to you

[@DrStructure]

@Mathematics with Inam Creating lectures on a mobile device is challenging, given the limited space and functionality of mobile apps. I am not aware of any mobile app for this purpose. There may be some useful apps for tablets such as iPad, but generally speaking, editing and producing videos becomes a cumbersome process on a small device.

[@INAMmathematicivilengineering]

@@DrStructure okay very thankful for replaying thank you ❤️🥰

[@OmarHadani]

Also do you have a separate video on calculatng reaction forces and drawing BMD?

[@DrStructure]

Yes, you can find more lectures on shear and moment diagrams (i.e., how to construct them) in our free online course. See the video description filed for the link.

[@fitnessnoob2271]

Thank you so much

[@gnidnoeled786]

In real world problem, when or how do we have the concentrated moment applied at beam's midpoint?

[@DrStructure]

We often isolate, and then analyze, parts of a larger mechanical/structural system. The isolated system, say a simple beam, could be subjected to an eccentric horizontal load causing a bending moment to develop at its midpoint (or any other point.)

[@gnidnoeled786]

Thanks.

[@ahmedbenlarbi2348]

Please how is calculated the following maximum bending moment : (WL)^2/24 . THANKS .

[@DrStructure]

Please provide the video time where your question originates from.

[@ahmedbenlarbi2348]

@@DrStructure it s not in this video . It s in a one from an other lecturer . I asked him the same question but he did n t replie . It s about designing a stairs . He just mentioned that the bending moment at the suport is as following : W(L squared) / 24 ( negative moment ) . Thank you so much .

[@DrStructure]

It is not clear what you are asking. The max bending moment is a function of the applied load and the nature of the beam itself. The max bending moment in a cantilever beam is different than the max moment in a simply supported beam, and they both vary depending on the location and magnitude of the load. What is the exact dimensions of the beam under consideration? How is the beam supported? What is the position and nature of the load? If you know the answers to the above questions, and the beam is statically determinate, you can analyze it, then draw its moment diagram in order to locate the position and magnitude of the maximum moment.

[@ahmedbenlarbi2348]

@@DrStructure it s a three spans beam . Thanks 🙏

[@DrStructure]

A three-span beam resting on pin/rollers is statically indeterminate. You need to analyze it and draw its moment diagram to determine the magnitude and location of max moment.

[@JesusandPals23]

How can V(4) = 6 kn and -4 kn at the same time? Wouldn't V(4) -4 kn where the 10 kn is applied? I suppose in reality point loads aren't really applied at exact points but I'm interested in understanding this in a bit more depth if you wouldn't mind helping me out?

[@DrStructure]

You are right, point loads are rarely applied at a single point. They are often distributed over a small area, for example, when the end of a beam rests on another beam (say, they are oriented 90 degrees apart), the weight of the former can be considered as a point load applied to the latter. But generally, the contact area is small enough so that we can assume the load is being transferred through a (dimensionless) point. Imagine a beam consisting of an infinite number of such points. Let’s number them 1 through N. Let’s suppose that the 10 kN force is being applied at point i. This point has two neighbors: Point i-1 and Point i+1. The vertical force between Points i and (i-1) constitutes the shear force just to the left of i, and the vertical force between i and (i+1) is the shear force just to the right of i. These two shear forces are not necessarily equal in magnitude. But, what must be true is the condition of equilibrium at i. That is the force being directly applied to i (in our case 10 kN), plus the shear force just to the left of it and the one just to the right of it must add up to zero. In this case, the shear force just to the left of it is an upward force of 6 kN, and the one just the right of it is upward 4 kN. Hence, their sum equals to zero. Viewed this way, we often say shear is not defined at i, rather, it is defined just to the left and just to the right of it. This makes sense if we view the point as dimensionless which cannot be subdivided. The smallest subdivision here is a single point, like i, i-1, i+1,… and if we define shear as the force between two neighboring points, then we have a shear force on either side of i, but not at the point.

[@JesusandPals23]

Thank you for the very detailed response!

[@BudhibirLoitong]

at around 7:6 where does 10x comes from

[@DrStructure]

10 is the point load magnitude. The load produces a moment about the cut point. The moment arm is (x-4) where x is the distance from the left end of the beam to the cut point and 4 is the distance from the left end of the beam to the point of application of the point load. So, the distance from the point of application of the point load to the cut point (the moment arm) is x-4.

[@Kidist-hb1mi]

Best ++++....

[@johnbrianescobio9243]

Where is the part 2?

[@DrStructure]

You can find an updated version of this and other related lectures in our free online course. See the video description field for the link.

[@minmax3160]

woke up my sleeping memory