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# Beam modal Analysis of using ANSYS APDL

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on In this blog, we will learn to find out the natural frequency of cantilever beam using Ansys as well as Matlab code.

Example: Obtain a natural frequency of a steel bar with a length of 0.45 m. The breadth and height of the beam are 0.02 m and 0.003 m. The modulus of elasticity of steel beam is 2.1*10^11 N/m^2. Along with analytical solutions verify it with Ansys software and Matlab code.

## Beam initial setting

Step1: Preferences > structural

Step2: Preprocessor > Element type > Add > Beam>2 noded 188 > ok

Step3: Preprocessor > material properties > material models > Structural > elastic > isotropic > 2.1e11 > ok > density > 7850 > ok > close

Step4: Preprocessor > modelling > create> keypoints > In active cs > keypoint number-1, location-0> Apply >> keypoint number-2, location-0.45 > ok

Step5: Preprocessor > modelling > create > line > In active coordinate system > ok

Step6: Preprocessor > section > beam > common section > b=0.02, h=0.003 > ok

Step7: Preprocessor > meshing > mesh tool > mesh5 > select element > ok

Step8: Preprocessor >loads > define loads>structural > select active point 1 > select all dof > ok

## Finding the solution

Step9: Solution > Analysis type > new analysis > modal

Step10: Solution > Analysis option > number of nodes to be extracted=3 > ok

Step 11: Solution > Solve >current LS > ok

Step12: General postprocessing > result summary

.Step 13: General Postprocesor > Plot results > deformed Shape

Step 14: General Postprocesor > Plot results > contour plot > nodal solution

## Matlab Code for cantilever beam

clc;
den=input(‘enter density’); % density value
E=input(‘Enter E’); % Modulus of elasticity
d=input(‘Enter d’); % depth
l=input(‘Enter L’); % length
I=bd^3/12; %inertia A=bd; %area
f1=sqrt((EI)/(denAl^4)); w1=((1.87^2)f1);
w2=((4.694^2)f1); w1=w1/(23.142);
w2=w2/(2*3.142);
fprintf(‘n first frequency of cantilever beam=%f’,w1);
fprintf(‘n second frequency of cantilever beam=%f’,w2);
% enter b0.02
% Enter d0.003
% Enter L0.45
%
% first frequency of cantilever beam=12.309133
% second frequency of cantilever beam=77.558681>>

## Conclusion:

The error was found out to be 0.001%.

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