Lecture 18

[Music] [Music] now we are in the second part of the second tutorial session we have been practicing certain numerical problems on the friction moment of inertia gravity in the first part here I will focus more on the stress strain and residual stresses just to recall certain relations and formula stress it is denoted by Sigma is force per unit area where f is force applied and a is the crosssectional area of the body your units of the stress are in Pascal or in Newton per met Square strain is deformation per unit length that is the change in length per unit the initial length where Delta is in the change in length and L not is the initial length units are dimensionless or sometimes it is expressed in percentage stream the percentage elongation or maybe in compression the percentage compression hooks law was also discussed that is the relationship between stress and strains we know Sigma is equal to e into Epsilon where e is the modulus of a material for which the units are Pascal residual stresses are the stresses that remain in a material it remain in a material after the original cause of the stress has been removed even so sources of the res stress are manufacturing processes such as welding Machining and heat treatment measurement of regual stress is often using the X-ray defraction or hold Drilling many Optical methods are there that helps us to find the residual stresses in the bodies and these residual stresses also determine the properties and the applications where the bodies are to be used let us now try to see some of the numerical problem all the relations which were discussed in the previous weeks this is a problem statement where it is given a square steel rod of 20 mm to 20 mm in section is to carry an actual load which is compressive of 100 k we need to calculate the shortening in a length of 50 mm I will try to show you the tensile strength test compression test and some other tests in the forthcoming weeks we will try to use these relations to find out what is your young modulus what is the tensile strength compressive strength impact strength of various materials so because it is a scare Rod so we need to find a shortening in a length that is we need to find the value of delta l so so here Delta L is to be determined the shortening in the length what we are given are area area is equal to uh it is given in mm so I convert the mm into meter that is 0.02 into 0.02 is = to 0. 0 4 m² and I have gotten length this is area length which I denote by L this is equal to 50 mm in me is equal to 0.05 M load that is given is I will denote it by P this is equal to 100 Kon e is also given that is R's modulus which is equal to 2.14 into 10^ 8 K per M Square now to find the shortening of the rod L we need more space shortening of Rod L that is Delta L I need to determine we know stress Sigma is = to P by a that is load per unit area this is equal to 100 is the load and area is 0.004 M Square 100 kilon is the node area is 0. 04 M squar so this turns out to be 25 Kon per M squar this is stress now strain we know that e is equal to stress by strain therefore strain is equal to stress by E which is equal to 25,000 by the value of V which is given as 2.14 into 10^ 8 so this Trin is Delta L by L which is 25,000 by 2.14 into 10^ 8 so therefore Delta L is equal to 25,000 by 2.14 into 10^ 8 into L length given is 0.05 this is equal to to 25,000 by 2.14 into 10^ 8 into 0.05 this is equal to 0 because 10 to^ 8 is here 0 0 0 0 5 84 meter it is better to put it in millimet so this into 10^ 3 this is equal to or I would write R this is equal to 0.0 584 mm so shortening of the rod that is Delta L that is shortening is equal to 0.584 4 mm this is answer I have a similar problem statement here where a Hollow cast iron cylinder 4 M long 300 mm outer diameter and thickness of metal 50 mm is subjected to a central load on the top when standing straight the stress produces 75,000 K per M Square assume modulus for cast tyan as 1.5 into 10^8 8 K per M square and find magnitude of the L longitudinal strain produced and total decrease in length this problem also I'll solve for you here it is given the outer Dia I'll put it as D is equal to 300 mm which is equal to 0.3 M thickness T is equal to 50 m mm which is equal to 0.05 M length L is equal to 4 M 4 M long road right now we have the stress that is produced that is given stress that is produced that is Sigma is equal to 75,000 Kon per M Cube and we have been given the Ang modulus here as well Ang modulus value is 1.5 into 10 to^ 8 Kon per M Square now inner diameter of the cylinder is to be determined indan diameter would be small D which is equal to out to diameter minus 2 * into thickness for instance this is a cylinder here this is outer di D this is inner D small D this is thickness thickness is here two times this is t t so this is D minus 2 of T this is equal to 0.3 - 2 or into 0. 