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The truss

How do I calculate the reactions in this truss?I would usually cut in the joint where beams 5,6,7 and 8 are and calculate reactions in the two trusses that are created.How do I do that when there is a force acting on the joint?

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  • $\begingroup$ Cut the joint between the left hand and middle sections, and forget about the load on it for now. Find the reactions at the support and the joint, to make the left section in equilibrium. Now you know now much load at the joint goes into the left section, you can do the same thing for the center section ignoring the load on the right hand joint. Finally solve the right hand section. $\endgroup$ – alephzero Feb 20 '19 at 1:42
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You need to realize that this isn't one large truss. In fact, it's three separate trusses, each supporting the other.

Here's your structure with some noteworthy points added and the three "sub-trusses" shown:

enter image description here

Start by looking at the left-most truss which ends at node B. If you deleted the vertical member at B, that truss would no longer be stable. This means this truss is basically supported by the rest of the structure at A.

Which means we can treat this truss as having a vertical support at B and solving it independently from the rest of the structure.

This is trivial, and we get that $R_A = R_B = 50\text{ kN}$.

Now, when looking at the rest of the structure, we can pretend that left-most truss doesn't exist and replace it with a concentrated downwards force of 50 kN at B.

enter image description here

However, looking at the central truss (which ends at node D), we can see the same thing applies: if it weren't for member 8 (the right-most diagonal), this truss would also be unstable. Therefore, we can once again consider node D to be a vertical support for this truss and solve for the reactions.

Now, you can choose whether to include the load applied at node D on this truss or on the next one. Since the load would be applied directly on the fictional support at D, that support will absorb the entirety of that load, which will then be applied when calculating the last truss. Or you can choose not to consider the load now, and then add the reaction you find for D to that load when calculating the last truss. It's exactly the same.

This one is still pretty straightforward though, and you get $R_C = 225\text{ kN}$ and $R_D = 25\text{ kN}$ (assuming you added the applied load now, otherwise it'd be -75 kN).

And then you move onto the last truss, replacing everything else with a downwards concentrated load of 25 kN at D (or with an upwards 75 kN but then adding in the applied load at the same location).

enter image description here

Unlike the other trusses, this one is self-standing, so you can just solve it as-is. This is also simple, and you get $R_E = 87.5\text{ kN}$ and $R_F = 37.5\text{ kN}$.


In image form, this is what we did:

enter image description here

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All the forces to the left of point at joint between member 6 and 8 add up to 0.25P down.

Assuming we ignore the weight of truss members, we get P/2 +P = 1.5P acting at the left of member 4. This will cause -0.75P at the joint between members 6 and 8. And adding this to 1P already shown will give 0.25P.

Taking moment about 1st support from right, $$ M= \frac{(0.25P*75 + P*25)}{50 } = 0.875P \ reaction\ at\ that\ support$$

$$ \Sigma F_y =0, \quad 0.8725P-(0.25P +P)= 0.375P \ reaction \ at \ rightmost\ post. $$ Note, the reaction at rightmost post is directed up. so I ignored minus sign.

And the rest of it I leave to you.

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  • $\begingroup$ What do you mean by "the loads [...] reduce to 0.75P"? $\endgroup$ – Wasabi Feb 21 '19 at 1:50
  • $\begingroup$ @Wasabi, I just added detailed calculations and corrected a numerical error. $\endgroup$ – kamran Feb 21 '19 at 4:24
  • $\begingroup$ I still don't understand what you mean by "we get P/2 +P = 1.5P acting at the left of member 4". Are you already including the reaction from the left-most support? As to how this "will cause -0.75P at the joint between members 6 and 8", I assume you got that by doing bending moment around the second support, but shouldn't you also consider the effect of the third and fourth supports and of the right-most force? $\endgroup$ – Wasabi Feb 21 '19 at 12:56
  • $\begingroup$ @Wasabi, no bending moment. take the free body diagram to the left of the intersection of 3 and 4. You get P/2 supported by the post and P/2 adding to force of P at this joint =1.5P. and now the FBD at 6 and 8 will show half of this force directed up and added to P= 0.25P at that point. now we get the moment about the right most support. $\endgroup$ – kamran Feb 21 '19 at 17:39
  • $\begingroup$ 3 and 4 don't intersect, though. And how do you go from "this at 3 and 4" to "that at 6 and 8"? Like, just accepting whatever you're doing at 3 and 4, how do you conclude that 6 and 8 will show "half of this force"? $\endgroup$ – Wasabi Feb 21 '19 at 17:44

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