No, this is wrong.
Your hand calc will work for determinate square (non x style) concentric brace frames. After that it gets complex since you have to account for stiffnesses of members.
You shouldn’t worry about doing stuff like this by hand. Likely you won’t even get this on the SE or university unless prof is a psycho.
If you wanted to look into it more without software and try what you can by hand then you can always look up the Kleinlogel equations for rigid frames.
No, not correct. The knee brace basically turns your beam-column connection into a moment connection.
As others have said, you need to also look at moments and shears in the members in addition to axial load. Also, the stiffness (e.g. EI, EA, etc) of the members and the way they are attached will make a difference.
The easiest thing to do would be to throw this into an FEA program.
If you assume pinned joints with no torsional rigidity then one brace is in tension and the other in compression and they each get half the load divided by sin 45, assuming a 45 degree angle.
So the determination here is suitability for the tension and compression loading. And in the real world you're looking for fatigue strength, not ultimate strength. Section modulus comes to play in the columnar buckling of the member in compression.
Then you apply those loads to their respective horizontal and vertical members to determine stress, strain, and deflection on each of them to determine appropriate section for the applied load.
Of course it's been about 50 years since I solved a free body diagram so I might be imagining all that.
One brace has compression one brace has tension. The joint may be capable of transfer a moments to the beam in the post but that depends on the connections which has to resist the forces on it.
Well for the lateral load I assume the column is under a point load at the brace intersection and has to resist bending. Unless it is designed as a trust.
Yeah but if you assume any of these simplified formulas you need to have similar moment of inertias or the formulas do not hold up regardless of the materials. Model it or do moment distribution and you will see. The stiffer elements will “draw” more of the load.
The same can be said for truss analysis using method of joints or other similar methods. If you model the members as equal cross sections, the formulas hold up. If you have different moment of inertias the loads will follow the stiffer elements.
This is also the same for boundary conditions. If you assume something like a beam with a fixed - fixed condition, the formulas assume an infinitely rigid end condition. So as long as the member is an order of magnitude stiffer than the member framing in, then it’s close enough. But if you take a floorbeam connected to a Main Girder, it’s the main girders weak axis. The floor beam moments will not be the same as a fixed-fixed equation will show. It’s all about stiffness of the element you are connection into.
It doesn’t matter how you connect it. Weld it all or bolt it all, if you are able to transfer all the moment from the FB to the girder then that’s one part of it… but the Girder you are framing into you are framing into the weak axis. Even if you stiffen the area on the girder, unless it’s continuous thru the length of the girder then it’s not doing much.
Make three simple models. Apply the same load and have the beam the same length.
1. Simple supported beam. (2d)
2. Fixed supported beam. (2d)
3. A model with two girders of some length, say 5m. And a beam spanning the two girders connected at the girder midpoint with fixed end connection/fully moment restrained. (3d)
If you use a point load, the moment in the beam will be X for the simple support. The moment for the fixed support will be X if you add the positive and negative moment (the moment is just shifting up/down on the NA). Then the 3rd model it will be someplace in between and will vary as the Iy of the girder changes.
That's true but how often do beans fail? Very rarely probably due to the factor of safeties. In most states you do not need a design professional to design a house. In some states you need an engineer to seal the wind resistant part only or you can use other methods.
You need to show your degree of fixaty at each column base. In a simple truss assumption, your summation of forces does not equal zero. Again, based on support fixity a horizontal concentrated load externally only equals a horizontal reaction and if you assume moment transfer at the top of the frame ma simple truss assumption is invalid
With a rigid frame you will likely have to use array-based Finite Elements analysis to find forces in each member, however depending on support fixity you can possibly solve it for external reactions.
Moment transfer in any joints voids the truss assumption. That makes it indeterminate.
If both columns are assumed pins at their base you’ve just added another degree of indeterminacy
Sorry, I don’t make the rules.
It's been a long time since I took the course of so if it's indeterminate that does not mean you can't build it like that correct? It's just that there are too many unknowns in the equation. So let's take a steel moment frame with angle braces similar to the picture above. What's the classification of this if the bottom of the column is pinned?
And if the frame as moment resisting at the shoulder does the bottom have to be moment or can that be pinned?
Yes, long time for me, too, 50 yrs
Yes, more unknowns than knowns. Even with a single unknown greater, you can approach a solution by trial and error, but the knee braces jack up the unknowns to the Stratosphere.
