Thanks for the feedback!

@@DrStructure you really deserved it

That is the moment arm (the distance from the center of the rectangle to Point A) for the rectangular load. The rectangular load is replaced by its equivalent concentrated load. The magnitude of this concentrated load equals to the area of the rectangle: 100(5) = 500. This concentrated load is located at the center of the rectangle which is (5/2) = 2.5 m away from Point A. This distance therefore is the moment arm for the concentrated load. The moment of the concentrated load about A therefore is its magnitude times the moment arm, or : 100(5)(2.5).

Dr. Structure Very helpful explanation. I didn't learn in class yet that the concentrated load is *located at the center of the rectangle,* in a circumstance like this. Many thanks.

Dr. Structure Understood! Thank you for the thorough explanation.

Moment arm is the distance between the load position (say, Point P) and the point we are taking the moment about (say, Point O), here the right end of the segment. How do we determine the distance between P and O? Generally, we assume the left end of the beam to be the origin of our coordinate system. Say, the load is positioned 3m away from the origin and the right end of the segment is 7m away from the origin. What is the distance between these two points? 7 - 3 = 4. So, if we are taking the moment of the load about the right end of the segment, the moment arm would be 4m. In our formulation however, the beam segment does not have a fixed length, we use x to denote the length of the segment. That is, x is the distance from the origin to the right end of the segment. Further, the first 5m of the segment is subjected to a distributed load. We are going to replace it with a concentrated load. The magnitude of the concentrated load is (5m)(100N/m), or 500 N. This load is placed at the geometric center of the distributed load which is located 2.5m to the right of the origin. So, the distance between the load and the right end of the segment is (x - 2.5). Another way of looking at this is to keep in mind that distance (the moment arm) is a positive value. This means we need to subtract the smaller value from the larger value when determining the moment arm. Since here x is the length of the entire segment, it is the larger of the two values.

+Dr. Structure Rock on! Mighty thanks for that explaination. Great video and I hope there's more to come :)

+Sands Kreative Verksted There is always an x^2 when a distributed load is present. One x is for calculating the area of the rectangle under the load (x is the width of the rectangle). The other x is for the distance from the center of the rectangle to the (cut) point. So, you need to multiply the first x (actually a multiple of it) by the second x to get the moment of the distributed load about the point.

Thank you very much!

Here, we are taking sum of the moments about the cut point. To take the moment of the uniformly distributed load about the cut point, first replace the load with its equivalent concentrated load (the area of the rectangle.) The rectangle has a base of x and a height of 100, therefore its area is 100 x. This concentrated load (of 100 x) is placed at the center of the rectangle which is x/2 from the cut point. Then, the moment of the concentrated load about the cut point is: (100 x) (x/2).

Dr. Structure thanks i get it

The correct equation indeed is Ay + Cy - 500 = 0.

@@DrStructure Might help "Students" if you show that correction on the video as to not confuse anyone.

It really is not the left- or right- hand rule. It is clockwise or counter-clockwise rule. It does not matter if you decide clockwise (or counter-clockwise) to be positive or negative, as long as the sign convention is used consistently throughout the equation.

No, the reaction force at the roller/pin support causes a discontinuity in shear and moment. You need to consider that as two separate segments each with its own shear and moment equations.

Dr. Structure Got it. Thanks a lot.

+Jody Lao The given equation is correct. Here, for writing the moment equation, clockwise assumed to be positive. The support reaction at the left end of the segment generates a clockwise moment about point O. This moment is 375x. The downward distributed load generates a counterclockwise moment about Point O, hence the negative sign. We write this as: -100(x)(x/2). And, since the internal bending moment (M) at Point O is drawn in the counterclockwise direction, it is shown as -M in the equation. So, the entire moment equation is written as: 375x - 100x(x/2) - M = 0. Simplifying and rearranging the equation, we get: M = 375x - 50x^2

***** Yes. In this example since we are taking the moments about point A (the left-most point in the beam), then any downward force would cause a clockwise moment. In general, a downward force causes a clockwise moment if the moment its being taken about a point to the left of the force. If the moment however is being taken about a point to the right of the downward force, the direction would be counter-clockwise.

Dr. Structure So if the Moment--taken from the right--is assigned positive in the ccw direction, then the downward force is assigned as positive?

