Hello friends here in this video we will see a problem on the unknown load and total elongation here we have a question determine the magnitude of P which is the unknown load here for equilibrium and total elongation of the bar as shown in figure this diagram is given as we can see in this diagram this bar is having different diameters and different lengths so it is called as a stepped bar here we have to take capital E that is Young’s modulus as 210 GPA now whatever is given here I’ll write that in the form of data 20 mm diameter I’ll write it as d1 here the length is 0.5 meter so I will convert it into mm so this will be 500 mm next d2 is 25mm length becomes 700 mm because it is 0.7 meter next diameter is 30 mm length is 900 mm now apart from this as we can see this system is subjected to different loads and here they are telling to calculate the magnitude of P for equilibrium and total elongation so magnitude of P is unknown and total elongation that is Delta L these are the two unknown quantities which we have to calculate so let us try to get the solution for this problem solution first I go on to find the value of unknown load P for this as we see in this problem all the forces are horizontal so for the equilibrium condition of this step bar I will say that sum of all horizontal forces are equal to zero since for equilibrium of stepped bar summation of FX is equal to zero where all right word forces I will take them as positive or leftward forces negative so therefore if we look into the diagram here we have minus 10 plus 5 minus 10 because it is acting towards left plus 5 since it is acting towards right minus P and plus 20 is equal to zero so from this if I calculate here I will be getting the value of P as 15 kilo Newton so this will be the first answer next after getting the value of unknown load here as the stepped bar is given in three different sections I will call this first section here we have the second section in there the third section as their diameters are given the first thing is we can calculate the area for all three sections so your I’ll write down since cross-sectional areas of all sections it will be given by the first section area I will denote it as a 1 so it is PI by 4 into the diameter is 20 20 square this comes out to be 314 0.16 mm square first area similarly the second cross sectional area diameter is 25 mm and my answer is 490 0.87 mm square at last the third area diameter is 30 mm so this area comes out to be 7-0 6.86 mm square now after getting the individual areas as we have to find out the total elongation so in this step bar there are different loads different values of load so we need to know how much load is acting on the first section on the second section and third section and for this purpose we have to take different sections separately and find the value of load always in such kind of problems when you want to find out the load of individual section try to find the load first on the extreme most section that is first we will find out the load on section 1 and section 3 at last we will go on to section two so starting with section one I’ll separate it this is section 1 having diameter 20mm and the length of this section is 500 mm 0.5 it is meter so it is 500 mm now your it is given that at section 1 we are having 10 kilonewton load towards left so here I’ll mock 10 kilo Newton load towards left to balance this 10 kilo Newton we have to add another load which should be towards right having the same value so that this section comes into equilibrium so here this is the first section and now after we have calculated this load for the first section next similarly what we can do here is that after getting the value of load here I can calculate the deflection of this individual section so you are I will say that therefore deflection in section 1 is given by Delta L 1 is equal to PL upon ei e for the first section so therefore deflection in the first section is P is tensile so we have to take it as positive 10 into 10 raised to 3 length of the section is 500 cross sectional area we have calculated for the first section and that was 314 0.16 next capital e the value was given in the problem as 210 GPA so it is 210 into 10 raised to 3 because 1 GPA is I’ll write it down here 1 Giga Pascal is equal to 10 raised to 3 Newton per mm square so in this problem they have given 210 Giga Pascal so it is 210 into 10 raised to 3 Newton per mm square now from this I will get deflection in the first section my answer is it is 0.07 6 mm and since this answer is positive it means there is elongation in the first section next similarly here after separating the first section now I will separate this third section because as I have told before we have to start with the first and the last section that is extreme sections and then the middle portion so here I’ll separate this portion which has a diameter of 30 mm and length is of 900 mm length is 900mm now as we can see in this problem in the third section we are having two loads that is first is 20 kilonewton and the other is load P but this load P will give us the value at the junction so it is better to find the load at only individual end so here I will start with this 20 kilo Newton which is acting towards right now here as I can see 20 kilo Newton is applied towards right so for the equilibrium condition of this beam here I have to add another 20 kilo Newton which is towards left so because of the action of these two forces there is tensile load acting on this beam or the section of the strip bar and it is subjected to tension so here I will calculate the deflection since deflection in section 3 is given by delta L 3 is equal to PL upon a e4 the third section so therefore Delta L 3 will be equal to here P is 20 kilo Newtons for 20 into 10 raise to 3 length is 900 mm area for the third section the value was 706 0.86 Young’s modulus it is 2 10 into 10 raised to 3 so from this the deflection in the third portion it comes out to be 0.1 to 1 mm again as this value is positive it means there is elongation next after finding the elongation in the first and the third section now we will separate the middle portion that middle portion is having a diameter of 25 mm and its length is 700 mm this is the second section for us now as we can see on this second section there are two load values one is 5 kilo Newton other is P so what is the value which we have to take now if we look into this at section 1 when we have separated it there was 10 kilo Newton load towards left 10 kilo Newton towards right now this 5 kilo Newton towards right is the load at the junction so now how I’ll explain this is that 10 kilo Newton load is there at this Junction it is we can say because of section 1 now because of section 2 there is also another value of load if you add 10 n that value of the load which is there on the section 2 that should give you an answer of 5 kilo Newton towards right so how we can complete this is that here we already have 10 kilo Newton so if you want the answer is 5 then we need to subtract 5 from this so if at this Junction I have another load which is a 5 kilo Newton towards left then by adding them my answer will become 5 kilo Newton which would be towards right so here we can complete this by adding a load of 5 kilo Newton towards left at 2nd section because here this is going towards left so minus 5 next at the same point we are having plus 10 so minus 5 plus 10 that gives us plus 5 kilo Newton and so here we are saying that this condition has been met or it is satisfied and to balance this we have another 5 kilo Newton load on to the other side so now if we can see the value of load P it was 15 kilo Newton 1 5 so we can calculate this at section 3 here when I have separated 20 kilonewton load is already there now there is another five kilo Newton load here so 20 is towards left so minus 20 and this is five which is plus so minus 20 plus 5 that gives me fifteen kilo Newton and that fifteen kilo Newton load would be acting here so the diagram is satisfied so here I can calculate the deflection for the second portion that since deflection in section three or section two I can say section three is over is given by Delta L 2 is equal to PL upon a II for the second section now here I will be putting the values therefore Delta L 2 is equal to P as we see second section it is subjected to tensile load so P has to be taken positive so positive 5 kilo Newton if I write into Newton it is 5 into 10 raised to 3 next its length is 700 mm divided by area of second section that was 490 point 8 7 into Young’s modulus is 2 10 into 10 raised to 3 so here I have this deflection as 0.03 4 mm and again since my answer is positive it means there is elongation so after reading these three deflections we can get the total deflection by adding them so here I can say that therefore total deflection is given by Delta L is equal to Delta L 1 plus Delta L 2 plus Delta L 3 so therefore total deflection is equal to Delta L 1 the value we have found out it was 0.076 Delta L 2 just now we have calculated point 0 3 for Delta L 3 it was zero point 1 2 1 mm if I add up these values my answer comes out to be zero point 2 3 1 mm and since this answer is positive second answer for us so here this means there is elongation or increase in length of the step bar so you’re in this video we have seen how to calculate the value of unknown load P which was the first question and the total deflection or we can say elongation of the step bar so both those questions have been solved and with this we complete the problem

