SFD and BMD – Problem 8 – Shear Force and Bending Moment Diagram – Strength of Materials


Let us take some problems on Shear force and Bending moment we are starting with question number one let’s take this example for the Beam loaded as shown in Figure here the diagram is given draw SFD and BMD so first of all whatever condition they have given based on that we would be writing the data now in the data first I will be drawing the beam which they have provided here there is 75 kilo Newton of point load at a distance of 3 meter from right hand support next at 6 meters distance from B we have UDL up to the point load of 75 kilo Newton intensity is 22 kilo Newton per meter now this is the diagram which we have and the question is draw SF DN BMD we have to draw the shear force diagram and the bending moment diagram so this much is a question part for us now we will start the solution for this here the first part is we have to calculate the support reactions as we see this is simply supported beam so there would be reactions at end ay a 10 B at a I will call the reaction as RA at B the reaction is RB so first we need to calculate values of RA and RB I will say that step number one is calculation of support reactions in support reaction calculation here I have to say that at first the beam is in equilibrium so I would be using the first condition that is summation of FY is equal to 0 where upward forces are positive and downward forces are negative so therefore here I can say ra n RB they are upward so RA plus RB 75 kilo Newton is downward so minus 75 22 kilo Newton per meter is the intensity of UDL acting at 3 meter distance so if I convert it into point load it would be 22 into 3 that gives me 66 and it would be in the downward direction so minus 66 is equal to 0 therefore RA plus RB will be equal to if I add these values and I shift it on to the other side I would be getting my answer as that is 75-66 if I add them I would be getting this as 141 keeping this as my first equation now after this I would be taking moments of all forces about point a so I will say that summation of moments about point a is equal to 0 where I will take clockwise moment positive and anti-clockwise moment negative therefore here I have RB which is acting upward if I take the moment at Point a it would be anti-clockwise so minus RB into 8 then 75 kilo Newton is acting downward so it would be producing clockwise moment plus 75 into 5 next this 22 into 3 it was 66 Salah Newton acting at half that is at 1.5 so from point A the distance would be 2 plus 1 point 5 that is 3 point 5 so here I have plus 66 into 3 point 5 is equal to 0 therefore if I calculate the values here that is this bracket 75 into 5 plus 66 into 3.5 shifted onto the other side then divided by 8 I would be getting my answer of RB as seventy five point seven five kilo Newton after getting our B I will say that put RB is equal to 75 point seven five kilo Newton in equation number one therefore here I have RA will be equal to 141 minus seventy five point seven five so my RA value comes out to be 65 point two five kilo Newton so these are my value of reactions that is the support reactions now this completes my step number one in step number two I would be calculating the value of shear forces at all the points so after this I would be writing step number two as sf calculations that is shear force calculations so here first I will be starting with shear force at Point a so it is SF at Point a that is equal to here my assumption is that the section is just at Point a or you can say just ahead of it very close to a this is my section and as we can see to the left of section there is our a reaction at a acting in the upward direction the value is sixty five point two five kilo Newton our B was seventy five point seven five kilo Newton so as per the sign Convention as I have told y’all that whenever you are calculating shear force first mark the sign convention now the sign convention is if we are taking a section to the left of section if there is upward force and to the right of section if there is downward force both have to be considered positive next if we have downward force to the left of section and upward force to the right of section they have to be taken negative so SF at Point a if I look here in the diagram that reaction at ay is in upward direction to the left of the section so it has to be taken positive section and to the left of section forces of put so here I have plus sixty five point two five kilo Newton next I will be calculating shear force at Point C because between A to D the shear force will remain constant as there is no load between A to D so after this I will say that SF at Point C is equal to if we look at Point C just before Point C we have UDL and at Point C we have point load so it is better to calculate two values at single point c1 which would be to the left of C and other section would be to the right of C so at a single Point C we would be having two values of shear force so first I will calculate SF at Point C to the left of section I will say that this SF at Point C will be equal to if I am taking the left section here I have RA to the left of section upward so it is positive 22 into 3 that is 66 in the downward direction so if to the section left to the section we have downward force that should be negative so here I have minus 66 this gives me a value as minus 0.75 kilo-newton next again I will be calculating shear force at Point C and that will be equal to sixty five point two five minus sixty six and we have to take even this seventy five kilo Newton because here I am taking the section to the right of C so it is minus seventy five therefore my answer comes out to be minus seventy five point seven five kilo Newton at Point C after this I will calculate SF at point B if I take the section just to the left of B and if I see here then to the right of this section I have our B which is upward so it is seventy five point seven five kilo Newton and if we have upward force to the right of section that is negative so here I have minus seventy five point seven five kilo Newton as the shear force at point B now as we have done in safe calculations here we are in a position to draw the shear force diagram so after this I will say that we are drawing the shear force diagram and for that shear force diagram first we should draw the given beam which they have provided this was the beam which they have given in the question now the first thing is for drawing SFD we need to project all the points downward that is starting from point A then point D at last point be now for drawing shear force diagram the length of the shear force diagram should be equal to the length of the beam at Point a I am getting the value as sixty five point two five kilo Newton then between A to D we don’t have any load so shear force diagram should be a straight line next at Point C we are having two values of shear force one is 0.75 and the other value is minus 7t five point seven five kilo Newton so these are the values which we have and even we can say between C to B since there is no load shear