# In Finite Element Analysis, what is the correct load vector for a linear triangular element with area Ae, thickness he and uniform body force vector f?

In Finite Element Analysis, what is the correct load vector for a linear triangular element with area Ae, thickness he and uniform body force vector f?

(a) $\frac{A_e h_e}{4}$f

(b) $\frac{A_e h_e}{3}$f

(c) $\frac{h_e}{3A_e}$f

(d) $\frac{h_e}{4A_e}$f

This question was posed to me in a job interview.

The doubt is from Plane Elasticity in portion Plane Elasticity of Finite Element Method

by (185k points)
selected by

Right answer is (b) $\frac{A_e h_e}{3}$f

The explanation: For a linear triangular (i.e., constant-strain triangle) element, for the case in which the body force is uniform and thus the body force components fx and fy are element-wise constant (say, equal to, $f_{x0}^e$ and $f_{y0}^e$ respectively), the load vector F has the form F=∫Ωche(ψ^e)^T$f_0^e$dx

=$\frac{A_e h_e}{4}\begin{bmatrix}f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\\f_{x0}^e\\f_{y0}^e\end{bmatrix}$. The internal load vector Q is computed only when the element falls on the boundary of the domain on which tractions are specified.

+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote