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Statics and trusses

A truss is a set of straight bars joined at their ends to form triangles, the shape that cannot deform without changing the length of its sides. It is used in bridges, roofs and towers because it carries large loads with little material. Analyzing it means finding the reactions and the force in each bar, and equilibrium is enough for that: it is the first big topic of statics.

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Assumptions and two-force members

The ideal-truss model assumes that the bars are joined by frictionless pins, that loads act only at the joints and that the weight of the bars is negligible (or shared out to the joints). With that, each bar takes forces only at its two ends and, to be in equilibrium, those two forces must be equal, opposite and along the bar: it is a two-force member, in tension or in compression.

Sign convention: a bar in tension pulls on its joints; in compression, it pushes them.

A trick that avoids sign errors: assume all bars in tension (arrows leaving the joint). Those that come out negative are in compression. That is what the simulator does.

Equilibrium and determinacy

Each joint is a point in equilibrium: in the plane it gives two equations ( $\sum F_{x} = 0$ , $\sum F_{y} = 0$ ), in space three. Comparing the unknowns (the $m$ bar forces and the $r$ reactions) with the equations ( $j$ joints) tells you, before calculating, whether statics is enough:

Plane	Space	Truss
$m + r = 2 j$	$m + r = 3 j$	statically determinate: equilibrium alone solves it
$m + r > 2 j$	$m + r > 3 j$	statically indeterminate: deformations are needed (structural analysis)
$m + r < 2 j$	$m + r < 3 j$	unstable: it is a mechanism and moves

The count is necessary but not sufficient: a truss with the right numbers can still be unstable if its bars or supports are badly arranged (for example, all rollers parallel). Supports provide what they prevent: a pin 2 reactions, a roller 1; in space, a ball joint 3.

Method of joints

Find the reactions from the equilibrium of the whole truss (moments about one support first), if you can.
Start at a joint with at most two unknowns and write $\sum F_{x} = 0$ and $\sum F_{y} = 0$ , with all bars in tension.
The components come from the geometry: a bar that rises 4 over a run of 3 has direction cosines $3/5$ and $4/5$ . You do not need to compute angles.
Move to the neighboring joint using the forces already found, and continue to the end. The equations left over at the end are a check.

Forces of example 1 (T = tension, C = compression).

Zero-force members

Before starting it pays to look for the bars that, under that loading, carry nothing. Two rules find them by inspection:

The two rules for zero-force members.

If a joint has only two non-collinear bars and no load or support, both are zero-force (each would have a component that nothing balances).
If a joint has three bars, two of them collinear, and no load or support, the third one is zero-force (and the two collinear ones carry the same force).

A bar that carries nothing under this load is not unnecessary: it stabilizes the truss, shortens the buckling length of the compressed bars and works under other loads (wind, a moving load).

Method of sections

For the force in a bar in the middle of a large truss, going joint by joint takes long. The method of sections goes straight there: cut the truss through the bar you want and at most two others, and write the equilibrium of one of the parts, which is a rigid body with three equations.

Take moments about the point where the other two cut bars cross: they drop out and the equation gives the force you want directly.
If the other two are parallel (they never cross), use $\sum F$ in the direction perpendicular to them.
Choose the part with fewer loads: either one gives the same result.

Cut of example 2 through bars IJ, ID and CD. ID and CD meet at O = D: moments about that point give F_IJ.

Space trusses

In three dimensions the tetrahedron plays the role of the triangle. Each joint gives three equations, and the force in each bar is written with its unit vector: for the bar from D to A, $u_{D A} = \frac{r _{A} - r _{D}}{∣ r _{A} - r _{D} ∣}$ . The equilibrium of joint D is then

i \sum F_{i} u_{i} + P = 0 (3 ecuaciones)

In space the method of joints is almost always used: a cut leaves up to six unknowns.

Common mistakes

Drawing the arrows of a bar at a joint in the «direction of the load» instead of assuming tension: the signs get mixed up.
Starting at a joint with three unknowns in the plane: it cannot be solved.
Using the length of the bar instead of its components when splitting the force.
In the method of sections, cutting four unknown bars, or taking moments about a point that eliminates none.
Removing the zero-force members from the design.