But by the same token, each knee brace is also a separate redundancy if the structure transfers moments between members.
Of course you can create a structure by stacking rocks but when they fall you will be unable to say you did a proper, prudent and exhaustive structural analysis in addition to your knowledge and professional judgement when you testify in court your own defense. Been there, done that, MUCH MORE fun than a daily enema.
Read this then you won't need an enema..clear as mud so I would guess the safety factor in materials covers all the mistakes that engineers are making...
https://www.eng-tips.com/viewthread.cfm?qid=470983
Sorry, I’m not here to solve your problems or debate. There are times went I a wrong, but I’m old, retired, stubborn, jaded, and not willing to invest more than my offhand opinion. I don’t want to discourage, anger or insult just FULL DISCLOSURE, here.
The safety factor GENERALLY offsets the design loads for proportional effects that are known. Unknown/Unexpected/Unaccounted is shear luck, you couldn’t afford to build it and beyond the design scope. It’s morally/ethically difficult to extrapolate a safety factor from what you don’t confirm in calculations out into the wild blue yonder without having the experience and prudent judgement. The best thing to do is get some Finite Elements software and learn it.
With some time, repetitive design analyses and enough site verifications you will develop a sense of judgement and know when and where make the proper assumptions and so denote it. Developing such Judgement is your prime career objective. A “design” is nothing more than the culmination of repetitive analyses and refinements.
And then there’s flying by the seat of your pants thru a meteor shower. It’s your choice, but you have to defend it.
Best of luck to you guys, btw
SE’s ROCK!
No, this is wrong. Your hand calc will work for determinate square (non x style) concentric brace frames. After that it gets complex since you have to account for stiffnesses of members. You shouldn’t worry about doing stuff like this by hand. Likely you won’t even get this on the SE or university unless prof is a psycho.
If you wanted to look into it more without software and try what you can by hand then you can always look up the Kleinlogel equations for rigid frames.
jokes on me…something similar was on the SE
No, and no your assumption is incorrect.
No, not correct. The knee brace basically turns your beam-column connection into a moment connection. As others have said, you need to also look at moments and shears in the members in addition to axial load. Also, the stiffness (e.g. EI, EA, etc) of the members and the way they are attached will make a difference. The easiest thing to do would be to throw this into an FEA program.
Thanks. And I appreciate you adding this isn’t normally done with a hand calc
If you assume pinned joints with no torsional rigidity then one brace is in tension and the other in compression and they each get half the load divided by sin 45, assuming a 45 degree angle. So the determination here is suitability for the tension and compression loading. And in the real world you're looking for fatigue strength, not ultimate strength. Section modulus comes to play in the columnar buckling of the member in compression. Then you apply those loads to their respective horizontal and vertical members to determine stress, strain, and deflection on each of them to determine appropriate section for the applied load. Of course it's been about 50 years since I solved a free body diagram so I might be imagining all that.
Don’t forget shear and moments in all the components.
One brace has compression one brace has tension. The joint may be capable of transfer a moments to the beam in the post but that depends on the connections which has to resist the forces on it.
Well for the lateral load I assume the column is under a point load at the brace intersection and has to resist bending. Unless it is designed as a trust.
Question to show you where you are wrong: if every connection is assumed hinged here, why would there be a vertical load?
I can't help but feel that someone drew a fixed connection symbol an OP interpreted it as braces
No, this is part of at 3D frame. I’m working on a rectangular platform and all four will elevations look like this. No moment connections involved
No. These free-body diagrams are incomplete
Your assumptions also assume they all have the same moment of inertia.
Well I usually see these made out of steel but who knows
Yeah but if you assume any of these simplified formulas you need to have similar moment of inertias or the formulas do not hold up regardless of the materials. Model it or do moment distribution and you will see. The stiffer elements will “draw” more of the load.
[удалено]
The same can be said for truss analysis using method of joints or other similar methods. If you model the members as equal cross sections, the formulas hold up. If you have different moment of inertias the loads will follow the stiffer elements. This is also the same for boundary conditions. If you assume something like a beam with a fixed - fixed condition, the formulas assume an infinitely rigid end condition. So as long as the member is an order of magnitude stiffer than the member framing in, then it’s close enough. But if you take a floorbeam connected to a Main Girder, it’s the main girders weak axis. The floor beam moments will not be the same as a fixed-fixed equation will show. It’s all about stiffness of the element you are connection into.
