 In last class, I have discussed about various foundation model, basically two parameter models and how the limitations of Winklam model can be removed by using two parameter model and then using non-linear soil parameters. Now, I have also discussed about the elastic continuum approach and elastic continuum model and then isotropic, elastic continuum, isotropic, elastic half pace and then anisotropic elastic continuum thing. Now, today I will discuss about beams on elastic foundation. Now, if we place a beam generally footing is can be idealized as a beam for the analysis purpose and then with the application of different types of load, different types of beams, some it is finite beam or infinite beam. Now, if we place beam on a soil medium, then what will be the response of that beam, how we can solve this, that beam equation. So, that thing will explain today. Now, if I plot that beams on elastic foundation, the beams of elastic foundation, suppose if this is the z axis or y axis, this is x axis and if we place a beam here and then we can apply different types of loading. Suppose, this is a u d l, we can put a concentrated load p, we can put a varying u d l here and then this is the position of the beam before deflection and then after that the beam will deflect and this will be the, again this is the concentration load, concentrated load p and then the loading which is acting here and this is the deflection line and here base of the beam, there will be the reaction of the soil. So, this would be the soil reaction, where we can analyze the soil reaction in this form. So, this will be the soil reaction and these are the applied load and this will be the deformation shape. So, this is this form is before deformation and this is the after deformation. Now, reaction force to be assumed to be acting vertically and opposing the deflection of the beam. So, that means this reaction force that is assumed to be acting, this is acting vertically and it is opposing the deflection of the beam. Hence, we have the deflection in directed downward direction. So, that means here the deflection in downward direction is taken as positive and deflection in the upward direction is taken as negative. So, although it is assumed that soil cannot take the tension, but here it is for the present analysis, it assume that it can take for the analysis purpose, but that is as treated as a heaving of the foundation soil. So, that means basically the direction or the deflection in the downward direction is taken as the positive deflection. Now, if we and this is the reaction, now if we take one beam, this is the beam and which is subjected to U D L Q or P and reaction or we can say this is Q reaction is P. So, that means this beam is placed on the ground and this is the thickness of the beam. So, now if we take one small component of this beam, suppose if I take this small element and then if we draw the various forces and moment on the beam. So, this is acting in the this thickness is taken as this small element which with of D X and whose Q is acting here. So, that means force this will be Q into D X. Now, these are the reaction P is the reaction force that is P into D X. So, this is reaction force or soil reaction then that Q the shear force acting in the left side is Q and one is one moment acting in the clockwise direction is M in the left side. Similarly, the moment will act in the anticlockwise direction in the right side that is M plus D M and Q will act another shear force that is Q plus D Q that will act in the downward. So, these are the forces and moments for the small segment of D X. So, this Q is the applied load P is the reaction soil reaction M is the moment capital Q is the shear force in the left side. Now, the sign convention that is used that upward acting shear force. So, here Q to the left cross section is considered as positive. So, the upward acting for shear force in the left side is considered as positive and the corresponding moment and the corresponding bending moment M in clockwise direction acting from left side on the element is taken as positive bending moment. So, we have the sign convention for the one for the deflection one for the shear force and one for the moment. So, for the in this analysis deflection in the downward direction in the downward direction is positive. Now, shear force acting upward direction in the left side is positive corresponding bending moment acting clockwise direction for on the left side of on the element is considered as positive. So, these are the sign convention where the clockwise moment in the left side is positive shear force upward direction left side positive and deflection downward direction is positive. So, next one that how to solve this beam expression. Suppose again if we draw the same element of d x this is the Q and here if I draw the this is x axis coordinate system this is y axis and then shear force in downward direction right side Q plus d Q and the