Problem on Shear Stress in I Section Beam – Shear Stress in Beams – Strength of Materials

Problem on Shear Stress in I Section Beam – Shear Stress in Beams – Strength of Materials


Hello friends here in this video we will see a problem on calculation of shear stress for an I-section here is the question I will write that in the form of data and I section I section beam 350 mm into 200 mm has a web thickness of 12.5 mm and flange thickness of 25 mm so these are the dimensions given it carries a shearing force of 200 kilo Newton shearing force is denoted by F it is 200 kilo Newton so 200 into 10 raised to 3 Newton at a section sketch shear stress distribution across the section so for that shear stress distribution diagram is what we have to draw now since it is given that we have an i-section we can start the problem by drawing the diagram now here is the I section I will mark the dimensions over that it is 350 mm into 200 mm so 350mm is the depth here into 200 mm next has a web thickness of 12.5 mm so web thickness is 12.5 mm then and a flange thickness of 25 mm so thickness of upper and bottom flange are same 25 mm now after knowing that I section which is given in the problem here I have marked all the dimensions we have to find draw the shear stress distribution diagram and for that purpose we need to know the values of shear stress at different locations so first I will start in this way that since the given PI section is symmetric so here is the y axis of the I section and its distance from the reference is X bar so as the given I section is symmetric therefore X bar will be half of 200 that is 100 mm next y bar will be equal to it is half of 350 that is the location of x axis this is also called as the neutral axis that is 175 mm now here is at this point we have the centroid which is at the intersection of X and Y axis now in this case after getting X bar Y bar we are going to find the moment of inertia about the neutral axis and also since here we have an eye section and we have to draw the stress distribution diagram so we require stresses at or we can say shear stresses at various locations so first we will calculate the shear stress at this location then here and then at last at the neutral axis so now after reaching here I will say that since mi about neutral axis is given by we know that the neutral axis is located at a distance of 175 mm from bottom so am I about this neutral axis that will be given by IX X is equal to from the complete section that is BD cube by 12 – that is what I am doing here is I am taking the complete rectangle which is having 200 mm width and 350 mm depth so from this complete rectangle I am going to subtract two small rectangles which is here so how I will get the moment of inertia is from the complete rectangle that is capital BD cube by 12 I am going to subtract the moment of inertia for these two small rectangles so for them I can write this as the value of width will be from 200 I will subtract 12.5 so that will give me the total width here so it is 200 minus 12.5 that gives us B and D is the depth of this small rectangle which is 350 – it is 25 plus 25 that is 50 so 350 minus 50 that is 300 so into 300 cube that is B D cube divided by 12 so here I will put the values of capital B and D capital B it is 200 capital D is 350 – the terms here which gives me an answer of this entire term as 421 point 8 7 into 10 raise to 6 and after I calculate this I will get the answer of I x6 as it is 292 0.7 1 into 10 raise to 6 mm raise to 4 so here I have got the value of I X X now I will say that since shear stress at the junction of flange and web is given by so now shear stress at the junction of flange and web is given by tau 1 is equal to F a y bar upon I into B 1 keeping this as equation number 1 now here I will say that F is given in the problem that was shearing force so it is 200 into 10 raised to 3 Newton next area here we are calculating shear stress at the junction of web and flange so it means we are calculating at this location so here the area is area of the flange that is 200 into 25 and that comes out to be 5,000 mm square then Y bar Y bar is since we want to calculate the shear stress at the junction which is at the flange so Y bar is the distance of the x-axis of the flange and the x-axis which is the neutral axis this distance is y bar and it can be calculated as since from bottom it is 175 mm from the top also this distance is 175 so from 175 I will subtract half of 25 that gives me this y bar distance so 175 minus 25 by 2 so that gives me an answer off it is 160 2.5 mm then only be one is left here B 1 is the width where we want to calculate the value of shear stress here the width is 200 mm so now once we have found out all values we can say that put all values in equation number 1 so I will put the values on to the next page so we have tau 1 is equal to F it is 200 into 10 raised to 3 a it is 5000 Y bar 1 62.5 upon I that is the moment of inertia we have found out this value 2 92.7 1 10 raise to 6 that is I into b1 which is 200 so on calculating this I will get the answer of tau 1 as 2.77 Newton per mm square now after getting tau 1 to calculate tau 2 here I’ll use a relation that using the relation of shear stress and width of the section so the relation is tau 1 upon tau 2 is equal to b2 by b1 that is it is inversely proportional because if we look into the formula of tau is equal to F a Y bar upon I be if I take all these terms as constants then tau is inversely proportional to B that is if B goes on decreasing shear