Combined Loading on an L-shaped beam

[morpheus343]

Homework Statement

Find Fx and Fy.
I tried calculating σα,σb,σc,σd using σ=εE for each point since i know the strains for each point. Then took the section from the top to points a,b. Fx will create a bending moment M and Fy is axial force, so for a: σα=-Fy/A - Mz/Iyy which has 2 uknowns the Fy and M. I do the same for point b: σb=-Fy/A +Mz/Iyy and solve a 2x2 system and find Fy and M. Not sure so far

Relevant Equations

stress

 

[erobz]

Homework Statement: Find Fx and Fy.
I tried calculating σα,σb,σc,σd using σ=εE for each point since i know the strains for each point. Then took the section from the top to points a,b. Fx will create a bending moment M and Fy is axial force, so for a: σα=-Fy/A - Mz/Iyy which has 2 uknowns the Fy and M. I do the same for point b: σb=-Fy/A +Mz/Iyy and solve a 2x2 system and find Fy and M. Not sure so far
Relevant Equations: stress

View attachment 332254

Can you break the diagram into two pictures so the text is larger?

 

[morpheus343]

Should be visible now

 

[erobz]

Should be visible now

Ok, go ahead and replace $F$ with its effective loading at the top of the column so we are clear.

Then proceed with finding the reaction forces at the base. It's a good idea to produce shear moment diagrams, as it should help you find an error (I believe) you are making.

 

[erobz]

View attachment 332257View attachment 332259

Missing a type of load in that diagram?

 

[morpheus343]

I need to add a moment at the top?

 

[erobz]

I'm assuming they have given the length of the beams?

 

[morpheus343]

No

 

[erobz]

No

Are they the same length? Does it say anything about their length?

 

[morpheus343]

No measurements given for any of them. It said the strains have been calculated with rosettes, not sure if that matters for the solution or it's just filler talk.

 

[erobz]

I think it's just to explain how you came to find said strains IMO. let's just call them both length $L$ for now ( I am now certainly unsure of how to get a numerical answer...), but lets continue so you can see the issue with what you were proposing.

 

[erobz]

I don't know if you go stuck or not, but have you had any luck finding reaction forces and/or producing a shear moment diagrams ( or just using the relevant equations to determine the internal moment ) ?

 

[erobz]

Furthermore, its preferable that you format your math using LaTeX Guide. It doesn't take long to learn, and it makes the communication process much smoother.

 

[morpheus343]

The internal moment at a random length x is M=Mtop+Fx*x ?

 

[erobz]

The internal moment at a random length x is M=Mtop+Fx*x ?

You have the right idea, but you are applying things incorrectly. What do you find for the reaction moments, forces at the base of the column. Please draw a diagram turning that portion horizontal.

Also, please follow the link I gave and reply with latex for math. It’s really not a difficult thing to learn. It is certainly not too much to ask given the fact that you are receiving free help.

 

[morpheus343]

 

[erobz]

View attachment 332272

What direction is the moment at B? And what is its value in terms the assumed beam length $L$?

 

[morpheus343]

counterclockwise?

 

[erobz]

counterclockwise?

It's the external effect of $F_y$ (applied at distance $L$) on the beam. Just like the forces. What direction is it?

 

[morpheus343]

you mean it should be clockwise to counteract the rotation that Fy creates?

 

[morpheus343]

Fx * when perpendicular

 

[erobz]

you mean it should be clockwise to counteract the rotation that Fy creates?

No it should be clockwise because it is the rotation $F_y$ creates. We are not looking at an internal force there. We are looking at external forces/loads at $B$. We are replacing the external loading condition out at the end where $F$ is applied, with an equivalent external loading condition at the column.

 

[erobz]

Fx * when perpendicular

These are perfect opportunities to use LaTeX Guide for math formatting. I'm not going to ask again.

 

[morpheus343]

I get it now

 

[morpheus343]

F_y , not sure why it's not showing correctly in preview i copied your LaTeX code

 

[erobz]

F_y , not sure why it's not showing correctly in preview i copied your LaTeX code

You don't have it in delimiters that lets it know to execute it as latex code.

##F_y##

makes inline Latex Code

Like this $F_y$

If you want an equation centered by itself use:

$$ F_y $$

Like this:

$$ F_y$$

 

[morpheus343]

$F_y$ okay thanks

 

[erobz]

$F_y$ okay thanks

Also there is a "glitch" ( side effect of trying to save processing power on the site) where if there is no latex in the thread, then latex will not show up in preview.

 

[erobz]

I get it now

So are you good on the whole problem, do you see the issue?

Your shear/moment diagrams suggest you might be having sign convention issues, and the internal moment should be $M(x) = M_A - F_x x$

$M_B$ is negative according to convention, so it has to cross the axis.

 

[morpheus343]

I am not really sure, so if i cut the beam at points $a$ and $b$ ,both $F_x$ and $F_y$ will create a bending stress and $F_y$ also a normal stress? Would this be the way to solve it, cutting 2 times once at $a$ , $b$, and once at $c$,$d$