SA06: Shear & Moment in Beams

[@RainbowPaint]

I love these videos, perfect for preparing for my Statics final. Thank you.

[@iwritechecksatthegrocerystore]

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

[@nasratullahamarkhail3217]

Greatest video ever!!!Easy to understand and good examples.....I LOVE u Dr. Structure!!!!!!!!!!!

[@rosannaubalde6492]

thank God for dr.structure!! answer to my prayers! great channel! just subscribed!!

[@racing_republic_cycle6458]

at 10.40 , when you calculate the moment of A, why is the 5 multiply with the 500(10). what does the 5 represent ?

[@DrStructure]

*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.

[@aveira.a4916]

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

[@DrStructure]

@@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

[@aveira.a4916]

@@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?

[@DrStructure]

@@aveira.a4916 Correct.

[@eminashindahouse]

Hi Dr..sorry to bother. Mind if i ask one thing. At 4:00, the downward V for the LHS and the upward V for the RHS does not mean that the shear force cancels out each other, am i right? This different in direction is only for us to identify on which side of the cross section do we use in calculation, am i right Dr?

[@DrStructure]

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.

[@eminashindahouse]

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

[@edg4246]

Hi. at 6:48 you are showing shear force V acting down. Why down? can we show it up?

[@DrStructure]

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.

[@kabirsharma4820]

Generally when calculating moment about a point we do not consider moment caused by a force passing through that point. For example, 12:47 we did not consider the reaction 5000N when finding out moment at point A. My question is, when we do not consider the force passing through the point A for calculating moment, how can we consider the moment 25000 which is applied at A?

[@DrStructure]

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.

[@aiswaryavnair3802]

@1.55, the moment is taken as clockwise which is created by the external load, but as we are determining the internal bending moments and shear forces, don't we have to write it as anticlockwise moment ?

[@DrStructure]

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.

[@aiswaryavnair3802]

Thank you :)

[@lebogangtsietsi3608]

Dr Structure you're the best.Thank you very much.

[@DrStructure]

Thanks for the note.

[@tahiraman6456]

a question plz.... at 12.07 u take -M which is not countering 25000 N-m . it is bothering me because i think it should be clockwise countering the effect of 25000 N-m

[@DrStructure]

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.

[@tahiraman6456]

Thanks a lot

[@trolllovindaddy]

At 7:42 wouldn't it be -4Vc since Vc is acting downwards?

[@DrStructure]

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.

[@fathereshark8421]

At 12:50 isn't the equation for M, M= Vx + 500x^2 - 25000, why in the video is it 5000x instead of Vx?

[@DrStructure]

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

[@CNsongs]

12:10 can you take moment about pint a and say just M? M is the moment about that right corner right? So if you take moments about m corner you can say the moment is M

[@DrStructure]

+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.

[@CNsongs]

ok thank you.

[@cookn.eatn.cravin]

thank you Dr, Structure! big help :)

[@shahedsiddiki1165]

i love dr structure .i and i believe on them who are explaining such a way easy way .

[@shaqboss14]

8:39 how did you get -1.5 vd an 8:52 md 17 ? @dr structure

[@DrStructure]

+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.

[@deenkz917]

on 12:20 you consider sum of moments about A point, just want to ask why do you include 25 000 Nm, if it has no distance to A, meaning why should we consider this moment if it s at point A?

[@DrStructure]

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.

[@СрбинПоносан]

at 7:40 how can Mc = 10 - m? What does m mean? Also when you make the equation for Ma why are you including the moment about C (a different point altogether) in it?

[@DrStructure]

+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.

[@EStreaks]

What sign convention fo you guys use for moments? Plz reply

[@DrStructure]

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.

[@MDNTRNC]

Easy to follow when you understand where numbers are coming from. My Strength of Materials professor wanted the class to use this video to help with an online quiz. But, I can't follow what is going on in the examples at times due to the lack of units on the numbers in the equations. To think that such a simple thing can completely mess up the learning experience.

[@honeydoringo5845]

at 6.13 where dis you get the equation for summation of M?

[@DrStructure]

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.

[@chineduochuba1813]

best explanation is in this video. thanks alot

[@vedprakashsingh7152]

how the direction of moment at C in free body diagram is anticlockwise

[@DrStructure]

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.

[@nikolaivacko5397]

Nice explanation :)

[@supertv2783]

Thank you Dr. Structure it is so great videos...

[@okjjjjjj]

hi dr. I got M= 5x -500x2 - 25 you got +25 because I did not calculated the moment at A . I did calculated it at the right side where the cut is. could you please explain

[@DrStructure]

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

[@nicholasoconnell5882]

HI 6.08 how did you get the .5 in Fy & Ma

[@DrStructure]

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

[@nicholasoconnell5882]

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

[@DrStructure]

+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

[@nicholasoconnell5882]

thank you so much

[@MDLegba]

Thanks very much Dr structure

[@sumansapkota9762]

At 11:20 u have written m = 25000 But after using equation, we got m=-25000 although we used same convention and figure Why?

[@DrStructure]

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.

[@mamii1759]

at 12.51 How do you go from Vx-M+500x(x/2)-25000=0 to M=5000x-250x^2-25000 where did the 5000x from?

[@DrStructure]

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.

[@basanthbabu]

PPT preparation is excellent. Feel like seeing in White Board in class room.

[@sinclairkamanya7481]

Just curious. What drawing tool (looks like a pen) were you using for these videos?

[@DrStructure]

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

[@sinclairkamanya7481]

Thanks greatly appreciated.

[@vedprakashsingh7152]

mam plzz tell me how to calculate bm at a distance 'x' from mid span of simply supported beams.

[@DrStructure]

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.

[@rohankatta7244]

all videos are excelent

[@Iraq4ever91]

God bless u dear it's really good and clear way to tech

[@oceanblue4818]

@15:11 Vx - M + ( 500x )( x/2 ) - 25,000 = 0 then M = 5000x - 250x^2 - 25,000 = 0 Why 250 is negative?

[@DrStructure]

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

[@liamgabrielullrich5742]

13:18 Shouldn't it be: M = 5000x + 250x^2 - 25000 Why is it + 250 rather than - 250?

[@DrStructure]

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.

[@liamgabrielullrich5742]

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.

[@DrStructure]

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

[@liamgabrielullrich5742]

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

[@RealGame101]

thank you so much for these videos. I'm struggling so badly to get the concept of this stuff

[@chandakalinso3705]

where did 5 come from?

[@DrStructure]

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

[@vinaymedapati8690]

This so amazing!

[@qasimsafdar3143]

Thank you so much

[@kurasha7683]

thanks for sharing

[@nicholasoconnell5882]

Excellent video btw

[@zahraarq6645]

Thank you 🫶🫶

[@chandakalinso3705]

The 500(10)(5) on the cantilever beam

[@DrStructure]

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).

[@chandakalinso3705]

Alright thank you

[@Jswa2024]

Thank you

[@salatshahid3752]

wow thanks a lot

[@accessuploads7834]

Useful

[@mariladdu]

if it is solved from right it will be more easier