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I'm having trouble with this problem:

problem

G and a are given, and we are asked to calculate A, B and C.

What I've Done:
The first thing I tried was a free body diagram cutting through A and C to create one body and C and B to create another.

Body 1

From here I calculated $A_y$ and $C_y$ as being equal to $-G\frac{\sqrt{2}}{4}$
Then I moved onto the second body:

Body 2

Here I ran into problems immediately. If I placed the moment at either B or C then it follows that the x component of the other must be zero, which it isn't (I have a numerical solution). So I've either made a mistake with my free body diagram or I'm doing something wrong with the second moment.

Here are my equalibrium calculations:
Body 1:
$$\Sigma M^{A}=C_y\cdot 2a+G\frac{\sqrt{2}}{2}=0 \\ \Sigma F_x=A_x-C_x+G\frac{\sqrt{2}}{2}=0 \\ \Sigma F_y=A_y+C_y+G\frac{\sqrt{2}}{2}=0$$

Body 2: $$\Sigma M^{B}=3a\cdot C_x=0 \\ \Sigma F_x=C_x+B_x=0 \\ \Sigma F_y=B_y-C_y=0$$

My results obviously make no sense. What am I doing wrong? I have also tried to calculate for the whole structure but there were too many unknows which I couldn't get rid of.

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  • $\begingroup$ I think I have just realised that in Body II I've forgotten to include the $3a\cdot G$ .. Maybe that's the answer $\endgroup$ – Andy Grey Dec 4 '16 at 13:47
  • $\begingroup$ Still no further.. Worse: I tried adding up all the horizontal forces as given in the solution and I can't make it come out as zero. Is there something fundamental which I'm missing here? $\endgroup$ – Andy Grey Dec 5 '16 at 6:24
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Let's start by transforming the load $G$ into the resultant forces applied to the structure. At point $R$, we have a vertical component equal to $G$ and an inclined component $G$. Since the inclined component is at 45°, we get a total force equal to $G\left(1 + \dfrac{1}{\sqrt2}\right)$ downwards and $\dfrac{G}{\sqrt2}$ to the left. Relatedly, at the end of the rope (between $A$ and $C$), the concentrated force has upwards vertical and rightward horizontal components both equal to $\dfrac{G}{\sqrt2}$.

The model is therefore equal to (for $G = 100$):

enter image description here

As you already found, we can isolate $AC$. It behaves like a simply supported beam with one horizontal constraint at $A$. Therefore, the horizontal load is fully absorbed by $A_x$. The vertical component is evenly split (due to being at the midspan of $AC$) between $A$ and $C$.

Now we have to deal with the forces at the top of the structure. The vertical force will be fully absorbed by $B$ (but is slightly offset by the vertical component at the end of the rope discussed above which is transmitted to $C$). The horizontal force, however, we can't yet know. For that, we need to balance out the moment around $B$ considering these forces and the horizontal force applied on/by $AC$.

$$\begin{gather}\sum M_B = -G\left(1 + \dfrac{1}{\sqrt2}\right)\cdot3a + \dfrac{G}{\sqrt2}\cdot7a - A_x'\cdot3a = 0 \\ \therefore A_x' = G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} \end{gather}$$

where $A_x'$ is the horizontal reaction in $A$ due to those forces, to be added to the reaction found previously of $\dfrac{G}{\sqrt2}$. So, $B_x = \dfrac{G}{\sqrt2} - A_x'$ (in this case $A_x'$ is negative, so this ends up being a sum).

