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I am reading a document about beam end release here:

An end release will allow either or both ends of a beam element to rotate about or translate along one or more of the local axes of the beam.

And the article continues with the following diagram:

enter image description here

(a) Fixed-fixed beam with a hinge point at 1 and 2.

enter image description here

(b) The theoretical rotation or slope of the beams. Note how the result is discontinuous at the hinge points.

The understanding I get from the above is that if we release at a beam end, then it will become hinge ( and hence the rotation at the end is not continuous)-- and that's it. Is it true?

If this is true, then I don't understand the moment released term in RISA software: enter image description here

I am not even sure whether the beam end release and the moment end release are connected, and if yes, how.

How are the beam end release and moment end release modeled mathematically?

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  • $\begingroup$ I don't quite know what it is about the end release dialog from RISA that you don't understand. I've posted an answer to the rest of your question; if that doesn't answer your RISA dialog question then please edit in additional information. $\endgroup$ – AndyT Jul 28 '16 at 8:41
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Mathematically, a release is achieved by giving a stiffness of zero.

A release in rotation is the same as a release in moment: the only way to ensure rotation continuity from one member to another is to transfer moment between them. Hence a moment end release is a stiffness of zero against rotation about the relevant axis.

Technically, you can release a beam end in any degree of freedom (hence the final option from RISA). In practice, the vast majority of beam end releases used are moment releases, hence your first source has conflated the two.

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    $\begingroup$ Andy, I think you haven't explain what does "release a beam end" mean? How is the "release in moment" or "release a beam end" represent mathematically? $\endgroup$ – Graviton Jul 28 '16 at 8:43
  • $\begingroup$ @Graviton - Updated to include. $\endgroup$ – AndyT Jul 28 '16 at 8:50
  • $\begingroup$ Andy, you are saying that if beam end is released, then the stiffness is 0, the beam stiffness is usually a 6x6 matrix=36 numbers, so which one should be set to 0? $\endgroup$ – Graviton Jul 28 '16 at 8:54
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    $\begingroup$ @Graviton "all of the stiffness terms in the corresponding row in the stiffness matrix must be set to 0" That looks wrong. You can't make a stiffness matrix unsymmetrical by just changing the terms in a row. The right way to do this is leave introduce a new variable for each "released" DOF. But since you are not usually interested in the value of that variable, in practice you eliminate the released variable from the element stiffness (which changes all the stiffness coefficients in the element) before you assemble the stiffness matrix for the complete structure. $\endgroup$ – alephzero Jul 28 '16 at 15:44
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    $\begingroup$ @alephzero - thanks for the contribution - I freely admit that I'm out of my depth with stiffness matrices, hence my actual answer was aimed at the user's viewpoint rather than the programmer's. It may be worth you writing a full answer covering the mathematical/stiffness matrix, if you have the time? $\endgroup$ – AndyT Jul 28 '16 at 16:04
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Beam End Release and Moment End Release are the same thing: it can be thought as converting a fixed beam end to a pivoted one (see later), relaxing the constraint on rotation while enforcing a new constraint on bending moment. The idea of applying constraints to beam ends is useful in understanding the mathematical implication in changing a fixed beam end to a pivoted one, and is discussed below.

Any continuous element of a beam, for small deflections, is governed by the following differential equation:

$$\frac{d^2}{dx^2}\left( E(x) I(x) \frac{d^2 u}{dx^2}\right) + q(x) = 0$$

Where $E(x)$ is the Young's Modulus of the beam, $I(x)$ is the Second Moment of Area about the axis of bending for the beam's cross section, $u(x)$ is the upward displacement of the beam, and $q(x)$ is the downward force per unit length acting on the beam. $x$ is a coordinate such that, for a beam of length $L$, $x=0$ at one end, and $x=L$ at the other end.

This equation is a forth order differential equation, and so it requires four boundary equations. This is done by applying two constraints at each end. There are three different types of beam end, each with different constraints to be applied:

enter image description here

FIXED END

This is where the end of the beam is rigidly clamped to e.g. a wall. This end allows both shear forces and bending moments to be transmitted from the beam to the wall. The end of the beam here cannot rotate nor vertically displace. This is like the beam ends shown in the first diagram of your question before releasing either end.

