Beam and Load: Understanding Uniformly Distributed Loads and Moments

[haruspex]

You are saying R1 and R2, R2 and R3 are separate problems?

Yes.

 

[Mohankpvk]

If the ends are fully moment connected (100% fixed).. why can't you determine the reaction forces and moments?

I think its because, there will be more unknowns than the number of static equilibrium equations that can be applied.If both ends of a beam are fixed, you have to find two reaction meoments(one at each end) and two reaction vertical forces(one at each end).But you have only two static equilibrium equations.(summation of vertical forces=0 and summation of moment about any point=0).

 

[dahoa]

anyway.. I have been trying to derive these for a month but couldn't... in the following:

How is R1=R3=5P/16 and R2= 11P/8 exactly derived?

 

[haruspex]

I think its because, there will be more unknowns than the number of static equilibrium equations that can be applied.If both ends of a beam are fixed, you have to find two reaction meoments(one at each end) and two reaction vertical forces(one at each end).But you have only two static equilibrium equations.(summation of vertical forces=0 and summation of moment about any point=0).

No, there's more to it than that.
Consider the simplest case: no load applied, weightless beam, ends at same height and fixed horizontally.
If this is statically indeterminate then there must be a nontrivial solution, i.e., one in which the end supports apply equal and opposite nonzero vertical forces. Correspondingly there will be nonzero torques so that equilibrium is maintained.
For a theoretical rigid beam that would all be possible, but this field of study considers beams to be flexible. The beam undergoes a deflection y=y(x).

Suppose the supports give a force F up at the left and down at the right, and anticlockwise torques T, T' respectively.
Taking moments at offset x from the left:
ky"=-T+Fx
Integrating, and using y(0)=0:
ky'=-Tx+Fx²/2
Integrating, and using y'(0)=0:
ky=-Tx²/2+Fx³/6
Now we use the same constraints at the right hand end (x=L):
0=-TL+FL²/2
0=-TL²/2+FL³/6
When the smoke clears:
T=F=0.

So it is not indeterminate after all.
(I was wrong to agree to that earlier without checking.)

 

[haruspex]

anyway.. I have been trying to derive these for a month but couldn't... in the following:

View attachment 228381

How is R1=R3=5P/16 and R2= 11P/8 exactly derived?

Try applying the procedure I posted at #22 and #34.

 

[Mohankpvk]

No, there's more to it than that.
Consider the simplest case: no load applied, weightless beam, ends at same height and fixed horizontally.
If this is statically indeterminate then there must be a nontrivial solution, i.e., one in which the end supports apply equal and opposite nonzero vertical forces. Correspondingly there will be nonzero torques so that equilibrium is maintained.
For a theoretical rigid beam that would all be possible, but this field of study considers beams to be flexible. The beam undergoes a deflection y=y(x).

Suppose the supports give a force F up at the left and down at the right, and anticlockwise torques T, T' respectively.
Taking moments at offset x from the left:
ky"=-T+Fx
Integrating, and using y(0)=0:
ky'=-Tx+Fx²/2
Integrating, and using y'(0)=0:
ky=-Tx²/2+Fx³/6
Now we use the same constraints at the right hand end (x=L):
0=-TL+FL²/2
0=-TL²/2+FL³/6
When the smoke clears:
T=F=0.

So it is not indeterminate after all.
(I was wrong to agree to that earlier without checking.)

Yes.I agree with the point that they are not totally indeterminate.
But I think they are called statically indeterminate because we can't find the reactions just by using the static equilibrium equations.
We have to exploit the cocept of deflection(deflection at the ends=0)to solve for the reactions.

 

[haruspex]

Yes.I agree with the point that they are not totally indeterminate.
But I think they are called statically indeterminate because we can't find the reactions just by using the static equilibrium equations.
We have to exploit the cocept of deflection(deflection at the ends=0)to solve for the reactions.

I'd put it a little differently: you have to use the information about the slopes at the ends.

 

[dahoa]

I'm using a structural software called ETABS or Extended 3D Analysis of Building Systems.

I'm manually verifying each computations and so far okay.. but I'm stuck on the determination of the reactions of the 3 points I mentioned earlier.

This is the input I made to the software.

I entered point live loads of 1 kN at each midspan. I didn't use dead load to avoid complicating with the concrete dead load.
The shear and moment diagram produced:

In the program, the fixed condition reactions at middle is about 60% and the ends 40%, whereas using pinned supports (supported by the formula), the reactions at middle is 11P/8 = 1.375 dividing by two is 0.6875 or 68.75% and end is 5P/16 = 0.3125 or 31.25%.

Using fixed supports, the ends have more contribution being 0.40 instead of 0.3125. Now how do you manually derive the 0.40 at fixed ends?

And what does the deflections have to do with it? There is a also a deflection output in the program.

Edit: I know deflections have a lot to do with the reactions but let's take first the concept of fixed vs pinned..

 

[dahoa]

about deflections I understood that

1) if an end deflects downwards, it will decrease the reaction at that end, increase the reaction in the middle, and decrease the reaction at the opposite end.

2) if an end deflects upwards, it will increase the reaction at that end, decrease the reaction in the middle, and increase the reaction at the opposite end.

3) if both ends and the middle support deflect the same amount, you're back to 5P/16 & 11P/8.

When an end is fixed.. how does it affect the deflections?

