Statics · Beams
Statics · Beams
What is inside the girder
The support reactions say what the crane runway girder passes on at its ends. They say nothing about how heavily it is loaded in the middle — and that is exactly what it is designed for. This module cuts the same girder open and develops the three internal forces normal force, shear force and bending moment: how to obtain them from equilibrium, how they run along the length, and how to spot their maximum.
To the exercises →Solve this chapter's task to keep reading.
The internal forces
On the right stands the crane runway girder from the module "Support reactions": length , self-weight as a distributed load , the trolley with at . The support forces are already known: and .
Now for the interior. Cutting at an arbitrary position and removing one half, the remaining part must be in equilibrium on its own. What the removed half transmitted before now appears as internal forces at the section:
- the normal force along the beam axis,
- the shear force transverse to it,
- the bending moment about the transverse axis.
No force acts along the axis of the crane runway girder — therefore everywhere. That leaves and , and both depend on . Drag the section position in the view on the right across the whole girder once.
Statics ends here and strength of materials begins exactly here: the bending stress follows from the bending moment and the cross-section, the shear stress from the shear force. Designing the girder therefore requires not the support forces but the largest bending moment and the position at which it occurs.
Sign convention
A cut produces two faces, and the internal forces act on both — oppositely, by actio = reactio. For a sign to mean anything, a fixed coordinate system is needed first. The textbook defines it thus: the axis coincides with the longitudinal beam axis and points to the right, while is directed downwards; the axis then points out of the plane of the drawing.
Which face is which is decided by the normal vector — it is perpendicular to the cut face and points outwards from the interior of the body. The face whose normal vector points in the positive direction is called the positive face; if it points in the negative direction, it is the negative face.
This single rule covers all three internal forces. The left part of the girder carries the positive face — there a positive points downwards, because points downwards. On the right part it is the other way round. That is precisely why both subsystems yield the same values although the arrows point in different directions.
In which direction does a positive shear force point at the negative face?
On beams the loads almost always act downwards. If points the same way, the usual loads are positive and fewer minus signs appear. The price is a coordinate system that differs from the rest of statics — which is why the axis cross appears in every drawing in this module: x to the right, z downwards.
Internal forces from equilibrium
The procedure is always the same: determine the support forces, cut at position , isolate one subsystem, write down the three equilibrium conditions. The internal forces are then the only unknowns.
The solution stepper does this for a section in the first field, that is for . There the left part carries only the support force and the part of the distributed load that reaches as far as the section.
Both lead to the same result — take the one with fewer forces. On the crane runway girder that is, for small the left part (only ), for large ones the right part (only ). Computing both supplies the check along the way.
Beam under point loads
Something characteristic happens where the trolley stands: the shear force jumps by the magnitude of the single force — before the load application , behind it ; the jump is exactly . The bending moment on the other hand stays continuous; it merely gets a kink.
That is no coincidence but a consequence of equilibrium: a force introduced over zero length can make a force quantity jump, but not a moment — because its lever arm is zero at the point of application itself.
Hands-on task: Set the section position so that the bending moment becomes largest. Careful — it is not the position of the trolley.
Once you have found the position, check here.
What does an externally applied single moment?
Relation between loading and internal forces
Cutting out an infinitesimal beam element of length and writing the equilibrium conditions for it yields a relation that shortens all the arithmetic. The textbook calls them the local equilibrium conditions:
Both statements can be read straight off the view on the right: where the distributed load is constant, falls linearly; and the slope of the line is at every position the value of . From this follows the practically most important consequence:
On the crane runway girder this occurs at — not under the trolley at 1.50 m. There .
For the support reactions a distributed load may be replaced by its resultant — the result is correct. For the internal forces this must not: the distributed load is not to be combined into a resultant before the section, otherwise precisely the dependence that matters disappears. First cut, then combine the share up to the section.
How is the position of the largest bending moment recognised?
Integration and boundary conditions
The local equilibrium conditions can also be used the other way round: if the distributed load is known, and by integration.
The two constants of integration follow from the boundary conditions — and those depend on the supports. A pinned support cannot carry a moment, so there . A free beam end transmits neither force nor moment, so and . A fixed support, by contrast, prescribes nothing: there and are exactly the support reactions being sought.
Both supports of the crane runway girder are pinned — which is why the line in the view on the right starts and ends at zero. These two zeros are at the same time the best check on any hand-computed moment diagram.
For a girder with few loads, cutting at position is faster. For continuously varying loads — a triangular or trapezoidal load — integration is the shorter route, because the resultant of the load share need not be re-determined at every section. Both are the same calculation in a different order.
Matching conditions with several fields
As soon as the loading changes abruptly — a single force, a single moment, the start or the end of a distributed load — different functions apply to the left and to the right of it, so the beam is divided into different functions apply. The beam is therefore divided into fields and separate constants of integration are determined in each field.
The fields are held together by the matching conditions at the field boundaries. They say exactly what chapter 04 showed on the stage:
- One single force makes jump by its magnitude, while stays continuous (kink).
- A single moment makes jump, while stays continuous.
- Where nothing is applied, both pass over continuously.
A special case are connecting elements. A hinge in the beam — as on the Gerber beam from the module "Support reactions" — cannot transmit a moment. There , and that is the additional equation that makes the Gerber beam solvable in the first place.
What holds at a moment hinge in the beam?
Föppl bracket
Dividing into fields is systematic but laborious: with three fields, six constants of integration have to be determined. The Föppl bracket writes the same function in one expression — then no division into several fields is needed any more.
The angle brackets therefore switch on only at position . Integration and differentiation work as with ordinary brackets:
With that, the bending moment of the crane runway girder can be written down over its whole length in one line. The stepper builds it up.
Hands-on task: Move the trolley in the view on the right so that both fields become equally long — then the bracket sits exactly at mid-span.
Once the trolley is at mid-span, check here.
From the loading to the moment diagram
Because and holds, the shape of the diagrams depends directly on the form of the loading. Every integration raises the degree by one:
| loading | shear force | bending moment |
|---|---|---|
| no | constant | linear |
| constant | linear | quadratic |
| linear (triangular load) | quadratic | cubic |
| single force | jump | kink |
| single moment | continuous | jump |
With this table, diagrams can be sketched without writing down a single function: enter the known values at the edges and at the load application points and connect them with a curve of the right degree. That is exactly what the second exercise practises.
Hands-on task: Set the distributed load to zero. What form does the moment diagram then have?
Once only the trolley acts, check here.
What degree does the moment diagram have under a triangular load?