Worked examples

Example 1Method of joints with an inclined load

Truss ABCD is pinned at A and rests on a roller at C. At D a load acts with components 6 kN to the right and 8 kN downward, and at B 12 kN downward. Find the reactions and the force in each bar.

Solution

Determinacy: $m + r = 5 + 3 = 8 = 2 \cdot 4$ , statically determinate.
Moments about A (counter-clockwise +): $6 C_{y} - 6 \cdot 4 - 8 \cdot 3 - 12 \cdot 3 = 0 \Rightarrow C_{y} = 14$ kN. Forces: $A_{x} = - 6$ kN (to the left) and $A_{y} = 20 - 14 = 6$ kN.
Joint A: $\sum F_{y}$ : $6 + \frac{4}{5} F_{A D} = 0 \Rightarrow F_{A D} = - 7.5$ kN (C). $\sum F_{x}$ : $- 6 + F_{A B} + \frac{3}{5} (- 7.5) = 0 \Rightarrow F_{A B} = 10.5$ kN (T).
Joint C: $\sum F_{y}$ : $14 + \frac{4}{5} F_{D C} = 0 \Rightarrow F_{D C} = - 17.5$ kN (C). $\sum F_{x}$ : $- F_{BC} - \frac{3}{5} (- 17.5) = 0 \Rightarrow F_{BC} = 10.5$ kN (T).
Joint B: $\sum F_{y}$ : $F_{B D} - 12 = 0 \Rightarrow F_{B D} = 12$ kN (T). Check at D: $\sum F_{x} = 6 - \frac{3}{5} (- 7.5) + \frac{3}{5} (- 17.5) = 0$ ✓.

A_{x} = - 6

A_{y} = 6

C_{y} = 14

kN; AB = BC = 10.5 T, AD = 7.5 C, DC = 17.5 C and BD = 12 T (kN).

Solve it in Simuveliq

In Engineering → Statics and trusses, truss type 2D · plane; in Examples, «Load truss» → Simple truss with an inclined load.
The drawing colors each bar (tension, compression) with its force; the cards give Determinacy («Determinate»), Largest tension 12 kN (BD) and Largest compression 17.5 kN (DC).
The «Step by step» writes the reactions and the method of joints (section 4) with these same equations, and the Bar forces and reactions table sums it all up.

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Example 2Method of sections on a Pratt bridge

A Pratt bridge with 6 panels of 3 m and a depth of 4 m carries 20 kN at each interior joint of the bottom chord (B to F), pinned at A and on a roller at G. Find the forces in IJ, ID and CD with a single cut.

Solution

By symmetry, $A_{y} = G_{y} = \frac{5 \cdot 20}{2} = 50$ kN.
Cut through IJ, ID and CD; left part (joints A, B, C, H, I), with the three bars in tension.
Moments about D (9, 0), where ID and CD meet (clockwise +): $50 \cdot 9 - 20 \cdot 6 - 20 \cdot 3 + 4 F_{I J} = 0 \Rightarrow F_{I J} = - 67.5$ kN (C).
Moments about I (6, 4), where IJ and ID meet (clockwise +): $50 \cdot 6 - 20 \cdot 3 - 4 F_{C D} = 0 \Rightarrow F_{C D} = 60$ kN (T).
Vertical forces: $50 - 20 - 20 - \frac{4}{5} F_{I D} = 0 \Rightarrow F_{I D} = 12.5$ kN (T). In a Pratt the diagonals work in tension and the verticals in compression.
Along the way: at J, IJ and JK arrive collinear together with the vertical DJ, with no load, so DJ is zero-force (rule 2).

F_{I J} = 67.5

kN (C),

F_{C D} = 60

kN (T) and

F_{I D} = 12.5

kN (T); DJ = 0.

