After finding n moments for all the segments, the next step is to find n shears for all the segments and note that the sign convention for n moments is positive for clockwise and negative for anticlockwise. So, I will draw free body diagram of all the segments one by one and find n shears for them. For segment AB at end A, n moment is 62.475. So, that is clockwise n moment and n moment at end B of segment AB is 125.25 that is also positive.
So that will also be clockwise. Let's say both the end shears are acting in upward direction. Let me call this Ra and this RB minus your note here I'm calling only Ra because I don't have anything on the left of A but there will be another segment at the right of B and the reaction for that I will denote by Rb plus.
So sigma m about A, this is end A and this is end B. So sigma m about A equals to 0 with clockwise is positive will give me the 62.475 is clockwise. So that will be positive.
Then this 125.25. is also clockwise so that will be positive then moment because of rb minus will be rb minus into lever arm that is 12 meters and moment because of rb minus about a will be in anti-clockwise sense so that will be negative rb minus into 12 equals to 0 so i can solve this to get rb minus equals to fifth 15.64 kN. Next, I will write sigma fy equals to 0 with upward direction is positive. So, this gives me Ra positive plus Rb minus and because value of Rb minus is 15.64. So, I will write that is 15.64 equals to 0. So, Ra will be minus 15.64 And on the segment BC at end B the end moment is minus 125.25 so that I will show as anticlockwise with plus sign and at end C of segment BC the end moment is 281.485 that is positive so that is clockwise and I will show the end shear says.
Rb plus and Rc minus now if I write Sigma m about b equals to 0 taking clockwise moments as positive so I get this 125.25 is anticlockwise so minus 125.25 moment because of the UDL of 20 kilo Newton per meter will be clockwise so plus 20 into 12 into the centroid of that will be midpoint of the beam so this distance will be 6 meters so into 6 lever arm then this 281.485 is clockwise so that will be plus 281.485 and then moment because of rc minus will be anticlockwise so that will be minus Rc minus into lever arm for that will be 12 meters. So, into 12 equals to 0. So, if I solve this, I get Rc minus equals to 133 kilonewtons. Next, I write sigma fy equals to 0 with upward forces is positive. So, here I have Rb plus in upward direction. So, Rb plus Then this 20 kN per meter in downward direction, so that will be minus 20 into 12. Then this Rc-is also in upward direction.
So plus and the value of Rc-is 133 kN. So plus 133 equals to 0. So this gives me Rb plus equals to 107 kN. Next I draw free body diagram. For segment CD, for CD the end moment at C is minus 281.485.
So that I will show as anticlockwise with 281.485 magnitude. At end D of segment CD the end moment is 234.25 with plus sign so that is 234.25 kNm and I show the end shears as Rc plus and just Rd because there is nothing on the right of D. Now I write Sigma m about c equals to 0 by taking clockwise moment is positive so this gives me this 281.485 is anti-clockwise so that will be minus 281.485 the moment because of 250 kilo newton will be clockwise with lever arm of 4 meter so that will be plus 2 5 0 into 4 then this 234.25 is clockwise so plus two three four 0.25 then moment because of Rd will be anti-clockwise so that will be minus Rd into lever arm in this case will be 8 meters so into 8 equals to 0 so this gives me Rd equals to 119.1 kilo Newton's and if I write Sigma f y equals to 0 with upward forces is positive so I get this Rc plus in upward direction so plus Rc plus then this 250 kilo Newton in downward direction so I get minus 250 and then this Rd in upward direction so I get plus and the value of Rd is 119.1 so 119.1 equals to 0 and this gives me Rc plus equals to 130.9 kN.
So I have found all the end shear forces for all the segments. So the diagram of all the segments with their end moments and shears and external load will be like this. Now using this I can draw shear force diagram for our beam and to draw shear force diagram I start moving from leftmost point of the beam and I move rightwards.
So as I move in the right direction, the first force that I encounter is 15.64 kN in downward direction. So my shear force diagram moves downward by magnitude 15.64 and because it is in downward direction, it will be with negative sign. Then there is no force till end of the segment AB. So my shear force diagram remains constant till then and then I encounter an upward force of 15.64 so there is a jump of magnitude 15.64 in upward direction so i reach zero value for shear force diagram then if i move further then i encounter 107 kilonewton force in upward direction so there will be an upward jump of magnitude 107 kilonewton in upward direction then i encounter a uniformly distributed load of 20 kN per meter for a length of 12 m.
