Macaulay’s Method – Slope and Deflection of Beams – Strength of Materials

Hello friends here in this video we will see what is Macaulay’s method and how it is to be used so let us get started Macaulay’s method it is also called as double integration method this is the other name of Macaulay’s method it is also called as the double integration method next in this method a section is taken extreme far away from the left hand support so in my college method we have to take a section at extreme far away from the left hand support of the beam then in this is systemic is systematic calculation of bending moment is done at the section next in my college method we use the double integration concept to get slope and deflection respectively so in my college method we use the double integration concept to get slope and deflection respectively next your let me explain that with the help of a diagram here I am considering a simply supported beam having length L then on this simply supported beam there are three loads I am just taking this example that there are three loads w1 w2 and w3 all three are point loads as we can see here next the distances of the load are w1 is located at distance a1 from the left support w2 is located at distance a 2 from the left sense of wood and w3 is located at distance a 3 from the left hand support now suppose for such a beam which I have drawn here we need to find out the slope and deflection at various points by using makalah’s method then the steps would be so I am going to write the steps in makalah’s method and the steps are first is calculation of support reactions that is first we are going to find the support reactions at a it is RA at B it is RB so first we will get the value of support reactions that is RA and RB next after getting the support reaction we will take a section X X at a distance small X which is small X from left hand support so here left hand support is point a so from point a we will consider the section X extreme far away from the left portion so suppose I am taking the section here that is extreme far away from a and just before B so this is the section xx located at a distance small X now when we are using the makalah’s method the first thing would be taking moments about section xx so here I will go I am going to take the moment at this section xx and for that whatever are the clockwise moment I will treat them as positive and anti-clockwise moment I take them as negative so taking the moments moment at xx will be equal to clockwise moment RA into X that is positive this is the first section next w1 will produce anti-clockwise moment now the distance between xx section and w1 is X minus a because X minus a 1 will give us the distance between W 1 and section xx so that is minus because it produces anti-clockwise moment W 1 into X minus a 1 the second section next W 2 is also producing anti-clockwise moment and the distance between xx section and W 2 is X minus a 2 so it is minus W 2 into X minus a 2 then after this W 3 is producing anti-clockwise moment so again negative minus W 3 into the distance between W 3 and XX section is X minus a 3 X minus a 3 so now once we get this bending moment equation this bending moment moment at xx can be made equal to e aí d square y by DX square because we can write the bending moment at any section by using this formula we can replace moment at xx by E I D square Y by DX square and this term I will copy it as it is now once I have reached here this year I have completed the third step now in the fourth step integrating the bending moment equation once we get so now integrating the bending moment equation we have here the bending moment equation I will call this as equation 1 now integrating this on the left hand side I have a I it is a constant I will keep it as it is a I is a constant next integration of d square Y by DX square that is dy by DX is equal to now RA will remain as it is because it is a constant value integration of X that will be X square by 2 plus c1 that is the constant of integration this will be our first term next minus w-1 will remain as it is constant next thing is this bracket will be integrated as a whole so X minus a 1 that will become X minus a 1 whole square divided by 2 the second term then minus w2 will remain as it is and X minus a 2 will become X minus a 2 whole square divided by 2 at last – w3 X minus a 3 square whole square divided by 2 so here I have integrated once I will keep this as equation number second next the fifth step is integrating equation two again we get now I am in case of my college method once we get dy by DX this dy by DX represents slope that is the angle by which the beam has been bent so once we get dy by DX equation which is from this we will be getting the slope equation now we have to integrate the slope equation to get deflection so I am integrating this equation number second so here we have e aí dy by DX integration is y ra will remain as it is this two will be written here X square integration becomes X cube by 3 plus C 1 into X plus another constant C 2 because I am integrating it twice next minus W 1 by 2 X minus a 1 whole square so integration becomes whole cube divided by 3 the next term minus W by 2 X minus a 2 whole square becomes whole cube divided by 3 minus W 3 by 2 X minus a 3 whole square integration becomes whole cube divided by 3 so therefore e I into y is equal to R a 2 into 3 that becomes 6 X cube plus C 1 X plus C 2 minus W 1 upon 6 X minus a 1 cube similarly W 2 by 6 X minus a 2 whole cube – w3 by 6x minus a 3 whole cube so now here also we have an equation I will call it as equation number third next if we look at equations again an equation third in both the equations we are having constraints in equation two we are having c1 in equation 3 we have c1 c2 both so we need to find this constant so the next step would be step number five in step number six we are going to find the constants by applying boundary conditions so by applying the boundary conditions we will have to find the constants and if I look into the diagram the boundary condition is when the distance X is zero here deflection Y is zero so the first boundary condition would be at X is equal to zero Y is zero and the second boundary condition is at X is equal to capital L at X is equal to L y is again zero so deflection is zero so after reaching here the next step would be we would be putting that is in equation seven applying boundary conditions in deflection equation one at a time to get constants c1 and c2 so once we know the boundary conditions we can apply them or put the values in the deflection equation which is equation number three to get the values of c1 c2 and once we know c1 c2 that is step number eight in makalah’s method once c1 and c2 are known then put their values in equations two and three respectively that is once we know c1 c2 values first we are going to put the c1 value in equation number second and that will give us the slope equation then c1 c2 values we will put in equation three that gives us Y that is deflection equation so one c1 and c2 unknown we can put their values and the in equation two entries respectively hence we will get the slope and deflection equation respectively and from them we can calculate slope and deflection at any point in a given beam so once we get after putting the constants c1 and c2 in equation 2 and 3 respectively from Equation 2 we will get slope equation that is dy by DX and once the constants are known we know the value of RA just we have to shift the distances and once we shift the values of X that is different values of X we will get the different values of slope similarly we can get different values of deflection from Equation 3 so here and this video we have seen how to use my college method what is meant by my college method and by using the double integration method we can get slope and deflection for a given beam