I know when it is moment connected at the end.. it has more reactions.. as shown by Etabs (see last message). I just wondered what formulas Etabs used to compute for it.

 

[Chestermiller]

Actually, both cases (simple support and fixed support) are statically indeterminate for this problem. If you do overall force- and moment balances on the beam in each of these two situations, you obtain the exact same relationship for the force balance as for the moment balance: $2R_1+R_2=2P$

So, in each of these cases, you are going to have to solve for the displacements using EIy""=M and employ the boundary conditions to solve for the unknown reaction force (and/or) moment at the ends.

In the case of simple support, the algebraic unknowns are the reaction force at the ends and the initial slope dy/dx at x = 0. In this case, you solve for these unknowns by requiring that the displacement y is zero at x = L, and the derivative of the displacement at x = L is minus the slope at x = 0.

In the case of fixed ends, the algebraic unknowns are the reaction force and the reaction moment at the ends, but you know that the slope at x = 0 is equal to zero: dy/dx = 0 at x = 0. You solve for the unknowns by requiring that the displacement and its derivative are zero at x = L.

So for both cases, you have enough information to solve for everything you want to determine. If you would like to solve either of these cases in detail, we will be glad to help.

 

[dahoa]

My thoughts before was that fixing the end can redistribute the moments and loads. But can't one make a general statement without computng for anything that fixing the ends versus just pinned can always produce more reactions for the fixed ends?

My reasoning is that making something stiffer can attract more load.

 

[Chestermiller]

My thoughts before was that fixing the end can redistribute the moments and loads. But can't one make a general statement without computng for anything that fixing the ends versus just pinned can always produce more reactions for the fixed ends?

My reasoning is that making something stiffer can attract more load.

I can't see it, but maybe you can. For me, it's easier to just solve the doggone thing already.

 

[dahoa]

After trying different beam sizes and columns in ETABs.. I found out that fixing the 2 ends lessen the moments at the center a bit.. so the reaction at center was lessened a bit and the end reactions increased a bit. I wonder if this is consistent with the concept of moment redistribution... that is.. by redistributing moments, you can redistribute the reactions, agree?

 

[haruspex]

fixing the 2 ends lessen the moments at the center a bit..

It is easy to see that this cannot be a general rule.
Suppose you have some arrangement where the ends are only pinned. There will be some gradient to the beam at each end. Now fix the ends at that gradient. Clearly nothing changes: no torque at the ends and no change to the verical reactions or moments anywhere.
Now consider fixing the gradients a little differently, up or down. If going one way increases a moment or reaction at some point then going the other way probably reduces it.

However, it may that what you say is the usual consequence for the typical arrangement: ends fixed horizontally with downward loads along the beam.

 

[dahoa]

It is easy to see that this cannot be a general rule.
Suppose you have some arrangement where the ends are only pinned. There will be some gradient to the beam at each end. Now fix the ends at that gradient. Clearly nothing changes: no torque at the ends and no change to the verical reactions or moments anywhere.
Now consider fixing the gradients a little differently, up or down. If going one way increases a moment or reaction at some point then going the other way probably reduces it.

However, it may that what you say is the usual consequence for the typical arrangement: ends fixed horizontally with downward loads along the beam.

But fixing the ends can change the deflections. Something about end bending moment reactions can change the deflections. What do you think of the following? The question was "Why does a continuous beam have less deflection than a pair of simply supported beams?"

 

[haruspex]

But fixing the ends can change the deflections.

Yes, of course, but the point of my previous post is that :
a) it does not necessarily change anything
b) depending on how the ends are fixed, the deflections, moments etc. can be increased or decreased.

 

[dahoa]

Yes, of course, but the point of my previous post is that :
a) it does not necessarily change anything
b) depending on how the ends are fixed, the deflections, moments etc. can be increased or decreased.

My concern is whether it can change the reactions at the ends.

If we have a single beam with 2 pinned or fixed supports.. it won't change the reactions because there is nothing to redistribute the weight.. but when you have 3 supports.. I was thinking if changing the ends from pinned to fixed can change the moment at center and shift the reactions too from center to end and back and forth.

 

[haruspex]

If we have a single beam with 2 pinned or fixed supports.. it won't change the reactions

Consider a beam with a central weight, pinned at each end. Now changed the left-hand support to fixed at the horizontal. That will shift some of the vertical reaction from the right-hand support to the left-hand.

 

[dahoa]

Consider a beam with a central weight, pinned at each end. Now changed the left-hand support to fixed at the horizontal. That will shift some of the vertical reaction from the right-hand support to the left-hand.

I see. So when the beam with 3 supports in message #33 is changed from pinned to fixed.. the reactions at the end will increase... this is agreed?

So the problem is we just don't know how to compute the increase in the reactions.. but I wonder how Etabs computed it. Also notice when I said "fixed".. of course I was referring to fixed horizontally.. were you imagining the fixing is from the top, hmm...

 

[haruspex]

we just don't know how to compute the increase in the reactions

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.

 

[dahoa]

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...

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..

 

[haruspex]

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.

 

[dahoa]

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.

 

[haruspex]

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.

 

[haruspex]

"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=0^z=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.

 

[dahoa]

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=0^z=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...

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.

 

[haruspex]

. 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.

 

[dahoa]

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"

 

[haruspex]

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.

 

[dahoa]

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?