 Hello friends, I am Prasanthi Shona Dhanshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Sulapur. Today, I am here to explain you about the share force and bending moment for a simply supported beam with point load at or UDL. The learning outcome of today's lecture is that at the end of this session, students will be able to understand the behavior of simply supported beam under different point loads or UDL. Then they can diagrammatically represent share force and bending moment of a simply supported beam. Now, here we will consider simply supported beam AB of length L with a point load at its center. So, as the load is symmetric, the reactions will be equal that is equals to total load divided by 2. Therefore, R A is equals to R B is equals to W by 2. Now, let us take a section xx at a distance of x from A. So, this will be section xx at distance of x from A and let fx be the share force at x and mx be the bending moment at x. Now, for the share force or bending moment, you have to consider either the left portion of the section or right portion of the section as we have seen in the earlier video. The resultant force acting on the left portion of the section is W by 2. So, here from this section in the left hand side, there is W by 2 which is acting upward and upward for the left of the section is treated as positive. So, fx is equals to W by 2. Now, the share force will be constant. Now, here we have to draw the share force above the baseline as it is positive value and now the share force is constant between these two loads that is the reaction R A and the load W. So, it is represented by horizontal straight line. Now, we will consider a section between C and B. Now, at this if we consider a section here, so there are two forces that is R A reaction and W. So, here fc, so you know that when there is a point load, there is a sudden change in the share force. So, here at this point fc is equals to W by 2 it is acting upward positive and W is acting downward that is minus W. So, it is equals to minus W by 2. So, here at C we are to draw minus W by 2 that is below the baseline and now again between C and B there is no load hence the share force is constant between C and B that is equals to minus W by 2 and we will connect this to the point B and this will be the share force diagram. At section C the share force changes from W by 2 to minus W by 2 as shown in the figure. Now, what will be the bending moment at A and B of a simply supported beam AB of length L? Here pause the video and try to write an answer on a paper. Now, it will be 0 as we know earlier that the bending moment at simply support and at the free end of the cantilever is always 0. Now, we will see the bending moment diagram for the beam. So, again reaction is there W by 2 load is W at the center. So, mx is equals to now taking a section xx at a distance of x from A again. So, the moment will be load into this perpendicular distance that is Ra into the x distance. So, here again the left of the section moment clockwise moment is positive or we can see that how the bending moment will occur. So, if the concavity is upward so, the bending will occur like this and the concavity will be at the top so, that is positive sign. So, therefore, now mx is equals to Ra into x that is positive Ra is W by 2 into x that is equation number 1. Then we will see the moment at various point at x is equals to 0. So, ma putting x value here we will get a moment at A that is W by 2 into 0 is 0 and at x is equals to L by 2. So, bending moment at C is equals to W by 2 will be as it is and instead of x we will place it as L by 2 so, it is equals to W L by 4. So, from the equation 1 it is clear that the bending moment varies according to the straight line law between A and C. So, the bending moment is 0 at A and it increases to W L by 4 at C. So, we have to draw the W L by 4 ordinate at C and connect this by a straight line. Now bending moment between a C and B so, again mx is equals to Ra into x now x will be the distance from A into this section the section is considered in CB portion. So, Ra into x minus so, this W into this distance it will be x minus L by 2 so, putting the value of Ra W by 2 into x minus W into bracket x minus L by 2 this is equation number 2 at x is equals to L by 2. So, putting the value of x is equals to L by 2 in this above equation 2 so, we will get mx so, mx or mc so, that is W by 2 into L by 2 minus W into L by 2 minus L by 2. So, this term will become 0 so, it is again W L by 4 so, the ordinate will be remain same as this is C dash and at x is equals to L putting the value of x is equals to L we will get moment at B. So, that W by 2 into L minus W into bracket L minus L by 2 so, it is equals to 0. So, the bending moment at C is W L by 4 and it decreases to 0 at B so, we will draw a straight line as per the equation of bending moment it will follow a straight line law. Now, the bending moment is maximum at the middle point C where shear force changes the sign we know that the shear force has changed the sign at C. Now, we will consider a next example which is having UDL throughout its length so, again a beam AB is there of span L with UDL over its entire length of intensity W per unit length. So, here again the load is symmetric hence the reactions are equal that is equals to total load divided by 2. Now, your total load will be so, W is for 1 meter length so, total load will be W into L. So, dividing it by 2 so, R A is equals to R B is equals to W L by 2. Now, again we will take a section x at a distance of x from and A between A and C so, this we will take a section. Now, fx will be the shear force at the section and mx will be bending moment at the section. Now, again considering the left portion of the section so, in the left portion of the section we are having 2 loads that is reaction which is acting upward and the UDL. So, again considering the left of the section upward as positive and downward as negative. The resulting force acting on the left portion of the section is R A minus W into x because this intensity is W and the length is x so, it is Wx. So, fx is equals to W L by 2 that is reaction minus W into x that is equation number 3 so, which shows that it again follows a straight line law. So, according to equation number 3 shear force varies to the straight line law so, we will find out the shear force at various point when x is equals to 0 putting the value of x is equals to 0 in this equation number 3 we will get shear force at A. So, fA is equals to W L by 2 minus W into 0 that is W L by 2 as it is having a positive value we are drawing this W L by 2 above the base line. Now, at x is equals to L, fB is equals to W L by 2 minus W into L that is minus W L by 2 so, this again here it is drawn below the base line and at x is equals to L by 2 so, fC is equals to W L by 2 minus W into L by 2 is equals to 0. So, now we have to join this all these 3 points that is W L by 2 here at A 0 at C and again minus W L by 2 at B. Now, we will find out the bending moment so, again considering the same section that is at a distance of x from A so, this clockwise will be positive anticlockwise will be negative. So, mx is equals to so, rA into x minus W into x and it is acting at x by 2 that is the center of this portion. So, putting the value of rA we will get this equation number 4 and seeing the equation now, it is having x square term so, the diagram or the bending moment varies according to the parabolic law. Now, bending moment at various points we can find out by putting the value of x at x is equals to 0 bending moment at A will be W L by 2 into 0 it is 0 then when at B at x is equals to L bending moment at B is W L by 2 into L minus W by 2 into L square that is again 0. Now, at C at x is equals to L by 2 bending moment at C is W L by 2 into L by 2 minus W by 2 into L by 2 bracket square so, solving this we will get mc is equals to W L square by 8. So, the bending moment increases according to the parabolic law from 0 at A to W L square by 8 at C and again it is 0 at B. So, these are my references which I have referred thank you thank you very much for watching my video.