Shear Force and Bending Diagram. Need help understanding.

[bob29]

Homework Statement

[PLAIN]http://img80.imageshack.us/img80/4476/bendingshearquestion.jpg

Homework Equations

$$\sum$$M = 0:
$$\sum$$Fy = 0:

The Attempt at a Solution

Revising for my finals and going over questions, and would like to understand more clearly how to get the following.

Solution is in the image.
[PLAIN]http://img249.imageshack.us/img249/7237/bendingshearsolution.jpg

When 2<x<5:
I do not understand how to obtain the graph.
Shouldn't the graph be from 0<x<5, since it includes the 2m and reaction force at A?
I also do not know how to obtain the (x-2).

Same when 5<x<7:
I do not understand how to obtain the (7-x) or (9-x) in the moment equilibrium equation.

Hope someone can help.
Cheers,

 

[nvn]

bob29: I drew on your diagram to show you how the (x - 2) and 0.5(x - 2) are obtained. The origin is at point A. The blue line is the section cut. The section cut is located at x, measured from the origin. The (x - 2) and 0.5(x - 2) are measured from the section cut, as I showed in the diagram.

Each diagram, such as 2 < x < 5 m, must show everything to the left-hand side of the section cut. If you omit part of the left-hand end of the beam, then you must add the force and moment at the left-hand-end section cut. Omitting part of the left-hand end of the structure perhaps makes it more confusing. And the diagram is inconsistent. Notice the first two domains in the diagram (0 < x < 2, and 2 < x < 5) show the entire structure to the left of the section cut (the blue line). But then the diagram changes its method, and omits a left-hand portion of the structure in the diagram domains 5 < x < 7, and 7 < x < 9.

You can draw the entire structure to the left-hand side of the section cut (the blue line) for each domain, if you wish. And it would make the solution more clear.

 

[bob29]

Thx much easier 2 understand. Should be able to ace this question if it pops up.

 

[MRMACKIE]

Sorry to dig up an old thread but i had a look at the workings posted by nvn and can see how most forces etc are achieved except the force of 2100N, can someone point out what I've missed. Thanks in Advance :-)

 

[DeeNos]

Dear student,

It seems like you are having some trouble understanding how to obtain the Shear Force and Bending Moment diagrams for this particular problem. Let me try to explain it to you in a clear and concise manner.

Firstly, let's start with the Shear Force diagram. The Shear Force diagram represents the change in shear force at different points along the beam. In order to obtain this diagram, you need to use the equation \sumFy = 0, which represents the sum of all the forces in the vertical direction must be equal to zero. This means that the sum of the vertical reactions at A and B must be equal to the sum of the vertical forces acting on the beam.

In this problem, we have two forces acting on the beam - the 20kN force and the 40kN force. At point A, we have a reaction force of 20kN acting upwards, and at point B we have a reaction force of 40kN acting downwards. Therefore, at point A, the shear force is 20kN and at point B, the shear force is 20kN - 40kN = -20kN. This means that the shear force diagram will start at 20kN at point A and decrease linearly until it reaches -20kN at point B.

Now, let's move on to the Bending Moment diagram. The Bending Moment diagram represents the change in bending moment at different points along the beam. In order to obtain this diagram, you need to use the equation \sumM = 0, which represents the sum of all the moments acting on the beam must be equal to zero. This means that the sum of the moments of the forces acting on the beam must be equal to the sum of the moments of the reactions at A and B.

In this problem, we have three forces acting on the beam - the 20kN force, the 40kN force, and the 60kN force. At point A, we have a reaction force of 20kN acting upwards, and at point B we have a reaction force of 40kN acting downwards. Therefore, at point A, the bending moment is 0 (since there is no distance from the point of rotation), and at point B, the bending moment is 20kN x 2m = 40kNm (since the distance from the point of rotation is