The BST binders · Trusses binder · Solution procedure — the order of steps · updated 27.9.2026

Trusses solution procedureMethod of joints — 9 steps

The solution procedure is the recipe. 9 steps — from the sketch to the summary table.

Static determinacy → Reactions → Zero-force members → Joints → Checks

In every step: goal, formula and order of work. A common mistake — only where it really happens, and where it falls in the flagship exercise.

Truss Straight members connected at joints, and the loads act only on the joints → every member is stretched or compressed along its length only.
Signs t = tonne (a unit of force). Moment: a force that turns about the point clockwise is +, counter-clockwise −. For example a downward force to the right of the point — clockwise.
Supports triangle = pinned support (AX, AY) · triangle on wheels = roller support (BY)

Flagship exercise: 6 joints, 9 members, vertical loads · scroll the drawing sideways

01
Sketch and labelling

Goal to mark joints, members, supports, loads and dimensions

Everything the calculation needs — written on the sketch.

Supports
Roller support One reaction — perpendicular to the surface
Pinned support Two reactions — horizontal and vertical
Labelling
Joints Letters: A, B, C…
Members Numbers: N1, N2, N3…

Order of work
1. A letter for every joint, a number for every member.
2. Reactions: at a pinned support AX and AY, at a roller support one reaction perpendicular to the surface. The direction is assumed — the sign of the result will correct it.
3. Every load — on its joint, with magnitude and direction.
4. Dimensions — only horizontal and vertical, as in the drawing.
02
Static determinacy check

Goal Make sure the truss is statically determinate

Every joint has two equations, ΣFx = 0 and ΣFy = 0. The unknowns: the force in every member and every reaction.

m + r = 2j
m
Number of members
r
Number of reactions
j
Number of joints

Order of work
1. Count m, r and j. Pinned support = 2 reactions, roller support = 1. The support joints are counted.
2. m + r = 2j → statically determinate, continue.
3. m + r < 2j → mechanism, the structure is unstable.
4. m + r > 2j → not statically determinate.
⚠ Common mistake Counting the pinned support as one reaction.
Why it is a mistake
You get m + r < 2j, and wrongly conclude that the structure is unstable.
How to avoid it
Pinned support = 2, roller support = 1.
In the flagship exercise
⁦r = 2 + 1 = 3⁩, so ⁦9 + 3 = 12 = 2 · 6⁩ — statically determinate. With ⁦r = 2⁩ you get ⁦11 < 12⁩.
03
Reactions

Goal to find the support reactions

From the equilibrium of the whole truss — before the joints.

ΣM_A = 0
ΣF_x = 0
ΣF_y = 0
A
The pinned support
lever arm
The perpendicular distance from the line of action of the force to A

Order of work
1. ΣM = 0 about the pinned support: one unknown is left — the reaction of the other support.
2. Every term = force × lever arm. For a vertical force, the lever arm is the horizontal distance from A.
3. Choose a positive direction of rotation and keep it through the whole equation.
4. ΣF_x = 0 → with vertical loads only, AX = 0.
5. ΣF_y = 0 → the vertical reaction of the pinned support.
6. Negative result → the reaction acts opposite to the assumption.
⚠ Common mistake Measuring some lever arms from A and some from B.
Why it is a mistake
A moment equation is written about one point — a lever arm from another point gives a wrong reaction.
How to avoid it
Write ΣM_A and measure all the lever arms from A.
In the flagship exercise
About A: ⁦4 t⁩ at lever arm ⁦2 m⁩, ⁦1 t⁩ at lever arm ⁦4 m⁩, so ⁦BY = 12 / 6 = 2 t⁩. Whoever takes the distance from B for ⁦1 t⁩ gets ⁦BY = 1.667 t⁩.
04
Angles and resolving the forces in the members

Goal to find the angle of every diagonal member and resolve the force in it

The force in a diagonal member is resolved into a horizontal component and a vertical component.

tan α = ΔyΔx
N_x = N · cos α
N_y = N · sin α
α
The angle of the member to the horizontal
Δx, Δy
The horizontal length and the height of the member

Order of work
1. tan α = Δy / Δx. Members with the same dimensions — the same angle.
2. Horizontal component = N · cos α, vertical component = N · sin α.
3. Each component points to the same side as the member's arrow — as in the table.
4. A horizontal or vertical member — no resolving.
Direction of the components
Down-rightF_x → rightF_y ↓ down
Down-leftF_x ← leftF_y ↓ down
Up-leftF_x ← leftF_y ↑ up
Up-rightF_x → rightF_y ↑ up

⚠ Common mistake Swapping sin and cos, or measuring α from the vertical.
Why it is a mistake
The horizontal component enters the vertical equation.
How to avoid it
α always from the horizontal: horizontal → cos, vertical → sin.
In the flagship exercise
In N1: ⁦Δx = Δy = 2 m⁩, so ⁦α = 45°⁩ and ⁦cos α = sin α = 0.707⁩. Here the swap does not show — at any other angle it changes the result.
05
Zero-force members

Goal Identify in advance the members in which the force is zero

A zero-force member carries no force under the given load. The rules — only at a joint with no load and no support.

