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I'm an engineering student, our professor assigned us this truss problem I found $A_y$ and $L_y$ by using the moment about l, and found ab, now what? It seems like I have too many unknowns to solve for after that...

  • $\begingroup$ Think about what else you can find by considering the equilibrium of the other joints, one at a time. For example, what can you say about F? $\endgroup$
    – alephzero
    Nov 1 '18 at 20:16
  • $\begingroup$ Well I 've already identified the zero force members... $\endgroup$
    – Max
    Nov 1 '18 at 23:51
  • $\begingroup$ guess I should have said that earlier... $\endgroup$
    – Max
    Nov 1 '18 at 23:51
  • $\begingroup$ I still haven't found a way to answer for the rest of the problem, though, could someone please help? $\endgroup$
    – Max
    Nov 4 '18 at 0:16

You say you've already identified the zero-force members, so I'll skip part (a) of the question. The structure then becomes (deleting all the zero-force members except for DE which is needed for structural stability):

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In this case, I prefer to work from the cantilever and then move in.

Since we know DE is zero-force, we therefore know $EG = 20\text{ kN (compression)}$.

Since the only other member with a vertical component on $G$ is $DG$, we can obtain that $$\begin{align} DG &= 20\times\dfrac{\sqrt{7^2+DG_v^2}}{DG_v} \\ \text{where }DG_v &= \dfrac{5}{3\times7}\times2\times7 = 3.33\text{ m} \\ \therefore DG &= 46.5\text{ kN (tension)} \\ \therefore GH &= 46.5\times\dfrac{7}{\sqrt{7^2+3.33^2}} = 42.0\text{ kN (compression)} \end{align}$$

Looking at $D$, we have the applied load, $CD$ (with horizontal and diagonal components), $DE$ (which we know is zero-force), $DG$ and $DH$. $CD$ is the only one which can absorb $DG$'s horizontal component (which is equal to the result found for $GH$, so we can find that

$$\begin{align} CD &= 42.0\times\dfrac{\sqrt{7^2+CD_v^2}}{7} \\ \text{where }CD_v &= 5 - 3.33 = 1.67\text{ m} \\ \therefore CD &= 43.2\text{ kN (tension)} \\ \therefore DH &= 43.2\times\dfrac{1.67}{\sqrt{7^2+1.67^2}} - 46.5\times\dfrac{3.33}{\sqrt{7^2+3.33^2}} + 40 = 50\text{ kN (compression)} \end{align}$$

Likewise, looking at $H$, only $CH$ can absorb $DH$'s vertical load, so by repeating the calculations above, we get that:

$$\begin{align} CH &= 86.0\text{ kN (tension)} \\ HI &= 112.0\text{ kN (compression)} \end{align}$$

And then looking at $C$, only $BC$ can absorb $CD$'s and $CH$'s horizontal components, so we then calculate that:

$$\begin{align} BC &= 169.3\text{ kN (tension)} \\ CI &= 195.3\text{ kN (compression)} \end{align}$$

Obviously $BI = HI = 112.0\text{ kN (compression)}$ and $IL = CI = 195.3\text{ kN (compression)}$.

Then looking at $A$, we use your previously calculated reactions to obtain

$$\begin{align} AB &= 84.8\text{ kN (tension)} \\ AL &= 10.1\text{ kN (compression)} \end{align}$$

And likewise with $L$ (or from $B$), we obtain the last remaining $BL = 23.8\text{ kN (tension)}$.

And we're done.

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