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Consistent load vector

A distributed load must be turned into forces at the nodes. The correct (consistent) way is to weight the load with the shape functions: $F^e_i = \int F_x\,\varphi_i\,dx$ . Then the nodal forces do the same work as the distributed load for every displacement the element can have.

F^e = \frac{qh}{2}\begin{pmatrix}1\\ 1\end{pmatrix}

uniform load, linear element

F^e = \frac{h}{6}\begin{pmatrix}2q_1 + q_2\\ q_1 + 2q_2\end{pmatrix}

linearly varying load

F^e = qh\begin{pmatrix}1/6\\ 2/3\\ 1/6\end{pmatrix}

uniform load, quadratic element

Symbols

$q$	distributed load	N/m
$h$	element length	m
$F^e_i$	nodal force	N

Example

Load from $q_1 = 0$ to $q_2 = 6$ kN/m over $h = 3$ m:

$F^e = \frac36(0 + 6,\ 0 + 12) = (3,\ 6)$ kN. The sum $9$ kN is the whole load $\tfrac12\cdot 6\cdot 3$ .

Check: the sum of the nodal forces must always equal the whole load on the element.

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