# I'm taking statics, and I'm stuck on this truss problem, how do do I solve this?

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...

• 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? Nov 1 '18 at 20:16
• Well I 've already identified the zero force members...
– Max
Nov 1 '18 at 23:51
• guess I should have said that earlier...
– Max
Nov 1 '18 at 23:51
– 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):

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.