sir send your Email id so i can send the problem

Thank u sir .aapke vdo se som ka concept bahut easy ho gya

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Sir this is similar to superposition principle . But I have a doubt in superposition principle that is"y that principle is not applicable for larger deformation s yy . That is in the starting only we assume that deformation on that body is small . Can u pls explain me how the text or by sending any link of the vedio . Pls reply me

sir u very good job do it …slute u…and thnk u so much sir …

thanks alot sir

a big respect to you sir I don't know how you makes complicated things like ABCD. love you sir # RESPECT

I still dont understand how you resolved forces in the middle segment

Great explanation

Awesome explanation sir it make us easy to understand

U concept ddnt clear enough

Wish had a teacher like you in our college

while taking na elongation of the first section, wouldn't the 10kN forces cancel out each other?

Sir kabhi esa ho sakta hain ki 1 bar ka elongation ho aur dusre ko length decrease ho

The middle section calculation is wrong it is compressive stress not tensile…

Can I get pdf file of stress stain notes

Email me to [email protected]

Sir pls give your link so that I can get new video of yours

Sir you are great . Once this problems were very here for me to understand but now I am able to understand this after seeing this video. Thank you very much for help us Sir๐๐ป๐๐ป

Sir deflection by conjugate beam method pr v vedio bnao

Sir when lengths is not given so can we assume the length

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Thankyou

Thanks you so much Sirji for your great explaination, and thanks for the making this video I understand all the problems of analysis of bars of varying sections..๐