force will be constrained as there is UDL between D to C then shear force diagram should be an inclined line next at Point C the shear force increases from zero point seven five to seventy five point seven five and then it remains constant and finally it reaches the value up to zero so here I have point a D C and B I have written the respective shear forces upward forces that is a forces which I have drawn above the line they should be positive and whichever are downward they should be negative so this is the shear force diagram which we have now after this we would be calculating bending moment at all points so here my step number three is BM calculations that is bending moment calculations as we can see in here that the beam is simply supported so at the supports bending moment will be zero I will write down moment at a is equal to moment at B and that is zero into the bracket I will mention since it is simply supported next I need to calculate moment at C for calculating moment at C first I will give you all the sign convention clockwise moment to the left and anti-clockwise moment to the right they are taken positive anti-clockwise moment to the left and clockwise moment to the right they are taken negative so here if we see when we are calculating moment at C then to the right of C we just have our B and this would be producing an anti-clockwise moment so if I have a section and then to the right of section there is anti-clockwise moment that has to be taken as positive so here I have moment at C is equal to R B which is 75 point 7 5 into 3 the distance is 3 meters from Point C from B to C that is 3 meters now the value of moment at sea it comes out to be two twenty seven point two five kilonewton meter next moment at D is equal to here if I am calculating moment at D to the left of D I have just RA and this would be producing clockwise moments so to the left of the section there is clockwise moment so it is positive so I have plus sixty five point two five multiplied by two and that value comes out to be one three zero one thirty point five kilo Newton meter so these are the values which I am getting of moment at C and moment at D now with these values we can draw the bending moment diagram so here we are drawing the bending moment diagram as we can see in this shear force diagram first of all that there is one location where shear force is changing at sign at that particular location shear force becomes zero so I will be projecting this point down now we have calculated the values of bending moment at all critical points but where shear force becomes 0 at that point bending moment will be maximum or in other words you can say where shear force is changing its sign at that location bending moments become maximum so here I would be having one point where I would be getting the bending moment M as maximum next we have calculated moment at C that is 227 Oh point two five kilo Newton meter moment at D is 1:30 0.5 kilo Newton meter and moment at a and B they are zero so here we would be drawing bending moment diagram after marking the values as we see between a 2d shear forces constant so whenever shear forces constant bending moment should be an inclined line next between D to C shear forces an inclined line so bending moment it should be a curve that is a parabola it is a parabola between C to B it is a straight line so bending moment should be an inclined line here I can say that we are having parabola from D to C and since all the values of bending moment they are positive I will mark plus sign here so this is the bending moment diagram which we have now after getting this bending moment diagram this problem is not completed yet here we need to calculate the value of maximum bending moment because we have calculated bending moment at a b c and d out of which we can see at Point C the bending moment is 227 point two five kilo Newton meter but we are having a volume which is more than 227 and that is the maximum bending moment that value we have to calculate so how to get that value I would be writing here calculation of maximum bending moment to calculate maximum bending moment first we should know the distance of this point where maximum bending moment is acting this distance we will find it out from point D I will call it as X if we are in a position to get this value of x we can easily calculate bending moment at this point so I will say that from the similarity of triangles here I would be magnifying these triangles that is here I have a bigger triangle whose base is X and here I have a smaller triangle so these two triangles I will be drawing here this distance is X between D to C it is 3 meters so from C to the location of this point that becomes 3 minus X so here I have distance as 3 minus X so from these two triangles I would be writing that the height of the bigger triangle that is 65 point 2 5 the height of the bigger triangle divided by the base of the bigger triangle which is X will be equal to the height of the smaller triangle that is 0.75 divided by the base of the smaller triangle that is 3 minus X so therefore 65 point 2 5 into 3 minus X is equal to 0.75 X therefore 65 point 2 5 into 3 minus 65 point 2 5 X is equal to 0.75 X so therefore if I calculate the values here that is if I shift 65 point 2 5 X on to the other side it would be added to 0.75 X and that becomes 66 X then if I calculate 65 point 2 5 into 3 the value is one ninety six point seven five and hence X will be equal to 196 point seven five divided by 66 the answer is it is two point nine 6 meter from point D so I have calculated the location from point B where shear force becomes zero and this answer of X is two point nine six meter now I will be calculating the bending moment at this point so I will say that taking moments at a point which is 2.96 meter from point D so here I’ll take the moments at this location where SF is changing its sign so if I calculate the moment here it would be first my section is here I would be transferring this point to the diagram of the beam here is my section and if I see to the left of section RA will be producing clockwise moment so that has to be taken positive so plus sixty five point two five and the distance is two plus two point nine six so it is maximum bending moment into four point nine six because this distance is two plus two point nine six that is four point nine six next this 22 kilo Newton per meter will be acting at two point nine six distance so it is minus twenty two in 22.96 acting at half of two point nine six so here if I calculate all the terms I will be getting my answer of maximum bending moment as two twenty seven point two six it is fractionally greater than moment at Point C that is at Point C I was getting the bending moment as two twenty seven point two five kilo Newton meter and my maximum bending moment is 227 point two six kilo Newton meter so just a fractional increase and we have given in this problem it was asked to draw the s FD and BMD and here we have solved the question and completed the shear force diagram and bending moment diagram with this the problem is completed

