Drawing Shear Force, Bending Moment Diagram

Posted by Jiro Doke, November 20, 2018

40 views (last 30 days) | 0 Likes | 0 comment

Jiro‘s pick this week is Shear Force Bending Moment by Lateef Adewale Kareem.

As an ex-mechanical engineer, I remember my statics course where I learned how to do shear force bending moment diagram. I remember really enjoying that topic, since the steps to drawing the diagram is very systematic. When things are very systematic, it’s suitable for automation. And this is what exactly Lateef did with this entry.

SFBM('Example',[20,5,20],{'CF',-2,0},{'M',10,8},{'DF',5,[1,3]},{'M',-10,12},{'DF',-4,[14,17]});

You call the function by specifying the length of the beam and the locations of the supports. Then you specify one by one the various types of loads, namely “point load”, “moment”, and “distributed load”, that are placed on the beam. Let’s decode the syntax above.

[20,5,20] – length of bar: 20m, support at 5m and 20m
{'CF',-2,0} – point load of -2KN at 0m
{'M',10,8} – moment of 10KNm at 8m
{'DF',5,[1,3]} – distributed load of 5KN/m from 1m to 3m
{'M',-10,12} – moment of -10KNm at 12m
{'DF',-4,[14,17]} – distributed load of -4KN/m from 14m to 17m

You can create an arbitrary linearly distributed load by simply changing the ‘DF’ parameter.

SFBM('Example',[20 5 20],{'DF',[5 30],[7 15]})

By increasing the number of break points, you can describe a non-linearly distributed load, expressed as a polynomial.

SFBM('Example',[20 5 20],{'DF',[0 25 30],[7 10 15]})

I also appreciate how active Lateef is in maintaining this File Exchange entry. As you can see from the comments section, he has active conversations with his users, and he has been updating the entry regularly. Thank you for your dedicated contribution to the community, Lateef!

Comments

Give it a try and let us know what you think here or leave a comment for Lateef.

Published with MATLAB® R2018b