moment this side M plus d M the force that will act is Q into d x and the reaction that will act P into d x. So, this is the force acting on the beam this is the soil reaction which is given on the beam and this shear force left side and this moment. Now, if I write the expression that is you can write Q minus Q plus d Q all the vertical forces. So, summation of vertical force that is 0. So, Q minus Q plus d Q plus d Q plus d Q plus d Q plus d Q plus d Q so upward direction that is plus P into d x minus Q into d x that is equal to 0. Now, here the soil P is soil reaction. So, P small p is soil reaction in early case now if this beam suppose this beam is placed on a soil where the soils are idealized as spring. So, this is spring with stiffness K or K is the modulus of sub grade reaction. So, on this beam Q is acting that is the UDL. So, these soils are giving the reaction and the soil this soil is giving the reaction and soil is idealized by spring. So, you can write that Q P will be equal to K into Y where K is the spring constant or modulus of sub grade reaction. K is modulus of sub grade reaction Y is the deformation. So, that is the principle of Winkler model that is any force which is proportional to the deflection and the K is the spring constant. So, finally, if I put this P in the final expression that will be plus K Y into d x minus Q into d x that is equal to 0. So, finally, so Q will be cancel out. So, d Q plus K Y d x minus Q d x equal to 0 where Q is the applied pressure and here it is a analysis. So, that means we can write the d Q by d x that will be equal to Q minus K Y. So, this is the reaction now if we do not have any applied pressure soil applied pressure load that means if Q is equal to 0 that means there is no force that means this Q is applied is 0 then the expression will be d Q by d x minus K into Y. Y is the deformation or deflection and K is the modulus of sub grade reaction. So, we have that expression final expression of the beam that d Q by d x equal to so that means here Q is the shear force M is the bending moment Q is the shear force M is the bending moment Y is the deflection. Now, finally, next step then how to calculate the other part of the when we will get the final expression of the beam then how we will calculate the other reaction or other forces that means from that beam how we will calculate the slope bending moment beam expression. So, that means final expression d Q by d x equal to Q minus K into Y. So, that means we will get this expression and then take the final expression that our Q is d M into d x. So, Q is the rate of moment that means d M by d x. So, we can write d Q by d x equal to that is equal to d square M by d x square. So, we will put this expression in the final equation that we will get that d Q by d x that is equal to K Y minus Q or we can take that is because here this negative term is there. So, we will get d K Q finally. So, that means finally from here we will get d Q by d x K Y into minus Q. So, we will replace this expression by d square M by d x square equal to K Y minus Q. So, in the previous expression if we get that expression so here there is a negative term. So, we can write this is minus Q plus K 1. So, this will be plus K 1. So, d Q by d x is K Y minus Q this will be the final expression of the because there is a negative term is there. So, we can write in this form. So, if there is no Q so there will be K into Y. So, finally from this K into this Y minus Q we put d square M d x square and again we know that E I d square Y d x square is equal to minus M where E I is the flexural rigidity rigidity of the beam. So, again we know this expression. So, finally for d square M by d x square we can if we differentiate this equation twice then we will get E I d to the power 4 Y d x to the power 4 that is equal to minus d square M d to the power d x square. Now, if you put this expression into here then we will get that E I d to the power 4 d x to the power 4 equal to minus K Y plus Q. So, again if Q is equal to 0 then there will be E I d to the power 4 d x to the power 4 will be minus K into Y. So, now we have that expression that E I d to the power 4 Y d x to the power 4 minus K Y plus Q. Now, we have to solve this expression for the different condition. Now, once we if we put take Y is equal to E to the power M x then we have the expression E I d to the power 4 Y d x to the power 4 equal to minus K into Y if Q is equal to 0. So, now if we differentiate this Y 4 times then we will get this d to the power 4 Y d x 4 that gives you E I d x to the power 4 minus M to the power 4 E M x. So, now if we put this expression here E I into M to the power 4 E to the power M x that is equal to K into Y and here Y is also E to the power M x. So, this E to the power E to the power M x will be cancel out. So, M to the power 4 is equal to minus K by E I. Now, this M which has two roots basically this M which has this 4 roots that is 1 is M 1 it is equal