ftool - doubt cutting diagram

[stef_design]

Hey, guys.
I'm using ftool to solve simple isostatics.
in image 1 you can see the starting isostatic.
image 2: cut pattern returned from the program.
in the section with distributed load, if adopted positive cut if timeShouldn't it be negative? therefore of value – 1 ? the distributed load should not rotate the beam in anticlockwise direction?

Thank you.

 

[SOLID]

Yes... but if "looks" on the left you must also consider other binding reactions.
instead if you put yourself in the middle of the beam where the distributed load is placed and "looks" to the right you see that the cut goes down and there are no other binding reactions in the same direction (you have a cart). therefore if it respects the convention of the signs of the elementary conical (right) is positive. . .

 

[Legs]

the rod you indicated has a distributed load requiring a binding reaction to the top that obviously has to pass all from the node to the left because the right bond (node a) allows free vertical shift.
This force is located on the left side of the auction and rotates it clockwise, that is precisely according to the convection of positive sign.

 

[stef_design]

Yes... but if "looks" on the left you must also consider other binding reactions.
instead if you put yourself in the middle of the beam where the distributed load is placed and "looks" to the right you see that the cut goes down and there are no other binding reactions in the same direction (you have a cart). therefore if it respects the convention of the signs of the elementary conical (right) is positive. . .

but if I leave from point to (carrello) and go left I only have the concentrated load that makes the beam rotate counterclockwise. No? Where am I wrong?

 

[Legs]

Without leaving the axial action of the auction, I give you the cutting force and the moment they have to be born in the extreme left to guarantee the balance:the cutting force at the extreme left (which is equal to the result of the load), as you see, is unequivocally such as to produce an hourly rotation of the rod.
Is it not that by chance you imagined a vertical binding reaction where the cart is? Be careful that that cart in a produces a reaction that is directed axially to the auction.

 

[SOLID]

if you put in to and look left you have to consider for cutting the q*l distributed load (of negative sign) plus the vertical binding reaction of the cart c. the sum must give you 0.

 

[stef_design]

Okay.

the mistake I made was to put me in and look right (verse a) and see what rotation did the distributed load q.
I thought it was anti-clockwise and therefore negative according to the convention.
but I was wrong as an approach because I didn't consider that in b I have an opposite balance force.

Hi.