# Solving a statically indeterminate truss

I'm trying to solve the statically indeterminate truss shown below and I'm having a little trouble.

$H$, $P$ and $\beta$ are given. The material is aluminum (density is 2700 ${kg/m^3}$) and has a cross section of 0.002 $m^2$. We are instructed to take the weight of each member into account by distributing $\frac12$ of each member's weight to its end joints.

I can easily solve for $f_1$, $f_3$, $f_4$, $f_5$ and the support reactions using statics (method of joints), but I can't figure out how to go about solving for the remaining members.

The main point of my question is, how do you solve statically indeterminate trusses? I assume I'll have to use some other property such as deformation. I've only posted the homework question to provide background as to why I am asking.

Thanks for any help.

• Hi edc1591, and welcome to Physics Stack Exchange! This is a site for conceptual physics questions, not general homework help. If you edit your question to focus on the specific physics concept that is confusing you, I'll be happy to reopen it. See our homework policy for more information. – David Z Nov 16 '12 at 0:12
• In the meantime, if you take all the element forces in a vector $f=(f_1,f_2,\ldots,f_{9})$ in terms of one unknown force (like $f_{10}$) then by minimizing the "energy" $U = \frac{1}{2 k} f^\top\,f$ with $$\frac{{\rm d} f^\top f}{{\rm d} f_{10}} = 0$$ will produce an answer for you. – John Alexiou Nov 16 '12 at 0:20
• That is a cool method indeed, @ja72. But you would probably need to use total strain energy, which is stress times deformation times length of the bar. So you would need to add a $\sqrt{2}$ to the $f^Tf$ terms for diagonal bars, wouldn't you? – Jaime Nov 16 '12 at 0:30
• No problem, it's fine as long as you improve the question (which you're doing). The edit helps, but still, it's the kind of question that would be answered by looking in your textbook or some other resource. But, for example, if you checked your textbook or whatever, and you found some methods but none of them seemed to work, then you could ask "I found methods X, Y, Z but X doesn't work because..., Y doesn't work because ..., Z doesn't work because .... So what else is there that I can use for this problem?" and that would be a perfectly fine question. – David Z Nov 16 '12 at 1:43
• @DavidZaslavsky OK, I'll do my best to make some more edits. However, I have no textbook to look in for help. This problem is for an applied math class where we are supposed to write a MATLAB function to solve the truss. The professor just assumes that we know how to do this. – edc1591 Nov 16 '12 at 2:23

Here is an idea, borrowed from linear elastic theory. Solve all the forces in terms of an unknown force (I chose $f_{10}$) and construct a long vector $f$

$$\boldsymbol f = \begin{bmatrix} f_1 & f_2 & \dots & f_9 \end{bmatrix}^\top$$

where each component is a function of $P$, $\beta$ and $f_{10}$.

Now assemble something resembling the total potential energy by doing this

$$U = \frac{L}{2\,E\,A} \, \boldsymbol f^\top \boldsymbol f$$

Now by minimizing this with

$$\frac{{\rm d}U}{{\rm d}f_{10}} = 0$$

will produce a result for $f_{10}$ and hence all of the values in $\boldsymbol{f}$.

This works because for each element $f_i = \frac{E_i,A_i}{L_i} \delta_i$ where $\delta_i$ is the deformation, and the energy is $U_i = \frac{1}{2} \left(\frac{E_i,A_i}{L_i}\right) \delta_i^2 = \frac{L_i}{2 E_i A_i} f_i^2$.

So $U = \sum U_i = \frac{L}{2\,E\,A} \, \boldsymbol f^\top \boldsymbol f$ except that not all the elements have the same length. So my method will produce an incorrect result. I just realized this.

To correct this you have to construct the total energy as

$$U = \sum_{i} \frac{L_i}{2 E_i A_i} f_i^2$$ and then minimize it with the derivative.

Illustrative Example

The force equilibrium on points A, B and C is

$$\begin{array}{cc} A_{y}-f_{1}=0\\ -P+f_{1}-f_{2}=0\\ C_{y}+f_{2}=0 \end{array}\biggr\}\begin{array}{cc} A_{y}=P+f_{2}\\ C_{y}=-f_{2}\\ f_{1}=P+f_{2} \end{array}$$ which is indeterminate. The total strain energy is $$U=\frac{L_{1}}{2EA}f_{1}^{2}+\frac{L_{2}}{2EA}f_{2}^{2} \\ =\frac{L_{1}}{2EA}\left(P+f_{2}\right)^{2}+\frac{L_{2}}{2EA}f_{2}^{2}$$
which is minimized by $$\frac{{\rm d}U}{{\rm d}f_{2}}=\frac{1}{2E\, A}\left[\frac{{\rm d}}{{\rm d}f_{2}}\left(L_{1}\left(P+f_{2}\right)^{2}\right)+\frac{{\rm d}}{{\rm d}f_{2}}\left(L_{2}f_{2}^{2}\right)\right]=0 \\ =\frac{1}{E\, A}\left(L_{1}\left(P+f_{2}\right)+L_{2}f_{2}\right)=0 \\ f_{2}=\mbox{-}\frac{L_{1}}{L_{1}+L_{2}}P$$ back substituting into $$A_y = \frac{L_2}{L_1+L_2} P \\ C_y = \frac{L_1}{L_1+L_2} P \\ f_1 = \frac{L_2}{L_1+L_2} P$$

Of course in your case, you need to add the member weights into the node equilibrium equations.

• It worked perfectly! Thanks so much for taking your time to help me! – edc1591 Nov 19 '12 at 3:02