Types of Stresses – Introduction to Design of Machine – Design of Machine

Types of Stresses – Introduction to Design of Machine – Design of Machine

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Hello Friends here in this video we will see various Types of stresses let us get started with it Types of stresses these Types of stresses will determine that how much or we can say what stress is acting on a particular material that means if we are able to identify how many Types of stresses are made then according to the duration of load given we can understand a material will be subjected to which kind of stress so let us get started with this for the types of stress we have the first one called as dangerous twist next compressive stress then we have shear stress after this we have Gausman shear stress next we have crushing stress then we have bearing stress and at last we have bending stress so now when we are studying design of machine elements or machine design in that case we should know these seven types of stresses because once we understand the types of stresses we can easily say by looking at the nature of the node like a material is subjected to which kind of stress so I will start the explanation one by one starting with the first one that is tensile stress know in order to explain tensile stress I will first draw a diagram yes I can say that we are having a lord now if I draw the axis for it here this is the axis which is passing through the world no I see that D is the diameter of this rod and now we’re not talking about tensile stress then the load is also 10 Zion ignition here I can say from this diagram that this angle shows the direction that this world is subjected to an upward force and you’re the road is subjected to downward force so this becomes an action of one type of loading so yes I can say this is the load denoted by B so the value of Lord is same on both the sides but their directions are different so because of the action of this what happens it is similar to this example like if I have a pain here which I can consider si Road and when I am applying two forces which will try to pull this in that is the force force will try to pull it upward in the second force will try to pull it downward so when I am pulling this there is tensile stress develop in this rod so the same thing happens here which I can write here tensile stress that because of prototype offloading the stress developed in the material is called as tensile stress so yeah I have an example that because of the pole type of loading pulling of rod in both the directions things I stresses develop in the rod next the second kind of stress that is compressive stress for compressive stress again I will draw this diagram of the same load having the same diameter D and length is in and again the load will be passing through this axis but now what will happen is that the direction of load it will be reversed here the arrow would be pointing towards the rod as I have shown here so now as we can see from this example that when there is a push type of load what happens is compressive stresses develop inside the material like for example destroyed I can consider it to be subjected to compressive stress and in that case what happens is when I applied two forces one from the upward and the other from the downward direction because of the action of these two forces this rod I am pushing this so it is subjected to compressive stress so I can see like because of push type of loading stress developed in the material is compressive stress now important thing to note here is that both in case of tensile and compressive stress the load is passing through the axis and the area around which the load is acting it is circular in section so this area is called as cross section area which I can write here if we have the rod of diameter D then this area of rod is PI by 4 d square and this area is subjected to stress so it is for this position Adium so this was regarding to insulate compressive stress next we will see shear stress for shear stress again I will draw a simple diagram here I have one plate next [Music] I have the second plate no they are connected with the help of pins or rivets now you only have two plates these two blades are connected by this joint which is at the center and it is called as a rivet having diameter I say that the diameter is d for this element no suppose on this top plate if force is acting towards left I will call it as P your this load on the top it is acting actually towards right it is not towards left towards left it is there for this plate which is at the bottom now we can see the directions bottom plate to that the road is acting towards left and for the top plate it is acting towards right now because of these two forces which are acting what can happen is that the top and bottom plates they can slide ok this since they are moving in the opposite direction so there is a possibility that the plates they can slide in that direction and because of that the rivet can break at the junction now when the remains they are breaking I draw the shape how it will look like when it breaks here I can show that item so we would be getting the rivets in two parts here like this this is the section at which they are breaking now it is very much clear from this diagram that because of the action of these two opposite forces the ribbon breaks into two halves like this and the action which is here it is called a shearing action and the load as I can say it is tangential to this area area is actually any of the serpent because here if I say that D is the diameter of the rivet that is on both the sides so here area which is resisting this load that is PI by 4 into d square so I can see that in case of shear stress the load is tangential to the cross-section area as we have seen here the load is parallel to this area so I can say either it is parallel or tangential now see shearing as I have seen this case it is of single shear I can write oh sure shearing may be single shear or double shear see if the rivet is breaking into two parts that becomes a case of double shear single shear if it is breaking into three parts then it becomes a case of double chien in that case we would be having four repairs of India and for that suppose I will draw the diagram for double shear here also we will be having number of plates here I have three plates say in this diagram we have three plates and the node which is acting for the topmost plate it is towards right and even for the bottom or straight it is towards right and for the middle plate the root is towards left now because of the action of these opposite forces there are chances of the rivet breaking first at this Junction and then it will break at this Junction so here we are getting one two and three three surfaces of the rivet that is it is divided into three parts and here we have two sets of areas so it is a case of double shear I hope shear stress is understood next is das consciousness the next kind of stress which we are seeing it is called as torsional shear stress for torsional shear stress what happens is that suppose we have a shaft which is rotating inside a bearing now here I have drawn he’s shaft and availing now this shaft would be subjected to rotation where is this bearing it is stationary now what may happen is that because of this rotation there are chances of the wheels – we can see separate or slip that is each layer is subjected to some kind of shear stress so what can happen is that suppose this first layer when the torque is acting this is the torque which causes the rotation so because of that there are