Bending and Shear

Bending Moments and Shear Force Envelopes

This program will compute the envelope of bending moment and shear force ofa beam segment subjected to various load conditions, as shown is the illustration below:

 

Program Notes

The maximum number of output points for this program has been set to 25. To modify this, change DIM statements as follows:

DIM X(N): DIM V(N,2N): DIM M(N,2)

DIM S(N): DIM B(N)

where N is the maximum desired.

 

Example

Given the beams below, compute the envelopes of bending moments and shear forces along the left spans for load multipliers of 1.0 in the first case and 1.25 in the second:

 

Downloads:

tBasic program - liberty basic - lbbmsfe.bas

Test Case:

BENDING MOMENT AND SHEAR FORCE ENVELOPES



**PROBLEM SPECIFICATION**

SPAN LENGTH = 10

NUMBER OF LOAD CASES = 2

**OUTPUT DEFINITION**

NUMBER OF SUBINTERVALS= 10

NUMBER OF ADDITIONAL

OUTPUT POINTS = 1

ENTER ADD.POINTS POSITION

POINT NO. 1
X = 3.5

**LOAD CASES SPECIFICATION**

****************1

> LOAD CASE 1

MULTIPLIER = 1

END MOMENTS

MA = 0
MB = 57.4054

DISTRIBUTED LOAD

WA = 2
WB = 2

NUMBER OF POINT LOADS = 1

LOAD 1

MAGNITUDE = 15
DISTANCE = 3.5
****************2

> LOAD CASE 2

MULTIPLIER = 1.25

END MOMENTS

MA = 0
MB = 43.3036

DISTRIBUTED LOAD

WA = 1
WB = 1

NUMBER OF POINT LOADS = 0
****************3

BENDING MOMENT ENVELOPE
***********************


POINT 1  AT X = 0

MAX M = 0
MIN M = 0

POINT 2  AT X = 1

MAX M = 13.00946
MIN M = 0

POINT 3  AT X = 2

MAX M = 24.01892
MIN M = -0.8259

POINT 4  AT X = 3

MAX M = 33.02838
MIN M = -3.11385

POINT 5  AT X = 3.5

MAX M = 36.78311
MIN M = -4.726575

POINT 6  AT X = 4

MAX M = 32.53784
MIN M = -6.6518

POINT 7  AT X = 5

MAX M = 22.5473
MIN M = -11.43975

POINT 8  AT X = 6

MAX M = 10.55676
MIN M = -17.4777

POINT 9  AT X = 7

MAX M = 0
MIN M = -24.76565

POINT 10 AT X = 8

MAX M = 0
MIN M = -33.3036

POINT 11 AT X = 9

MAX M = 0
MIN M = -43.09155

POINT 12 AT X = 10

MAX M = 0
MIN M = -57.4054

IS IT OK TO CONTINUE (Y/N) ?Y


SHEAR FORCE ENVELOPE
********************


POINT 1  AT X = 0

MAX V = 14.00946
MIN V : 0

POINT 2  AT X = 1

MAX V = 12.00946
MIN V : -0.41295

POINT 3  AT X = 2

MAX V = 10.00946
MIN V : -1.66295

POINT 4  AT X = 3

MAX V = 8.00946
MIN V : -2.91295

POINT 5  AT X = 3.5

MAX V = 0
MIN V : -7.99054

POINT 6  AT X = 4

MAX V = 0
MIN V : -8.99054

POINT 7  AT X = 5

MAX V = 0
MIN V : -10.99054

POINT 8  AT X = 6

MAX V = 0
MIN V : -12.99054

POINT 9  AT X = 7

MAX V = 0
MIN V : -14.99054

POINT 10 AT X = 8

MAX V = 0
MIN V : -16.99054

POINT 11 AT X = 9

MAX V = 0
MIN V : -18.99054

POINT 12 AT X = 10

MAX V = 0
MIN V : -20.99054