***** If I understand your question correctly, the answer is no. The direction (sign) of the force is not really related to that of the moment. 1. By convention we assume an upward force to be positive and a downward force to be negative. If you wish, you can choose a different sign convention as long as you do it consistently. For example, you can take the downward direction to be positive and upward direction to be negative. 2. When it comes to a bending moment, it either is causing a clockwise rotation or a counter-clockwise rotation. Again, by convention we assume clockwise to be positive. So let's say we go wit the convention: upward force is positive and clockwise moment is positive. Now say we have a point and an upward (positive) force. Depending on the location of the point, the force could cause a clockwise (positive) moment or a counterclockwise (negative) moment about the point. Simply put, a positive moment is not necessarily caused by a positive force. The best way to determine the direction of the moment that a force causes, imagine a circle where its center is at the point and its radius is the distance from the center to the head of the force. Now make the force orbit around circle in the direction of the force's arrow. Is it moving in the clockwise direction or counter-clockwise direction? That is the direction of the moment.

Dr. Structure After watching your videos and others, I think I finally understand. [Going with what you said; BM in the cw direction is positive (+) while BM in the ccw direction is negative (-)] and I will be using the example at 6:28 to illustrate my understanding. Let's say I'm taking the Summation of Moment at B.If I assigned ccw for Moment at B to be negative (or which is the same as saying cw at Moment of B is postitve), then: M = -By(0m) - 10kN(6m) + Ay(10m); The rest is solvable. What I wanted to show was that I assigned By(0m) & 10kN(6m) to both be negative because they produce ccw motion at B and since I already assigned ccw at B to be negative means that they have to be negative. Ay(10m) is positive because it produces a cw at B & cw at B have been assigned as positive. Hopefully, this assessment was right.

***** Yes, your analysis is correct.

Since shear and moment equations cover the entire beam, then x needs to be defined form 0 to L, where 0 is the start point (often the left end of the beam) and L is the end point (usually the right end of the beam). Obviously shear and moment diagrams could be piecewise continuous, meaning we can have different equations for different segments in the beam. So, x could be defined for each beam segment. This is demonstrated in the video. You can also use different coordinate system, say, x going from left to right for the left segment, but changing from right to left for the right segment. But this is not advised as it often leads to confusion. For the sake of consistency, it is best to pick one origin (usually the left end of the beam) and measure x from that origin in every interval (segment) in the beam.

Imagine a simply supported beam of length L subjected to a uniformly distributed load of w. The vertical reaction force at the left end of the beam is wL/2, and the bending moment equation for the entire beam is: M(x) = (wL/2)x - w x^2/2 This is a continuous function with no jump in the moment value. For example, at x = L/2, we get a moment value of wL^2/8. So, bending moment just to the left of midpoint as well as bending moment just to the right of the mid point would be wL^2/8. Now, suppose there is a concentrated moment (M) at the mid point. This means, if bending moment just to the left of the mid point is m1, then moment at the right end of the point would be equal to m1 + M, since going from left to right, all of the sudden M appears on the beam. This all of the sudden appearance of a moment on the beam means the difference between bending moment at the left end and the right end of the midpoint equals M, hence the drop, or rise (depending on the direction of the moment) in the moment value. Depending on the moment value at the left end of the midpoint, the drop/rise in the moment may not cause a change in the sign. For example, if moment at the left end of the midpoint is 100 kN-m, and the concentrated moment is 80 kN-m, the moment at the right end of the point would become 20 kN-m . Here, there is no change in sign.

Thank you :)

anytime!

I guess that I am wrong, I just checked it. Dr structure, are you using counterclockwise as -ve in your sign convention?

The sign convention for writing the moment equilibrium equation is rather arbitrary. We cannot do wrong as long as we are adding all the clockwise moments together, all the counterclockwise moments together, then subtracting one sum from the other and setting it equal to zero.

We first cut the beam to the left of the concentrated load, drawing the free-body diagram for the left segment of the beam, and use that diagram to write the shear and moment equations for the segment. We are not using the information from the right segment of the beam when writing the equilibrium equations and generating the shear and moment equations. We then cut the beam a second time, this time somewhere to the right of the concentrated load. Since we want to write the moment and shear equations in terms of x where x is the distance from the origin of the coordinate system to the cut point, and the origin is located at the left end of the beam, we take the beam segment from the origin to the cut point as our free-body diagram.

