Problem on Shear Stress for T Section – Shear Stress in Beams – Strength of Materials

Problem on Shear Stress for T Section – Shear Stress in Beams – Strength of Materials


Hello friends here in this video we will see a problem on calculation of shear stress for a T section and for that here we have a question whatever is given here I will write that in the form of data now a beam a be supported at its ends has a span of 2 meter so span means the length of the beam is given it is 2 meter so 2000 mm and it carries a UDL of 200 kilo Newton per meter over the entire span so it means that here if I can explain it with a diagram the beam is supported at its two ends I will call it as end A and B and then it carries a UDL over this beam there is a uniformly distributed load whose intensity is given as 200 kilo Newton per meter over the entire span means over this entire length then the cross section of the beam is T section so if we look from the side then the cross section is T and for that T section the dimensions are given having flange width 125 mm so flange width is 125 and thickness is 25 mm similarly web thickness is 25 and overall depth is 200 mm so all these things I’ll denote it directly onto the diagram calculate maximum shear stress that is tau max and maximum shear stress in the beam and draw shears distribution diagram so this is the question we have now let us try to get the solution to this problem now in the solution part first I will say that since the beam or I’ll say the given beam carries UDL overheads entire span so therefore reaction at each end will be same means when we are having UDL over the entire span then that UDL would be converted into point load and that point load would be shared equally by each of the supports as we can see here at a I will call it the reaction as an RA at B the reaction as RB so here I will say that therefore reaction RA will be equal to RB and that will be W into L divided by 2 because W is the intensity of UDL when we multiplied by the capital L that is the span we converted into point load and that we would be dividing into two half that is the load is shared equally by each of the supports so therefore RA is equal to RB and that will be W value is 200 it is given kilo Newton per meter into the span is 2 meter divided by 2 so therefore RA is equal to RB and it comes out to be 200 kilo Newton now I will say that since in this case here we are solving the problem on shear stress so in shear stress this force which we are getting are ARB these are called as shear force so therefore shear force it can be denoted by s or it can be denoted by F that is nothing but the reaction which is 200 kilo Newton now after getting the shear force the next thing is since it is given that we are having a T section here so I’ll draw the diagram for that T section No here I have drawn the dissection I’ll mark the dimensions over this here is the flange here we have web so the cross-section of the beam is T section having flange width 125 mm so the width of the flange here is 125 mm next flange thickness is 25 mm this is the flange thickness 25 mm then web thickness is 25 mm and overall depth is 200 mm so this is the overall depth 200 mm now first of all once I have marked all the dimensions in the T section I need to get the value of cent idea that is centroid needs to be located as I can see this T section is symmetric about the vertical axis or Y axis so centroid will lie on this axis this is the y axis so therefore the distance of this by X is from the origin that will be nothing but X bar so I will say that therefore X bar is directly half of 125 so therefore here we have X bar as 6 to 2.5 mm next we need to calculate y bar here so I will say that after this therefore Y bar is equal to a 1 y 1 plus a 2 y 2 upon a 1 plus a 2 keeping this as equation 1 now this web portion I will keep it as section 1 for us flange will be the second section so area 1 will be there for area 1 that will be here the width is 25 so it is 25 into the depth of this web is 200 minus 25 this is the depth of the web which is 175 mm so 25 into 175 so area 1 it comes out to be 4375 mm squared so now after getting area 1 area 2 will be equal to 4 the second section the width is 125 depth is 25 so this area comes out to be 3000 125 M square now after getting these areas y1 will be now y1 is the location of centroid for the first section so it will be at half of 175 and it is 87 point 5 mm next y2 is the location of centroid for the second section and its centroid is located at the x axis which is passing here so the distance of this x axis from the bottom will be 175 plus half of 25 so it is 187 0.5 mm next I will say that therefore put all values in equation number 1 so I will put all the values here so we have Y bar is equal to area 1 was for 375 y1 87 point 5 plus area 2 three one two five into y21 87 point 5 upon area 1 plus area 2 so therefore y bar value comes out to be it is one twenty nine point one seven mmm point one seven mmm this is from bottom now after getting the location of y bar I locate the x axis for the complete section this is the x axis of the complete section which distance is one twenty nine point one seven mmm from the bottom now once we have X and y axis here is the centroid for the complete section now if we look into the question here we have to find out the maximum shear stress and for that maximum shear stress we even require the moment of inertia the moment of inertia about the centroidal