So, my shear force diagram will drop in linear fashion by magnitude 20 x 12 kN over the length of 12 m. So, at the end of 12 m, my shear force diagram will be 107-20 x 12 that is, 133 kN. So, Here this value will be minus 133 kN.
Then I encounter a force of 133 kN in upward direction. I'll move back to zero. Then I encounter a force of 130.9 kN in upward direction.
So my shear force diagram will go up by 130.9 kN. Then there is no force till this 250 kN in downward direction. So my shear force diagram will remain constant up to that point and then there will be a jump of 250 kN in downward direction. So I will come down by 130.9 minus 250 that is minus 119.1 kN.
So come down 2. minus 119.1 kN then there is no force so shear force diagram will remain constant and then I encounter a force of 119.1 kN in upward direction so the shear force diagram will go back to zero so this is my shear force diagram next step is to draw bending moment diagram using free body diagrams of all the segments and shear force diagram so we start from left most end of our beam at the leftmost end we have 64.475 kilonewton meter bending moment in clockwise sense and we know that clockwise moment on the left end is of sagging nature so that is positive so my bending moment diagram will start with 64.475 positive moment then the shear force diagram is negative and constant with value minus 15.64 so my bending moment diagram will be linearly decreasing and the total decrease from point a up to point b will be equals to area of the shear force diagram between a and b and that area is equals to minus 15.64 into base that is 12 and that has value equals to minus 187 point 68 kNm so the drop 64.475 minus 187.68 will be minus 125.21 kNm. Ideally we should get this value to be equals to this value but because of truncation error there is slight difference. So at point B the value of bending moment will be 125.21.
or we can say 125.25. Now from this value next we have shear force diagram which is triangular and this is positive triangular. So the bending moment diagram will be increasing in parabolic way and the increase from this point will be equals to area of this triangle and to find area of this triangle we need this distance.
Let's call this x and this distance will be 12-x. and here we can use the two similar triangles to find x and 12 minus x so for that i can write 107 upon x equals to 133 upon 12 minus x here note that i am not writing this negative sign because i am only considering distances here to find x and this gives me x equals to 5.35 meters and this 12-x will become 6.65 meters. Now I can find area of this triangle. Here height is 107 and base is 5.35 and that area will be called that A2 that area A2 will be 286.225. So from minus 125 I will go up to minus 125 plus 286.225 and that value will be 160.975 kNm so I'll go up to 160.975 kNm then from here there will be a parabolic decrease in my bending moment diagram because shear force diagram is linearly decreasing and the total decrease will be equals to area of this triangle it's called that A3 will be equals to minus 442 point.
So bending moment diagram will be linear and because it is positive constant, it will be increasing linear and the increment will be equals to area of this rectangle and that area if I call that A4 that area will be equals to 523.6. So from minus 281.25 the increment will be 523.6 so the value will be minus 281.25 plus 523.6 and that will be 242.35 after that the shear force diagram is constant with negative value so my bending moment diagram will decrease linearly and that decrement will be equals to area of this rectangle and if I call that A5 the value of A5 will be equals to minus 476.4 so from 242.35 it will go down up to 242.35 minus 476.4 that is minus 234.05 kilonewton meter and again Ideally, this value should be equals to this value, but there is slight difference because of truncation error. So here we have our bending moment diagram.
Now if we also want to find support reactions, then they are easy to find at the end supports, the bending moments and shear forces at the end supports are equals to the support reactions. So here this 15.64 kN is equals to Ra. that is in downward direction this 64.475 kNm is equals to MA and that is in clockwise sense similarly 119.1 kN is equals to RD and it is in upward direction and 234.25 kNm is MD and that is also in clockwise sense the support reaction at intermediate supports is slightly tricky to find but that we can find by finding the jump in shear force diagram so here the jump in shear force diagram is equals to Rb and that jump is 107 plus 15.64 that is equals to 1.22.64 kilo newton and the support reaction at C will be jump in shear force diagram at C R C will be equals to 130.9 plus 133 that is equals to 263.9 kN. And note that the support moments at the internal hinges is 0. So, MB will be 0 and MC will also be 0. And also note that these support moments are different from values of bending moment diagram at B and C which are normally non-zero for internal hinge supports but they are zero at external hinge supports. So don't confuse these bending moment values at B and C with the support moments at B and C.