100 Replies to “Macaulay’s Method – Slope and Deflection of Beams – Strength of Materials”

  1. sir, i don't know why and where we use this method can't we solve this example by simple method? and is there some cases where Macaulay's method fail and we can't use it?

  2. Your videos are very helpful to gate preparation. Thanks. I want problems on magnitude of slope in cantilever beam

  3. To Get Full Course Of Strength Of Materials, Click

  4. nice video, but would be better if you write down the notes before hand, and not during the video itself~ keep up the good work

  5. thanku sir
    but your taken three loads and get only two constants but if we taken more loads and getting more constants it is very cum

  6. Awesome learning video.
    Thank you sir.
    My problem is solved after saw your video.
    Good YouTube channel

  7. Can we find support reaction by taking different signs conventions than the ones which we use for actual macualays method??

  8. Sir Please 9 may se exam hai so aap moment area method and conjugate beam method aur daldo please humble request hai


  9. Good video maybe add numbers and you have the moments negative and positive the wrong way good video made me understand

  10. So this works for multiple point loads on a beam but how would you adapt the same method to work for a spread load across a beam?

  11. sir apke som videpo lecture me moment area method ka question nhi mil rha h..sir please upload this lecture fastly sir lecture achha study ke liye useful sabit ho rha h..

  12. Thank You Sir,very nice explaination. Just had one doubt, when You were taking the moments, why didn't You take the moment due to the reaction from the right support?

  13. What is the most suitable method to find out deflection of beam. Please comment everyone and give your ideas. Thank you

  14. Your channel is to much helpfull for me… it is too good and esay way to learn but…. didnt got the solution of sfd and bmd for frame …..i think u haven't made or deleted …plz could u make that with in 2 3 days ….because my esam is comming near …plz

  15. Sir u the best πŸ™πŸ™πŸ™
    Before watching your video I don't understand the whole process about that…..
    I had watched your video previous day of my exam…….
    Then I have done hard all problems regarding to this one and also the shear force -bending moment problems after watching your video…….
    Your videos helps me a lot to capture the main basic knowledge…..
    Thanks sir for support us.😁😁

  16. It's irritating to see ur potrait everytime u talk just remove it u might get more views πŸ˜‘πŸ˜‘πŸ˜‘πŸ˜‘

  17. words will be very less for thanking you, i'm thanking with you all of my heart thank you very much sir , you re awesome

  18. Some one tell him the difference between macaulis method and dobule integration method:(don't teach if you don't know. By your knowledge I messed up my paper

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