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Statics · Concurrent forces

Statics · Concurrent forces

All forces through one point

If several forces act at the same point — or if their lines of action pass through a common point — they form a concurrent force system. Two questions suffice for them: that of the resultant and that of equilibrium. This module addresses both using one recurring example (called the “reference example” here): a Wall bracket made of two bars, with a load hanging at its node. The path there is the workflow of every statics problem: isolate, draw the free-body diagram, set up the equilibrium conditions, solve.

01 · Guiding system

The wall bracket — two bars carry a load

Here, the bracket means the entire wall support formed by the two bars. On the right, a load at the node of a wall bracket: Bar 1 runs horizontally to the wall, Bar 2 braces diagonally at the angle downwards. At their ends, both bars are pin-jointed and loaded only there — such a bar is called a two-force member and can only transmit forces along its axis : tension or compression, nothing else. It passes this force on unchanged from one end to the other.

The bracket is designed via its dimensions: the Projection — how far the node projects from the wall — and the vertical distance between the two wall connections. The angle is then no longer a chosen value but a Result of the geometry:

Adjust , and with the sliders: the dimension lines in the view on the right move along, and the cell shows the resulting angle. The value box also gives the two bar forces — as readings to experiment with; derived they are derived in chapter 05.

Fundamentals · Concurrent force system

A force group is called concurrent if the lines of action of all forces involved pass through a common point. With respect to that point none of the forces has a lever arm — no moments arise, and for equilibrium the Force conditions suffice; moments follow in the next module. Whether and where the wall bracket meets this condition is something you find out yourself in chapter 02.

The projection and the distance between the wall connections are both doubled. The load stays the same. How do the member forces change?

02 · Isolating

The cut makes the forces visible

Now comes the most important step in statics: Isolating the body. Cutting free means isolating a piece of the structure and replacing every severed connection by the force it transmits — previously invisible internal forces become external forces on the isolated part.

Where you cut is a decision, and it determines everything that follows: at every cut the force appears that the removed part exerts on the remaining one. A cut in which the unknowns do not appear at all is of no help — which spot is the right one for the question of the two bar forces is something you decide yourself in a moment.

Fundamentals · Tension and compression

A tension bar is stretched; its internal force pulls outwards at both cut faces — the sign is positive. A compression bar is compressed; the force presses on the cut faces — under the tension assumption its value becomes negative. So the sign is no arithmetic whim but the answer to the question: does the bar pull or does it brace?

Hands-on task: Practise isolating a joint of the wall bracket in the view on the right. Choose the point where both member forces and the load form a concurrent force system. Isolate this joint from both members and the load suspension, then draw the three forces. A complete free-body diagram obtained by cutting at the wall connections is also a possible approach; here, we specifically practise isolating the joint. For the member forces the sense of direction is free: you may assume tension (pointing away from the node) or compression (pointing towards it) — they only have to lie on the member axis. Which assumption was right is revealed later by the sign. Only the load is fixed.

Convention from here on: The usual choice is the tension assumption — bar forces are assumed to be tensile, so the arrows point away from the cut. This is not compulsory; the assumption only has to be applied consistently. If a negative value comes out, the force acts opposite to the assumption: the bar is in compression. The view on the right uses this convention from now on.

03 · Composition

Composition of forces in the plane

Three forces now act at the isolated joint. Pick two of them — the load and the bar force : which a force could replace both?

The answer is given by the parallelogram law of forces: if two forces act at a point and act at one point, they can be replaced equivalently by a single force. The resultant results in magnitude and direction as the diagonal of the parallelogram spanned by and spanned parallelogram. Determining the resultant is also called reduction — the force group is reduced to a single equivalent force.

For the drawing, half the half parallelogram suffices, that is a force triangle. This does lose the fact that the lines of action pass through one point — in exchange the construction extends to any number of forces: in the the forces are chained in any order, and points from the start point the end point of this force polygon (force polygon).

Analytically the problem is solved from a sketch of the force triangle. Known are the magnitudes , and the angle is between them; the law of cosines gives the magnitude of the resultant, the law of sines the angle of its line of action relative to . Mind the notation: is the angle between two forces — remains, in this module, the inclination of bar 2. At this point the textbook writes , because no structure sits next to it there:

With many forces one calculates instead of drawing: all x components and all y components are added separately, and the two sums give magnitude and direction of the resultant.

