@Ali Alghamdi CAME HERE WONDERING THE SAME!?!?!?

Thanks for the note.

In beams, a positive bending moment is one that bends the beam segment concave up. Consider a beam segment. If we want it to bend concave up, we are going to apply a clockwise moment at the left end of the segment and an anti clockwise moment at the right end of the segment. Hold on to the ends of an (imaginary) flexible ruler with your left and right hands. Now gently force the ruler to bend downward (concave up). Notice how you are turning your left hand clockwise and right hand anti clockwise to bend the ruler. You are applying a clockwise moment to the left end of the ruler (beam) and an anti clockwise moment to the right end of the beam. This is considered a positive (pair of) moments. So, if we are asked to draw the free-body diagram of a beam segment, we place a clockwise moment at the left end of the segment and an anti clockwise moment at the right end of the segment. Here of course the right end of the segment is free so there is no need to show an anti clockwise moment at that end.

Thank you :)

*That is the distance from the center of the distributed load to Point A called the moment arm. *We are taking the moment of the distributed load about A. *Bending moment is the product of a (concentrated) force by a distance (moment arm). *The concentrated force due to the distributed load is the area of the rectangle: (500)(10). *This equivalent concentrated force is placed at the center of the rectangle, 5 meters away from Point A. *Therefore, the moment of the force about A is (500)(10) times the distance 5 m.

@@DrStructure please how did you input your values into the equilibrium equation @5:49 4(8)+2(12)-16By

@@aveira.a4916 Three forces cause a moment about A. The applied load of 4 kN is 8 m away from A, hence it create a clockwise moment of 4(8) = 32 kN.m about A. The applied load of 2 kN is 12 meters away from A, its clockwise moment about A, therefore, is: 2(12) = 24 kN.m The support reaction at B is 16 m away from A, causing a counterclockwise moment of 16 By. Assuming clockwise is positive, and counterclockwise is negative, the total moment about A becomes: 4(8) + 2(12) - 16 By. Setting this sum equal to zero and solving for By, we get: By = 3.5 kN

@@DrStructure thank you, the equilibrium equation @5:50 Efy= Ay+By-4-2=0, the 4 and 2 are the forces acting on the beam right?

@@aveira.a4916 Correct.

Bending moment is the product of a force and a distance. If the force is applied at point A and we are taking the moment at A, then the distance between the two points (here A and A) is zero. Therefore, the moment = (force)(0) = 0. A concentrated moment is a product of a force and s distance already. There is no point in multiplying this product by a distance again. So, when writing the moment equilibrium equation about any point on the beam, the applied concentrated moment appears by itself without any distance multiplier.

+Europa Erwacht In Mc = 10 kN-m, the unit of bending moment is kiloNewton-meter (kN-m). This represent the product of a force (having kN as its unit) by a distance (having meter as its unit). When a beam is cut, the internal forces at the cut point must be drawn on the free-body diagram and accounted for in the equilibrium equations. The internal forces in the beam at point C include a shear force (V) and a bending moment (Mc). Both these forces which appear on the free-body diagram must be included in the equilibrium equations.

The sign convention being used here is attached to the moment equilibrium equation, not to the moment itself. A bending moment is either clockwise or counterclockwise, not positive or negative. When writing the moment equation, we can assume either clockwise or counterclockwise to be positive. The purpose of this sign convention is to make sure we add up all the clockwise moments (say, the sum is CM), and add up all the counterclockwise moments (say, CCM), then subtract one from the other. Whether we write: CM - CCM = 0 or CCM - CM = 0 does not really matter. They both represent the same equation. This is different than the beam sign convention for bending moment. There, the pair of end moments that causes the beam segment to deflect concave up is considered positive. That means, given a beam segment, a clockwise moment at the left end of the segment coupled with a counterclockwise moment at the right end of the segment represents positive moment in the segment.

+Roger Carter Starting at 12:10, the second equation is obtained by taking sum of the moments about point A (the left end of the segment). Alternatively we can take sum of the moment about the right end of the segment. Assuming counter-clockwise moment being positive, we get: 25000 - 5000(x) + 500(x)(x/2) + M = 0. Either way, the resulting moment equation is: M = 5000x - 250x^2 - 25000.

ok thank you.

