I am an electrical engineering student trying to teach myself Finite Element Methods (FEM) through a couple of textbooks and independent study. While I believe that I understand the basic ideas behind interpolation and shape functions, I am struggling to understand load distribution due to surface tractions.

For a 3 node bar element (quadratic bar), how do I compute the equivalent nodal forces for all nodes. Note that nodes (1, 2, 3) belong to bar element 1 and nodes (3, 4, 5) belong to bar element 2. There is a uniformly distributed axial pressure being applied resulting in forces in the $$z$$ direction.

I understand the nodal forces $$\{f_s\}$$ for a given surface traction $$\{\Phi_s\}$$ can be found $$\{f_s\} = \int\limits_{-L}^{L} [N_s]^T \{\Phi_s\}dS$$ In the case of a constant thickness ($$t$$) and pressure ($$p$$), this becomes $$\{f_s\} = pt\int\limits_{-L}^{L} [N_s]^Tdx$$

For element one, I compute the shape (interpolation) functions as follows (one shape function per node in the element): $$A = \begin{bmatrix} 1 & -L & L^2 \\ 1 & -\frac{L}{2} & \frac{L^2}{4} \\ 1 & 0 & 0 \end{bmatrix}, A^{-1} = \begin{bmatrix}0 & 0 & 1 \\ \frac{1}{L} & -\frac{4}{L} & \frac{3}{L} \\ \frac{2}{L^2} & \frac{-4}{L^2} & \frac{2}{L^2}\end{bmatrix}, \text{where } N = \left(A^{-1}\right)^T \begin{bmatrix} 1 \\ x \\ x^2 \end{bmatrix}$$ Thus, for nodes $$1, 2, 3$$ in element $$1$$ the shape functions are found: \begin{align*} N_1 &= \frac{x}{L} + \frac{2x^2}{L^2} \\ N_2 &= \frac{-4x}{L} - \frac{4x^2}{L^2} \\ N_3 &= 1 + \frac{3x}{L} + \frac{2x^2}{L^2} \\ \end{align*}

Integrating over the range for element 1 $$(-L, 0)$$ gives: \begin{align*} f_{s,1} &= pt \int\limits_{-L}^{0} N_1dx &= \frac{Lpt}{6} \\ f_{s,2} &= pt \int\limits_{-L}^{0} N_2dx &= \frac{2Lpt}{3} \\ f_{s,3} &= pt \int\limits_{-L}^{0} N_3dx &= \frac{Lpt}{6} \\ \end{align*} If I follow the same process for element $$2$$, I find the resulting: \begin{align*} f_{s,3} &= \underbrace{\frac{Lpt}{6}}_{\text{element 1}} + \underbrace{\frac{Lpt}{6}}_{\text{element 2}} = \frac{Lpt}{3}\\ f_{s,4} &= \frac{2Lpt}{3} \\ f_{s,5} &= \frac{Lpt}{6} \\ \end{align*} Intuitively, this doesn't seem correct since there is less force on node 3 than there is nodes 2 or 4. It appears like the force dips in between elements.

Does anybody know where my mistake(s) could be? I can certainly clarify any parts if need be. Unfortunately, I don't have the intuition to tell if this result makes sense.

• The result seems strange to picture, but I didn't retain much of the FEM that I learned; this could be expected for how simplified the model is. It seems like a consistent result with how the forces are acting on node 1 and 5 though, when considering the elements separately as you are. You have it treated as two independent bars that just share the force on a node, so it makes sense that both have the same distribution across.
– JMac
Feb 28, 2019 at 1:49
• If you don't get a good/satisfactory answer (or one at all) in a few days, you may want to try over at Computational Science (though you may have to adjust the format of the question to suit the site policies). Feb 28, 2019 at 11:07
• Thank you for your replies. I have been spending a lot of time thinking about this problem and I'm wondering if my intuition is wrong. These are the node equivalent forces, which means they have to represent the force distribution and consider the shape of the interpolation function. I wonder if my results are actually correct mathematically, but my understanding of what these forces actually represent is off?
– Mike
Feb 28, 2019 at 23:31
• I do think that Computational Science is better suited for this question. Aug 6, 2021 at 19:01