Fundamentals · The parallelogram law of forces

The parallelogram law of forces is not a calculation rule but an empirical fact — an axiom: the effect of two forces acting at one point is equivalent to the effect of a single force following from the parallelogram construction. Geometrically this corresponds to vector addition. The magnitudes may only be added when both forces point the same way.

Seven steps on the guiding system: parallelogram, diagonal, force triangle, law of cosines, numerical value and finally the direction via the law of sines.

Two equally large forces act at 90° to each other. How large is the resultant?

04 · Decomposition

Decomposition of forces, component representation

Just as forces can be composed, they can also be decompose: a force is replaced by two forces with prescribed lines of action — one of the two directions is drawn through the start and the end point respectively. The two forces are called Components with respect to these directions. Most commonly the coordinate axes are used; the inclined member force decomposes at the angle in

The cosine belongs to the adjacent side (horizontal), the sine to the opposite side (vertical) — with which sign the components enter the balance follows from the sense of direction of the arrow in the free-body diagram. The derivation below builds up the decomposition step by step — the view on the right draws each component as soon as its line appears. Afterwards adjust and watch how the two components shift against each other.

Four steps from the force arrow to the component pair — each line also appears in the view on the right.

The angle gets smaller, bar 2 lies flatter. What happens to the magnitude of the bar force ?

05 · Equilibrium

Equilibrium in the plane

A concurrent force group is in equilibrium when the vector sum of all forces — that is, the resultant — is zero; such a group is called an equilibrium group. Geometrically this means: the force polygon is closed, start and end point coincide. The chain of , and closes. On the right, the scaffold is ready: the load to scale and the two lines of action from the free-body diagram. Their position follows from the chaining — the line of runs through the tip of , because that is where the next force attaches, and the one of through the start, because the chain has to arrive back there. The derivation below fills both lines with length, line by line. In terms of calculation, equilibrium of the central planar force group reads:

These two scalar equilibrium conditions are equivalent to the vector equation. With them at most two unknowns can be determined — here exactly the two member forces. Such problems are called statically determinate; if more unknowns occur, the problem is statically indeterminate.

Seven steps from the tension assumption through both balances to the closed force polygon — symbolic first, then with the example values α = 35° and G = 2.0 kN.

Hands-on task: Build the bracket via and so that the magnitude of the member force becomes exactly twice as large as the load .

06 · Space

Concurrent force systems in space

In space a third direction is added — the principle stays the same. The view on the right shows the same joint, now held by three bars:

Three equations for up to three unknowns. The only new part is the resolution: the rule of thumb "cosine to the adjacent side, sine to the opposite side" no longer carries in space, because a bar axis is no longer fixed by a single angle. Instead the direction is read off the coordinates. If the bar points from joint to point , then is the direction vector, and the force follows from its magnitude times the unit vector:

The three factors are called direction cosines — the cosines of the angles between the force and the three axes. They are not independent of one another but always satisfy

The procedure is then the same as in the plane: cut free, resolve each force into components via its unit vector, balance component by component, solve. Problem sheet 2 of the lecture trains exactly this case.

Fundamentals · Static determinacy

A plane concurrent node provides exactly two force balances — so it can carry at most two unknown bar forces, three in space. Provided the bar axes are not parallel: if two lines of action coincide, both balances say the same thing and the problem stays unsolvable despite the matching count. If a further bar is added, the joint is statically indeterminate: equilibrium alone no longer distributes the forces uniquely — only the deformation of the members decides.

A spatial node connects four bars under one load. Why can the bar forces no longer be determined uniquely from force equilibrium alone?

Control accounts

All results for the example values of the guiding system — for checking by hand.

Bar force 2

Negative under the tension assumption: bar 2 braces — it is in compression.

Bar force 1

Positive: bar 1 pulls the node towards the wall — it carries in tension.

Check in the force polygon

The three arrows form a closed triangle — equilibrium.

Sources: Gross/Hauger/Schröder/Wall, Technische Mechanik 1 (Springer) · Lecture Technische Mechanik I (HSO, summer term 2026).

On to the three exercises — from concept check to the full solution path: build the free-body diagram, set up the equations, derive the formula.

To the exercises →