to minus M 3 that is equal to K 4 E I 1 plus this is lambda 1 plus I. Another one M 2 is minus M 4 that is equal to 4 K 4 E I minus 1 E I minus 1 E I plus I. So, that is equal to lambda minus 1 plus I where lambda is 4 K 4 to the power E I 4 root by K by 4 E I. So, in general solution of this equation solution will be Y equal to M 1 E to the power M 1 x plus A 2 E to the power M 2 x plus A 3 E to the power M 3 x plus A 4 E to the power M 4 x. So, Y A 1 E to the power M 1 y is not equal to this. So, M x, a 2 e to the power m 2 x, a 3 e to the power m 3 x and a 4 e to the power m 4 x. Finally, so we will get this expression, now using e to the power i lambda x is equal to cos lambda x plus i sin lambda x and e to the power minus i lambda x, this is equal to cos lambda x minus i sin lambda x. So, again if we take that a 1 plus a 4 is equal to c 1 and i a 1 minus a 4 is equal to c 2, similarly a 2 plus a 3 that is equal to c 3 i a 2 minus a 3 is equal to c 4. Then the final expression, general equation will be y is equal to e to the power lambda x c 1 cos lambda x plus c 2 sin lambda x plus e to the power e to the power minus lambda x c 3 cos lambda x plus c 4 sin lambda x. So, this will give us the final expression of this beam where lambda is given for root k 4 e i. So, this lambda is equal to c 1 c 2 c 3 and c 4. Now, we have to determine next step is that we have to determine these 4 constants. Then how we will determine these 4 constants? This is the final expression of the beam and then how we will determine these 4 constants of this expression that will depend the to use the boundary condition to solve these 4 constants. Now, before we solve these 4 constants, then we have another value that how to determine the bending moment slope from this expression, because we have this expression final expression this is the settlement expression that e to the power lambda x into c 1 cos lambda x plus c 2 sin lambda x plus e to the power minus lambda x into cos c 3 cos lambda x plus c 4 sin lambda x. Now, the slope expression will be 1 by lambda d y t x that is equal to e to the power lambda x into c 1 cos lambda x minus c 2 sin lambda x. So, sin lambda x plus c 2 cos lambda x plus sin lambda x minus e to the power minus lambda x into c 3 cos lambda x plus sin lambda x minus c 4 cos lambda x minus c 4 cos lambda x minus sin lambda x. So, this is just differentiate this expression this once we will get the slope expression. So, I differentiate it twice we will get the bending moment expression in the shear force expression. So, this is del square y del x square that is equal to minus e to the power lambda x and differentiate this expression once and then rearrange that you will get the equation second equation. This is the one equation this is b equation and then the differentiate this expression again b expression again and rearrange it will get the third expression that is c 1 sin lambda x minus c 2 sin lambda x minus c 2 sin lambda x cos lambda x plus e to the power lambda x c 3 sin lambda x minus c 4 cos lambda x. So, this will give the moment expression then once again differentiate that c expression will get 2 lambda square sorry lambda cube. So, initially this is lambda square. So, this again expression 1 by 2 lambda square this is 1 by 2 lambda cube then d 3 y d x 3 that will equal to e to the power minus e to the power lambda x c 1 cos lambda x plus sin cos lambda x plus sin lambda x minus c 2 cos lambda x minus sin lambda x plus e to the power minus lambda x c 3 cos lambda x minus sin lambda x plus c 4 cos lambda x plus c 1 sin lambda x plus sin lambda x. This is the shear force expression D. So, here we can write d y d x is equal to tan theta, which is the slope. Then the next one E i d 2 square y d x 2 d 2 square y d x 2 d 2 y d x 2 is equal to m minus E i d 2 y d x 2 is equal to m and minus E i d 3 y d x 3 that is equal to q. So, that is shear force. So, these are the shear force bending moment and slope expression. This B is A is the deflection expression, B is the slope expression, C is the bending moment expression and D is the shear force expression. So, 3 next one is how to determine this C 1, C 2 and C 3 and C 4 value. So, D is the slope expression depending upon the using the boundary condition of the beam. Then how we will calculate this unknown factor that we will determine here. So, suppose if I place beam with a beam infinite beam subjected to concentrated load. Suppose you have a beam, this is x, this is the y direction, this is the center where beam is subjected to a concentrated load P. So, this is the concentrated load, this is the infinite beam. Now, the deflection equation that we have here, here analyze only the half portion of the beam as this is a symmetric case. So, the final deflection equation of the beam say y is equal to e to the power lambda x C 1 cos lambda x plus C 2 sin lambda x plus e to the power minus lambda x C 3 cos lambda x plus C 4 sin lambda x. So, this is the