Beam and load question

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  • #51
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I have shown you how to compute them.
But you seem to be looking for a simple rule of thumb, and that may or may not exist.
I thought some commented here it was statistically indeterminate.. meaning uncomputable... but asking some engineers.. their initial comments was that if all the supports are fixed.. then the moments would be that of simple point load with 2 fixed supports...

ZQd2Bf.jpg



I wonder if he was right. I don't know how to compute for the reactions when the 3 supports beam were changed from pinned to fixed.. some say here statistically indeterminate.. so last favor.. if you can compute all the reactions for 3 fixed supports beam.. please share them.. thanks a lot..
 

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  • #52
haruspex
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statistically indeterminate.. meaning uncomputable
You mean "statically indeterminate", which means that you cannot deduce the reactions from the usual equilibrium equations alone, i.e. an external view of the torques and forces on the beam as a whole.
But you can often solve these questions when you take into account the known constraints on the shape of the beam. E.g. if it is a fixed support at one end then the torque is an extra unknown, but if you know it is fixing the slope there at a certain angle then that gives you an extra equation too.
 
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  • #53
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You mean "statically indeterminate", which means that you cannot deduce the reactions from the usual equilibrium equations alone, i.e. an external view of the torques and forces on the beam as a whole.
But you can often solve these questions when you take into account the known constraints on the shape of the beam. E.g. if it is a fixed support at one end then the torque is an extra unknown, but if you know it is fixing the slope there at a certain angle then that gives you an extra equation too.
In my structural books such as Design of Concrete Structures 14th Edition by Dolan & Co. "Torque" and "Slope" were not mentioned.. only Torsion and Deflections.. maybe that's because they didn't derive the formulas.. Do you have other synonyms for Torque and Slope that I can research so I can understand your computations? Thanks.
 
  • #54
haruspex
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if you can compute all the reactions for 3 fixed supports beam
As I already explained, if there is a fixed support part the way along the beam you can treat it as two separate problems.
The most general support system you may have to deal with is:
- at the ends, fixed, pinned or one of each
- some number of pinned supports at intermediate points
In principle, each of these could be at a different height, and each fixed end at any prescribed angle.
But even for the simplest cases the manual steps are quite involved, as you can see from my posts.
 
  • #55
haruspex
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"Torque" and "Slope" were not mentioned.. only Torsion and Deflections
Not sure how they would be using "torsion". In everyday English it can mean either torque (i.e. an applied twisting load) or the twist that results from it.
If the deflection is given by some function y=y(x) then the slope at some point is the first derivative, y'(x), and the twist the second derivative, y"(x).
If the upward force density on the beam is f(x) and the anticlockwise torque density t(x) [could use the Dirac delta function to represent point loads and torques] then the basic equation is:
ky"(x)=∫z=0z=x(f(z)(x-z)-t(z)).dz
The reasoning is this: if at the beam element dz the locally applied force is f(z) then that is distance (x-z) from the point x, so exerts a clockwise torque f(z).(x-z)dz about the point x, and this leads to an upward curvature in y. Similarly for the applied torque t(z).dz, but in that case the effect at x is just t(z).dz, not depending on the distance x-z.
 
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  • #56
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This is one method I found out at net about how to solve for
Not sure how they would be using "torsion". In everyday English it can mean either torque (i.e. an applied twisting load) or the twist that results from it.
If the deflection is given by some function y=y(x) then the slope at some point is the first derivative, y'(x), and the twist the second derivative, y"(x).
If the upward force density on the beam is f(x) and the anticlockwise torque density t(x) [could use the Dirac delta function to represent point loads and torques] then the basic equation is:
ky"(x)=∫z=0z=x(f(z)(x-z)-t(z)).dz
The reasoning is this: if at the beam element dz the locally applied force is f(z) then that is distance (x-z) from the point x, so exerts a clockwise torque f(z).(x-z)dz about the point x, and this leads to an upward curvature in y. Similarly for the applied torque t(z).dz, but in that case the effect at x is just t(z).dz, not depending on the distance x-z.
In structural books. They don't derive those formulas.. the torsion or torque I was referring to was about twisting of the concrete beam...

