# Finding all members using method of joints

Well, I'm having a hard time finding the forces through summation of moments in figure 4.14, the problem is, there's 3 hinges, which means there's 4 unknown forces. It's kinda hard to find the forces.

I tried researching about it but I can't find anything similar to this one.
Do you have any advice or do you know the right process to do so?

I really don't need the specific answer to the image below, I just want to know how to find the 4 unknown forces.

• What have you done so far? If you want anyone on this site to help, so what work you've done.
– Fred
Commented Dec 8, 2016 at 7:17
• Welcome to Engineering! This looks like a homework question. In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information.
– Wasabi
Commented Dec 8, 2016 at 10:08
• Okay! I'm sorry about that, I'll edit the post and will add details on what I have tried so far, thanks for the response! I'll update it as soon as I arrive home! Commented Dec 8, 2016 at 10:55
• thank you so much guys! I edited the problem now and I tried my best to be specific about it! Commented Dec 8, 2016 at 12:10
• I've been thinking about this question for some time now and it really is quite tough. I can't figure out how to do it elegantly (without simply dumping a computer model's answer or using a monstrous stiffness matrix).
– Wasabi
Commented Dec 10, 2016 at 12:00

This question is far simpler than it looks. Since it has four supports (RX and RY at A plus RY at D and E), it would seem to be statically indeterminate, but it actually isn't.

To see this, let's count the number of unknowns and the number of equations we have:

• 1 axial force for 10 bars = 10 unknowns
• 4 reactions ($A_x$, $A_y$, $D_y$, $E_y$) = 4 unknowns
• 2 equations ($\sum F_x = 0$, $\sum F_y = 0$) for each of 7 nodes = 14 equations
• Number of equations = number of unknowns, therefore we have an isostatic (statically determinate structure).

One way of solving this would be to create a system of 14 equations and solving it. This would work, but ain't nobody got time for that.

A much simpler way of solving this is by observing that this is actually two trusses, with the one to the left being supported by the one to the right by the node at $C$ (much like a Gerber beam).

So let's start by solving the truss to the left, placing a support at $C$ representing the effect of the other truss. Also, since we know that the horizontal force at $G$ will be fully absorbed by $A_x$, let's apply that force at $C$. Due to the Gerber-beam-esque behavior of this truss, we don't have to worry about the bending moment generated by shifting the force this way. After all, that bending moment wouldn't have crossed over the hinge at $C$ and the force itself would indeed have been transmitted via $C$, so there's no issue.

This is very straightforward, so I won't go into how to calculate the internal forces in each bar, but just calculate the reactions. It is clear that the results are $A_x = -50\text{ kN}$, $A_y = 60\text{ kN}$ and $C_y = 180\text{ kN}$.

We now move on to the rightmost truss, where we apply the fictitious value of $C_y$ (and notice that the horizontal force at $G$ is still there!):

This is also easy to solve for the reactions:

\begin{align} \sum M_E &= -4D_y + 8\cdot180 - 3\cdot50 = 0 \\ \therefore D_y &= 322.5\text{ kN} \\ \sum F_y &= D_y + E_y - 180 = 0 \\ \therefore D_y &= -142.5\text{ kN} \end{align}

Given the reactions, the structure is also easy to solve for the internal values, so that's left as an exercise for the reader.

Here's a computer model of the structure, as well as of the individual truss models. Notice I had to place a fictitious RX at $E$ for the isolated rightmost truss to be stable to horizontal forces, so ignore its horizontal reaction. This also alters the axial compression values for the bottom chord, so just subtract 50 kN from the compression in that model to make it align with the original structure.

Assuming positive $x$ to be to the right, positive $y$ to be upwards, and all the bars to be in compression, we can write the force balance for each joint. In the end if a force turns out negative it means that the reaction force is in the negative $x$ or $y$ direction, or the bar is in tension.

Joint A: $$A_x-F_{\text{AB}}-F_{\text{AF}} \cos (\theta ) = 0$$ $$A_y-F_{\text{AF}} \sin (\theta ) = 0$$ Joint B: $$F_{\text{AB}}-F_{\text{BC}} = 0$$ $$-F_{\text{FB}}-120000 = 0$$ Joint F: $$F_{\text{AF}} \cos (\theta )-F_{\text{FC}} \cos (\theta ) = 0$$ $$F_{\text{AF}} \sin (\theta )+F_{\text{FB}}+F_{\text{FC}} \sin (\theta ) = 0$$ Joint C: $$F_{\text{BC}}-F_{\text{CD}}-F_{\text{CG}} \cos (\theta )+F_{\text{FC}} \cos (\theta ) = 0$$ $$F_{\text{CG}} (-\sin (\theta ))-F_{\text{FC}} \sin (\theta )-120000 = 0$$ Joint D: $$F_{\text{CD}}-F_{\text{DE}} =0$$ $$D_y-F_{\text{GD}} = 0$$ Joint G: $$F_{\text{CG}} \cos (\theta )-F_{\text{GE}} \cos (\theta ) +50000 = 0$$ $$F_{\text{CG}} \sin (\theta )+F_{\text{GD}}+F_{\text{GE}} \sin (\theta ) = 0$$ Joint E: $$F_{\text{DE}}+F_{\text{GE}} \cos (\theta ) = 0$$ $$E_y-F_{\text{GE}} \sin (\theta ) = 0$$

I concur that nobody got time for solving these equations, so all my computations were done in Mathematica to get the following results: $$A_x=-50kN, \ \ \ A_y=60kN, \ \ \ D_y=322.5kN, \ \ \ E_y=-142.5kN$$ $$F_{\text{AB}}=-130kN, \ \ \ F_{\text{AF}}=100kN, \ \ \ F_{\text{BC}}=-130kN, \ \ \ F_{\text{CD}}=190kN, \ \ \ F_{\text{CG}}=-300kN$$ $$F_{\text{DE}}=190kN, \ \ \ F_{\text{FB}}=-120kN, \ \ \ F_{\text{FC}}=100kN, \ \ \ F_{\text{GD}}=322.5kN, \ \ \ F_{\text{GE}}=-237.5kN$$

1. Consider the whole system: $$\sum x = H_A+50=0$$ Thus: $$H_A=-50$$
2. Consider the left part ABCF: $$\sum m_C=V_A\times 8-120\times 4=0$$ Thus: $$V_A=60$$
3. Consider the right part CDEG: $$\sum m_D=V_E\times 4+(120+60)\times 4-50\times 3=0$$ Thus: $$V_E=-142.5$$
4. Consider the whole system again:$$\sum y=V_D+V_A+V_E-120-120=0$$ Thus: $$V_D=322.5$$
5. Do you only want to find the 4 unknown reactions? If yes, the 4 steps above might help.

What's going on here is that you have a what looks like Hyperstatic structure or Statically indeterminate structure, but isn't. Since you have a hinge in the middle, you get and extra equation. Cut your structure at the hinge and solve for your reactions. Here is an example video from youtube. Start by the left hand side triangular truss and make your way right.