SFBM

所属分类:matlab编程
开发工具:WINDOWS
文件大小:73KB
下载次数:3
上传日期:2018-11-16 16:28:10
上 传 者sm359
说明:  Shear force bending moment calculation

文件列表:
SFBM\ConvertCoordinates.m (249, 2018-11-15)
SFBM\Diagrams.m (14312, 2018-11-15)
SFBM\DrawArrow.m (325, 2018-11-15)
SFBM\Interpolate.m (368, 2018-11-15)
SFBM\MomentArrow.m (584, 2018-11-15)
SFBM\rotateXLabels.m (18506, 2018-11-15)
SFBM\Round.m (48, 2018-11-15)
SFBM\SFBM.asv (5696, 2018-11-15)
SFBM\SFBM.m (5738, 2018-11-15)
SFBM\shaft.png (62712, 2018-11-15)
SFBM\SpecialPoints.asv (2459, 2018-11-15)
SFBM\SpecialPoints.m (2473, 2018-11-15)
license.txt (1490, 2018-11-15)

This file explains how to use the SFBM.m code. as well as supporting accessories. SFBM means Shear Force & Bending Moment. This program calculates the shear force and bending moment profiles, draw the free body, shear force and bending moment diagrams of the problem. Under the free body diagram, the equations of each section is clearly written with Latex To use this program, you call the function placing the arguments in cells with keywords at the beginning of each cell except for the first 2 arguments. First Argument The first argument is the name of the problem as a string e.g.: 'PROB 1'. Second Argument -Simply supported beam The second argument is a row vector containing length of the beam and location of the supports, for example, if the length of the beam is 20m and has 2 supports, one at 3m and the other at 17m, the second argument will thus be: [20, 3, 17] -Cantilever If the problem is a cantilever problem, then you have only one clamped support, at the beginning or end of the beam. In such a case, the number is second argument contains 2 elements instead of three. For instance, fir a cantilever of length 20m, supported at the beginning, the second argument would be [20,0], and if supported at the end, we have [20,20]. -Beam on the floor Its possible to have a problem in which the body is lying on the floor without any point support. In such scenario, the second argument will just be the length of the beam Third argument and on From the third argument and onward, we use cells. The first element of the cell contains a keyword describing what type of load is inside the argument. The second element is the magnitude of the load while, the third element of a cell argument is its location. Keywords: Point Load = 'CF' Moment = 'M' Distributed Load = 'DF' To add a downward point load of magnitude 5N at location 4m, the argument would be {'CF',-5,4}. Note the negative sign. If the force is acting upward the argument would be {'CF',5,4}; To add a clockwise moment of magnitude 10N-m at location 14m, the argument would be {'M',-10,14}. Note the negative sign. If the moment is anticlockwise the argument would be {'M',10,14}; To add distributed load we need to describe all of them with the minimum number of point required to describe the profile with the highest complexity. For example, a linear profile can be described as {'DF',[5,5],[2,10]} meaning uniform force per unit length of 5N/m from point 2m to 10m. If the values of the profile were given at 3 points, the code will automatically assume it to be quadratic. If profile is uniform, the coefficient of the second and first degrees would be zero.Hence describing the constant 5N/m from 2m to 10m as {'DF',5,[2,10]}, {'DF',[5,5],[2,10]}, {'DF',[5,5,5],[2,8,10]} will make no difference. But in case where the values in the force vector are different, SFBM will generate a polynomial fit for the forces as a function of position. For instance {'DF',[1,5,5],[2,8,10]} will generate a quadratic function, while {'DF',[1,4,5],[2,8,10]} will generate a linear expression and {'DF',[5,5,5],[2,8,10]}will generate a degree zero expression There is no limit to the number degree of polynomial that can be used. its is important that all concentrated loads and torques are listed in the order of locations For example: SFBM('Prob 200',[20,5,20],{'CF',-2,0},{'M',10,8},{'DF',5,[1,3]},{'M',-10,12},{'DF',-4,[14,17]}) Name : Prob 200 Length: 20m Supports: 5m and 20m Point Load: 2N at point 0 Distributed Load: Constant 5N/m from 1m to 3m and Constant -4N/m from 14m to 17m Moment: ACW 10Nm at point 8m and CW 10Nm at point 12 Solution: The Freebody (with Legend), Shear Force and Bending moment diagrams are generated and saved in picture format titles Prob 200.png.

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