Trusses: method of joints

By the end of this topic you will be able to

State the assumptions behind the ideal truss and why members carry only axial force.
Apply ΣFx = 0 and ΣFy = 0 at a joint to find two unknown member forces.
Decide whether each member is in tension or compression and describe the load path.

The ideal truss

A truss is a frame of straight members joined at their ends by pins. If loads are applied only at the joints, and the weight of members is ignored, each member is a two-force member: it is pushed or pulled along its own length and carries no bending. So every member is either in

tension (being stretched, force pulls away from the joint), or
compression (being squeezed, force pushes into the joint).

Roof structures, bridges, towers and cranes use trusses because straight axial members use material efficiently. Triangles are rigid; most trusses are built from them.

Is the truss solvable by statics?

A truss with $j$ joints has $2j$ equilibrium equations (two per joint). The unknowns are the $m$ member forces plus the $r$ support reactions. If

m + r = 2j

(and the arrangement is stable), the truss is statically determinate. The preset trusses in the model satisfy this.

Method of joints

Find the support reactions by treating the whole truss as one rigid body.
Start at a joint with no more than two unknown member forces (usually a support).
Draw the joint with all forces. Assume every unknown member is in tension (arrow pulling away from the joint). A negative answer then means compression.
Solve $\sum F_x = 0$ and $\sum F_y = 0$ .
Move to the next joint that now has at most two unknowns, using the forces already found (reversing the direction at the other end of each member).
Use the last joint, or an overall check, to verify.

Interactive modelChoose a truss, move the loads and tap a member to read its force.

This interactive model needs JavaScript. The theory, worked example and problems on this page work without it.

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Worked example: a simple triangle

A truss has pins at A (0, 0) and B (4, 0) with a roller at B, and apex C at (2, 2). A 10 kN downward load acts at C.

Reactions. By symmetry $R_A = R_B = 5$ kN upward.

Joint A. Member AC makes $45^\circ$ with the horizontal. $\sum F_y = 0$ : $5 + F_{AC}\sin45^\circ = 0$ , so $F_{AC} = -7.07$ kN (compression). $\sum F_x = 0$ : $F_{AB} + F_{AC}\cos45^\circ = 0$ , so $F_{AB} = +5.0$ kN (tension).

Joint C. By symmetry $F_{BC} = -7.07$ kN (compression). Check: vertical forces at C are $2\times7.07\sin45^\circ = 10$ kN, balancing the load. ✓

In the model, choose the Triangle preset with a 10 kN load at C to see the same values.

Worked example: a Warren truss

In the Warren truss preset (8 m span, 1.8 m deep) a 20 kN load acts at the middle bottom joint C, so $R_A = R_E = 10$ kN. At joint A the diagonal AF has direction cosines (0.486, 0.874). $\sum F_y = 0$ : $10 + 0.874F_{AF} = 0$ , so $F_{AF} = -11.44$ kN (compression), and then $F_{AB} = 0.486\times11.44 = 5.56$ kN (tension). Working along gives the top chord GH $= -22.2$ kN (compression) and bottom chord BC $= +16.7$ kN (tension), the greatest forces. Check them in the model.

Practice problems

A truss has 7 members and 3 support reactions. How many joints are needed for it to be determinate?

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$m + r = 2j$ : $7 + 3 = 10 = 2j$ , so $j = 5$ joints.
At a joint a 20 kN vertical load is balanced by two symmetrical members at $30^\circ$ above the horizontal. Find the member force.

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$2F\sin30^\circ = 20$ , so $F = 20$ kN (each member pushes, i.e. compression, if they are below the load).
Why is a member that is in compression made thicker or braced?

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A slender compression member can buckle at a load far below the one that would yield it; tension members cannot buckle.
In the Warren-truss model, moving the 20 kN load to joint B changes the reactions. Predict $R_A$ and check.

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Taking moments about E: $R_A \times 8 = 20\times6$ , so $R_A = 15$ kN and $R_E = 5$ kN.

Key formulae

Joint equilibrium: $\sum F_x = 0, \quad \sum F_y = 0$
Determinacy check (m members, r reactions, j joints): $m + r = 2j$
Force components of a member: $F_x = F\cos\theta, \quad F_y = F\sin\theta$

Lecturer notes: suggested plan, activity and assessment

Suggested plan (50 min). 10 min: types of truss with photos (roof, Warren bridge, Pratt railway); 10 min: assumptions; 15 min: worked triangle on the board; 15 min: Warren truss in the model, students predict the sign of each member before tapping it.

Activity. Play "tension or compression?": students call each member, then the model shows the colour.

Discussion. What happens in a truss if a load is applied away from a joint? (The member bends: the ideal truss no longer applies.)

Assessment idea. Solve a five-joint truss by joints and verify using the model.