Validating Beam Deflection Equations for Pinned and Fixed Supports

[fonseh]

Homework Statement

In this question , is figure 11-2 a and 11-2 b the same beam ? Why in a , the support is pinned end , it change to fixed end in figure b ?

Homework Equations

The Attempt at a Solution

In equation 11-9 , i gtet different ans with the author ..
Here's what i gt :
MN = 2EK( 2 θN + θF - 3 Φ ) + (FEM)N
So,
0= 2EK( 2θA + θB - 3 Φ )

At here θA = 0 , and (FEM)N = 0 , far end = B ,near end = A

MN = 2EK( 2 (0) + θB - 3 Φ ) +(FEM)N ----(1)
0= 2EK( 2θB + 0 - 3 Φ ) ------(2)

By applying the same method as author ( Multiply equation 1 by 2 , and subtract equation 2 ) , i gt
MAB = 3EK(-Φ)

 

[haruspex]

is figure 11-2 a and 11-2 b the same beam ?

No. They are the two different cases of "far end not fixed" (i.e. pin or roller). The near end is fixed or not fixed.

I cannot follow your algebra. You seem to be setting some things to zero without cause. I get the same equation as the author.

 

[fonseh]

No. They are the two different cases of "far end not fixed" (i.e. pin or roller). The near end is fixed or not fixed.

I cannot follow your algebra. You seem to be setting some things to zero without cause. I get the same equation as the author.

why in case 11-8(a) , the far end angular dispalcement theta _B doesn't have to be determined ? I only learned that when the end is fixed support , then , the the angular displacement is 0

 

[haruspex]

why in case 11-8(a) , the far end angular dispalcement theta _B doesn't have to be determined ? I only learned that when the end is fixed support , then , the the angular displacement is 0

θ_B (=θ_F) does not need to be determined because it was possible to combine the two equations so as to eliminate it and produce an equation without it. There is no suggestion that is zero.

By the way, I find the image you posted hard to read. It is very small. When I blow it up large enough to read it becomes quite fuzzy.

 

[fonseh]

θ_B (=θ_F) does not need to be determined because it was possible to combine the two equations so as to eliminate it and produce an equation without it. There is no suggestion that is zero.

By the way, I find the image you posted hard to read. It is very small. When I blow it up large enough to read it becomes quite fuzzy.

In 11-8 a, we could see that at the left end , it's pin supoorted , why at 11-8b , it will become fixed supported ?

 

[fonseh]

θ_B (=θ_F) does not need to be determined because it was possible to combine the two equations so as to eliminate it and produce an equation without it. There is no suggestion that is zero.

By the way, I find the image you posted hard to read. It is very small. When I blow it up large enough to read it becomes quite fuzzy.

Can you explain how to get 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? (Just below equation 11-9)

 

[haruspex]

In 11-8 a, we could see that at the left end , it's pin supoorted , why at 11-8b , it will become fixed supported ?

The left end is the "near end". The situation being considered is "far end not fixed". There are two cases of that, near end fixed or near end not fixed. Hence the two diagrams.

 

[haruspex]

Can you explain how to get 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? (Just below equation 11-9)

Apparently this comes from Eq 11-7 and/or 11-8, but those are not in your post.

 

[fonseh]

Apparently this comes from Eq 11-7 and/or 11-8, but those are not in your post.

Here is it . Can you help to explain ? 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? Does the first 0 represents MN , and the last 0 represent (FEM) N ?

 

[haruspex]

Here is it . Can you help to explain ? 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 ? Does the first 0 represents MN , and the last 0 represent (FEM) N ?

Yes, but remember this is applying eq 11-8 at B in fig 11-8b, so in the context of eq 11-8, B is the near end and A is the far end (which is fixed in fig 11-8b). (The unfortunate coincidence of similar reference numbers in eqs and figures is confusing.)

 

[fonseh]

Yes, but remember this is applying eq 11-8 at B in fig 11-8b, so in the context of eq 11-8, B is the near end and A is the far end (which is fixed in fig 11-8b). (The unfortunate coincidence of similar reference numbers in eqs and figures is confusing.)

then , how about MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N ? What does it represent ? 11-8a ? Which is near end ? and which is far end ?

 

[haruspex]

then , how about MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N ? What does it represent ? 11-8a ? Which is near end ? and which is far end ?

That is the other way round, A is the near end here.

 

[fonseh]

That is the other way round, A is the near end here.

So , both MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N and 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 describes about case in 11-8b ?

 

[haruspex]

So , both MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N and 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 describes about case in 11-8b ?

No. Compare your second equation above with the second equation of 11-9. The angles are swapped around.

I think what is confusing you is the near-far notation in eqs 11-9. In both of them, the near end is A and the far end is B, so for clarity rewrite them replacing N with A and F with B everywhere.

