Q6. A 10m long beam ABCDEF carries 20KN load at its free...

Q6. A 10m long beam ABCDEF carries 20KN load at its free left end A. It is simply supported at B and D, 2m and 6m from A respectively. It carries a uniformly distributed load of 30KN/m between B and D, and a point load of 40KN at E which is 8m from A. The remaining 2m length which terminates at F carry no loads. Let point C be 4m from A at the centre of span BD. Draw the Shear Force Diagram and Bending Moment Diagram for the beam and indicate the values at intervals of 2m. (25 marks)

Transcript

00:01 Okay, so this is the bar that we've got, and we want to draw shear force and bending moment diagrams.

00:06 So to start off with, we need to figure out what these reaction forces rb and rd are.

00:10 So just by balancing vertical and horizontal forces, we know that the forces going up, which is rb plus rd, has to equal the sum of the forces going down, which is 20 plus 40 plus 30 times 4, because that 30 kilonewtons is distributed evenly over the 4 metres.

00:28 Is well it's 30 kilonewtons per meter for four meters so that's 30 times four in total and so we can see that is 180 kilonewtons and then if we take moments about point b we are going to find that the anti -clockwise moment is 4rd plus 2 times the 20 kilonewtons on the left and then the clockwise moment is going to be the evenly just the uniform load the strength of which is 30 times 4 and we take it to act in the middle of its load so that's 2 meters from b and then you've also got the 40 kilonewtons 6 meters from b and we can find that this means that rd is equal to 110 kilonewtons and therefore subbing that into here we can find that rb is equal to 70 kilonewtons.

01:35 So for the shear force and bending moment diagrams then, so if we do shear force to start off with, then we have 20 kilonewtons straight away, so if we have the forcing kilonewtons at the side.

01:53 Then we have 20 kilonewtons until we get to the reaction, the first reaction support at point a, at point b, sorry, where we, well actually we should start on, we should really start on minus 20 because it's going downwards.

02:15 And then when we get to point b, we add on the force of rb, which is 70 newtons, 70 kilonewtons, which gets us up to 50 plus 50 kilonewtons and then we're going to the load is uniformly applied so the force is going to linearly take us into the negatives until we get to our d where we'll have added on the 120 kilonewtons downwards which will have taken us to minus 70 so we've got minus 70 down here and then we've added on the 110 which will take us back up to plus 40 and then from there we remain constant until we get to point e where the 40 newtons is added back downwards and then like this.

03:34 So that's our shear force diagram and the bending moment one on.

03:40 So if again we draw on points a, b, d, e and f, it's the moment needed to balance it at any point...

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