If the end at $x=0$ was a fixed end, the following two constraints are applied:

No vertical displacement $u(0) = 0$

No rotation $\theta(0) = \frac{du(0)}{dx} = 0$

PIVOTED END

This is your moment-released end: the beam still may not vertical displace at the end, but it can rotate. However, since it may rotate, the beam can no longer transmit bending moments to the support, so a bending moment of zero must be set at the end. Hence moment-released.

If the end at $x=0$ was pivoted, the following constraints are applied:

No vertical displacement $u(0) = 0$

No bending moment $M(0) = 0$

Where $M(x)=-E(x)I(x) \frac{d^2 u}{dx^2}$

FREE END

This type of beam end is connected to nothing: it is free to displace and rotate at the ends. However, no bending moments or shear forces can be transmitted from the end since there is nothing attached.

If the end at $x=0$ is free, the following constraints are applied:

No bending moment $M(0) = 0$

No shear force $S(0) =0$

Where $S(x) = -\frac{d}{dx} \left(E(x)I(x) \frac{d^2 u}{dx^2} \right)$

Note that a cantilever is a beam with one fixed end and one free end, and a simply supported beam has both ends pivoted.

Upon setting two constraints to each end, the general solution to the differential equation can be obtained, allowing the vertical displacement, rotation, curvature, bending moments and shear forces to be determined.

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  • $\begingroup$ While a good primer on structural analysis, I don't see how this actually answers the question. $\endgroup$ – Wasabi Jul 29 '16 at 10:50
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    $\begingroup$ The edit improved the answer, but it still seems to only discuss supports, which end-releases can be internal hinges not associated to any supports (where $u \neq 0$). $\endgroup$ – Wasabi Jul 29 '16 at 18:16
  • $\begingroup$ The "pivot end" you mention above doesn't explain why there are discontinuity in rotation ( and hence, the hinge). Pin and hinge behave different structurally. $\endgroup$ – Graviton Aug 3 '16 at 5:54
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A beam could move in 3 axis x,y,z and bend in 3 axis and rotate along the x axis (torsion) . In all and every one of these cases the support can act in a variety of ways, allow free movement, allow no movement, allow resisted movement, impart pre configured hard constraint such as is the case for pre/post tensions re-bars or none flexible pre designed displacement to load the member in a desired setting.

Releasing one end means to release one or several or all of these constrains. I have worked with Risa it is a good FEM software but 25 years ago the version i had did not have a good menu of release. Such as spring loaded constrain, or pre set hard displacement.

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    $\begingroup$ release one or several or all of these constrains-- this is not explaining anything at all. What do you mean by releasing constraints? What is the boundary conditions/mathematical formula for this? $\endgroup$ – Graviton Aug 3 '16 at 0:27
  • $\begingroup$ @Graviton you may have to go back to study some beginners statics to be prepared to at least comprehend the basics of joint constrains. It is assumed that the questioner has the minimum knowledge to be familiar with the basic ideas. Here is just a few examples of joint release: Total release, removal of support. Moment release, joint is pin connected. Shear release, the joint is on rollers in the direction where moment is allowed, like a drawer on rollers which can't tip over. torsion release means member can roll at joint but is constrained otherwise. Like a car antenna's base connection. $\endgroup$ – kamran Aug 3 '16 at 1:15
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    $\begingroup$ @kamran - great answers within the StackExchange arena are able to draw in the less experienced along with confirming things with those who understand more of the subject matter. It appears that the OP is working through those statics that you mentioned, and your answer could afford to go into more detail to provide more explanation. $\endgroup$ – user16 Aug 3 '16 at 2:02
  • $\begingroup$ Also, perhaps you should enlightened me whether beam end release/moment end release are the same or not the same thing. $\endgroup$ – Graviton Aug 3 '16 at 5:52
  • $\begingroup$ Software companies use the terms as it suits the application, not the practice standards. From the image you have it seem they use 'moment release' as pin joint . Let's say the beam was twisted a bit in the shipping. So the support will have moment and shear in both y-y and z-z directions. I think what they say by moment end release is the Moments are set to zero, not carried to the joint but the shear is. In practice end release is removing the support at the joint. In some methods you release the end like a cantilever and then apply forces to bring it back to geometry of the joint constrain. $\endgroup$ – kamran Aug 5 '16 at 21:20

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