5 is equal to 0.2 M this is my inner di now let us start solving the problems that it has asked magnitude of the load magnitude of the load I need to determine I'll try to solve it here magude of load that is if I know that Sigma is equal to P by a that is load per unit area so I need to find p isal = to Sigma into a that is stress into area which is equal to 75,000 into the area now this area is < by 4 because this is a cylender this is D ² minus D ² that is outer D minus smaller d right so this is equal to 75,000 into < by 4 4 outer D is 0.3 squ minus inner D is 0.2 s here I calculate the value of P = to 29 45.2 K but this is first part magnitude of the load right second part is longitudinal strain produced now second part that is longitudinal state for this I need to find longital strain produced I denote it as e here e is equal to stress by Ang modulus which is equal = to 75,000 by the value of young modulus 1.5 into 10 Power 8 which is equal to 0. 00 five there are no units strain is dimensionless now decrease in length is the third part that is Delta L like the previous numerical to find the value of delta L I know strain is Delta L by L right so and I know the value of L here that is 4 M I know the value of this strain here so that means the strain that you have just calculated as 0.005 is equal to Delta L by 4 which gives me Delta L is = 4 into 0.00005 this is equal to 0.2 meter so this met if I convert into CM this is multip 10^ 3 this turns out to be 2 mm this is decrease in length right so this is all solved now now there the numericals in the similar fashion that I have here for instance there is a safe stress for a holel column which carries an actual load of 2.1 into 10^ 3 k and the safe stress is 125 Mega Newton per M Square the external diameter of the column is 30 cm determine the internal diameter you will use a similar relation as I did in the previous numerical so this is again a hollow steel column so we have external Dia we have internal Dia like we had it here external and the internal diameters right so you will try to solve this numerical you will find the solution in the lecture notes and try to compare how have you determined how have you calculated it we might have to use some relationships which have been given the theory part of the lectures covered in week two and three and then you have to find the solution to this Nole problem similarly I've given you another numerical statement here find the Jung's modulus of a brass Rod the diameter is 25 mm and length is 250 mm which is subjected to a tensile load of 50 Kon when extension of the road is equal to 0.3 mm for this we have to find the J modulus you know the relationship of J modulus that has been used in the previous numericals as well so here you might have to first find uh the uh area of the rod the diameter is given the area of the rod is to be taken so uh Dia is given area would be question mark then you have to find the tensile load tensile load is already given so the length is already given you would will find the stress stress you will find based upon the stress then you will find the strain strain you will find strain e then based upon the stress and strain you'll try to find the value of young modulus that is stress for unit strain that you try to find here and try to compare the answer that you get with the answer which I give you in the elotes then there is another problem statement where a hydraulic press exerts a total load of 3.5 Mega Newton this load is carried by two steel rods supporting the upper head of the press if the safe stress is 85 megapascal and zang's modulus is uh 210 Kon per mm squar find number one diameter of the rods number two extension in each rod in a length of 2.5 M to find the the diameter of the rod you first have to find area over which the load is being applied so load carried by the two rods has to be determined that is value of p is to determined So based upon the value of the P then calculate the total area stress is given you will finally find the diameter of the rods it is D you will find accordingly second part is extension in the rod once you know the diameter of the rod so then you will use the Young's modulus relation that is stress by strain so you will put the value of stress here and St here that is p by a over Delta L by L this relation you will use for young modulus to find finally the value of delta l so these problems you will solve now another thing that we discussed in the previous uh weeks are hooks law hooks law H modulus these were discussed the elasticity of the material is given here young modulus is the measure of material stiffness the units is pcal is already discussed next is poison ratio the definition is ratio of transas strain to actual strain so transas strain to actual strain ratio is question ratio typical values are 0.25 to 0.35 for most metal let us try to see some problem statements on this a steel rod of cross-sectional area 2 cm is subjected to a tensile force of this calculate the stress in the rod this is a kids play we have gotten the area a is equal to 2 cm² which is equal to 2 into 10^ -4 m here force is equal to 10,000 nton so stress is equal to force per unit area this is equal to 10,000 by 2 into 10^ - 4 which is equal to 50 map Pascal this was just again a warmup similarly there the small very short