- #1
elbarto
- 32
- 0
Hi,
I have been working on a MATLAB program for solving the member forces in a 2d frame structure which many people have helped me out with on this forum. I have finally got the script to a point where it is working ok and have shown the command window output for a solving a simple frame.
My question is, is there a way you can "rollback" on a user input? My program requires the user make many inputs and it is very easy to make a mistake, which is frustrating because you must then enter the 20+ previous inputs again. So, if there is an option to rollback to the previous input it would be extreamly usefull.
Thank You
Elbarto
<<<<<<<<<<<<<<<<<<Enter Node Geometry, Number of members>>>>>>>>>>>>>>>>>>
Enter Number Of Nodes:3
Enter X Co-Ordinate of Node 1:0
Enter Y Co-Ordinate of Node 1:0
Enter X Co-Ordinate of Node 2:100
Enter Y Co-Ordinate of Node 2:0
Enter X Co-Ordinate of Node 3:200
Enter Y Co-Ordinate of Node 3:-75
Enter Number Of Members:2
Enter Node i for Member 1:1
Enter Node j for Member 1:2
Enter Node i for Member 2:2
Enter Node j for Member 2:3
<<<<<<<<<<<<<<<<<<Section properties>>>>>>>>>>>>>>>>>>
Enter E for material {enter 0 for steel}:1e4
Do you wish to use more then 1 section [y/n]?:n
Enter Area m^2 for all members:10
Enter Second Moment of Area (m^4) for all Members:1e3
<<<<<<<<<<<<<<<<<<Member Releases>>>>>>>>>>>>>>>>>>
Would you like to release the Moment (Mz) of a member [y/n]?:n
<<<<<<<<<<<<<<<<<<Enter in Restraint Data>>>>>>>>>>>>>>>>>>
Enter Restraint Code For Node1 {"1 1 1" is fixed}:1 1 1
Enter Restraint Code For Node2 {"1 1 1" is fixed}:0 0 0
Enter Restraint Code For Node3 {"1 1 1" is fixed}:1 1 1
<<<<<<<<<<<<<<<<<<Enter in apllied load data>>>>>>>>>>>>>>>>>>
Enter Applied load vector {R} For Node1 {"Fx Fy M"}:0 0 0
Enter Applied load vector {R} For Node2 {"Fx Fy M"}:0 -10 -1000
Enter Applied load vector {R} For Node3 {"Fx Fy M"}:0 0 0
<<<<<<<<<<<<<<<<<<FIXED END ACTIONS>>>>>>>>>>>>>>>>>>
Are there any member loads (ie Fixed end actions) [y/n]?:y
Enter members with applied member loads ie (# # #):1 2
Enter point loads at mid span for member_1 (Fx Fy) GLOBAL:0 0
Enter point loads at mid span for member_1 (fx fy) LOCAL:0 0
Enter Distributed load for member_1 (w) LOCAL:-0.2
Enter point loads at mid span for member_2 (Fx Fy) GLOBAL:0 -20
Enter point loads at mid span for member_2 (fx fy) LOCAL:0 0
Enter Distributed load for member_2 (w) LOCAL:0
............
RESULTS
............
Nodal Displacments {rf}
r2x = -0.019
r2y = -0.09408
theata_2 = -0.001826
............
Support Reactions {Rs}
R1x = 19
R1y = 10.33
M1 = 365.8
R3x = -19
R3y = 39.67
M3 = -875
............
Basic Member Forces {S}
Mi1 = 365.8
Mj1 = -332.8
F1 = -19
.
Mi2 = -667.2
Mj2 = -875
F2 = -39
.
............
I have been working on a MATLAB program for solving the member forces in a 2d frame structure which many people have helped me out with on this forum. I have finally got the script to a point where it is working ok and have shown the command window output for a solving a simple frame.
My question is, is there a way you can "rollback" on a user input? My program requires the user make many inputs and it is very easy to make a mistake, which is frustrating because you must then enter the 20+ previous inputs again. So, if there is an option to rollback to the previous input it would be extreamly usefull.
Thank You
Elbarto
<<<<<<<<<<<<<<<<<<Enter Node Geometry, Number of members>>>>>>>>>>>>>>>>>>
Enter Number Of Nodes:3
Enter X Co-Ordinate of Node 1:0
Enter Y Co-Ordinate of Node 1:0
Enter X Co-Ordinate of Node 2:100
Enter Y Co-Ordinate of Node 2:0
Enter X Co-Ordinate of Node 3:200
Enter Y Co-Ordinate of Node 3:-75
Enter Number Of Members:2
Enter Node i for Member 1:1
Enter Node j for Member 1:2
Enter Node i for Member 2:2
Enter Node j for Member 2:3
<<<<<<<<<<<<<<<<<<Section properties>>>>>>>>>>>>>>>>>>
Enter E for material {enter 0 for steel}:1e4
Do you wish to use more then 1 section [y/n]?:n
Enter Area m^2 for all members:10
Enter Second Moment of Area (m^4) for all Members:1e3
<<<<<<<<<<<<<<<<<<Member Releases>>>>>>>>>>>>>>>>>>
Would you like to release the Moment (Mz) of a member [y/n]?:n
<<<<<<<<<<<<<<<<<<Enter in Restraint Data>>>>>>>>>>>>>>>>>>
Enter Restraint Code For Node1 {"1 1 1" is fixed}:1 1 1
Enter Restraint Code For Node2 {"1 1 1" is fixed}:0 0 0
Enter Restraint Code For Node3 {"1 1 1" is fixed}:1 1 1
<<<<<<<<<<<<<<<<<<Enter in apllied load data>>>>>>>>>>>>>>>>>>
Enter Applied load vector {R} For Node1 {"Fx Fy M"}:0 0 0
Enter Applied load vector {R} For Node2 {"Fx Fy M"}:0 -10 -1000
Enter Applied load vector {R} For Node3 {"Fx Fy M"}:0 0 0
<<<<<<<<<<<<<<<<<<FIXED END ACTIONS>>>>>>>>>>>>>>>>>>
Are there any member loads (ie Fixed end actions) [y/n]?:y
Enter members with applied member loads ie (# # #):1 2
Enter point loads at mid span for member_1 (Fx Fy) GLOBAL:0 0
Enter point loads at mid span for member_1 (fx fy) LOCAL:0 0
Enter Distributed load for member_1 (w) LOCAL:-0.2
Enter point loads at mid span for member_2 (Fx Fy) GLOBAL:0 -20
Enter point loads at mid span for member_2 (fx fy) LOCAL:0 0
Enter Distributed load for member_2 (w) LOCAL:0
............
RESULTS
............
Nodal Displacments {rf}
r2x = -0.019
r2y = -0.09408
theata_2 = -0.001826
............
Support Reactions {Rs}
R1x = 19
R1y = 10.33
M1 = 365.8
R3x = -19
R3y = 39.67
M3 = -875
............
Basic Member Forces {S}
Mi1 = 365.8
Mj1 = -332.8
F1 = -19
.
Mi2 = -667.2
Mj2 = -875
F2 = -39
.
............
