 Today we will see plastic analysis of a simply supported beam subjected to a point load. Now the learning outcome of this session is the students will be able to understand the concept of plastic collapse that is for simply supported beam subjected to point load. This is a simply supported beam subjected to one point load at the center that is L by 2 from this end and L by 2 from this end. Figure a represent the loading on the beam figure b represent the stress variation due to step by step loading. This is the cross section of the beam b is the width and d is the depth. This is the stress variation within the elastic limit. This is at the at elastic limit the stress strain variation. This is above the elastic limit the stress strain variation. This is at the plastic limit the stress strain variation. Don't focus on this figure just now we will see afterwards. Here the same beam is represented now at the center of the plastic moment will develop which is equal to Me. This is the length of plastic hinge and this is Me. We will see all these in detail further in the further slides. This is the details of the figure. So, the stress diagram 1 it will show the elastic stress distribution for working or the service load. Basically here the maximum bending stress which is Fb at the extreme fiber is allowed to reach the permissible stress F permissible stress which is equal to 0.66 Fy. So, this Fb is less than 0.66 Fy or we can also say that Fb is less than Fy. This is known as the elastic stress distribution and this second diagram represents the limit of elasticity at this stage Fb is equal to Fy and this figure is known as the limit of elasticity. Now elastic moment of resistance at this stage at the second stage by the Euler Bernoulli equation Me is equal to Fy into I by y. Now I by y is nothing but our section modulus which is equal to Ze I is equal to Bd cube by 12 Bd cube by 12 and y is equal to D by 2. So, Ze is equal to Bd square by 6 Ze is the elastic section modulus as per Euler's Bernoulli equation. If we want to prove all the stuff we have discussed in the previous slide. So, we can prove it here the total compressive force is equal to C and total tensile force is equal to T. First of all we will take C force is equal to average stress multiplied by area in compression. The compression is this upper triangular portion. So, the stress multiplied by the area of upper triangular portion. So, stress is nothing but our average stress. Average stress is calculated by this here is the stress is equal to 0 here is the stress is equal to Fy. So, 0 plus Fy divided by 2 therefore it is equal to Fy by 2. Now area is equal to B multiplied by D by 2. So, B multiplied by D by 2 which is equal to Fy into Bd by 4. This is a total compressive force in a similar way we can calculate the total tensile force which is equal to average stress multiplied by area B into D by 2 due to symmetry. And this will act at a distance of 1 by 3 from D by 2 from top as well as from bottom. So, from 1 by 3 from top end as well as 1 by 3 from the bottom end it will act at that distance. And for equilibrium we will equate C equal to T and we will calculate the lever arm is equal to 2 by 3 of D and elastic moment of resistance which is equal to C into A or T into A. Moment is equal to force multiplied by distance this is this will form a couple C into A or T into A because C is equal to T we have taken C into A or T into A. Just now we have calculated C which is equal to Fy into Bd by 4. Here we have calculated A which is equal to 2 by 3 of D. So, we will multiply with A we will get Fy into Bd square by 6 Fy which is equal to Fy into Ze. Now we will go a step further. Further increase in load does not produce increase in stress Fy here the ultimate stress is Fy only it will not get increased. As per it remains constant as per idealized stress strain graph. But the strain deformation increases the stress in the fiber immediately below the extreme fiber now reaches the value Fy. So, this is the extreme fiber the stress in the fiber which is present below the extreme fiber it will reach the stress value of Fy that is yielding penetrates inside the cross section. So, the yielding will penetrates inside this section this stage this complete stage is known as the elasto-plastic condition as some extreme area of the section undergoes in plastic condition while the central part remains the elastic this is the central part it this central part will remains in the elastic condition. Now further will increase the load and yielding will penetrates inside the cross section. So, this figure is known as fully plastic section and the movement develop at this section is called as fully plastic movement which is equal to C into A or T into A C is the compressive force T is the tensile force C is equal to average stress for rectangle Fy plus Fy divided by 2 which is equal to Fy multiplied by area the area which is present in this section above this axis. So, B into D by 2 now this lever arm can be calculated the total depth we know that is D upper depth is D by 4 and lower depth is D by 4. So, D minus D by 4 minus D by 4. So, this will equal to D by 2 which is the lever arm now plastic movement will be equal to average stress multiplied by area which is equal to force multiplied by lever arm will get the movement equating this Fy into BD square by 4 and this BD square by 4 is equal to Zp which is known as plastic section modulus. Now we will see in detail this figure this section EE is called as the limiting case or it is a case of fully elastic range this is ME is the elastic moment. Section DD is the elasto plastic stage some part is in plastic and some part is in elastic stage the central part will be in elastic and extreme fibers will be in plastic stage and section DD which is directly below the load for simply supported beam subjected to point load it is a case of fully plastic stage. Now we will see some theory so at section CC now when yielding of a section takes place over the entire depth section CC behaves similar to that of mechanical hinge however there is a difference in mechanical hinge and hinge developed due to plasticity of section which is commonly known as plastic hinge. Now at mechanical hinge movement will be equal to zero but while in case of plastic hinge movement will be equal to MP excessive rotation of cross section starts at the plastic hinge now the part AB and CB will no longer act as a continuous continuously as one beam that is ACB and excessive rotation at C due to the formation of plastic hinge at C makes a beam a mechanism and at mechanism the load the beam will not carry any load and collapse will occur. The hinge formation at CC takes place when the central load W is reached the value WP that is the plastic collapse load bending moment MP is equal to WL by 4 plastic moment of resistance developed at CC which is equal to MP which is equal to FI into ZB just now we have seen which is equal to FI into BD square by 4 here we have here we will see this zone AB and CB it will not behave as a single beam this part will behave somewhat different it will behave in a straight line as a straight line and it will behave as a straight line the part CB next is plastic collapse of the beam so WP that is the collapse load equal to 4 MP by L will calculate the value of ME which is equal to WE L by 4 which is equal to FI into ZD which is equal to FI into BD square by 6 we will take a ratio of plastic movement to that of elastic movement which comes out to be 1.5 that is equal to shape factor S what is mean by that 1.5 so at the plastic stage rectangular section has 50 percent more strength in flexure as compared to that of elastic stage 50 percent more strength that is 1.5 and this ratio is known as the shape factor so in the discussion of plastic collapse of simply supported beam under central load WP we have seen that in the progressive loading plastic hinge is developed at C we have just now seen that at the section CC plastic hinge will get developed here the maximum bending moment will occur as there is only one section CC of maximum bending moment possible formation of plastic hinge is only at CC so hinge will form only at CC it will not form on any other location and simply supported beam it is a determinate beam and the static degree of redundancy will be equal to zero so the number of hinges required for plastic collapse by mechanism is equal to r plus one r equal to zero therefore one only one hinge will form for collapse now during collapse process as soon as the plastic hinge is developed by the load WP flexural deformation of the part AC and CB vanishes and they become a straight line this is due to typical mechanism characteristics the mechanism cannot resist any external load and does not develop any deformation so it will not resist any external load the mechanism so we will have some questions just now whatever we have seen the shape factor for the rectangular cross section is just now we have seen a rectangular cross section of a simply supported beam the shape factor will be equal to you can pause the slide pause the video and answer the questions so the answer for this question is the shape factor for rectangular cross section is 1.5 and the second question is in the plastic analysis of the beam which of the following assumptions are made the plane section under bending remain plane at all the stages of bending stress strain relation is bilinear consisting of two straight lines shear deformations are neglected select the correct options from this below and the answer to this question is all the three are correct these are my references thank you