Question:
How is the load distributed to the members in the angular axis method if the members are excluded from the load application?
Answer:
An area load of 1 kN/m² delimited by Node 1 to Node 4 is only applied to Member 3 (Image 01).
The entries carried out in the load generator are shown in Image 02. There is no correction of the distribution according to the moment equilibrium (Image 03).
The generated member load is shown in Image 04. This is calculated as follows:
q = 1.00 kN/m² (area load)
h-1 = 4.00 m
h-2 = 6.00 m
b-tot = 12.00 m
$\mathrm\alpha\;=\;\arctan\left(\frac}\;=\;\frac{1.00\;\cdot\;1.333}{12.166}\;=\;0.110\;\mathrm{kN}/\mathrm m$ (constant load component on the loaded member)
${\mathrm q}_2\;=\:{\mathrm q}_{\mathrm c}\;+\;{\mathrm l}_1\;\cdot\;\mathrm q\;=\;\:0.110\;+\;4.055\;\cdot\;1.000\;=\;4.165\;\mathrm{kN}/\mathrm m$ (member load Node 2)
${\mathrm q}_5\;=\:{\mathrm q}_{\mathrm c}\;+\;{\mathrm l}_2\;\cdot\;\mathrm q\;=\;\:0.110\;+\;5.918\;\cdot\;1.000\;=\;6.028\;\mathrm{kN}/\mathrm m$ (member load Node 5)
q-4 = q-c = 0.110 kN/m (member load, Node 4)