Solve it in Simuveliq

In Examples, «Load truss» → Pratt bridge (6 panels).
With the Move tool, click bar IJ: the «5. Method of sections · bar IJ» step draws the cut, takes moments about O = (9, 0) and gives $F_{I J} = - 67.5$ kN, and also $F_{C D}$ and $F_{I D}$ .
The «3. Zero-force bars» step points out DJ, and the Zero-force bars card counts it.

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Example 3Cantilever truss: zero-force members

The cantilever truss is pinned to a wall at A and B and carries 15 kN at F and 10 kN at D. Find the zero-force members and the forces in CF and DF.

Solution

Determinacy: 8 bars and 4 reactions (two pins) against $2 \cdot 6 = 12$ equations: statically determinate, although the reactions cannot come from global equilibrium alone (there are 4) and are found at the end.
Joint E: only CE (horizontal) and EF (vertical) arrive, with no load or support: by rule 1, $F_{CE} = F_{EF} = 0$ .
Joint F, with EF = 0: $\sum F_{y}$ : $\frac{1}{2} F_{CF} - 15 = 0 \Rightarrow F_{CF} = 21.21$ kN (T). $\sum F_{x}$ : $- F_{D F} - \frac{1}{2} F_{CF} = 0 \Rightarrow F_{D F} = - 15$ kN (C).
Continuing through C, D, B and A gives the rest: BC = 15 T, CD = 15 C, BD = 35.36 T and AD = 40 C.

CE and EF are zero-force; CF = 21.21 kN (T) and DF = 15 kN (C).

Solve it in Simuveliq

In Examples, «Load truss» → Cantilever truss (wall).
Bars CE and EF appear as zero-force in the drawing; the «3. Zero-force bars» step explains the rule, and the table marks them with ★.
Step «2» warns that with 4 reactions global equilibrium is not enough and that they come out of the method of joints.

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Forces of example 3 in kN: the dashed lines are the zero-force members.

Example 4Tripod: a space truss

A tripod has its legs on ball joints at A (0, 0, 0), B (4, 0, 0) and C (2, 3, 0) m, and its head at D (2, 1, 4) m. At D a load of 5 kN in

+ x

and 30 kN downward acts. Find the force in each leg.

Solution

Unit vectors from D: $u_{D A} = \frac{( - 2 , - 1 , - 4 )}{21}$ , $u_{D B} = \frac{( 2 , - 1 , - 4 )}{21}$ , $u_{D C} = \frac{( 0 , 2 , - 4 )}{20}$ .
Equations at D: $x$ : $\frac{2 ( F _{B} - F _{A} )}{21} + 5 = 0$ ; $y$ : $- \frac{F _{A} + F _{B}}{21} + \frac{2 F _{C}}{20} = 0$ ; $z$ : $- \frac{4 ( F _{A} + F _{B} )}{21} - \frac{4 F _{C}}{20} - 30 = 0$ .
Substituting the $y$ equation into the $z$ one: $- \frac{12 F _{C}}{20} = 30 \Rightarrow F_{C} = - 11.18$ kN. Then $F_{A} + F_{B} = - 22.91$ and $F_{B} - F_{A} = - 11.46$ .
$F_{A} = - 5.728$ kN and $F_{B} = - 17.18$ kN: all three legs in compression, the most loaded one on the side the horizontal load pushes toward.

AD = 5.728 kN (C), BD = 17.18 kN (C) and CD = 11.18 kN (C).

Solve it in Simuveliq

Truss type 3D · space; in Examples, «Load truss» → Tripod (3 ball joints).
The Determinacy card shows $m + r = 3 + 9 = 12 = 3 j$ , and Largest compression 17.18 kN in BD. Rotate the view by dragging and select each leg to see its force.

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Statics and trusses

Assumptions and two-force members

Equilibrium and determinacy

Method of joints

Zero-force members

Method of sections

Space trusses

Common mistakes

Worked examples

Example 1Method of joints with an inclined load

Solution

Solve it in Simuveliq

Example 2Method of sections on a Pratt bridge

Solution

Solve it in Simuveliq

Example 3Cantilever truss: zero-force members

Solution

Solve it in Simuveliq

Example 4Tripod: a space truss

Solution

Solve it in Simuveliq

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