SA09: Drawing Shear & Moment Diagrams without the use of Equations

[@doctorp286]

With the movement to distance learning during the virus crisis, you are a godsend. Thank you.

[@jeanapp_]

After the first example I was able to draw the diagrams of the second example before looking at the final answer. Your explanation is fantastic as it doesn't have many confusing numbers! For me it is much easier to understand if we try to solve the questions via concepts than via numbers. Thank you so much for the great video!

[@hightopnikes104]

I like this method, it's a very logical approach. A lot of the other ones I have come across skip a lot of what you talked about. Thanks!

[@devvv4616]

fr fr, that's when students get lost, when instructors assume that students can follow what they're saying and shortcuts everything. Everything should be told when teaching a new concept!! even the obvious ones, because they're most likely not obvious for beginners!

[@abhaytalathi2929]

GREAT !!! GREAT !! GREAT! quality videos i never heard before. thanks a lot

[@saeedullahyasinzai7715]

Superb way of expression , please continue in future too.

[@2011iryna]

Thank you so much! I spent ages to find a good video as was stuck at method for drawing moments diagram. Delighted -as i now understand it!!! Thank you.

[@DrStructure]

You are quite welcome.

[@rikdevghosh7324]

@@DrStructure sir can you please make a similar kind of video for uniformly varying load

[@DrStructure]

A generalized load function most often forces us to actually formulate shear and moment equations before we can graph them. The approach described in this lecture works best when we have concentrated or uniformly distributed loads, the kind of loads that result in linear shear diagrams. If the shear diagram is nonlinear, it would be difficult to determine the point that it crosses the x-axis (the point of maximum moment) without having the shear equation.

[@rikdevghosh7324]

@@DrStructure ok sir thank you.. Sir your teaching methods are tooo good and lucid

[@ravindrassk]

excellent videos

[@jamesakaso3614]

Great video, thank you

[@official_ebro]

well understood

[@supertv2783]

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

[@jithinprimz2972]

quality lecture

[@momenmohamed8337]

مش فاهم11:49 i dont understand this part,sir.

[@Kebab_with_extra_garlic_mayo]

a bit late but maybe if you're still confused, looking left to right, at A we have an upwards shear of 6, this stays constant till B wherein shear drops by 10kN culminating at -4kN, at C shear drops again by 4kN making a culminative shear of -8kN returning to 0 at segment D due to the +8kN force.

[@ssssss5389]

Hello, st 12:11 what is shear under B. From a previous reply I learned that shear at a point is undefined rather just left or right of a point. So considering that if I want design the beam for shear at B where point load is 10 kN , what should be my design shear 10 kN or 6 kN(max of 6 and 4 kN)? Thanks in advance doc.

[@DrStructure]

For design purposes, maximum shear is 6 kN, it is not 10 kN. If provisions need to be made for negative shear, then a value of 4 kN would suffice.

[@00iy0]

Thank you so much! :)

[@venkatesh2285]

May I know the name of the font you are using in the video? It is really looking a real handwriting effect.

[@DrStructure]

The text was traced using a stylus on an ipad, so it is hand-written.

[@AhmedKhaled-lz5dc]

thanks dr ,we want more videos about frames and moment calculation

[@ruchiawasthi3664]

If there was no support at the end and rather an overhang at the end then how would we initiate the graph?

[@DrStructure]

At the free end of the overhang both shear and moment are zero. Say we have a cantilever beam subjected to a uniformly distributed load. The beam is fixed at the left end and is free at the right end. Assuming the length of the beam is L and the intensity of the distributed load is w, then at the fixed end of the beam (at the left end) vertical reaction is wL and bending moment is (wL^2)/2 (counterclockwise). The starting value for shear (at the left end) is wL, and the ending value, at the free end, is zero. This gives us a right triangle as our shear diagram. The starting value for moment is negative (wL^2)/2 and the ending value is zero. Since the degree of moment curve is one higher than that of shear, we end up with a nonlinear (quadratic) curve for moment diagram. The curve is going to be concave down.

