I How Do Boundary Conditions Change When a Beam Is Fixed at x=L Instead of x=0?

[papanatas]

Hi! I have a question related to boundary condition in a one dimensional beam subject to compression and traction efforts.

In my class notes I have the following: If we consider a 1D beam of length L which is fixed at x=0 and subject to an effort F at x=0 we have the following boundary condition depending if the effort is a compression or a traction effort:

u(0)=0 ; SE(L) \frac{du}{dx}(L)=F if F is a traction effortu(0)=0 ; SE(L) \frac{du}{dx}(L)=-F if F is a traction effort

Where S is the beam section and E is the Young constant.My question is the following: what happens with boundary conditions if the beam is fixed not in x=0 but in x=L? In my opinion it would be the following:

u(L)=0 ; SE(0) \frac{du}{dx}(0)=-F if F is a traction effortu(L)=0 ; SE(0) \frac{du}{dx}(0)=F if F is a traction effort

What I did was simply put "0" instead of "L" and change the F sign: in this situation F would be positive if we have a compression effort and negative if we have a traction one. Is this correct?

Thank you very much and sorry for my English...

 

[PhanthomJay]

Hi! I have a question related to boundary condition in a one dimensional beam subject to compression and traction

the proper term here in English is tension, not traction.

In my class notes I have the following: If we consider a 1D beam of length L which is fixed at x=0 and subject to an effort F at x=0 we have the following boundary condition depending if the effort is a compression or a traction tension[/color] effort:

u(0)=0 ; SE(L) \frac{du}{dx}(L)=F if F is a traction tension[/color] effortu(0)=0 ; SE(L) \frac{du}{dx}(L)=-F if F is a traction compression[/color] effort

Where S is the beam section and E is the Young constant.

S is the area of the beam's cross section, usually called 'A'.

My question is the following: what happens with boundary conditions if the beam is fixed not in x=0 but in x=L? In my opinion it would be the following:

u(L)=0 ; SE(0) \frac{du}{dx}(0)=-F if F is a traction tension effortu(L)=0 ; SE(0) \frac{du}{dx}(0)=F if F is a tractioncompression effort

What I did was simply put "0" instead of "L" and change the F sign: in this situation F would be positive if we have a compression effort and negative if we have a tractiontension one. Is this correct?

assuming forces pointing right are positive and left negative, then, yes. Other conventions assume that tension forces are positive and compression forces are negative. That's why it's best to show the direction of the forces in a sketch.

Thank you very much and sorry for my English...

No apology necessary.