numericals 1 M Long alenium Bar is tetched by 0.2 mm under Sil canel The Strain strain is change in length over original length it could be directly determined it is 0.2 / 1 M that is 0.02% strain would come as an answer I've given very short numericals here because some problems might also come in your exams which will very short to get marks in your exam these problems would be easy for you to score now comes another problem statement where we need to determine the E stress ultimate tensile stress percentage reduction in area percentage elongation when we are given a m Steel Rod of 12 mm diameter which was tested for tensile strength with a Gorge length of 60 mm following observations were recorded final length is 80 mm final diameter is 7 mm yield load is is 3.4 K and ultimate load is 6.1 K now we need to determine the following parameters heeld stress ultimate tensile stress percentage reduction in area percentage elongation so we have been given here D diameter is equal to 12 mm we have been given length is equal to 60 mm and uh recorded observations the final length that we have gotten final length I will put it as capital L that is 80 mm final diameter is D isal to 7 mm E load is 3.4 K I'll put it as W yield is equal to 3.4 4 Kon which I convert into Newton that is 3400 Newton and I have gotten ultimate load that is W ultimate is equal to 6.1 K which is equal to 6100 Newton these are all the given values now I need to find the eeld stress first to find that let me first find the original area of the rod original area that is [Music] AAL < by 4 d ² is equal to < by 4 into 12 s is equal to 113 I keep it in millim only mm squ then I have gotten the final area here in the calculations final area small a is equal to < by 4 the final diameter which is smaller one that is small D this is is equal to > by 4 small D is 7 Square this is equal 38.5 mm squar now let us try to find the solutions to the questions which are given first is Eid stress eeld stress we know eeld stress is equal to load that is the E load per unit initial area this is equal to 3400 by 113 this turns out to be 30.1 Newton per mm squ or I can put it as this is equal to to 30.1 megapascal Second point is Ultimate tensile stress that is altimate tensile stress so this is again calculated in the similar fashion we have ultimate load per unit area this is = to 6100 by 113 this is equal to 54 newon per mm squ or this is equal to 54 MEAP Pascal next point that is asked is percentage reduction in area so percentage reduction in area it is so simple we have the initial area and the reduction in area that is a minus a this is in percentage we'll calculate that is 113 minus final area is 38.5 that we have calculated by the initial area itself this is equal to 0.66 in percentages we can call it 66% right fourth point it has asked is percentage elongation percentage elongation is again so simple we have we have the final length L and change in length so change in length is 80 - 60 the final length is 80 this is equal to 0.25 or I can say this is 25% right these kind of problem statements keep coming when we try to talk about stress strain curve or elasticity we'll also try to solve something in potion ratio now so in potion ratio you see this numerical statement that is there a 2.0 mm long metal wire is loaded resulting in a 4 mm elongation find the change in diameter of wire when elongated if diameter of wire is 1.5 mm and potion ratio of wire is 0.24 just to recall once again oion ratio is the transverse strain by exal strain this is my portion ratio now to calculate the position ratio what we have here is the given values are the length of wire as uh 2.0 M only the change in length is also given 4 mm so if I turn it to M it is equal to 0.04 M diameter of Y is given diameter of Y is 1.5 mm so so I have diameter of wire D is = 1.5 mm and I have been given position ratio that is 0.24 the longitudinal strain find the change in diameter wire when elongated by this diameter right so longitudinal stain let me determine first right this is equal to change in length by original length this is equal to 0.4 by 2.0 this is equal to 0.2 so longal stain I have gotten right now to find the change in diameter of wire right I need to find the change diameter so let me try to see here using poision ratio we have mu is equal to lateral strain by longitudinal strain it is actual strain or longitudinal Str right so that means 0.24 is equal to lateral strain by 0 002 it is the longitudinal strain given so this gives me the value of lateral strain or the transm strain as 0.24 into 0.2 which is 0.48 using this I will find the change in diameter so lateral strain is this so literal strain this is also equal to change in diameter per unit original diameter so that is is equal to 0.0048 is equal to change in diameter by original diameter is 1.5 mm so I got the change in the diameter as 1.5 into 0.0048 which is equal to 0.0072 so this is my answer right so using the potion ratio uh I found the ratio between the lateral stra and longitudinal strain that is the ratio between the transverse strain and actual strain using