x axis so the next thing would be I’ll calculate the moment of inertia so after reaching here I will say that now since mi about x axis that is moment of inertia about x axis 480 section that is given by I xx is equal to ixx 1 + i6 – keeping this as the second equation now I will calculate separately I xx 1 i6 – I xx 1 will be I 1 plus area 1 into H 1 square now what is I 1 your I 1 is the x-axis for the first section which will be at half of 175 so this is the x-axis I will call it as X is 1 1 located at distance y1 which is nothing but 87 point 5 mm from bottom so about this axis which is 1 1 the moment of inertia would be since it is horizontal so that will be I 1 is equal to BD cube by 12 plus area 1 is B into D H 1 is the distance between the two x axis that is x-axis of section 1 and x axis for the complete section this distance is nothing but H 1 here we have h1 distance so that will be 120 9.17 minus 87 point 5 whole square so putting the values here for the first section with B is 25 and depth is 175 area we have found out it was for 375 into this bracket so here I am getting the value of I xx 1 it comes out to be 18 point 7 6 into 10 raise to 6 mm raise to 4 now similarly I will calculate I xx 2 that will be I do plus area 2 into H 2 square now if we look into the problem the x axis for complete t section it is passing at x x from this x x axis and the x axis for section 2 is passing here so there is some distance between the 2 axis and this distance is H 2 so here we need to know what is this H 2 H 2 will be here as we can see from 200 if I hear this is 120 9.17 mm from bottom so I will say that therefore X 6 2 will be first of all I 2 is nothing but BD cube by 12 area 2 is B into D and here we have h 2 square and this is by parallel axis theorem as the two x axis they are not passing through the same line so we are using parallel axis theorem as we have used in the first case as well now therefore X 6 2 will be equal to B and D values for the T section here for this flange B is 125 D is 25 so 125 into 25 cube divided by 12 plus 125 into 25 next H 2 is nothing but 187 0.5 minus one twenty nine point one seven square so from this I will get the answer of IX x2 and that comes out to be 10 point 7 9 5 into 10 raise to 6 mm raised to 4 so here I have the value of x x2 now after getting the value of IX X to the next thing would be putting these values in equation number second to get IX X so therefore put IX X 1 and X 6 2 in equation number second so I’ll add the values therefore IX X is equal to X 6 1 18 point 7 6 into 10 raise to 6 plus is 6 to 10 point 7 9 5 into 10 raise to 6 so therefore I xx value will be it comes out to be 29 point 5 6 into 10 raise to 6 mm raise to 4 and this is nothing but I now after getting this here in the question it was mentioned to calculate maximum shear stress so I will say that for given P section maximum shear stress will be at the neutral axis so now for given t-section maximum shear stress will be there at the neutral axis and that maximum shear stress is denoted by tau max formula is f a y bar upon I be now sometimes this F which is called as shear force can also be denoted by capital s capital S or F both are same here so next now tau max since it is there at the neutral axis so to calculate that we require the area from the bottom only up to the neutral axis because we want the stress at neutral axis so we have to consider the area below the neutral axis as we are taking the reference from bottom so the area will be one twenty nine point one seven that is height into width as 25 so here I will say that keeping this as equation number third I will say that capital a is one twenty nine point one seven into twenty five that is the area of the web which is below the neutral axis and that comes out to be three thousand two hundred and twenty nine point two five mm square next y bar is nothing but the half of the height height is one twenty nine point one seven so half of that that is y bar and it is sixty four point five nine mm then I will put all the values in equation three and for that I is the value your B since we want to calculate the stress at the neutral axis so with B will be the width of the web because at the neutral axis we have web so this much with we have to take which is 25mm so I will say that putting all values in equation number third so we have tau max is equal to now f is nothing but the shear force we have calculated at the start of the problem that shear force was 200 kilo Newton so that becomes 200 into 10 raised to 3 Newton next area Capital a it here we have found out so I will put the value then y bar y bar value is 64 point 5 9 upon IB is 29 point 5 6 into 10 raise to 6 and B is 25 the width of the web so on calculation I will get tau max value as it is 56 point 4 4 Newton per mm square this is the answer I will call it as our first answer now if we look into the question here the question was to calculate maximum shear stress we have found out the value in the beam and draw shear stress distribution diagram now for this shear stress distribution diagram we require stress at other points as well so after this I will say that therefore the first thing would be once I have calculated the maximum shear stress now I will find out shear stress at the junction of web and flange so to calculate the shear stress at the junction of web and flange I will call it as tau 1 and the formula will remain same fa y bar upon I B now in this case since I am calculating