20 November 2020

004829

How is the load distributed to the members in the angular axis method if 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)

Write Comment...

• Views 150x

If not, contact us via our free e-mail, chat, or forum support, or send us your question via the online form.

RFEM | Basics

Online Training 29 January 2021 8:30 AM - 12:30 PM CET

RFEM for Students | USA

Online Training 3 February 2021 1:00 PM - 4:00 PM EST

The Most Common User Errors With RFEM and RSTAB

Webinar 4 February 2021 2:00 PM - 3:00 PM CET

RFEM | Steel | USA

Online Training 16 February 2021 9:00 AM - 12:00 PM EST

Eurocode 2 | Concrete structures according to DIN EN 1992-1-1

Online Training 19 February 2021 8:30 AM - 12:30 PM CET

RFEM | Structural dynamics and earthquake design according to EC 8

Online Training 24 February 2021 8:30 AM - 12:30 PM CET

Eurocode 5 | Timber structures according to EN 1995-1-1

Online Training 17 March 2021 8:30 AM - 12:30 PM CET

Eurocode 3 | Steel structures according to DIN EN 1993-1-1

Online Training 18 March 2021 8:30 AM - 12:30 PM CET

RFEM | Dynamics | USA

Online Training 23 March 2021 1:00 PM - 4:00 PM EST

RFEM | Basics

Online Training 23 April 2021 8:30 AM - 12:30 PM

Eurocode 3 | Steel structures according to DIN EN 1993-1-1

Online Training 6 May 2021 8:30 AM - 12:30 PM

Eurocode 2 | Concrete structures according to DIN EN 1992-1-1

Online Training 11 May 2021 8:30 AM - 12:30 PM

Eurocode 5 | Timber structures according to DIN EN 1995-1-1

Online Training 20 May 2021 8:30 AM - 12:30 PM

RFEM | Structural dynamics and earthquake design according to EC 8

Online Training 2 June 2021 8:30 AM - 12:30 PM

Webinar 19 January 2021 2:00 PM - 3:00 PM EST

Dlubal Info Day Online | 15 December 2020

Webinar 15 December 2020 9:00 AM - 4:00 PM CET

Length 0:54 min

Length 1:48 min

Length 0:46 min

Length 1:28 min

Length 0:40 min

Length 0:30 min

Length 0:25 min

Length 1:02 min

Length 0:35 min

Length 0:33 min

Length 0:52 min

Length 0:47 min