Notes

Simple Loading Beam 1:

For the simple loading beam 1, I worked out all missing forces that are acting on the beam, this would be Va and Vb. To do this, I put all forces acting upwards equal to all the forces acting downwards, this however did not allow me to work out the missing forces. After that, I worked out all the forces acting in moments – clockwise and anticlockwise. By using one point as an anchor, I can see all the forces that are acting around that point. By using this I was able to work out that Vb is 61KN.

To figure out Va, I put replaced Vb with 61 and put it back into the equation of upwards is equal to downwards and I manage to figure out Va which is 49KN.

The Shear Force diagram is drawn by using the beam, if there is a force acting upwards, the force would be shown on the shear force diagram by going up the amount that was on the beam. On the beam, there is a 49KN force upwards, so the shear force diagram has an upward force of 49KN from 0 metres. It stays at 49KN for 2M before dropping down by 20KN putting it at 29KN. It stays at 29KN for 3M before dropping down by 30KN into –1M and stays like that for 2M. After that it drops by 60KN for 3M before going up by 61KN which ends up at 0, this shows that the beam is in equilibrium.

The Bending Moment diagram is drawn by multiplying the force on the shear force diagram by the length. For the first section, it starts at 0 since anything multiplied by 0 is equal to 0. The shear force diagram then has a 49KN force for 2 metres, so I multiplied 49 by 2 and reached 98 and is drawn diagonally on bending moment diagram. I then multiplied 29 by 3 and added that on the previous force on the bending moment diagram which got me to 185. I then did –1 multiplied by 2 and added that on the diagram which got me to 183 and I Finally did –61 multiplied by 3 which put my diagram to 0.

Simple Loading Beam 2:

For the second simple loading beam, I worked out the missing forces again which were R1 and R2 by using the same method as in the first simple loading beam. Using that method, I worked out R1 which is 5KN and R2 which is 7KN.

Like the first beam, the Shear Force diagram was drawn and worked out by using the beam and drawing where the forces acted on the beam. There is a 5KN force at 0M going up and stays like that for 5M before dropping down 2KN for another 5M and then drops again by 4KN for another 5M before going back by 7KN to 0KN.

For the Bending Moment diagram, I multiplied the force by the length again to work out the bending moments, 5KN multiplied by 5M which got me to 25KN, I then did 3KN multiplied by 5M and added the answer back into 25KN which got me to 40KN, I then did –2Kn multiplied by 5M and added the answer back to 40KN which got me to 35 KN and then I did –7 multiplied by 5M and added the answer back to 35KN which got me to 0KN.

Simple Loading Beam 3:

For the third simple loading beam, I did not have to work out the clockwise and anticlockwise forces as the only force that would be acting on a cantilever beam would be Ra. To work out Ra I did forces going upwards equal to forces going downwards, and after working out the equation Ra was 1600N

To draw the Shear Force diagram, I drew a line going up to 1600N as there is a 1600N force going up from 0M. It stays like that for 0.5M before drawing another line down 300N which puts the shear force at 1300N, and it stays like that for 0.7M before drawing another line down by another 500N to 800N on the diagram and it stays like that for 0.8M before going down another 800N to 0N.

The Bending Moment diagram is a little bit different for this cantilever beam as I would start from the right side and work out the points in sections. I would start at 0 from the right side as the distance from the right side would be 0. to work out the first section, I multiplied the force on the beam by the distance until the next force (800 multiplied by 0.8 which got 640 and since the force is acting downwards the bending moment line is also going downwards.) For the next section I would multiply the force by the length, so 500 multiplied by 0.7 while also adding 800 multiplied by 1.5 as the distance has increased since I am working out the second section. After adding the two answers I got –1550 and for the final section I multiplied 300N by 0.5M, 500N by 1.2M and 800N by 2M. The last point was –2350 which I drew on the bending moment diagram.