Fixed End Moments and Stiffness Coefficients for Beams
This program computes the fixed end forces and stiffness properties for elastic straight beams, subjected to arbitrary loading.. In order to use this program you will need to idealize your data into a series of segments such that:

* The cross-section moment of inertia varies approximately linearly

* The distributed load on the segments varies approximately linearly

* There are no concentrated loads within each segment span

Program Notes

The program requires as inputs:

* The modulus of elasticity

* The total length of the beam

* The number of segments

* Segment data (either the moment ofinertia or the depth of the end cross sections)

The maximum number of segments the program can handle is set at 20, but this can be altered by changing the DIM statements as follows:
DIM F1(N), F2(N), F3(N),Z(N)

where N is the maximum number of segments.
 

Example
Compute the fixed-end moments and stiffness coefficients for the beam shown in the illustration below:

Downloads:

Basic program - liberty basic - lbfemscb.bas

Test Case:

FIXED END FORCES AND STIFFNESS
COEFFICIENTS FOR STRAIGHT ELASTIC BEAMS

ENTER GENERAL DATA

ELASTIC MODULUS = 1

TOTAL LENGTH = 15

NUMBER OF SEGMENTS = 2

CROSS SECTI0N GE0METRY
SPECIFICATI0N TYPES =

TYPE 1 = MOMENT OF INERTIA
TYPE 2 = DEPTH

TYPE =2

ENTER SEGMENT DATA

SEGMENT 1

LENGTH = 3

END SECTI0NS DEPTHS
D0 = 3
D1= 1.5

DISTRIBUTED LOAD
W0 = 1
W1 = 1

NODAL LOADS
FORCE =0
MOMENT = 0

SEGMENT 2

LENGTH = 12

END SECTI0NS DEPTHS
D0 = 1.5
D1= 1.5

DISTRIBUTED LOAD
W0 = 1
W1 = 1

NODAL LOADS
FORCE =0
MOMENT = 0

SOLUTI0N
********
FIXED END FORCES

VA = 8.38077739
MA = -27.9172495

VB = -6.61922261
MB = 14.7055886

STIFFNESS COEFFICIENTS

KAA = 0.13178686
KBB = 0.08559940
KAB = 0.61933016e-1