[@mangeshratolikar8772]

Is there any other graphical or analytical method to calculate bending moment apart from the area under SFD?

[@DrStructure]

Yes, we can write an algebraic equation for bending moment without resorting to shear diagram. See Lecture SA08.

[@adarshkumarh7576]

Sir for cantelever beams how to calculate?

[@andreasandre4756]

Perfect

[@venkatesh2285]

Sir do you have videos on in relation between loading , shear force and bending moment.?? I am unable to find it...

[@DrStructure]

No, currently we don't have a lecture that specifically discusses the relationship between load, shear and moment.

[@juliaradius4846]

Dear Dr. Structure, At 3.04 you've drawn a sheer force diagram. Why is the value at B postive? Shouldn't wL/2 be -wL/2. thanks in advance

[@DrStructure]

Please keep in mind that the sign convention we use for reaction forces is different than the sign convention used for shear and moment. Shear at B is not shown as positive, it is correctly drawn as a negative shear on the shear diagram. The reaction at B however, since upward, is considered positive and drawn as an upward force on the free-body diagram.

[@viveksimon6714]

is this technique applicable only for simply supported beams or can also apply for cantilever & overhanging beams Thanks in advance

[@DrStructure]

It works for any beam.

[@faisalelnady1354]

what would be the shear if there is only one reaction acts at the middle of the beam assuming a uniformly distributed load along the beam ?

[@DrStructure]

A beam with only one support reaction is unstable, therefore not subject to analysis.

[@faisalelnady1354]

Dr. Structure thank you for caring, but i got one more question, does the shear becomes zero at the point where it turns from positive to negative?

[@DrStructure]

No. Shear changes sign at interior supports. At one side of the support shear is positive, then it jumps to a negative value at the other side of the support. Shear is considered undefined at the support, even though it appears as zero on the shear diagram.

[@raviprajapati6981]

Please do this for frames as well

[@kabirsharma4820]

Shear force is defined as a force which acts parallel to surface. Then why is it always represented perpendicular to beam in SFD?

[@DrStructure]

If we examine an infinitesimal element in a solid body, in addition to a normal force, we see shear force components acting along two planes: the parallel (horizontal) plane and the perpendicular (vertical) plane. We cannot have one without the other, else static equilibrium of the element cannot be maintained. In beams we focus on the shear force along the vertical plane since it is directly related to the bending moment in the beam, and the vertical plane gives us the geometric properties needed for calculating shear strength.

[@kabirsharma4820]

Thank you very much for replying.

[@asmaulhusna732]

@@DrStructure the answer was enlightening. Please give us more guidance about it. This question is an enquiry of mine too- how we can conceptualize shear force .. As a parallel of plane force or something that comes to resist falling of the beam portion when a section is cut? Which concept is suitable where, that is, in different cases? Please make us guided in details. Thank you.

[@DrStructure]

@@asmaulhusna732 Take a look at ST08 lecture (kzbin.info/www/bejne/mn7RoI13qMyXadU) In the middle of the lecture, there is a conceptual explanation for shear. May be that could help for better wrapping your mind around the concept.

[@kabirsharma4820]

Could you please explain the BMD of pure moment.

[@arnoldchristian9501]

Why is shear at a negative? The equation is only balanced if ShearA is negative as that would make ShearB-(-ShearA) = Total Load AB. Can you please explain? Shear A and Shear B should both be positive as the reaction forces are both acting upwards.

[@DrStructure]

By definition, a positive shear consists of two forces, applied at the ends of the beam segment, that tend to rotate the segment in the clockwise direction. Similarly, a positive moment refers to two end moments that want to bend the segment concave up. Therefore, it is important to keep in mind that for positive shear, we have two forces that collectively are trying to cause a clockwise rotation in the segment. That is, when we talk of a positive shear, we are talking about an upward shear force at the left end of the segment AND a downward force at the right end of the segment. So, any shear force that is drawn downward at the right end of a beam segment is considered positive. The above sign convention is not the same as the one we use for writing the equilibrium equations, for balancing the forces. In summing the forces in the y direction, up is considered positive and down is considered negative.

