Engineering Combined Loading on an L-shaped beam

AI Thread Summary
The discussion focuses on solving for the axial force (Fy) and bending moment (M) on an L-shaped beam under combined loading. The user attempts to calculate stresses at various points using strain data but encounters difficulties due to the presence of multiple unknowns. Suggestions include finding reaction forces at the base and creating shear and moment diagrams to clarify the problem. The importance of using LaTeX for mathematical formatting is emphasized to enhance communication. The conversation concludes with the user acknowledging the complexity of the problem and the challenge of having too many unknowns relative to the available equations.
morpheus343
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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
Screenshot 2023-09-20 155607.png
Screenshot 2023-09-20 155716.png
Screenshot 2023-09-20 155716.png
 
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morpheus343 said:
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?
 
Should be visible now
 
morpheus343 said:
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.
 
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I need to add a moment at the top?
 
I'm assuming they have given the length of the beams?
 
No
 
morpheus343 said:
No
:wideeyed:

Are they the same length? Does it say anything about their length?
 
  • #10
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.
 
  • #11
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.
 
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  • #12
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 ) ?
 
  • #13
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.
 
  • #14
The internal moment at a random length x is M=Mtop+Fx*x ?
 
  • #15
morpheus343 said:
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.
 
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  • #16
1695229943631.png
 
  • #17
morpheus343 said:
What direction is the moment at B? And what is its value in terms the assumed beam length ##L##?
 
  • #18
counterclockwise?
 
  • #19
morpheus343 said:
counterclockwise?
It's the external effect of ##F_y## (applied at distance ##L##) on the beam. Just like the forces. What direction is it?
 
  • #20
you mean it should be clockwise to counteract the rotation that Fy creates?
 
  • #21
Fx * when perpendicular
 
  • #22
morpheus343 said:
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.
 
  • #23
morpheus343 said:
Fx * when perpendicular
These are perfect opportunities to use LaTeX Guide for math formatting. I'm not going to ask again.
 
  • #24
I get it now
 
  • #25
F_y , not sure why it's not showing correctly in preview i copied your LaTeX code
 
  • #26
morpheus343 said:
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.

Code:
##F_y##

makes inline Latex Code

Like this ##F_y##

If you want an equation centered by itself use:

Code:
$$ F_y $$

Like this:

$$ F_y$$
 
  • #27
##F_y## okay thanks
 
  • #28
morpheus343 said:
##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.
 
  • #29
morpheus343 said:
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.
 
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  • #30
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##
 
  • #31
In order to find the bending moment at each point
 
  • #32
morpheus343 said:
In order to find the bending moment at each point
Yes, the moment at section ##a-b## and ##c-d## are defined by ##M(x)## which is a function of ##F_x, F_y##.

Before you were applying the same moment at point ##b## and ##d##. But they are ##10 ~ \rm{cm}## apart. So they don't have the same internal moment.

You have two other points to consider, so maybe you can even solve for ##L##, I don't know...You'll have to try that out.

To me it appears as though section ##a-b## is at ##\frac{L}{2}## but that isn't specified anywhere - I'm skeptical, but who knows.

Also, it’s not clear whether or not the height of the I beam ( ##20 \rm{cm}## )is to be neglected against ##L##, that would factor into your external moment.
 
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  • #33
I see, i will work the algebra and see if i can solve it, but i think there are too many unknowns.(##F_x##, ##F_y##, ##M_B##, ##L##, ##x## distance). And i only have 4 equations, one of each of the combined stress at each point. Thank you for your time and help
 

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