deflection equation. Now, you have to solve this equation for different condition. So, now if this is the infinite beam. So, if x tends to infinity then y will be equal to 0, because here we are applying this concentrated load at 0 0 position. Now, if it is the infinite beam x is infinity then the effect of this concentrated load will be negligible. So, there is no effect. So, that means y will be 0 and this is possible only if the terms related to e to the power lambda x is 0. So, if e to the power lambda x and we just terms which are attached with e to the power lambda x is 0 then it is only possible. So, in that condition you can write that and this is again it is possible if C 1 and C 2 both are 0. So, you can write C 1 is equal to C 2 is equal to 0 then only you will get this condition. So, the expression will be e to the power minus lambda x C 3 cos lambda x plus C 4 sin lambda x. So, we have e to the power minus lambda x and then this expression is further reduced to this expression. Now, the next condition so now we have to apply the another boundary condition or the next condition that is that will be at our slope at this center is 0. That means d y d x at x equal to 0 is 0 because it is a symmetric problem. So, the slope at the center will be 0. Now, if we differentiate this expression 2 so therefore, this is expression 1 this is expression 2 once we differentiate this expression 2 and then put this condition the slope is 0 at x equal to 0 then you will get this form of expression that minus C 3 minus C 4 is equal to 0. So, then you will get another condition that is C 3 is equal to C 4 is equal to C. So, from one first conditions infinite beam then you put this condition then it is only possible if C 1 is equal to C 2 and the next condition the slope at the center of the beam is 0 by putting that condition will get that C 3 is equal to C 4 is equal to C. So, the equation 2 will be further reduced to e to the power y C e to the power minus lambda x cos lambda x plus sin lambda x. So, that is expression 3 because C 3 is equal to C 4 is equal to C. Now, if I write the expression 3 again that that is y is equal to C e to the power minus lambda x cos lambda x plus sin lambda x. Now, from the equilibrium of the forces that is the sum of the reaction forces next third condition the first condition is infinite beam second condition is slope due to the symmetric slope at the center is 0. The third condition the sum of the reaction forces will be forces will keep equilibrium with the total load. So, that means the total sum of the reaction that will be equal to the force external force which is applied that is p p. So, that means this thing will be in the equilibrium condition when only that the sum of the reaction force will be equal to the applied force the p. So, in that condition this will be twice 0 to infinity k into y into dx that will be equal to p. So, twice means the we are analyzing the only one half. So, this is the two halves or we can write that is 0 to infinity k y dx that is equal to half of the total external load because we are analysis on the half portion. This is 0 to infinity and the soil reaction is k into y. So, that k into y term that is the soil reaction. So, finally, once we use once I this is solved then if I put this expression that is 2 k will be in the outside and 0 to infinity y is c e to the power minus lambda x cos lambda x. Plus sin lambda x into dx that is equal to. So, if I take this expression left side left hand side that is 2 k will be in the outside. Then if I put the y value here that is expression 3 here that is c e to the power lambda x. And then if I take that c e to the power c also outside that is 2 k c 0 to infinity e to the power lambda x cos lambda x plus sin lambda x that is dx. Then we will get this form of expression that is 2 k c into 1 by lambda. So, we will get this expression after solving this after integrating this expression equation. So, we will get the expression that 2 k c 1 by lambda that is equal to p because from this expression that is equal to p. So, finally, the third or the final constant that is c will be equal to p lambda divided by 2 k, where k is the modulus of sub grade reaction p is the concentrated load and l is the fracture rigidity of the beam. So, now we will put the final expression will be y is equal to p lambda 2 k e to the power minus lambda x cos lambda x plus sin lambda x and that is for x equal to greater than 0. So, this will give us the final expression of the deformation in this form. So, that means we will get the final expression y is equal to p lambda 2 k e to the power minus lambda x cos delta x plus sin delta x this is for the deflection equation. So, for the slope expression dy dx which is equal to theta that is equal to minus p del square by k e to the power minus del k del x into sin del x. So, slope will be in this form. So, next expression that is the bending moment expression and the shear force