No, there is not.

ok thank you so much keep uploading i really appreciate how you are making the things easy

In frames, we draw shear and moment diagrams member by member, then assemble them into a single diagram. For example, if a frame member is vertical, you can rotate it (align it horizontally), draw its shear moment diagram, then rotate it (along with the diagrams) back to its actual orientation.

Now I get your technique for solving frames. Thank you for helping me out.

375x - 100(5)(x-2.5) - M = 0 M = 375x - 100(5)(x-2.5) M = 375x - 500(x-2.5) M = 375x - 500x + 1250 M = -125x + 1250

Thank you!

+Nitin E P No, 6 m is the distance between the applied load and the right support.

This particular lecture was prepared by (1) tracing the lecture notes using an stylus on an iPad and saving the trace as an SVG file, (2) loading and running the SVG file through VideoScribe online tool to generate a video file, and (3) loading the video file into Camtasia Studio and synchronizing the video with the audio file which was recorded separately.

We do not have a pdf version of this specific lecture. But we do have the pdf for a different version of the lecture. You can find it here: lab101.space/Course-Statics.asp

+David Gallo Here, since the uniformly distributed load is applied to the half of the beam, we have two segments: the (left) segment subjected to the uniformly distributed load and the right segment with no load on it. Assuming the origin is located at the left end of the beam, then the left beam segment is between 0 and 5m, and the right segment is between 5m and 10m. Since we are measuring x from the origin, then for the left segment we have 0 < x < 5 and for the right segment we have 5 < x < 10.

+Dr. Structure Thank you!

I cannot follow your line of reasoning here. Are you saying Cy should be half of the total load? Half of (5)(100)? That would have been the case only if the load was distributed over the entire length of the beam. Here, the load covers half of the beam only, and since the load is closer to A, then Ay carries a larger portion of the load. Now on to the math behind this. The last equilibrium equation @10:30 reads: 100(5)(2.5) - 10 Cy = 0. This gives us Cy = 125. To show that Cy is anything but 125, you need to prove that either the equation is incorrect or the calculation is done incorrectly.

XD I see the eq could be also written as 500*2.5-10*Cy=0 right )?

Yes, the distributed load can be replaced with its equivalent concentrated load. The magnitude of the concentrated load is (5)(100) = 500. The location of it is at the center of the load rectangle which is 2.5 m away from point A. So, the moment of the load about A can be written as (500)(2.5). The entire equilibrium equation then becomes: (500)(2.5) - 10 Cy=0.

The magnitude of the distributed load is 100 N/m, it is distributed over 5 m. Let's replace it with a concentrated load. What would be the magnitude of the concentrated load replacing the distributed load? (100)(5). That is the area of the rectangle. Where would we place that load on the beam? At the center of the rectangle, at 2.5 m away from point A. So, what would be the moment of the concentrated load about point A? Its magnitude time its distance from A: (100)(5)(2.5) The other term in the equation is the moment of the right support reaction about A. The two moment terms differ in sign, one is positive the other negative, because one has a clockwise direction and the other has a counterclockwise direction.

oh my god you reply so fast, I havnt gone through it yet but this definitely helps more than my college professor.

great i got it now, thank you so much!!!! you dont know how this mean to me esp i am going to have the exam right tmr

You are welcome!

Imagine two beam segments each having a different length and is subjected to a different load. For each beam segment, we define the origin of the coordinate system to be located at the very left end of the segment. We then define variable x to be the distance from the origin to an arbitrary point on the beam, and use x (a variable, not a constant) to formulate the shear and moment equation in that beam segment. This reference system does work. Feel free to elaborate why you think it does not work, what flaw do you see in the formulation. The more specific you are, the better we can address your concern.

Yes, the equation should have been written as Ay + Cy - 500 = 0

Dr. Structure thank you very much

The sign of shear in a beam segment is always decided using a pair of forces: a shear forces at the right end of the segment, and one at the left end of the segment. These forces act in opposite directions. By convention, if they tend to cause a clockwise rotation, shear is considered positive. This means, shear is considered positive, if the force at the left end of the segment is upward and the force at the right end of the segment is downward causing the segment to turn clockwise.