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enter image description here

So in class we went over the different kinds of supports and in this problem I have to determine the reactions at A and E.

When I drew my force diagram, I drew (using components) a total of 7 forces. A and E are clearly pins that much I know.

However apparently at B there is nothing going on even though I drew 2 forces for B, because it looks like a pin (it's even shaded in gray like A and E). So are there only 5 forces?

We have covered the importance of determining the supports and I am absolutely confused as to how I can accurately distinguish what the supports are in these problems. How is B not a support? It looks like a pin.

So if the problem instead said determine the reactions how would I know it's only at A and E and not B?

I feel like I'm just not understanding the engineering portion of this. And how can I solve for this just using components without using triangles?

I would greatly appreciate a drawing for your force diagram. Also these metallic looking boomerangs structures seem like common problems so I need help understanding what's going on in the spots where they connect and why there is no reaction there.

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  • $\begingroup$ If B is a pin, isn't it simply a connection to the arm connecting to E? If you consider that the force applied on D is being "vectored" via support A, it's a force applied to a diagonal arm, rather than a vertical one (compression), yes? Why the restriction of no triangles? $\endgroup$
    – fred_dot_u
    Commented Feb 10, 2022 at 22:45
  • $\begingroup$ USUALLY support for these types of questions will be on the ground or wall. The concepts are generally how the forces get from the applied point through a series of members to those points. B is a pin connection for the member leading to support E. The important part to note here is that the C to A member looks to be continuous to me. This means that moment cannot be transferred to the BE member, only vertical and horizontal loads. A moment will transfer through point B along the CA member. $\endgroup$
    – Forward Ed
    Commented Feb 10, 2022 at 23:28
  • $\begingroup$ @fred_dot_u Can you elaborate a little more on what you mean by vectored via support A? Earlier I finally figured out that the forces from the arm ED or reaction E is the same at E and B correct? Like a wire connecting the ground and a traffic pole those endpoints have the same forces. So I set up my sum of forces as -200 +Ay + Ey = 0 and -Ax + Ex =0 clearly the horizontal x forces are equal to each other. I drew a line connecting E and B like it was a cable and got the correct answer but I'm not sure what I would have done if I started with A instead for a triangle. $\endgroup$
    – r2d2
    Commented Feb 10, 2022 at 23:53
  • $\begingroup$ @fred_dot_u and i want to know how to do it using component forces because that's what I'm more used to and would like that option. Although I'm still not sure how I would set up a triangle to get the component forces if I tried drawing a triangle using A I would have used the horizontal leg to be 5 inches and the vertical one to be 10 in. This is wrong though and I'm not sure why. Can you explain? $\endgroup$
    – r2d2
    Commented Feb 10, 2022 at 23:57
  • $\begingroup$ @ForwardEd by continuous you mean it's a separate "arm" right? I'm just confused on how to solve this using component forces instead of the fancy triangles. I finally understand that the arm BE is the same forces on B and E ... I just want to know how to algebraically determine the reactions now $\endgroup$
    – r2d2
    Commented Feb 11, 2022 at 0:00

4 Answers 4

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The misunderstanding here is quite simple: there is no support at B.

You are absolutely correct that there's clearly something there, as indicated by the grey circle. However, it isn't a support.

The way to realize this is simply to remember what a support is: a means of transferring the forces applied on the structure to the "ground". This is quite visible when you look at A and E: there's those chunky grey "things" around the beam which clearly connect the beam to the "ground." Obviously the beam is transferring its load to those grey things and they're transferring it to the ground.

But do we have anything like that at B? No. There's just that grey circle, but it's clearly something that's merely "internal" to the beam, not something which allows it to transfer forces to the outside world.

So ok, the only supports are at A and E. But then what is that thing on B? It's an internal hinge/pin. That is, a pin which merely serves to connect two beams.

As you correctly identified, at A and E we have pinned supports: these are supports which restrict the vertical and horizontal deflection of the beam (therefore generating vertical and horizontal reactions) but allow for the beam's rotation (therefore not generating any moment reaction).

At B, however, we simply have a pin: a device which ensures that two separate beams have the same vertical and horizontal deflections (therefore transferring axial and transversal forces between the beams), but allowing them to rotate relative to each other (therefore not transferring bending moment between them, which implies zero bending moment at the pin for the element BE).

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    $\begingroup$ Since the beam AC is continuous isn't there really a potential moment at B? Its just that the moment is not transferred to BE which is connected via the hinge? Everything you say seems to support this other then the _"...at the pin)" part....maybe FROM the pin is better wording? $\endgroup$
    – Forward Ed
    Commented Feb 11, 2022 at 2:44
  • $\begingroup$ @ForwardEd: Oh yes, very well caught! Yikes. Edited. $\endgroup$
    – Wasabi
    Commented Feb 11, 2022 at 11:37
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The 200 lbs load will produce a vertical load of 200 lbs on joint "B" along with a clockwise moment 1000 lbs-in moment. Since there is no horizontal load in the system, if the deformation is ignored, you can treat A-B-E as a rigid body that resists the load, and solve the reactions at A & E through simple static.

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I just give you the answer, you work the solution. the vrtical reaction of

$R_yA = R_yE= 100lbs$

$ R_xA =250lbs= -R_xE$

Think of it as a rigid truss. And take moments about A.

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Youe bending moment diagram should look like this:

enter image description here

and your shear diagram should look like this:

enter image description here

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