Derivation of Flexural Formula for Pure Bending – Stresses in Beams – Strength of Materials

Derivation of Flexural Formula for Pure Bending – Stresses in Beams – Strength of Materials


Hello friends here in this video we will see a derivation of flexural formula for pure bending and for that purpose here I’ll draw the diagram that considering a beam which is simply supported at its two ends here we have the horizontal beam which is supported I’ll say these are ends A and B or I can say ends A dash and B dash now when we are deriving the flexural formula it is the bending formula and for that here I’ll assume on this beam there is a point load W acting at the center so the distance is L by 2 from a and L by 2 from B so now under the action of this load W the beam a dash B dash it is going to bend and the nature of bending will be in this way so your this arc which I am getting it is the bending curve of a beam so now we have to derive the formula which is called as the flexural formula and flexural here it means bending we are going to derive the bending formula here for the case which I have considered by taking an example of point load but even if we apply the uniformly distributed load then also the beam will be bending in this manner now what I will do I will say that consider the beam section before bending as shown below now here I will consider the section of this beam before bending so when the beam is not bending it means it is completely straight then the diagram will be in this manner now after this here I am considering a portion of a beam on this I will take two sections one is section a B and the other is section C D now here I have taken a part of that beam and in that part I am taking two vertical sections next Here I am drawing the centreline for this beam now on this center line here I will mock a layer joining a B and C D and this layer is called as EF and here the central axis which is passing through the beam it is also called as neutral axis denoted by n suffix a so I have taken the section of this beam which is before bending now after this here I can say that the length of this beam which I am considering that is between section a B and section C D the length of the beam I am considering is DX next after this here I will say that consider consider another layer gh at a distance small Y from neutral axis n a now you’re from this neutral axis I am going to consider another layer which is layer gh and its distance from the neutral axis is y now what I have done up till here is that I have considered the bending of a beam and then taken a section of the beam which is before bending and in that I have taken two vertical sections that is a B and C D between them the length of the beam is DX and I have taken the center of the beam which indicates the neutral axis and on that I have taken the layer which is called as layer EF also called as the neutral layer next from this EF at a distance small Y I have taken another layer which is called as gh now this was considering before bending now after bending what happens is I will see that considering the section of the beam after bending now after bending how it will look like first I will draw the diagram for that now when the beam is subjected to loading as shown in the diagram this beam section which was straight before loading now when the load is acting that beam will Bend in this way so what will happen is these points ABCD all would be shifted so here I’ll mark those points so here is 4a when it gets shifted I’ll make a dash B as B dash then C Dash and D dash joining them up to the center where they will intersect there we will get the center now at this intersection I am getting 0.0 so here what has happened that after bending a has been shifted to a dash B has been shifted to B dash similarly C – Andy – next after this we have layer EF so that layer EF will be e dash F dash that is the neutral layer and here we have the neutral axis similarly the layer GH it will also shift so we have G dash and H dash this is the layer now the distance from the neutral axis to the lair g – h – that was why we have taken next i’ll assume that from point O up to the neutral layer that is layer II – f – this radius I am going to assume it as capital R so now I will say that for this diagram let the beam may Bend as shown in above figure so now after this I will say that let the beam may Bend as shown in figure here I will say that capital R is equal to radius of curvature of layer G – f – that is the layer here II – f – its radius with respect to the center is capital R next after this I will say that Y which is the distance from neutral axis up to layer g – h – next i will see that the angle here I’ll take it as theta so theta is angle subtended by arc at the center that is theta now after giving the notations the next thing would be since we can see that previously this layer gh it was straight but after the load has been applied that layer is bent so the shape of this layer has changed it means when the shape has changed there is strain in that layer so I will say that therefore strain in layer gh due to bending now that strain will be denoted by E is equal to and the formula of strain is change in length upon original length so here the change in length is it is g – h – that is the final length – initial length was gh also what is the original length so here change in length is g – h – minus gh which was the original length divided by G H that is the original length now but I can say that as it is clear from the diagram before bending gh is equal to EF so here I will say that therefore E is equal to g – h – will remain as it is gh is equal to EF and here I’ll mention since gh is equal to EF now after this reaching at this stage I will call this strain equation as equation 1 next now since length of Arc E – f – or EF is given by now that length of Arc is nothing but EF or we can say e – f – both are same because at the neutral axis the length of Arc remains same before and after bending so therefore it will become EF is equal to now to calculate length of Arc since E – f – here I am getting the radius of it as capital R so capital R radius into theta that will give me the length of our Kia so EF is equal to R into theta that is the length of arc which I am getting next this length of arc which I am getting it is for EF now similarly I will calculate the length of arc for G – H – I want to know the length of this arc so I will say that similarly length of Arc g – h – is given by g – h – is equal to now for this layer the distance or the radius of this layer is capital R plus y into theta so capital R plus y into theta so hence here g – h – will be R into theta plus y into theta now after reaching at this stage therefore I will say that in equation number one g – h – minus EF that represents the change in length so I will say that therefore change in length which is denoted as Delta L is equal to g – h – minus EF so it is equal to R theta plus y t de – EF was again our data so R theta next from this plus R theta – our data it gets cancelled out so we have therefore deflection Delta L is equal to Y into theta now after reaching here the next thing would be here I will put the value of change in length and original length in equation 1 so therefore put all values in equation number 1 so in equation 1 we have strain small E is equal to now g g – h – minus EF that is change in length change in length is Delta L we have found out it is y Tita upon the original length which is EF it was R theta so from this