When writing the moment equation, we usually show the internal shear and moment in the positive direction. So, here we get 25000 +.....+ M = 0. When we solve for M, we get: M = -25000 +... A negative value for M indicates that the correct direction for the moment is opposite to what was assumed in the free-body diagram.

Thanks a lot

The cantilever beam has a vertical support reaction of 5000 N. The reaction force results in a clockwise moment of 5000x at the cut point.

Yes, it means exactly that-- the two forces cancel out each other, their algebraic sum is zero. Whenever we cut a beam, the internal forces (i.e., shear and moment) at the two sides of the cut add up to zero. So, the shear forces act in opposite directions with the same magnitude; so do the bending moments.

@@DrStructure i see. Very helpful Dr. Thanks a lot

If the 25000 was a vertical force, then its moment about point A would have been zero. But, the 25000 is not a vertical force, it is a concentrated moment (view it as the product of a force by a distance). Therefore, we would not/should not multiply it by a distance again. We simply write the concentrated moment in the equilibrium equation, be it about point A or any other point.

I am not sure which moment equation you are referencing. For the cantilever beam, if we take point A to be the origin of the coordinate system, meaning x is measured from left to right, the moment equation can be written as: M = 5000 x - 250 x^2 - 25000. If however we place the origin of the coordinate system at the right end of the beam, and define the x-axis going from right to left, then when we cut the beam at point P, we can take the right segment of the beam for writing the shear and moment equation. In such a scenario the shear equation becomes: V = 500x and the moment equation can be written as M = -250x^2

Cut the beam at x, draw the free body diagram of the left segment of the beam (from the left support to x), show the distances correctly on the diagram (for example, if x is to the right of mid-span, and the beam's length is L, then the distance from the left end to the cut point would be (L/2)+x. Then write the equilibrium equations for the left segment and calculate the moment at the cut point.

The sign convention for writing the equilibrium equations is completely arbitrary. There is no right or wrong here. Think of it this way, suppose our moment equation (assuming clockwise is positive) comes out to be: a + b - c = 0. The same equilibrium equation, but using the opposite sign convention, is going to be: -a - b + c = 0 Note that algebraically we can rewrite the second equation as: -(a + b - c) = 0, which differs from the first equation by a factor of -1 on the left side. But, since the right side of the equation is zero, that negative sign does not effect the results. The second equation can be turned into the first equation by multiplying both side of it by -1.

The shear force is drawn in the positive direction (according to the established beam sign convention). Although you can draw the force upward, it would be best to follow the sign convention when showing internal forces in beams.

+Sinclair Mupupa The pen-action simulation is done in VideoScribe. The actual writing/drawing is done using iDraw on an iPad.

Thanks greatly appreciated.

+Tull shaquille cooper Not sure if I understand your question. The two values are calculated using the equilibrium equations which are obtained from the free-body diagram.

The moment equation is the result of adding three moment terms, all taken about point A. The vertical load of 4 kN has a moment arm of 8 m about A, therefore, its moment about A is 4(8) = 32. The vertical load of 2 kN, with a moment arm of 12 m about point A, results in a moment of 2(12) = 24. Both these moments are in the clockwise direction. The reaction force By, having a moment arm of 16 m, causes a counterclockwise moment of 16 By about point A. Therefore, the total moment about A, assuming clockwise is positive, comes out to be 4(8) + 2(12) - 16 By = 0.

When dealing with bending in beams, we shouldn't think clockwise or counterclockwise, we should think concave up or concave down. By convention, we assume the moment direction(s) that cause the beam segment to bend concave up as positive. A counterclockwise moment at the left end of the segment coupled with a clockwise moment at the right end of the segment forces it to have a concave up deflection. Therefore, these end moments are assumed to be drawn in the positive direction.

When writing a moment equation, we can assume either the clockwise or the counterclockwise direction to be positive. If we assume the latter, the equation can be written as: 4Vc - Mc = 0. If make the alternative assumption, the moment equation becomes: -4Vc + Mc = 0. These two equation mean the same thing and yield the same results.

