I've been solving something for a month as well as reading half a dozen structural engineering books. Maybe you can help.

Pinned and fixed supports can affect the moments and reactions.. for example.. a simple support pinned and fixed has the following moments...

pinned supports: fixed supports: You will noticed fixed support have more moments at the supports and less moments at midspan..

In structuring engineering.. you get the loads of the beam by the tributary areas of the floors or slabs. In 3 pinned supports.. the formulas for the reactions are: However, I can't get reference what would be the reactions if all the supports were fixed instead of pinned. The ends (A & C) would be have reactions and contributions to the load carrying if fixed than if they were all pinned. Any idea how to derive formulas for the fixed support cases?

Thank you.

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haruspex
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what would be the reactions if all the supports were fixed instead of pinned.
If an intermediate support is fixed then what happens on side of it cannot affect what happens on the other side, so the problem splits into two separate problems.

If an intermediate support is fixed then what happens on side of it cannot affect what happens on the other side, so the problem splits into two separate problems.
If R2 which is in the middle is fixed.. and R1 and R3 are pinned.. there are more reactions at R2. Whereas if R1 and R3 are fixed.. R1 and R3 can take more reactions. You are saying R1 and R2, R2 and R3 are separate problems? Hmm... Then let's just take R1 and R2.. what happened if R1 is fixed instead of Pinned?

But then the case is about 2 point loads at either sides and R1 formula is 5P/16 or 0.3125P, whereas R2 formula is 11P/8 or 1.375P. Now if all are fixed.. R1 and R3 should take more loads maybe 0.65P or sorta. I want formula to compute it.

Chestermiller
Mentor
Is it clear to you that, if the ends are pinned, the reaction forces are statically determinate? Is it also clear to you that, if the ends are "built in," the reaction forces and moments are statically indeterminate? Do you know the difference?

Is it clear to you that, if the ends are pinned, the reaction forces are statically determinate? Is it also clear to you that, if the ends are "built in," the reaction forces and moments are statically indeterminate? Do you know the difference?
If the ends are fully moment connected (100% fixed).. why can't you determine the reaction forces and moments?

haruspex
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You are saying R1 and R2, R2 and R3 are separate problems?
Yes.

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).

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?

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haruspex
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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.)

• dahoa and Mohankpvk
haruspex
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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.

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 cant 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
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Yes.I agree with the point that they are not totally indeterminate.
But I think they are called statically indeterminate because we cant 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.

• Mohankpvk
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. 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..

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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
Mentor
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.

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Chestermiller
Mentor
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.

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
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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.

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?" #### Attachments

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haruspex
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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.

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
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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.

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