stress goes on increasing that is at minimum width shear stress is maximum and where width is maximum shear stress will be minimum so tau 1 is the shear stress at width of 200 mm now we want tau 2 and that will be at a width of 12.5 mm that is here at this location I will say this is 0.1 this is 0.2 so we have found out tau 1 now tau 2 is at 12.5 mm so using this special relation tau 2 will be equal to tau 1 into B 1 by B 2 therefore tau 1 is 2.7 7 into B 1 and B 2 values B 1 is 200 B 2 is 12.5 so on calculation I will get the value of tau 2 and it comes out to be 44 point 3 2 Newton per mm square now we have found out the shear stress at section 1 now at section 2 next I will calculate the shear stress at the neutral axis so I will write the formula now since I let down maximum shear stress maximum shear stress because it will there it will be there at the neutral axis maximum shear stress at the neutral axis is given by we can directly write the formula that is tau max is equal to it is f into a 1 by 1 plus a 2 y 2 upon I into B now your a 1 and a 2 values a 1 is the area of the flange so I will write down a 1 is equal to 200 into 25 which is five thousand mm square next after this area 2 a 2 will be equal to since I am considering the neutral axis so I have to consider the area above the neutral axis so flange was 200 225 web it is 12.5 mm width and height of the web is it is 175 from total that is from 175 mm if we subtract 25 that will give us this height so it is 12.5 into 175 minus 25 that gives us area 2 and the answer is 1 8 7 5 mm square so getting this as area 2 now y1 and y2 values therefore y1 will be it is nothing but the distance between x axis of the flange and the neutral axis so this distance is 162 we have already calculated 162 point 5 this is 1 6 2 2.5 mm then y2 y2 is now we have this web and the height of the web is above the neutral axis this is 150 mm so taking half of this will give us Y 2 that is 75 mm now I can say that therefore tau max will be here in the formula we have F F was 200 into 10 raised to 3 area 1 5,000 into y1 a 1 by 1 plus a 2 by 2 upon IB so I will put go on putting the values area 2 1 8 7 5 into y2 upon I value that is 292 0.7 1 into 10 raise to 6 and B is nothing but the width of the web which is 12.5 so on calculation of this I am getting the answer of tau max as 52 point zero nine Newton per mm square this is the maximum value of shear stress so now after getting the values of shear stresses at different locations I will plot the shear stress distribution diagram in this shear stress distribution diagram first I row the I section which was given in the problem here I will mark the neutral axis projecting the neutral axis light now for the shear stress distribution diagram first I will draw the vertical line and here shear stress is zero at the extreme ends next here I had given this as 0.1 so tau one value was two point seven seven so it will be somewhere here tau 1 as two point seven seven Newton per mm square next at section two here the area is decreasing so stress will increase and it will increase up to a value of tau 2 and tau 2 was forty four point three two Newton per mm square in a similar manner since it is an i-section here I’ll mock tau 1 and tau 2 Webb the bottom as well so the shear stress variation will be in the form of an arc from Tao it is equal to zero so tau one next at the neutral axis we are getting the maximum value of shear stress and that is fifty two point zero nine Newton per mm square so at the neutral axis we are getting tau max fifty two point zero nine Newton per mm scale now so this is the shear stress variation for an i-section or we can say the shear stress distribution diagram where the shear stress is zero at the extreme fibers at section one that is at the junction of where the flange is connected to web it is two point seven seven Newton per mm square where the area decreases that it is section two the stress is forty four point three two Newton per mm square at last at the neutral axis it is maximum shear stress and here it will go up to a maximum value similarly here the shear stress will go on increasing up to a maximum value and with this we complete the problem

48 comments

  1. Can we apply the same procedure for solving unsymmetric I section…???What will be the change in shear stress distribution diagram??

  2. I think , as the dimensions 350mm x 200mm the first number show's the base I.e 350mm and height of 200mm please checkout

  3. @ekeeda , here 350 mm*200 mm it means horizontal dimension 350 mm and vertical dimension 200 mm . But you showed it inverse in figure

  4. (Tau1/Tau2) = b1/b2

    Please explain it.

    If A is constant, b will also be constant right?

    And if you've considered b as variable then how have you considered A as constant?

    Is E-learning an easy way to avoid contradictional doubts by students?

  5. Hello, actually in the book of Rk bansal,the Sol for I section is completely different..there using formula for finding max shear stress at flange and web,which methodology is preferable to solve for this problem…can u explain??

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  7. A small mistake in the problem solving ..u have to substaract two rectangles moment of inertia from the entire rectangle but u did only for one.. U r videos are really helping so many students and from next time onwards please go indepth of the concept so that it makes some difference from others..thank you sir.

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