[@arnoldchristian9501]

Dr. Structure woah thank you so much for such a detailed response. Your videos have not only helped me but my entire class and we have a Facebook group where I found your videos. So once again thanks a lot for your effort, I subscribed a while ago :)

[@DrStructure]

I am glad the videos have been helpful. Thanks for the feedback.

[@andrey6161-l8l]

Hi, sorry for my English and thank you for your lectures. I have logically misunderstanding about shear diagram, then mathematically I can solve them. Why shear diagram is not symmetrical and don't have a "V" shape, then all of the load is symmetrical? Thanks.

[@DrStructure]

Be being the shape of a V, you mean, shear at the midpoint is zero and everywhere else is positive? That would not work. A positive shear means that if we cut the beam (taking the left segment) and examine the shear force at the cut point, we would have a downward shear force. To maintain equilibrium, if the beam is cut to the left of the mid-point, shear needs to be acting downward (positive), but if we cut the beam to the right of the midpoint, shear has to be upward to maintain equilibrium (to ensure that the sum of the forces in the y direction equals zero). That upward shear, by convention, is a negative shear. This means, the shear diagram to the left of mid-point has to be above the x-axis indicating a positive shear, and to the right of the mid-point, below the x-axis indicating a negative shear.

[@andrey6161-l8l]

@@DrStructure Thank you ;)

[@kabirsharma4820]

We get maximum bending moment when shear force is zero. what does zero bending moment imply?

[@DrStructure]

The relationship between shear and moment is similar to the relationship between moment and the slope of the elastic curve (deflected shape of the beam). Moment is maximum/minimum when shear is zero. Slope of the elastic curve is maximum/minimum when moment is zero.

[@damithsandaruwan5558]

​@@DrStructurewhen slope is minimum; moment should be maximum

[@AmiraMohamed-dx4kc]

is the area under curve represent the direvative of the function +dr_structure

[@DrStructure]

The area under a curve represents the integral of the function, not its derivative.

[@AmiraMohamed-dx4kc]

thank u

[@accessuploads7834]

Thnaks dr. structure

[@DrStructure]

You are welcome!

[@kabirsharma4820]

Could you please answer this as well. If area under SFD gives BM, then what does area under BMD give? AND if slope of BM gives SF and slope of SF gives load. Does slope of load give some thing? Thank you.

[@DrStructure]

The integral of bending moment (area under the moment diagram) is related to the slope of the elastic curve. More precisely, the slope of the elastic curve equals to the integral of M/EI where M is bending moment, E is modulus of elasticity and I is moment of inertia. No, the slope of the load has no significance with regard to internal forces in the beam.

[@kabirsharma4820]

Kindly pardon my ignorance, but what is elastic curve? and does area under bending moment diagram give some important property or something? Does area under BMD geometrically mean something? and similarly what is geometrical meaning of slope of load ? What does slope of load imply?

[@DrStructure]

Elastic curve is the deflected shape of the beam. The curve represents how the beam deforms under the load. The area under the moment diagram represents the slope of elastic curve. Say, you dump a pile of sand on a bridge ( beam). That is your distributed load on the beam. How does the pile look like geometrically? Did you try to flatten the pile? Or, did you leave it as a cone? Depending on how you distributed the sand in the pile, it has a geometric shape (which has a slope.) So, if you flatten the pile, the load has a zero slope. If you leave it in the form of a cone, the slope is constant,...

[@kabirsharma4820]

I appreciate you for replying. I express my gratitude for that. Truly a "Dr" indeed.