Substitute the expression for V (it is 5000 - 500 x) in the moment equation, then simplify. The result is: M = 5000 x - 250 x^2 - 25000

V = 5000 - 500x Then, V x - M + (500x)(x/2) - 25000 = 0 Or, (5000 - 500 x) x - M + 250x^2 - 25000 = 0 Or, 5000x - 500x^2 - M + 250x^2 - 25000 = 0 Or, 5000x -250x^2 - M - 25000 = 0 Or, M = 5000x - 250x^2 - 25000

When writing the equilibrium equation, we assumed the moment at A to act in the clockwise direction. That equation then gave us a value of -25000 for the moment. The negative sign here indicates that the moment, in actuality, is acting opposite to the assumed direction. So, we can show the moment on the diagram in one of two ways: 1. Either draw it in the clockwise (assumed) direction and show its magnitude as -25000, or 2. Draw the moment in the counterclockwise (the correct) direction and show its magnitude as 25000.

+Nicholas O'Connell I am not certain what you are asking, please elaborate.

sorry Ay = 2.5kn By = 3.5kn how did you get

+Nicholas O'Connell We get By from the third equilibrium equation: 4(8) + 2(12) - 16 By = 0 gives By = 3.5 Then, substitute 3.5 for By in the second equation to get Ay: Ay + (3.5) - 4 - 2 = 0 gives Ay = 2.5

thank you so much

That is the moment of the distributed load about Point A. (500)(10) =5000 is the total equivalent concentrated load. We can replace the distributed load with this concentrated load. But where, on the beam, would we place this concentrated load? at the midpoint of the beam (which is the geometric center of the rectangle representing the distributed load.) Since the midpoint is 5m away from A, the moment of the concentrated load about A becomes (5000)(5).

Alright thank you

You need to be more specific. What 5? where?

There are three terms in the moment equation. Taking sum of the moments about a point located at distance x from the left support, we get one clockwise moment and two counterclockwise moment terms. The moment of the support reaction about the cut-point is clockwise (5000)(x). The moment of the distributed load about the cut-point is counterclockwise (500)(x)(x/2). The concentrated moment of 25000 N-m is also acting in the counterclockwise direction. If we assume clockwise to be positive, we get: M(x) = 5000x - 250x^2 - 25000. If we assume counterclockwise to be positive, we get: M(x) = -5000x + 250x^2 + 25000. Either formulation is correct.

Thank you very much. That clears things up a lot! :) Just to confirm, the concentrated moment at A is only affecting joint at A, and therefore we don't incorporate it into the equation? Also, I think the confusing part was at 12:01, where it was said that "sum of the moments about point A must be 0" - assuming then that the moments were being taken about point A and not a random cut to the right of A.

Although the concentrated moment is applied at A, it does affect the rest of the beam. Say, we cut the beam at some distance (x) from A. If we draw the free-body diagram of the left segment (from A to x), we need to show the vertical reaction force at A, the concentrated moment at A, the distributed load between A and the cut point as well as internal bending moment and shear force at the cut point. This means five forces are shown on the diagram. So, when we write the equilibrium equations for the beam segment, we need to include all 5 forces/moments. Therefore, the moment equilibrium equation about the cut point should have five terms in it, one term for each load/moment appearing on the free-body diagram. Here are the five terms (assume clockwise direction being positive): 1. The moment term due to the vertical reaction at A is: + 5000(x) 2. The moment due to the concentrated moment is: -25000 (note that here we are not multiplying the value by a distance since 25000 is already a moment, it has a unit of moment, it is not a vertical force). As a general rule, when a concentrated moment appears on the free-body diagram, it affect every point in the beam, so it appear in the moment equilibrium equation as is. 3. The moment term due to the distributed load: -500(x)(x/2). 4. The moment term due to the shear force at the cut point: V (0) = 0. Note that since the shear force passes through the cut point, its moment arm is zero. So, the force does not produce any moment about the point. 5. The moment term due to the bending moment at the cut point: -M. Here I am assuming the free-body diagram shows a counterclockwise moment at the cut point labeled M. Now, summing the above terms and setting it equal to zero, we get: (5000)(x) - 25000 - 500(x)(x/2) - 0 - M = 0 Or, M(x) = 5000x - 25000 - 250 x^2

Thank you so much for all of that. That helps so much. It all works now perfectly. You saved me - thank you! :D