the reation I found the change in diameter when I was given the length of the wire the resulting elongation that was given and also the diameter of the wire was given and the potion ratio was given so using the potion ratio I calculated the changing Di diameter the similar problems could be taken like another one when a brass Rod of 6 mm is subjected to a tension of 5 into 10 3 Newton the diameter changes by 3.6 into 10us 4 calculate the longitudinal strain and positions ratio of for brass given that Y for the brass is 9 into 10^ 10 Newton per met square that is elastic modulus is also given so we have been given here diameter we have been given here tension we need to find the longitudinal strain again using the PO ratio relation similar to the pattern that we did in the previous numericals you can try to solve this numerical right and find the answer you can compare your answer again with the solution that I give you in the lecture notes okay there's another problem statement where metal wire of length 1.5 m is loaded and elongation is 2 mm if the diameter of wire is 1 mm find the change in the diameter of the wire when it is elongated we have been given here that the position ratio is 0.24 this is almost exactly similar to what I just did here right this also you can try to solve so there is another problem statement in which which Young's modulus is given as 110 gons per met squ this here modulus is 42 gons per M squ find the bulk modulus and lateral contraction of a round bar of 37.5 mm diameter and 2.4 M length when stretched 2.5 mm this is an interesting problem where we have been given value of e as 110 G newon per M Square D modulus we have been given the shear modulus Shear modulus I denote it as C that is 42 G Newtons per M Square we have been given also the diameter 37.5 that is diameter of bar is 37.5 millimet so because all the things are in meter I will also convert this into meter this will this is equal to 0.0375 M length of the round bar is also given that is L is equal to 2.4 M I have to find bulk modulus and lateral contraction right this is one this is two so bulk modulus k for this relation K is bulk modulus where K is equal to M into e by 3 into m - 2 here sheer modulus is also given for Shear modulus also there's a relation that also will put here Shear modulus for that the relation is e = 2 c 1 + 1 by m right let us try to First find the value of M using the shear modulus right this is relation one this is relation two so using two I just put the values e it is 110 gonon is given convert this into Newton into 10^ 9 this is equal to 2 into c c is 42 gonon that is 42 into 10^ 9 into 1 + 1 by m 1 + 1 by m right so this gives me the value of M as if you solve this equation it will give you m is 1 by 0.31 which is equal to 3.22 now substituting the value of M substituting in one relation one right I got it here K is equal to value of M that is 3.22 into e value of e is 11 0 into 10^ 9 by 3 into M - 2 that is 3.22 - - 2 this is equal to 9677 gons per M Cub so I've gotten the value of the bulk modulus first part is over next comes the lateral contraction for the lateral contraction that is change in the diameter Delta D here I have also been given the value of the Delta L that is Delta L is given here that is equal to 2.5 mm this is equal to 0.25 M so for lateral contraction I come back here I first find the longitudinal strain longitudinal strain that is Delta L by L is = 0.0 25 by the length that is 2.4 this is equal to 0.104 so lateral strain is equal to longitudinal strain into 1 by m lateral strain is equal to longitudinal strain that is 0 00104 into 1 by m this turns out to be 0.104 by 3.22 this is equal to 0.0 0323 so this gives me the value of the Delta D now lateral contraction Delta D is equal to lateral strain into D that is 0.00 323 into d d is 37.5 which is equal to 0.121 mm right so this is uh one where we have calculated the bulk modulus and also we have used the relationship for the shear modulus as well in this problem so the last problem statement in this lecture is where we have been given the steel wire of 100 cm length and cross-sectional area of 0.03 cm squ and load that is applied on this is 100 Newton and this is slower load that is being applied the Young of the steel is also given that is 2 into 10^ 11 Newton per M squ we need to calculate the work done calculate the work done in stretching right that means we need to calculate the energy stored so I will just put it here work done is equal to energy stored right this you will calculate by yourself and try to compare the result with this I'm ending this tutorial and I will meet you further we'll take uh YouTube further practice sessions and also I'll discuss about the um laboratory sessions laboratory sessions means I will show you the videos on how different tests are being conducted in laboratory and also we'll try to see some simulations on the results on different tests which are there in the forthcoming lectures and further we will also try to talk about certain mechanisms in the forthcoming lectures in the fourth weeks thank you [Music] [Music] [Music]