11/20/2020

004829

# How is the load distributed to the members in the angular axis method if the members are excluded from the load application?

A surface load of 1 kN/m² delimited by nodes 1 to 4 is only applied to member 3 (Figure 1).

The entries made in the load generator are shown in Figure 02. There is no correction of the distribution according to the moment equilibrium (Figure 3).

The generated member load is shown in Figure 4. This is calculated as follows:

q = 1.00 kN/m² (area load)

h1 = 4.00 m

h2 = 6.00 m

btot = 12.00 m

$\mathrm\alpha\;=\;\arctan\left(\frac{{\mathrm h}_2\;-\;{\mathrm h}_1}{{\mathrm b}_{\mathrm{ges}}}\right)\;=\;\arctan\left(\frac{6,000\;-\;4,000}{12,000}\right)\;=\;9,46^\circ$

${\mathrm b}_1\;=\;\tan\left(\mathrm\alpha\right)\;\cdot\;{\mathrm h}_1\;=\;\tan\left(9,46^\circ\right)\;\cdot\;4,000\;=\;0,667\;\mathrm m$

${\mathrm l}_1\;=\;\sqrt{{\mathrm b}_1^2\;+\;{\mathrm h}_1^2}\;=\;\sqrt{0,667^2\;+\;4,000^2}\;=\;4,055\;\mathrm m$

${\mathrm l}_2\;=\;\cos\left(\mathrm\alpha\right)\;\cdot\;{\mathrm h}_2\;=\;\cos\left(9,46^\circ\right)\;\cdot\;6,000\;=\;5,918\;\mathrm m$

${\mathrm A} _ {\mathrm R}\; =\frac {\; {\mathrm b} _1\;\cdot\; {\mathrm h} _1} 2\; =\;\frac {\; 0.667\;\cdot\; 4,000} 2\; =\; 1.335\;\mathrm m ^ 2$ (remaining area marked in red in Figure 4)

${\mathrm l}_{\mathrm{ges}}\;=\;\sqrt{{\mathrm b}_{\mathrm{ges}}^2\;+\;\left({\mathrm h}_2\;-\;{\mathrm h}_1\right)^2}\;=\;\sqrt{12,000^2\;+\;\left(6,000\;-\;4,000\right)^2}\;=\;12,166\;\mathrm m$

${\mathrm q} _ {\mathrm c}\; =\:\frac {\mathrm q\;\cdot\; {\mathrm A} _ {\mathrm R}} {{\mathrm l} _ {\mathrm {ges}}}\; =\;\frac {1.00\;\cdot\; 1.333} {12.166}\; =\; 0.110\;\mathrm {kN}/\mathrm m$ (constant load component on 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)

q4 = qc = 0.110 kN/m (member load node 4)

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• Updated 12/01/2020

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