Eq 11-8 is for far-end-fixed. The author doesn't write it out but states that if, in addition, the near end is not fixed we can simply substitute M_N=(FEM)_N=0 in eq 11-8. Call this eq 11-8₀.
In fig 11-8b, we have A fixed, B not fixed. This means we can use eq 11-8₀ with A as F and B as N. This gives the second eq of 11-9. We can also use eq 11-8 but with A as N and B as F. This gives the first eq of 11-9.

 

[fonseh]

think what is confusing you is the near-far notation in eqs 11-9. In both of them, the near end is A and the far end is B, so for clarity rewrite them replacing N with A and F with B everywhere.

so , do you mean both MN = 2Ek( 2 theta_N + theta_F -3phi) + (FEM)N and 0 = 2Ek( 2 theta_N + theta_F -3phi) + 0 describes about the near end is A and the far end is B in case 11-8(b) ?

 

[fonseh]

The author doesn't write it out but states that if, in addition, the near end is not fixed we can simply substitute MN=(FEM)N=0

why if the near end is not fixed , we could use MN=(FEM)N=0 ??

 

[fonseh]

We can notice that the case in 11-8b yield equation (11-10) , so , is it possible that the equation 11-10 to be used and applied in case 11-8(a) in which both ends are roller / pin supported ?

 

[haruspex]

why if the near end is not fixed , we could use MN=(FEM)N=0 ??

Because a pin or roller cannot exert a torque.

 

[fonseh]

Because a pin or roller cannot exert a torque.

post #17 , can you try to reply ?
We can notice that the case in 11-8b yield equation (11-10) , so , is it possible that the equation 11-10 to be used and applied in case 11-8(a) in which both ends are roller / pin supported ?

 

[haruspex]

We can notice that the case in 11-8b yield equation (11-10) , so , is it possible that the equation 11-10 to be used and applied in case 11-8(a) in which both ends are roller / pin supported ?

I think so, just setting M_N=(FEM)_N=0. That would lead to θ_N=ψ. Does that make sense? I'm unsure how ψ is defined. Is it something to do with the conjugate beam?

By the way, I think I found your text online at http://mrkhademi.com/chapter11.pdf.

 

[fonseh]

I think so, just setting MN=(FEM)N=0. That would lead to θN=ψ. Does that make sense? I'm unsure how ψ is defined. Is it something to do with the conjugate beam?

so , for both case 11-8(a) and 11-8(b) , we can use equation 11-10 ?

 

[haruspex]

so , for both case 11-8(a) and 11-8(b) , we can use equation 11-10 ?

I can only repeat my post #20. That looks right, but in the case of neither end fixed M_N and (FEM)_N would both be zero, right? That would imply θ_N=ψ. Does that equality make sense? I cannot tell because I do not understand how ψ is defined. Can you explain that to me?

 

[fonseh]

I can only repeat my post #20. That looks right, but in the case of neither end fixed M_N and (FEM)_N would both be zero, right? That would imply θ_N=ψ. Does that equality make sense? I cannot tell because I do not understand how ψ is defined. Can you explain that to me?

ψ is the settlement of the support , which can be ignored since there's no statement stated that the support undergo settlement

so , for both case 11-8(a) and 11-8(b) , we can use equation 11-10 ?

 

[haruspex]

ψ is the settlement of the support , which can be ignored since there's no statement stated that the support undergo settlement

so , for both case 11-8(a) and 11-8(b) , we can use equation 11-10 ?

No, I'm not happy with the implication. Having read the online doc, I see that you are right about ψ; we can take it as zero for present purposes.
As I understand it, M_N and (FEM)_N are the moments exerted by the supports at the near and far ends respectively. (Can you confirm that?). So with both ends pinned, these should be zero. Putting that in the equation leads to θ_N=ψ=0, which makes no sense.

I will try reading and digesting the whole online doc, but it will take some time. Other than trying to help you with this subject, I have no background in it.

 

[fonseh]

As I understand it, MN and (FEM)N are the moments exerted by the supports at the near and far ends respectively. (Can you confirm that?).

I think so

 

[haruspex]

As a first step, I have calculated some simple cases from first principles.
In each, there is a load P on a uniform beam length L, coefficient E. The deflection angle at a pinned end is cPL²/E for some constant c.
Loads on supports are F_A, F_B; moment at A is M_A if A fixed.
1. Ends pinned, load central: c = 1/16. F_A=F_B=P/2.
2. Ends pinned, load distributed uniformly: c=1/24, etc.
3. End A fixed horizontal, end B pinned: c=1/32, F_A=11/16 P, F_B = 5/16 P, M_A=3/16 PL.

I hope to be able to use these to validate various equations in the text.