Rule 1
2 members Not on one line → both are zero-force members
Rule 2
3 members Two on one line → the third is a zero-force member

Order of work
1. Find the joints that have no load and no support.
2. Apply the matching rule and mark the zero-force member on the sketch.
3. In the next joints the zero-force member enters as 0.
⚠ Common mistake Applying a rule to a joint that has a load or a support.
Why it is a mistake
The load or the reaction balances the members — they do not have to be zero.
How to avoid it
First ask: is there a load at the joint? A support? If so — do not apply.
In the flagship exercise
At C there is no load and no support, and N2, N6 are on one line — so ⁦N3 = 0⁩. At B two members are not on one line, but there is a support: ⁦N8 = −2.828 t⁩ and ⁦N9 = 2 t⁩.
06
Choosing a joint

Goal Choose which joint to start from and where to continue

Every joint has two equations — so at most two unknowns.

Unknowns at a joint ≤ 2

Order of work
1. Start at a support joint with at most 2 unknowns.
2. A force that was found is also known at the joint on the other end of the member.
3. Continue to a neighbouring joint with up to 2 unknowns left.
⚠ Common mistake Entering a joint with 3 unknown members.
Why it is a mistake
Two equations do not solve three unknowns.
How to avoid it
Count the unknowns before entering a joint.
In the flagship exercise
At first D has four unknowns and A has two — start at A. The order: ⁦A → C → D → E → B⁩

Supports
Roller support	One reaction — perpendicular to the surface
Pinned support	Two reactions — horizontal and vertical

Labelling
Joints	Letters: A, B, C…
Members	Numbers: N1, N2, N3…

Rule 1
2 members	Not on one line → both are zero-force members

Rule 2
3 members	Two on one line → the third is a zero-force member

Free body and joint equations

Goal Solve the joint with two equilibrium equations

Isolate a joint and draw all the forces on it.

ΣF_x = 0

ΣF_y = 0

Right and up are positive

Direction of a member's arrow at the joint
Unknown member	Assume tension — the arrow points away from the joint
Known, in tension	The arrow points out; substitute the magnitude
Known, in compression	The arrow points in; substitute the magnitude, without a minus

Sign cards — a member between two joints

+ tensionOn the member: the arrows point inwardOn the joint: the arrow goes out

− compressionOn the member: the arrows point outwardOn the joint: the arrow goes in

Order of work

Draw the joint with the loads, the reactions and all its members — the arrows as in the table.
Resolve a diagonal member: horizontal N · cos α, vertical N · sin α.
First solve the axis that has one unknown, and substitute the result into the other axis with the sign that came out.

⚠ Common mistake Drawing the arrow of an unknown member into the joint, because 'it looks compressed'.

Why it is a mistake: Then a positive result is compression — the opposite of the convention, and at the next joint it is unclear which sign to substitute.
How to avoid it: Every unknown member — arrow out. The sign of the result determines the state.
In the flagship exercise: At A, assume tension in N1 and get ⁦N1 = −4.243 t⁩ — the member is in compression.

08
Sign of the result and moving to the next joint

Goal Read the state of the member and carry it forward

Sign convention
- +Tension — at the joint the arrow points out; on the member the arrows point inward
- −Compression — at the joint the arrow points in; on the member the arrows point outward
- 0Zero-force member
Order of work
1. Write the magnitude and the state next to the member — for example 4.243 t compression. From now on it is a given.
2. At the next joint: compressed member — arrow in and the magnitude without a minus; tensioned member — arrow out and the magnitude.
3. Repeat the free body and the equations until all members are known.
⚠ Common mistake Compressed member: drawing an arrow in and also substituting a minus.
Why it is a mistake
The sign is counted twice — the force enters reversed, and everything after it is wrong.
How to avoid it
Arrow in → positive magnitude. A minus only when the arrow is still out (within the same joint).
In the flagship exercise
At D: N1 is compressed, so substitute ⁦+(4.243 · 0.707)⁩ and get ⁦N5 = −1.414 t⁩ and ⁦N4 = −2 t⁩. With a double minus you get ⁦N5 = −9.9 t⁩.
09
Checks and summary

Goal make sure the truss is in equilibrium and summarise all members

The equations we did not use — they are the check.

Allowed deviation ≤ 0.1
in units of t

Checks checklist
1. Sum of the vertical reactions = sum of the loads.
2. At the remaining joints substitute all the forces: ΣF_x = 0 and ΣF_y = 0.
3. Not 0 → check the reactions, the angles and the sign of a compressed member.
4. Summary table: for each member the magnitude and state, for each support the reactions.

Flagship exercise solved — N3 dashed: zero-force member

Summary table: AY = 3 t, BY = 2 t, AX = 0
Member	Between joints	Force	State
N1	A–D	4.243 t	Compression
N2	A–C	3 t	Tension
N3	C–D	0	Zero-force member
N4	D–F	2 t	Compression
N5	D–E	1.414 t	Compression
N6	C–E	3 t	Tension
N7	E–F	1 t	Tension
N8	F–B	2.828 t	Compression
N9	E–B	2 t	Tension

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Trusses binder · BST · beamsolvertool.com/en/learn/sdp/trusses/

Solve a truss in BST BST guides you by the same procedure — joint after joint, or waits for you to solve on your own

All pages as a print-ready PDF — in the Binders to download of BST PRO

Trusses solution procedureMethod of joints — 9 steps

Sketch and labelling

Static determinacy check

Reactions

Angles and resolving the forces in the members

Zero-force members

Choosing a joint

Free body and joint equations

Sign of the result and moving to the next joint

Checks and summary