the shear stress at the junction of flange and web so when the flange in web they are meeting here is the junction and even here I can say that this would would be the area so area I have to take it as the area of the flange and not the area of web so I will say that your a is equal to 125 into 25 which is 3 1 to 5 mm square 125 into 25 here I have got the value of area I is constant B we have to take the width as the width of the flange which is 125 mm I will call that as B 1 and B 1 is 125 mm which is your next y bar after this the Y bar value will be I will calculate Y bar here itself Y bar will be now Y bar means here we want to know the distance between the neutral axis and the x axis for the second rectangle that is this height H 2 and it is 129 it was from 200 first I’ll subtract 120 9.17 so 200 minus one twenty nine point one seven so once I subtract from 200 this distance 129 0.17 I am going to get this total distance and from this total I will subtract half of 25 so that I get this distance which is h2 so here we have after this – from this 25 by 2 so therefore y bar value it comes out to be fifty eight point three three mmm next I will put all the values here so calling this as equation four so therefore put all values in equation number four hence we have tau one as equal to now if as we know it is 200 into 10 raised to 3 area we have found out it is three one two five into y bar is 58 point three three upon I into b1 I was twenty nine point five six into 10 raise to 6 and b1 value is 125 so from this I will get the answer of tau one and it comes out to be nine point eight seven Newton per mm square next after this I will calculate another value of shear stress and that value will be at this location previously the shear stress which I had calculated it was here now I am calculating the shear stress at this location so for that I use a special relation that now using the relation of shear stress and corresponding Wed so using that relation so now here I am using the relation that tau 1 upon tau 2 is equal to inverse of the ratio of width that is b2 by b1 here a very important conclusion I am using that shear stress are inversely proportional to the width that is where the width is minimum shear stress will be maximum so from this ratio I can say that therefore tau 2 will be tau 1 into B 1 upon B 2 so tau 1 is 9.8 7 B 1 is the initial width which was 125 B 2 is the width at the junction which is nothing but with of web 25 so from this I will get tau 2 therefore tau 2 value comes out to be 49 point 3 3 Newton per mm square now once we have got all the values now we can draw the stress distribution diagram as it is asked in the question so here I’ll write down shear stress distribution diagram first I’ll draw the T section which was given in the question now here the D section was given and we have found out that the neutral axis was located at a distance of it was 120 9.17 mm so I’m marking all the dimensions here first no after marking the dimensions here now I will plot the shear stress variation and for that variation first I need to draw a vertical line then the neutral axis would be extended by a light line after that the junction of flange and wave would be projected light now at the extreme corners the shear stress value is zero so tau is equal to zero then at the neutral axis the shear stress is maximum so here I’ll write down tau max value was fifty six point four four Newton per mm square next now at the junction where the flange width is 125 mm there means you’re at the junction you’re the stress value we have got it as tau 1 which was 49 point it was 9 point 8 7 Newton per mm square tau 1 so the nature is such that here I will get the value tau 1 as 9.8 7 Newton per mm square and joining them so from 0 the shear stress will go on increasing and it will reach up to 9 point 8 7 Newton per mm square then after this these points we had calculated the shear stress at the inner junction taking the width of the web which was 25mm so you’re I to with web is equal to 25mm the stress was forty nine point three three so when area decreases as we can see here stress will increase so from nine point eight seven it will go on increasing up to a value which is forty nine point three three Newton per mm square and then it will reach a maximum value here and after reaching the maximum value again it will go on decreasing such that it becomes zero so this is the shear stress variation as we can see here and with this we complete the question

43 Replies to “Problem on Shear Stress for T Section – Shear Stress in Beams – Strength of Materials”

  1. Sir, why we didn't considered the full rectangle first and then subtract the excess portion like previous I section method?

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  3. How did u know initially knw that for maximum shear stress we have to take the lower section?? Plse explain

  4. sir ,I think there is a mistake in Ixx1…..it should be 175×25^3….plz make it clear to me….we should commonly cut the flange and web, isn't it?

  5. For calculating the Shear Force, in this case both reaction forces are equal. What if there is unequal reaction forces? Do we use the highest reaction force as the shear force in this case?

  6. Nice explanation thanks sir it was helpful vediou for me today we start this topic I got so many doughts but after seeing this vediou I got some confidence so thanks u so much sir🙏🙏🙏🙏

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