[@vitrovogan9635]

How can we draw bending moment diagram with the help of known direction of bending moment.

[@DrStructure]

The direction (positive/negative) of bending moment across the beam tells us whether the diagram is above or below the x-axis, but it does not help us draw the diagram itself. To draw the moment diagram, generally we need to have the moment equation. Although under certain loading patterns (for example, when the beam is subjected to concentrated or uniformly distributed load only) we can draw the diagram without needing to explicitly write the moment equation. But, the sign/direction of moment does not provide enough information for drawing the diagram.

[@vitrovogan9635]

@@DrStructure Do we take counter clockwise direction as positive and if we take it as positive then should bending moment be drawn above x- axis or not?

[@vitrovogan9635]

@@DrStructure thank you for for reply 😎

[@DrStructure]

With regard to the sign of bending moment in beams, we always consider two (not one) bending moments , those at the ends of a typical (infinitesimal) beam segment. That is, we always need a pair of moments (one at the right end and one at the left end of a typical beam segment) in order to bend the segment. The pair of moments either wants to bend the segment in the concave up direction (we refer to this as positive moment), or in the concave down direction (negative moment). If the beam is being bent concave up (if the moment is positive), then it means the moment at the left end of the segment is acting in the clockwise direction and the moment at the right end of the segment is in the counterclockwise direction. Conversely, if the segment is bent concave down, the moment at its left end is counterclockwise and the one at the right end is clockwise in direction.

[@brixmonton3577]

I love your videos Dr. Structure!!!!!! reply pls ;_;

[@kabirsharma4820]

We know, a material will fail if applied load exceeds the ultimate strength of that materials. But, even if we load a member below its ultimate strength for a longer duration it will fail, but how to compute that duration?? We mostly talk about deflection, slope, bmd, sfd in a member, rather we are talking w.r.t to dimension of the member, but how to calculate afore mention parameters w.r.t time, like after 10 minutes how much will be the deflection or sf or bm

[@DrStructure]

Yes, cyclic loading causes fatigue which could lead to structural failure. Fatigue is a design consideration which involves keeping stresses in the member below the fatigue strength. Depending on the type of design methodology being used, we may still be able to assume linear elastic behavior for the structural material. This means stresses and deflection can be assumed to be linear and independent of time. Otherwise, we may consider time-variant nonlinear behavior for the material, and use advanced analysis techniques for our stress calculations.

[@yassertalib3957]

Thank you

[@레인스톤]

Thank you for your lecture. I am not good at English hearing. Subtitles please~.

[@jawadahmadi1483]

In the following beam if the shear force at B is -250N and if we add the 0N at the other end of the right segment it will remain -250N (-250+0), while it should be zero ( from equation) 900N(the only force) | ------------------------------- A (2m) B (1m) |..........650N | | |--------------------...........(0N)------------------------------------------------------(x) | |-250N ......... | |

[@jawadahmadi1483]

I meant to say sometimes the addition rules doesn't work.

[@DrStructure]

I am not sure if I understand your diagram, but I'll give it a try. From the numbers shown on the diagram, one can infer a simply supported beam subjected to a downward load of 900 N at 5 m point from the left support. The length of the beam is 18m. This makes the left support reaction 650 N and the right support reaction 250N. For this beam, shear in the left segment of the beam (from the left support all the way to the left of the applied load) is constant positive 650N. In the right segment of the beam (from just to the right of the applied load to the right support) shear is constant negative 250 N. There is no zero shear point in this beam. If we examine the point of the application of the load, shear just to the left of it is positive 600N, then there is a drop of 900 N in the shear value giving us a negative shear of 250N just to the right of the applied load.