100 Replies to “SFD and BMD – Problem 8 – Shear Force and Bending Moment Diagram – Strength of Materials”

  1. I appreciate for helping the helpless ones before exam. Sir i subscribed u .
    Please make such videos they help a lot

  2. That's ok..
    but I'm studding Civil Engineering.
    Already I have completed my Engineering Machanics & Solid Machanics courses.. During examination time we don't get more time. If I calculates like this way, I mean step by step, examination will be incomplete..

  3. Sir why u have taken shear force at c on both the sides while for other points only for one point. Pls reply fast tomorrow is my paper.

  4. Just wondering, does it matter if you set clockwise to positive verses negative? Because I was taught that clockwise was negative.

  5. sir while finding the moment at c you took the right side for finding the moment not the left side why?

  6. Sir how to watch deleted videos of yours that is ., Some of them are concept on thick cylinders…?

  7. For taking shear force at B the section should be taken at right side of B…Right? well actually doesn't matter , cause answer is still the same . But am i right?

  8. Rattebaaz…. You should tell why Inclined SF line will come parabolic line in BM diagram … And so … Dont just copy method from book and make videos

  9. Sir, there is UDL on DC 22kN/m = 66kN acting on the middle of DC, so that from point A to the point where these 66kN, the distance is 3.5m. Why do we have to take in consideration that 66kN are on point C, when calculating the SF?

  10. my professor was talking about this subject for a week and I had no idea what the hell he was doing. Thanks to you I know what I'm doing know!

  11. Sir I can not understand 0.75kN in SFD below the lower line how it possible plz tell me I mean 0.75kN is the positive value it can upper m the main line

  12. Dear sir,
    I got the BM at D as 261KNM.and at C as 454.5KNM . These both reading are exactly double the asnwer.how it is possible.i have tried as below

    BM AT D:
    RA*2-22*3*1.5-75*3+75.75*6=261KNM

    BM AT C:
    65.25*5-22*3*1.5+75.75*3= 454.5KNM

  13. Sir 0.75 KN in SFD should be pointed above the line .. the line denotes 0 KN and 0.75 is a higher value .. please clear the doubt . Thank you

  14. Hello sir m a diploma mechanical second year student…can i pass my som subject by watching this video's or you have special diploma subject vidoes….sir i actually want diploma som subject vidoes…send me the link sir.

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  16. Sir,We have to get -65*3.5 in calculating moments because it is in anti-clock wise but sir you have given +66*3.5 how it would me positive

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