Statics · Beams
Condition: The theoretical fundamentals develop the three internal forces, the sign convention via the normal vector, the local equilibrium conditions and the Föppl bracket on the crane runway girder.
Open theoretical foundations →First sketch every girder on paper: isolate it, determine the support forces, mark off the fields, enter the diagrams. The part steps here are your check.
Exercise 1 · Concept check
A beam is cut. At the negative face, a positive bending moment points …
The rule reads, in one sentence: Positive internal forces point, at the positive (negative) face, in the positive (negative) coordinate direction. Which face is present is decided by the normal vector .
That is exactly why the arrows on the two faces of a section point oppositely — and still give the same numerical value.
Assign to each loading the form of the corresponding moment diagram.
runs linearly
runs quadratically
runs cubically
jumps
Every integration raises the degree by one: . No loading yields a constant shear force and a linear moment diagram, a constant load a quadratic one, a linear load a cubic one.
Single force and single moment fall outside this series — they produce jumps: the force in , the moment in .
Where does the largest bending moment lie on a girder under several loads?
From it follows: has an extreme value where . On the crane runway girder of the theory page this position is at 2.50 m — the trolley stands at 1.50 m.
For a jump of the shear force through zero, the maximum lies at the jump itself; that too follows from the same relation.
A girder on two supports carries the support force and a single force for . Which expression describes over the whole length?
The ordinary bracket in a) would also contribute to the left of — but there does not act yet. The angle bracket switches on only at there.
The exponent must be be: jumps from 0 to 1 and thereby describes the shear forcejump, not the kink in the moment.
Conclusion: three rules that carry everything
Sign convention, and — nothing more is needed. The forms of the diagrams, the position of the maximum and the jump rules at the field boundaries all follow from the two local equilibrium conditions.
Exercise 2 · Travelling load across a bridge
Across a bridge of span drives a crane of self-weight . The front axle is loaded with , the rear axle with the remainder; the axle spacing is . Considered is the position in which the front axle at . The bridge rests on a pinned support at the left and a roller support at the right.
From it follows that .
The support closer to the crane carries more — a good first plausibility check.
Draw the shear force diagram . Place points by tapping, drag them into position; a double tap deletes a point.
Without a distributed load, field by field constant and jumps only at the axles:
At the right-hand edge, — the second good check.
Draw the bending moment diagram . Without a distributed load it consists of straight pieces.
The slope of each piece is the value of in this field:
At both pinned supports, — if your diagram starts or ends elsewhere, the error is in the support forces.
— at the position where the shear force jumps from on so that it passes through zero.
For , under the front axle — not at mid-span and not under the rear axle. The change of sign of decides, not the position of the loads by itself.
No distributed load acts between the axles — so is constant there and straight.
Start the Q diagram with and end with . In between it jumps downwards at each axle by that axle's load.
For the M diagram you need only four points: both supports (zero there) and both axles. The values in between follow from the areas under the Q diagram.
Conclusion: the Q diagram is the derivative of the M diagram
Putting both diagrams side by side shows directly: where is positive, ; where jumps, rises; and where changes sign lies the apex.
With a travelling crane this changes with every position. For design one therefore looks for the position in which becomes largest — that is the task behind the task.
Exercise 3 · Cantilever under a triangular load
A cantilever of length is fixed at the right end; the left end is free. It carries a triangular load that starts at zero at the free end and reaches at the fixed end. The coordinate is counted from the free end , so .
The area under the load line is a triangle: .
The centroid of a triangular area lies at two thirds from the apex: from the free end — that is in front of the fixed end.
That is exactly the tool from the module “Centroid”, applied here to a load area.
Magnitude of the shear force at position , symbolically.
Allowed symbols: q0, x, l
Notation help: power with ^, fraction with /, brackets allowed.
The shear force carries the load share between the free end and the section, that is the triangular area up to :
Check at the fixed end:
Magnitude of the bending moment at position , symbolically.
Allowed symbols: q0, x, l
Notation help: power with ^, fraction with /, brackets allowed.
The resultant of the load share acts one third of in front of the cut:
That matches the table: the load is linear, so is quadratic and cubic.
As a check by hand: total load times lever arm gives the same.
Combine the distributed load only after the cut — otherwise the dependence on .
The load share up to is again a triangle: height , width . Its centroid lies one third of in front of the cut.
Conclusion: one formula instead of many sections
On a cantilever no support is needed first — the free end supplies the boundary conditions and for free. That is why it pays to count from there.
Both expressions hold for any length and any load magnitude. Once you have them, the fixed-end moment of any cantilever under a triangular load can be read off directly.
Report
The completion status of all three tasks is exported as a PDF report with a verification token — checkable via the platform's report verification.
| Exercise | Status |
|---|---|
| Exercise 1: Concept check | Open |
| Exercise 2: Travelling load | Open |
| Exercise 3: Triangular load | Open |