9RR9mP.jpg


Anyway. About deflections. Is deflection just like the concept of moments where it can occur without any material properties just like I originally asked in my first message? In structural books. This wasn't made clear so most of us think moments and deflections are only related to the elements.
 

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  • #57
haruspex
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. Is deflection just like the concept of moments where it can occur without any material properties
No. The moment is the cause, and only depends on the load distribution. The deflection is the result, and depends on both the load and the properties of the beam.
 
  • #58
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No. The moment is the cause, and only depends on the load distribution. The deflection is the result, and depends on both the load and the properties of the beam.
Well in the following derivations of 11P/8 and 5P/16 in the 3 supports pinned case.. there is no material properties of the beam so how did he derive using the deflection that has no material properties?

"one solution method:

Remove R2 (R2 is located at the centerline)

Deflection at R2 for the simple condition:
Pa/24EI * (3*l^2 - 4*a^2)
where: a = L/4
Pl/96EI * (3*l^2 - l^2/4)
Pl/96EI * (12*l^2/4 - l^2/4)
Pl/96EI * (11*l^2/4)
11*Pl^3/384EI

To restore that deflection

consider a single point load at R2:
R2 l^3 / 48*EI

set those two equations equal to each other and solve for R2
R2 = 11*P/8

from here it is simple statics to determine R1 and R3
Sum Fy = 0 -> R1 + R2 + R3 = 2P
Sum M1 = 0 -> R2*l/2 + R3 l - P*l/4 - P*3l/4 = 0 -> R3 = P - R2/2 = P - 11*P/16 = 16*P /16 - 11*P/16 = 5P/16

R1 + 11*P/8 + 5*P/16 = 2P
R1 + 27*P/16 = 2P
R1 = 2P - 27*P/16
R1 = 32*P/16 - 27*P/16
R1 = 5P/16"
 
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haruspex
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Well in the following derivations of 11P/8 and 5P/16 in the 3 supports pinned case.. there is no material properties of the beam so how did he derive using the deflection that has no material properties?
Ok, I can see why you are getting confused.

If all the forces and moments acting on the beam are known then the deflection depends on those and the properties of the beam.
In these problems, only some of the applied forces are known initially. The remaining reactions will depend on how the beam reacts to those.
If the beam is known to be uniform then we can figure out the shape of the deflection without knowing its extent. Halving the EI will simply double all the deflections. This allows us to calculate all the reactions without knowing the extent of the deflection.
 
  • #60
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Ok, I can see why you are getting confused.

If all the forces and moments acting on the beam are known then the deflection depends on those and the properties of the beam.
In these problems, only some of the applied forces are known initially. The remaining reactions will depend on how the beam reacts to those.
If the beam is known to be uniform then we can figure out the shape of the deflection without knowing its extent. Halving the EI will simply double all the deflections. This allows us to calculate all the reactions without knowing the extent of the deflection.
can you give example of a beam that is not uniform? do you mean not uniform in thickness or the rebars inside? so a uniform beam is one with identical rebars ratio and resistances in the whole length of the beam?
 
  • #61
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a uniform beam is one with identical rebars ratio and resistances in the whole length of the beam?
Yes.
 
  • #62
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Yes.
By the way.. when the ends are moment connected.. for example, the column and beam can transfer moments to each other.. so the moment in the column is equal to the moment in the beam at edge connections and rotation occurs at the joint... do you consider it fixed?

Some use definition of fixed where there is no moment rotation. If the above is not fixed.. then what kind of real or actual connections would make it fixed? How can you make column-beam fixed with no moment rotations?

And when we described pinned.. do you mean something that is just put on top of the support?
 

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