[@jawadahmadi1483]

sorry I could not show what I meant, I was arguing about the shear force of last part of the beam ( right-hand side). the beam has the following properties; -The length of the bar is 3m - support A is at the very left side of the bar with a reaction force of 650 N upward -The support B is 2 metres away from support A ( in right-hand side) with reaction force of 650N - there is a 900 N force downward in the middle of A and B -the rest of the beam (1m) is at the right-hand side of support B without any reaction or forces. *I know that the shear force at the point B ( support B) is -250 and for the rest of the beam (the last one metre at right-hand side) if we want to draw the shear force diagram using addition (-250+0) the shear force will remain -250 as at the very right-hand side of the beam (end of beam shear is zero).

[@DrStructure]

Okay, let's go over the problem. We have a beam of length 3m. From left to right, we have: at x = 0, a pin support. We call this point A. at x = 1m, a downward load of 900 N. We call this point C. at x = 2m, we have a roller support. This is point B. at x = 3m, we have a free end. This is point D. The upward support reaction at A and B each has a magnitude of 450 N. Since the load is applied at the mid-point of segment AB, the reaction forcess are equal in magnitude. This gives us a positive shear of 450 N in segment AC, a negative shear of 450 N in segment CB, and a zero shear in segment BD. Keeping in mind that shear at the support points and the point of application of concentrated load is undefined, this is how you want to think about shear and the way it changes in the beam: At A (x = 0), shear is undefined. Why? because there is a support reaction here. at C (x = 1), shear is undefined. Why? because there is a concentrated load here. at B (x = 2), shear is undefined. Why? because there is a support reaction here. * A positive shear develops just to the right of A with a magnitude of 450 N. * Shear remains constant +450 N from just to the right of A to just to the left of C. * Since there is a downward load of 900 N at C, we are going to have a drop in shear value, from +450 just to the left of C to -450 just to the right of C. * Shear remains constant -450 N from just to the right C to just to the left of B. * At B, since there is an upward reaction force of 450 N, there would be a jump in shear value by 450. This means shear changes from -450 N just to the left of B to 0 just to right of B. * Shear remains zero from just to the right of B to point D.

[@bil-alak3382]

Dr. Structure,At max point of S.F the B.M is zero AND vice versa, Do you have any logical explanation. not mathematical explanation only logic, why this is so - via YTPak(.com)

[@DrStructure]

+Bilal Khan Thank you for the interesting question. It is true that when shear in a beam is zero bending moment is maximum, but the converse is not true. That is, bending moment at maximum shear is not necessarily zero. For some problems (specially problems involving simply supported beams) it may look that way, but not because of the relationship between shear and moment, but due to the imposed boundary conditions. To answer the question why bending moment is maximum when shear is zero, it is necessary to keep the relationship between shear and moment in mind. Of course, the relationship is mathematical in nature (so there is no escaping mathematics here): By definition: Shear is Change in Bending Moment. Shear = 0 means Change in Bending Moment = 0. Take a very very tiny (infinitesimal) slice of a beam. The slice has a shear and a bending moment in it. More precisely, there is a moment (say clockwise) at the left end of the slice and a counter-clockwise moment at the other end of the slice (a pair of end moments in the slice is what we call bending moment in the slice). If shear in the slice is zero then the two moments must have exactly the same value (there is no change between the end values). Simply put, when change in bending moment is zero, shear is zero. But under what condition change in bending moment be zero? When bending moment is maximum. If we have an algebraic function (f), how do we determine its point of maximum, or minimum? We take the derivative of f and set it equal to zero. The derivative is change in f. So f has its maximum value when change in f is zero. Since shear is the derivative of moment, shear is change in moment. When moment is maximum, when change in moment is zero, shear is zero. I know you wanted a purely non-mathematical explanation, but I am afraid the question cannot be answered effectively without examining the mathematical relationship between shear and moment.

[@bil-alak3382]

+Dr. Structure Thanks. - via YTPak(.com)

[@kabirsharma4820]

very beautifully explained .

[@sangameshsannakki689]

I didn't understand moment

[@DrStructure]

Please elaborate. We would be more than happy to answer any specific questions that may help you better understand the topic.