numerator and denominator theta will get cancelled out so therefore we have strain is equal to Y upon R but I can say that from Hookes law as we know that in hoops law stress is directly proportional to strain when we remove the proportionality sign we have a constant which is called as Young’s modulus so strain will be stress upon Young’s modulus so I’ll use this relation and therefore instead of small e I’ll write down stress upon Young’s modulus is equal to Y upon our next I will shape the terms so therefore Sigma I will give it as it is why I will shift it in the denominator here II will go into the numerator divided by our keeping this as the second equation now after this I will see that therefore from Equation number ii bending stress acting on the layer is given by I will keep Sigma on one side because this Sigma is nothing but Sigma suffix V bending stress and I will shift Y onto the other side so it becomes e by r into y so now here i am getting this bending stress equation i will call it as equation number three now as we can see from Equation three capital e and capital R that is Young’s modulus and the radius of curvature which is at the neutral axis this capital R D it is a constant so I will say that from equation number three we see that Young’s modulus capital e and radius of curvature capital R are constants so therefore Sigma B is directly proportional to Y Sigma B becomes directly proportional to Y now after reaching at this stage I will say that consider the cross section of the beam as rectangular and in that consider an elemental strip having area da located at distance small Y from neutral axis as shown in figure so now Here I am going to consider the cross section of the beam as a rectangle here we have the neutral axis in that I am considering a small area or a small strip now this small strip has area da and it is located at distance Y from the neutral axis now the next thing would be I will calculate how much is the force acting on this elemental strip because of bending so therefore force acting on the elemental strip because of bending now that force is nothing but small amount of force DF because it is acting on small area da and that force will be equal to bending stress into area here I can write down since stress is equal to force upon area so therefore force is equal to stress into area so I have used the similar concept here that the force acting on the elemental strip is small D F and that is equal to bending stress into area now here we have found out the bending stress in equation number three it was capital e by r into y into y into da next now after getting this force I will take the moment of this force that is Here I am getting the force which is acting on this elemental strip which is DF next I take the moment of this force about the neutral axis so therefore taking moment of force about neutral axis so we have that moment as small DM that is equal to small amount of force into distance Y and here DF is equal to e by r into y into da into y here i put the value of DF so next DM will be e upon r y into y will become Y square da now this much is the value of the moment which is acting for that elemental strip the moment which I have taken for that small strip now if I consider the entire beam then the moment will be so therefore total moment of the beam about neutral axis now for that total moment here I will have to integrate this equation so your integration of d-m will become capital M next after this your ebuy r I will keep it as it is because it is a constant now one thing interesting here is this is area and distance square so area into distance square that is called as the second moment of area because distance is y and when I am taking the distance twice it becomes Y Square + y square into area that is called as the second moment of area and second moment of area is nothing but I which is the moment of inertia so I will write down where I is equal to second moment of area or it is called as moment of inertia about the neutral axis so here I have replaced integral Y square da by I which is the second moment of area next year also I will shift the terms so M this I will be shifted in the denominator on the right hand side we have a upon R keeping this as equation number four so here as we can see from Equation number second we were having Sigma Y upon Sigma upon Y is equal to e by r and this Sigma is nothing but Sigma suffix B that is bending stress so I will say that from equation numbers 2nd and 4th Sigma B upon y is equal to here it is e by r and similarly in equation 4 we have e by r so e by r is also equal to M upon I so from both the equations I am getting this form next I will write it as therefore M upon I is equal to Sigma B upon y is equal to e by r I have written it in this form and this equation which I am getting here I will give this as equation number 5 and this is nothing but the flexural equation so therefore from Equation number 5 we are getting the flexural formula or it is also called as bending formula or bending equation so now here in equation 5 we are getting the bending formula or it is also called as the flexural formula and I will define all the terms where capital M is called as bending moment and since it is bending moment the unit is either it can be Newton mm or it can be Newton meter it can be kilo Newton meter and so on because it is force into distance next I it is called as second moment of area it’s unit is either mm square mm raise to 4 centimeter is to 4 or it can be made is to four then we have Sigma B which is called as bending stress and the unit is Newton per mm square or it can be kilo Newton per mm square or kilonewton per meter square so on next y is the distance of layer in which bending is considered from neutral axis so this is distance so it can be in mm it can be in centimeter or meter so Y is the distance of the layer in which bending is considered next capital e it is called as Young’s modulus or modulus of elasticity for the beam material and young’s modulus unit is Newton per mm square at last we have capital R which is called as radius of curvature and that radius of curvature of beam it is since radius so in terms of mm centimeter or meter and that radius radius of curvature of the beam it will be in the terms of mm centimeter meter and it is the arc which is formed when the beam gets bent so here in this video we have seen what is the flexural formula and how it is T right

30 Replies to “Derivation of Flexural Formula for Pure Bending – Stresses in Beams – Strength of Materials”

  1. Good but method aisa bataye jo students ko simple lage ap to book ka ditto utaar dete ho

    Jab ratna hi hoga to video ki kya jarurat hai

  2. Super sir thank you very much this type of videos u are excellent explainer upload more videos on mohrs circle sir my exams on next week please upload thoroughly tq u sir all the best

  3. Thank you so much sir
    Good job
    It's easy to understand sir I really appreciate of your work ! it's very important for engineering students

  4. Hello Friends,

    Watch Complete Video Series of Subject Strength of Materials only on Ekeeda Application.

    Use Coupon Code "NEWUSER" and access any one Subject free for 3 days!

    Download Ekeeda Application and take advantage of this offer.

    Android:- https://play.google.com/store/apps/details?id=student.ekeeda.com.ekeeda_student&hl=en

    iOS:- https://itunes.apple.com/tt/app/ekeeda/id1442131224

Leave a Reply

Your email address will not be published. Required fields are marked *