Computational Engineering / Finite-element building blocks

Restraints and releases

How the model is held to the world, and what happens inside it — with the moments and deflections for every case.

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Why this matters

Support restraints and member releases determine which actions a model can transfer. Compare their definitions with the intended connection behaviour, and use simple load cases to check the resulting moments and displacements.

By the end of this lesson you should be able to

Distinguish a restraint from a release
Compute the moments and deflections for three sets of end conditions
Show that a release is a spring of zero stiffness
Recognise which restraint mistakes stop the model and which do not

Two different things

A restraint joins the model to the world. It is a statement about what is outside the model, and it produces a reaction.

A release describes what happens inside the model, between two members at a joint. It produces no reaction and transfers no force in the released direction.

Confusing them is common because both are set on the same screen in most software. The test is whether the thing being described is outside the model or inside it.

What end conditions do

One 8 m span, 12 kN/m, EI = 21 000 kN·m²:

End conditions	Support moment magnitude	Midspan moment	Midspan deflection
Simply supported	0	96.0 kN·m	30.48 mm
Propped cantilever	96.0 kN·m	48.0 kN·m	12.19 mm
Fixed both ends	64.0 kN·m	32.0 kN·m	6.10 mm

Three numbers worth carrying:

Fixing both ends divides the deflection by exactly five — 5wL⁴/384EI against wL⁴/384EI.
The peak moment moves from midspan to the supports, and it falls, from wL²/8 to wL²/12.
The propped cantilever's fixed end takes wL²/8 — more than either end of the fully fixed beam.

That last one surprises people. Removing restraint from one end increases the moment at the other.

A release is a spring of zero

It is tempting to treat 'released', 'pinned' and 'sprung' as three different settings. They are one parameter.

A rotational spring of stiffness s between the member end and the node reduces exactly to a release as s → 0, and to full continuity as s → ∞. There is no discontinuity anywhere between them, and real connections live in the middle. Module 8 sweeps that parameter and finds the middle is where all the action is.

One consequence is worth stating plainly. A release should be implemented by condensing the freedom out, not by inserting a very small stiffness. A condensed release carries exactly zero moment; a small-stiffness release carries a small moment, and 'small' is relative to numbers you have not checked.

Four ways to get the supports wrong

The same 8 m beam, meant to be simply supported:

Model	Deflection	What happens
Pin and roller	30.48 mm	Correct
Pinned at both ends	30.48 mm	Solves. Identical in a linear analysis
Rotation held at both ends	6.10 mm	Solves. Exactly one fifth of the deflection
Vertical only at both ends	—	Mechanism. Will not solve

Three of the four solve. Only one of them is caught.

The third is the dangerous one. 'Fully restrained' is a phrase that means laterally restrained to a designer and rotation held to a modelling dialogue, and the consequence is a beam five times stiffer than the one being built. Nothing warns, and 6 mm is an entirely plausible deflection for an 8 m beam.

The second is worth understanding too. Adding horizontal restraint at both ends changes nothing at all in a linear analysis, because a transversely loaded beam develops no axial force in linear theory. Under the stated transverse-only linear load case the extra axial restraint has no effect. Temperature or imposed-strain loading can reveal it in a linear analysis; large-deflection effects require the appropriate nonlinear formulation.

The cost of a wrong end condition

\delta_{\text{simple}} = \frac{5wL^4}{384EI}, \qquad \delta_{\text{fixed}} = \frac{wL^4}{384EI} \;\Rightarrow\; \frac{\delta_{\text{simple}}}{\delta_{\text{fixed}}} = 5

What it calculates: The factor by which a model over-states a beam's stiffness when rotation is held at both ends by mistake

w: uniformly distributed load (kN/m)
L: span (m)
EI: flexural rigidity (kN·m²)

This assumes

A prismatic member under a uniform load, linear elastic

In plain terms: The factor is exactly five because both expressions share wL⁴/384EI and differ only by the 5. It is worth carrying as the size of the error a mis-set restraint produces.

Try it

Support and release laboratory

Three sets of end conditions, and four ways to get the supports wrong. Only one of the four refuses to run.

Span (m)

Load (kN/m)

EI = 21 000 kN·m² throughout.

End conditions
Ends	Support M	Midspan M	Deflection
simple	0.00	96.00	30.48 mm
propped	96.00	48.00	12.19 mm
fixed	64.00	32.00	6.10 mm

Moments in kN·m. Fixing both ends divides the deflection by exactly five.

Four ways to set the supports, all meant to be simply supported
Model	Deflection	vs correct	Solves?
Pin and roller	30.48 mm	× 1.000	yes
Pinned at both ends	30.48 mm	× 1.000	yes
Rotation held at both ends	6.10 mm	× 0.200	yes
Vertical only at both ends	—	—	no — mechanism

Three of the four solve. The one that reports a fifth of the true deflection raises nothing at all, and ‘fully restrained’ means one thing to a designer and another to a dialogue box.

Freedoms the model gave no stiffness

simple: L rz, R rz
propped: R rz
fixed: none

What this shows: Fixing both ends divides the deflection by exactly five, and three of the four common restraint mistakes solve cleanly.

Worked example

The cost of 'fully restrained'

Given

An 8 m secondary beam, 12 kN/m, EI = 21 000 kN·m²
Meant to be simply supported; the modeller ticks 'fully restrained' at both ends

Find

What the model reports and what is actually built

Practice

A uniformly loaded span deflects 30.48 mm when simply supported. What does it deflect when both ends are fully fixed?

Practice

An 8 m span carries 12 kN/m. What is the fixed-end moment at the built-in end if the other end is propped?

Predict first

A simply supported beam model has horizontal restraint accidentally left on at both ends. What does a linear analysis report?

Summary

A restraint joins the model to the world; a release acts inside it
Fixing both ends divides the deflection by exactly five
A propped cantilever's fixed end takes wL²/8, more than a fully fixed beam's wL²/12
A release is a spring of zero stiffness — condense it, do not fake it with a small number
Three of the four common restraint mistakes solve cleanly
'Fully restrained' means two different things to a designer and a dialogue box