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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

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

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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.

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