 Hello friends, I am Prashant Vishwanath Dinshati, Assistant Professor, Department of Civil Engineering from Walchand Institute of Technology, Singapore. So, today I am here to explain you about the share force and bending movement for a simply supported beam carrying inclined load and kappa. So, the learning outcome of today's lecture is at the end of this session student will be able to understand the behavior of simply supported beam under inclined load. Then they can diagrammatically represent the share force and bending movement of a simply supported beam with inclined load and kappa. Now, share force and bending movement for a simply supported beam carrying inclined load. So, when a beam carries inclined load then these inclined loads are resolved in vertical and horizontal components. The vertical components will cause share force and bending movement whereas the horizontal components of the inclined loads will introduce axial force or thrust in the beam. The variation of axial force for all the section of the beam can be shown by a diagram known as thrust diagram or axial force diagram. Now, in the most of the cases one end of the beam is hinged and other end is supported on rollers. The rollers support cannot provide any horizontal reaction hence only the hinge end will provide horizontal reaction. So, the roller loads will allow to move the beam horizontally so it will not provide horizontal reaction. So we will consider a example of horizontal beam AB of length 4 meter is hinged at A and supported on rollers at B. So, it is hinged at A and supported at roller on B. The beam carries inclined loads of 100 newton, 200 newton and 300 newton inclined at 60 degree, 45 degree and 30 degree to the horizontal as shown in figure. So we have to draw the shear force and bending movement diagram for this beam. So what is our first step? So first step is to resolve this inclined load into vertical component and horizontal component. So the point load at the position and here these are the vertical component and horizontal component. So at point C so this is 100 newton acting at angle of 60 degree with the horizontal. So the vertical component here it will be 100 sin 60. So 100 sin 60 will be 86.6 newton will be the vertical load. And horizontal load it will be acting towards the left end. So it will be 100 cos 60 that is 50 newton. So here the horizontal load is acting horizontally towards left. At point D the vertical load will be 200 sin 45 that is 141.4 newton. So this will be the vertical load and horizontal load will be 200 cos 45 that is 141.4 newton. Then at point E again resolving is this 300 sin 30 it is 150 newton it is acting downward and towards the left again it is 300 cos 30 that is 259.8 newton. Now a beam is supported on a ruler at B. Hence will not provide any horizontal reaction. The horizontal reaction will be provided at hinged end that is A. So the horizontal reaction HA. So this HA will be resisting force so it will be acting towards right. And this all forces are acting towards left. So this will be the summation of all horizontal components of a inclined load. So 50 plus 141.4 plus 259.8 it will be 451.2 newton. So this RA this will be the reaction 451.8. And to find the reaction RA and BB RB that is vertical reactions. So taking moment about A of all the forces. So here horizontal forces will pass through the same point it will be 0. So therefore RB into 4 this is anticlockwise is equals to and all these three forces will have a clockwise movement. So first one will be 86.6 into the distance is 1 meter. Then the second load that is 141.4 into distance will be 2. Then 150 into 3. So solving this we will get RB is equals to 204.85 newton. So 204.85 newton will be RB. Now RA is equals to total vertical downward force minus vertical force. So the total vertical downward force minus vertical force that is so these three are the vertical downward force that is 86.6 plus 141.4 plus 150 minus this upward force is 204.85 that is RB. So therefore RA will be 173.15. Now let fx is the shear force at x and mx is the bending moment at x. So shear force due to the vertical loads only including reaction therefore shear force at A. So we will consider a section x at a distance of x from A. So this vertical shear force will be RA and it is acting upward so it is plus 173.15. And this shear force remains constant between A and C as there is no another load. And the shear force suddenly changes its sign at C due to this vertical load. So the shear force at C is plus 173.15 minus 86.6 that is this it is acting downward. So this will be 86.55. So here you can see this is suddenly downward. Then the shear force again remains constant between C and D so this is a horizontal line. Then again shear force at D now again due to point load there will be sudden change. So if we consider all these forces what we will get plus 173.15 minus 86.6 and again minus 141.4. So it will be at D minus 54.85 Newton. So this is minus 54 point so it is below the baseline. So there is a sudden change. Now again this will be constant between D and E as there is no load and at E the shear force at E is again you considering all the forces on the left hand side. So plus 173.15 minus 86.6 minus 141.4 minus 150 it will be minus 204.85. So this is 204.85 so it is the sudden change and it will be constant up to B and it will be again 0 at this point. So this is the shear force diagram. Now let Mx is the bending moment at any section X and it is the distance X will be the distance from A. So the bending moment at A it will be 0 because we know at simply support the bending moment is 0. Now bending moment at C so this bending moment considering the left section only one force is there and it is having a clockwise rotation. So bending moment at C is Ra into 1 that is 173.15 into 1 that is 173.15. So we have to plot this 173.15 on the top side of the baseline and we will connect this with the straight line. Then bending moment at D point so at D point we are at two vertical loads. So one will have clockwise movement other will have anticlockwise movement. So this Ra into distance between this is 2 Ra into 2 minus 86.6 into 1 is equals 2. So it will we will get here 259.7 Newton meter. So the 259.7 Newton meter so here it will be 259.7 Newton meter. Then bending moment at E so bending moment at E again we will consider these three forces so Ra into 3 minus 86.6 into 2 now this will be the distance 2 and this will be having distance 1 that is 141.4 into 1. So solving this we will get here again the bending moment 204.85 Newton meter 204.85 Newton meter here. So if we join all this we will get this bending moment diagram. The diagram due to horizontal component inclined horizontal reaction beam is known as here pause the video and try to write answer on a paper. So it is the thrust or axial force diagram. Now for the thrust diagram or axial force diagram the thrust diagram is due to the horizontal component including horizontal reaction. So the axial force at A is equals to plus HA so this is HA so we will consider right side as positive so here it will be plus that is 451.2 the axial force remain constant between A and C and at C it is HA minus 50 so you will get here 401 so here it will be downward again 401.2 Newton the shear force at D HA is again HA minus 50 minus 141.4 so here we will get 259.8 so here it will be 259.8 the axial force remain constant between D and E so then again axial force at E is equals to HA minus 50 minus 141.4 minus 259.8 so this will be 0 and this is constant between E and B so this is the thrust diagram. So these are the references which I have referred thank you thank you very much for watching my video.