Frame element, equivalent nodal loads

Results.

Example	Reference	Records	Largest deviation
Uniformly distributed load over the full span	Published source	2	0%
Point load at mid-span	Published source	1	0%
Point load away from mid-span	Published source	1	0%
Triangular distributed load	Published source	2	0%
Trapezoidal distributed load	Published source	1	0%
Partial distributed load, starting at end i	Published source	1	0%
Concentrated moment on the element	Published source	1	0%

Problem. A load on the element is transferred to the nodes as equivalent end forces. The end forces the element produces are compared with the fixed-end force tables without solving a system of equations.

Reference document. AISC Steel Construction Manual Table 3-22b and the fixed-end moment table of Hibbeler, Structural Analysis, are used.

Under a uniformly distributed load over the full span the equivalent end force and end moment take these values.

f_i = \dfrac{wL}{2},\qquad m_i = \dfrac{wL^2}{12}

Under a point load at mid-span the equivalent end force and end moment take these values.

f_i = \dfrac{P}{2},\qquad m_i = \dfrac{PL}{8}

The equivalent nodal load is the integral of the shape functions along the load.

\mathbf{f}_e = \int_0^L \mathbf{N}(x)^{\mathsf{T}}\, w(x)\, \mathrm{d}x

The equivalent nodal loads are computed with the Hermite shape functions for the transverse translation and rotation of end i.

N_2(\xi) = 1 - 3\xi^2 + 2\xi^3,\qquad N_3(\xi) = L\left(\xi - 2\xi^2 + \xi^3\right),\qquad \xi = x/L

Under a uniform load the end force is found by integrating the load against the shape function $N_2$ .

f_i = \int_0^L N_2\, w\, \mathrm{d}x = wL \int_0^1 \left(1 - 3\xi^2 + 2\xi^3\right) \mathrm{d}\xi = wL\left(1 - 1 + \tfrac{1}{2}\right) = \dfrac{wL}{2}

Under a uniform load the end moment is found by integrating the load against the shape function $N_3$ .

m_i = \int_0^L N_3\, w\, \mathrm{d}x = wL^2 \int_0^1 \left(\xi - 2\xi^2 + \xi^3\right) \mathrm{d}\xi = wL^2\left(\tfrac{1}{2} - \tfrac{2}{3} + \tfrac{1}{4}\right) = \dfrac{wL^2}{12}

For a point load the shape function is evaluated at the load point.

f_i = P\,N_2(\tfrac{1}{2}) = \dfrac{P}{2},\qquad m_i = P\,N_3(\tfrac{1}{2}) = \dfrac{PL}{8}

Under a triangular load the end force is found from the same integral with the load intensity taken as $w\xi$ .

f_i = wL \int_0^1 \xi\left(1 - 3\xi^2 + 2\xi^3\right) \mathrm{d}\xi = wL\left(\tfrac{1}{2} - \tfrac{3}{4} + \tfrac{2}{5}\right) = \dfrac{3wL}{20}

Under a triangular load the end moment is found from the same integral.

m_i = wL^2 \int_0^1 \xi\left(\xi - 2\xi^2 + \xi^3\right) \mathrm{d}\xi = wL^2\left(\tfrac{1}{3} - \tfrac{1}{2} + \tfrac{1}{5}\right) = \dfrac{wL^2}{30}

A trapezoidal load is the sum of a uniform and a triangular load.

m_i = \dfrac{w_1 L^2}{12} + \dfrac{(w_2 - w_1)L^2}{30} = \dfrac{L^2\left(3w_1 + 2w_2\right)}{60}

For a partial load the upper limit of the integral becomes $c/L$ .

m_i = wL^2 \int_0^{c/L} \left(\xi - 2\xi^2 + \xi^3\right) \mathrm{d}\xi = \dfrac{wc^2}{12L^2}\left(6L^2 - 8cL + 3c^2\right)

For a point moment the derivative of the shape function at the load point is used.

m_i = M\,N_3'\!\left(\dfrac{a}{L}\right), \qquad N_3'(\xi) = 1 - 4\xi + 3\xi^2

Numerical model. A single frame element is used. The equivalent load is the integral of the shape functions along the load, and for linearly varying loads the four-point Gauss rule evaluates it exactly. When a shear area is given, shear deformation is included in this calculation.

Uniformly distributed load over the full span

Parameters. L = 6 · w = 12

Published source. Table of fixed-end forces (AISC Steel Construction Manual, T.3-22b)

Quantity	Reference	Computed	Deviation
End force $f_i$	36	36	0%
End moment $m_i$	36	36	0%