# Finding the displacement in a truss using energy method Question

I've been trying to revise for an upcoming final so I am solving problems from the previous chapters.I've been wracking my head on this for a while (about 1-2hrs) but I just can't seem to get it. For some stupid reason I can't figure out why Member CE is 0 in the virtual loading, shouldn't that pin at C have a vertical reaction of 1. That would mean the summation of the vertical forces at joint C would be something like this Fy=1-FCEsin(26.565)=0. Failing at method of joints is very depressing to think about with the upcoming finals :/. Of course, the solution manual is no help it already assumes any scrub can solve method of joints. Can you help me out :|

Consider the geometry. Horizontal forces on the bearings can not be determined with forces and moment equilibrium, but you can determine the vertical forces!

From there, calculate those forces in the beams, which you are able to calculate.
Note: Those little green triangles you see are the forces in the truss members, split up to horizontal and vertical. That way you can avoid using nasty decimal point numbers. (To get the absolute value, you can just use our fellah Pythagoras' formula)

The numbers in orange specify the order in which my calculations were made:
0. Calculate horizontal forces on bearing (moment equilibrium)
1. $AE_V=1$, as $AB$ cannot take the vertical force attacking at $A$. Therefore $AE_V=2$ (geometry)
2. Therefore, the $AB_H=-1$ (only has a horizontal part)
3. External horizontal force $=2$. Therefore $DE_H=2$, thus $DE_V=1$
4. $DE$ and $AE$ cancel each other out at node $E$, so $CE$ and $BE$ must do so as well. As only $CE$ has a horizontal part, both $CE=BE=0$.
5. see 4.
6. only horizontal part, so $BC=2$
7. from point 3 you can determine, that $D_V=1$ (see the triangle), and consequently $C_V=0$

conventions:
1. $AB_H$ is the horizontal part of the force in member AB, $AB_V$ the vertical one...
2. $-$: compression, $+$: tension

• I understand what you're doing to an extent, rather than dealing with the resultant forces and pesky angles you turned them into horizontal and vertical forces and used gemtery to simplify the process. You then proceeded to solve joint A,D,E,C sure I get that but I don't know how to solve it if I went to solve say joint C directly and get member CE, I would falsely assume it is equal to AE since I would assume that pin on C has a vertical force of 1 when Cy=0 how did you realize that? – Khaloodxp May 12 '18 at 12:18
• In this specific case you're just getting closer to it with every step. But without knowing all bearing forces, you can't really solve pin C right away (as Cv is missing). That's why I think this triangle method is quite handy. See, for example at step 3: You get 2 as horizontal, and by geometry can conclude that vertical is 1. This leads to an upwards Dv=1. Now you could already see, that Cv=0, because Dv+Cv=1 (sum of all bearing forces = sum of external forces) Did it become any clearer now? – Andrew May 12 '18 at 15:39
• I get it now it really does help to move away from a problem and work on something else to get perspective,"(as Cv is missing)" my mind didn't even register that CV was missing and solved as if it were there, my mind glossed over that point completely how can I solve with 1 equation and 2 unknowns! Thanks :o – Khaloodxp May 12 '18 at 16:56
• Yeah it took me some time as well to figure that out. You're welcome. – Andrew May 13 '18 at 10:35

You're asking about the virtual case which only has the 1 kip load, so that's what I'll answer.

Let's start at node A. It has beams AB and AE. Only AE is capable of supporting the vertical load and we know that AE's slope is 1/2, so the horizontal component is equal to 2 kip, for a resultant of $\sqrt{1^2+2^2}=2.236\text{ kip}$.

Since AE's horizontal component is 2 kip, we know that AB is also 2 kip.

Now for node B. This has beams AB, BC and BE. We already know AB is 2 kip (horizontal), and since BE is vertical, it can't absorb any of this load (it therefore suffers 0 kip), leaving all of it for BC, which therefore is also 2 kip.

Now for node E. This has beams AE, BE, CE and DE. AE we know has a resultant of 2.236 kip and BE is 0 kip. We therefore need to figure out how to split AE's force between CE and DE.

However, this is actually also trivial. Look back at what we did for node B: we saw that one of the beams (BE) could only resist vertical forces, of which there were none, and we could therefore conclude that it suffers 0 kip. This case is identical, other than that it is rotated. Imagine for this case that the X-axis follows the line A-D, and the Y-axis is perpendicular to that. What you'll see is that AE and DE only work on the X-axis, while CE has a Y component. Since it is the only one with this component (since we already know BE is 0 kip), it can't be cancelled out. Therefore, for this node to be in equilibrium, this component (and therefore CE's entire resultant force) must be equal to 0 kip.

• oh yes I know it can be solved with ease by using zero force members, I like to train myself and try different techniques. Plus I like to poke and see if there are other ways to solve a problem I can easily procastinate on one problem for 2 hrs. A comment actually helped me figure it out in my initial statement I blindly assumed Cy=1 and Dy=0 which is stupid because a) the initial problem has a 1.5 kip load b) why would the vertical reaction only be for C that would let ED=0 – Khaloodxp May 12 '18 at 16:50