chances to the first layer suppose it mislead like this and similarly the other layers they are also they would be subjected to stress because of that they would be skipping so that is for this torsional shear stress so you are let all the important points that torsional shear stress is seen in members which are rotating so torsional shear stress will be deadly those members which are rotating and because of that rotation there are chances of each of the vales for slipping so this is a case of torsional shear stress next after this we have crushing stress now in crushing stress what happens is that suppose for this also I draw a diagram like of a limit which we have seen in case of shear stress also now again we have two plates here I could draw a section for them now here there are two plates which are subjected to loading in opposite direction now suppose as we have seen in case of shear stress instead of the rivet ability then can be a chance well because of the load applied this limit it can just crush crush means it is like we can say compression it can crush here so first of all crushing stress it is a localized stress acting on a member so crushing stress it is a localized stress acting on a member irrespective of the time of loading so here I am written for crushing stress that it is a localized stress means it will be acting only on those members like for example which are pins prevents corner joints then crushing takes place more frequently so here I can say that crushing is also called as local compression and because of crushing what can happen is the material it cannot break but it will be crushed into fine powders like for example for a limit I can say that if I draw it separately here now this rivet it will not break but its surface will be converted into fine powders like this so this is an example of crushing like we have a concrete and if we apply compressive force over concrete and if that concrete it just converts into fine powders that is called as local machine so here the area which undergoes crushing is the projected area projected area means here the outer diameter and then I can see that if the diameter of the rivet it is T and here if the length of the rivet is n so therefore the area which is crushed that crushing area is capital e is equal to it would be PI D that is circumference multiplied by Lin so this was regarding thrashing stress next we have after crushing we have barristers now bearing stress it is somewhat similar to crushing stress but the difference is in case of crushing the members are stationary like they are pin that is caught and we can save in bolts nuts they are subjected to crushing but in case of burying the member one of the mobilization product or they are circular in cross section so I can give you the example like suppose we have a shock and over that if we have a wish like the diagram and right now I can see that this is a shop and here we have a bush now push is that member which acts as a medium to connect shaft with the bearing it means first we have a bearing inside that we will be inserting a bush and in that bush we will be inserting Nisha now what happens when this shaft is rotating that is when the torque we provide there are chances of this bush to get worn out because the material of the bush is mostly soft like example it can be rubber it can be brass gunmetal Bush or we can say bronze etc so these are some of the common materials for the bush whether its shaft it is mostly made up of steel which is harder as compared to the bush so there are chances that because of continuous rotation of the shaft this bush may get worn out and the failure which is it it is for this baling failure the stress which is developed in the bush that would be called asked bearing stress so I can see that bearing stress it is most frequently seen in those members which are continuously rotating like here I have given the example of a shark rotating in a bush next like crushing since it was localized stress crushing stress even bearing stress is applicable only to the area which is in contact with the bush and shock so therefore barring India capital a it will be fine D that is if I say D is the diameter of the shaft that the small D is the diameter of the shaft so PI D is the circumference over which this bush is in contact and if the length of this bush is L so PI D into L is the bearing area and therefore bearing stress will be it will be equal to bearing load which I can denote it by W upon bearing area which is a so this was regarding very stressed now we have the last kind of stress that is called as bending stress now bending stresses also simple for this suppose we have a beam which is simply supported at its two ends I can call them as a and B length of the beam is capital L now because of the action of the central load which is w this beam it can bend in such a way like this it can bend in this manner similarly if I have a cantilever beam that has a beam which is fixed at one end and which is free at the other end like this example again if a load is acting at the free end of this cantilever beam its length is capital L now if the load is acting this beam will be deflected downwards like this so here we can see that both the beam they are subjected to waiting and therefore the stress which is developed because of bending in the wheel that would be called as bending stress so here I can say that bending stress it is the stress developed in a member because of bending load next bending stress is Sigma B therefore bending stress it is denoted by Sigma B and it can be calculated it can be calculated by using like general formula and that general formula is nothing but M upon I is equal to Sigma P of 1y where m is the bending moment I is the moment of inertia of the material of the beam Sigma V is the bending stress in via the distance at which maximum bending others so here in this video we have seen various kinds of stresses we had started with enzyme stress then we had seen first property tensile stress that was completed next we had seen compressive stress then shear stress we have seen torsional shear stress crushing stress bearing stress and bending stress so these seven kinds of stresses are necessary to study the subject of design of machine elements without their knowledge it is very much difficult to analyze how the failure takes place what is the area and how much is the stress stored in the memory so I hope everything is clear in this video

54 thoughts on “Types of Stresses – Introduction to Design of Machine – Design of Machine”

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  7. Hello Sir, Am following classes regularly it was good and nice explanation… But my doubt is, can we say Tensile and Compressive stress are applicable for single part?

    As shown in video, you considered single part as a example for Tensile and Compressive stresses.
    and for shear stress Two parts with joints.

    Please let me know your opinion.

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  9. idon,t understand why the area of the rod is divide by 4. normal area should be pie r square, but i see you have divide it by 4 why?

  10. Built up column consist of three rolled steel beam section ISWD 450 @ 0.794 KN/m
    CONNECTED EFFECTIVELY TO ACT AS ONE COLUMN
    Length is 4.25m
    DETERMINE THE SAFE LOAD CARRYING CAPACITY OF SECTION

    Sir can you please have a video on sums related to steel sections like this…

  11. In the double shear when p and p acts on one side then the opposite side reaction will be 2p..but u have mentioned as p only.

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