So we end up with: $$\begin{alignat}{3} A_x &= -\dfrac{G}{\sqrt2} + G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{1}{3\sqrt2} - 1\right) &&\approx -0.764G \\ A_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\ B_x &= \dfrac{G}{\sqrt2} - G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(1 - \dfrac{1}{3\sqrt2}\right) &&\approx 0.764G \\ B_y &= G\left(1 + \dfrac{1}{\sqrt2}\right) - \dfrac{G}{2\sqrt2} &&= G\left(1 + \dfrac{1}{2\sqrt2}\right) &&\approx 1.354G \\ C_x &= G\dfrac{\dfrac{7}{\sqrt2} - 3\left(1 + \dfrac{1}{\sqrt2}\right)}{3} &&= G\left(\dfrac{4}{3\sqrt2} - 1\right) = &&\approx -0.057G \\ C_y &= -\dfrac{G}{2\sqrt2} &&&&\approx -0.354G \\ \end{alignat}$$

And now, to check our work:

enter image description here


Alternatively, using your method, what you forgot to do was to consider the bending moment due to the forces at the top of the structure in your second cut ($BC$). That moment is equal to $G\left(1 + \dfrac{1}{\sqrt2}\right)3a - \dfrac{G}{\sqrt2}\cdot4a = Ga\left(3 - \dfrac{1}{\sqrt2}\right)$, which can only be balanced by the force binary generated by $A_x$ and $B_x$. Dividing the moment by the $3a$ lever arm between $A_x$ and $B_x$, we get that each of those forces must be equal to $\pm\dfrac{Ga\left(3 - \dfrac{1}{\sqrt2}\right)}{3a} = \pm G\left(1 - \dfrac{1}{3\sqrt2}\right)$, which is precisely what we got for $A_x$ and $B_x$ above. Knowing this, you can then solve for the other variables as well.

All diagrams given by Ftool, a free 2D frame analysis tool.

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  • $\begingroup$ That's great! Thanks again. I am a little confused about C though. I haven't had a chance to work though your solution yet, but it seems as if you're saying that $C_x = 0$, is that right? $\endgroup$ – Andy Grey Dec 5 '16 at 21:07
  • $\begingroup$ @AndyGrey: Well, $C$ isn't actually a support since it's a part of the structure, so describing $C_x$ is sort of meaningless. But the segment from the end of the rope to $C$ is under a tension equal to $A_x'$. $\endgroup$ – Wasabi Dec 5 '16 at 21:12
  • $\begingroup$ Ok, i understand that, but we have to calculate it anyway, as Suba has below. Splitting the rope into horizontal and vertical sections was the thing I missed, I think. $\endgroup$ – Andy Grey Dec 6 '16 at 9:26
  • $\begingroup$ @AndyGrey: Ah, sorry, I missed that in your question. I've added the values for $C$ to my answer. $\endgroup$ – Wasabi Dec 6 '16 at 10:07
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The balance equations would be the following

$$\Sigma F_x = A_x-C_x+\frac{G}{\sqrt{2}}=0$$ $$\Sigma F_y = A_y+C_y+\frac{G}{\sqrt{2}}=0$$ $$\Sigma M_P = A_y a-C_y a=0$$ $$\Sigma F_x = B_x+C_x-\frac{G}{\sqrt{2}}=0$$ $$\Sigma F_y = B_y-C_y-\left(\frac{G}{\sqrt{2}}+G\right)=0$$ $$\Sigma M_B = C_x (3 a)-\frac{G}{\sqrt{2}}(7 a)+ \left(\frac{G}{\sqrt{2}}+G\right)(3 a)=0$$

And the solution turns out as $$ A_x=\frac{1}{6} \left(\sqrt{2} G-6 G\right) \ \ A_y=-\frac{G}{2 \sqrt{2}}$$ $$B_x=\frac{1}{6} \left(6 G-\sqrt{2} G\right)\ \ B_y=\frac{1}{4} \left(\sqrt{2} G+4 G\right)$$ $$C_x=\frac{1}{3} \left(2 \sqrt{2} G-3 G\right) \ \ C_y=-\frac{G}{2 \sqrt{2}}$$

The free-body diagrams

enter image description here

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  • $\begingroup$ Giving the answer without explaining your approach isn't likely to help other readers solve similar problems. $\endgroup$ – Air Dec 5 '16 at 17:12
  • $\begingroup$ What have I not 'explained'? $\endgroup$ – Suba Thomas Dec 5 '16 at 17:31
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    $\begingroup$ I found this to be concise and useful - for me at least $\endgroup$ – Andy Grey Dec 5 '16 at 21:07

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