expression, because there will be bending moment expression and the shear force expression for these two cases. So, next we will put the final expression y is equal to p lambda 2 k e to the power minus lambda force expression that minus E i d to the d square y dx square that is equal to m that is equal to p by 4 lambda e to the power minus lambda x into cos lambda x minus sin lambda x. So, we have this expression into sin minus lambda x. Then the further if I differentiate this expression then we will get the shear force expression d 3 y dx 3 that is equal to q. So, this will be minus p by 2 e to the power minus lambda x cos lambda x. So, this is expression 1 that is for the deformation this is for the slope this is for the bending moment 4 equation number that is for shear force. Now, if I use some different symbol for this case first one if I use that e to the power minus lambda x and cos lambda x plus sin lambda x is equal to a lambda x and e to the power minus lambda x into sin lambda x that is equal to p lambda x. Now, e to the power minus lambda x cos lambda x minus sin lambda x equal to c times e to the power minus lambda x cos d lambda x and e to the power minus lambda x cos lambda x is equal to d lambda x. Then these are the symbols then all the expression will change in this form that y will be equal to p by lambda into 2 k divided by 2 k into a lambda x theta e to the power minus lambda will be equal to minus p lambda square divided by k into b lambda x m will be equal to p by 4 lambda c lambda x and q will be equal to minus p by 2 d lambda x. So, these are the expression of these 4 quantities that is deformation, slope, bending moment and shear force. Now you have to draw the shape of these 4 quantities then how this can this shape we can draw so that we can identify here. So, first we will get that y is equal to p lambda 2 k a lambda x theta is equal to minus p lambda square by k b lambda x bending moment m is equal to p by 4 lambda c lambda x and shear force q is minus p by 2 d lambda x. So, these are the 4 expression. Now, first one that we will get suppose this is the beam and the concentrated load is applied here this is x direction and here it is y direction this is 0 0. Now first case this will be deformation. So, that expression of this curve is y is equal to p by lambda 2 k into a lambda x because this is the expression of that form. Now here the distance from this point to this point where the this is 0 is 3 by 4 into pi by lambda. Next one that is for the slope expression this is a straight line then this will be the slope and slope is 0 at the centre it is a boundary condition. So, this is the slope expression so this is the slope expression. So, this is positive this is negative this is negative this one is positive this is negative. So, this expression where this is touching that point actually it is in this form here also. So, this distance where from here to here this distance is pi by 4. So, this will be minus theta is equal to minus p lambda square by k b lambda x. Next one is the bending moment which we can draw in this form. We will start here then it will go down then it will go up and again 0 because the centre bending moment is maximum. So, we will have the bending moment this is negative negative positive. So, this expression m is equal to p by 4 lambda c lambda x and this expression will form here to here that up to the end where it is again this is the negative part is pi by 4 and this part is one fourth by pi by 4. Next one is the shear force where at the centre then we have this expression this type of variation then it will go down and then it will follow this path. So, q will be equal to minus p by 2 d lambda and this will be x. So, this is positive this is negative. So, there is this part is given by that is this part where this value is given by half pi by 4. So, we have this is the four diagrams or the this is the deformation diagram this is 3 by 4. This is pi by lambda and this is sorry this is not pi by 4 this is all a pi by lambda here also this is pi by lambda this is pi by lambda. So, you have this is this distance where the deformation is the changing from positive to negative that distance from the centre is 3 4 pi by lambda and for the slope here the slope centre is 0 and from centre to another point where it is slope is 0 that distance is pi by lambda. Then bending moment that will also follow this expression which is derived where this is where the point is going positive to negative that is 1 by 4 pi by lambda then from negative to the 0 point that distance is pi by lambda. And then again from this point where this shear force is also changing sign that is half pi by lambda. So, these are the different diagram of the one case that I have discussed that is for the infinite beam subjected to concentrated load. In the next class I will discuss very other loading condition for the infinite beam and as well as if possible for the finite beam also. Thank you.