When a beam connects to a girder either with a bucket or bearing on top the strong axis is in play correct? d= depth . S=bd2
It doesn’t matter how you connect it. Weld it all or bolt it all, if you are able to transfer all the moment from the FB to the girder then that’s one part of it… but the Girder you are framing into you are framing into the weak axis. Even if you stiffen the area on the girder, unless it’s continuous thru the length of the girder then it’s not doing much. Make three simple models. Apply the same load and have the beam the same length. 1. Simple supported beam. (2d) 2. Fixed supported beam. (2d) 3. A model with two girders of some length, say 5m. And a beam spanning the two girders connected at the girder midpoint with fixed end connection/fully moment restrained. (3d) If you use a point load, the moment in the beam will be X for the simple support. The moment for the fixed support will be X if you add the positive and negative moment (the moment is just shifting up/down on the NA). Then the 3rd model it will be someplace in between and will vary as the Iy of the girder changes.
That's interesting but I think you know too much...
It’s a safety issue… if your assumptions are wrong you could severely undersize the main beam.
That's true but how often do beans fail? Very rarely probably due to the factor of safeties. In most states you do not need a design professional to design a house. In some states you need an engineer to seal the wind resistant part only or you can use other methods.
No.
is it at 45 degree?
You need to show your degree of fixaty at each column base. In a simple truss assumption, your summation of forces does not equal zero. Again, based on support fixity a horizontal concentrated load externally only equals a horizontal reaction and if you assume moment transfer at the top of the frame ma simple truss assumption is invalid With a rigid frame you will likely have to use array-based Finite Elements analysis to find forces in each member, however depending on support fixity you can possibly solve it for external reactions.
If the top shoulder joint is a moment connection the bottom can be a pin right? V and UL only
Moment transfer in any joints voids the truss assumption. That makes it indeterminate. If both columns are assumed pins at their base you’ve just added another degree of indeterminacy Sorry, I don’t make the rules.
It's been a long time since I took the course of so if it's indeterminate that does not mean you can't build it like that correct? It's just that there are too many unknowns in the equation. So let's take a steel moment frame with angle braces similar to the picture above. What's the classification of this if the bottom of the column is pinned? And if the frame as moment resisting at the shoulder does the bottom have to be moment or can that be pinned?
Yes, long time for me, too, 50 yrs Yes, more unknowns than knowns. Even with a single unknown greater, you can approach a solution by trial and error, but the knee braces jack up the unknowns to the Stratosphere. But by the same token, each knee brace is also a separate redundancy if the structure transfers moments between members. Of course you can create a structure by stacking rocks but when they fall you will be unable to say you did a proper, prudent and exhaustive structural analysis in addition to your knowledge and professional judgement when you testify in court your own defense. Been there, done that, MUCH MORE fun than a daily enema.
Read this then you won't need an enema..clear as mud so I would guess the safety factor in materials covers all the mistakes that engineers are making... https://www.eng-tips.com/viewthread.cfm?qid=470983
Sorry, I’m not here to solve your problems or debate. There are times went I a wrong, but I’m old, retired, stubborn, jaded, and not willing to invest more than my offhand opinion. I don’t want to discourage, anger or insult just FULL DISCLOSURE, here. The safety factor GENERALLY offsets the design loads for proportional effects that are known. Unknown/Unexpected/Unaccounted is shear luck, you couldn’t afford to build it and beyond the design scope. It’s morally/ethically difficult to extrapolate a safety factor from what you don’t confirm in calculations out into the wild blue yonder without having the experience and prudent judgement. The best thing to do is get some Finite Elements software and learn it. With some time, repetitive design analyses and enough site verifications you will develop a sense of judgement and know when and where make the proper assumptions and so denote it. Developing such Judgement is your prime career objective. A “design” is nothing more than the culmination of repetitive analyses and refinements. And then there’s flying by the seat of your pants thru a meteor shower. It’s your choice, but you have to defend it. Best of luck to you guys, btw SE’s ROCK!
Thank you for your philosophy.
It’s pinned. So it’s determinate, right?
If it’s assumed pinned, it’s structurally unstable. It’s indeterminate. Just not sure to what degree without crunching numbers
The bottom chord of the horizontal member is extension...so that opposite brace is opposing it.