Method of sections: cut a truss through three members and solve

Method of sections on the 8 m truss

On the method of joints page, two joints had to be solved before the force in the top chord D-E appeared. The method of sections gets it in one line.

The truss is the same: pin A (0, 0), roller C (8, 0), middle bottom joint B (4, 0), top joints D (2, 3) and E (6, 3), members A-B, B-C, D-E, A-D, D-B, B-E and E-C, with 12 kN down at B. The reactions are 6 kN at A and 6 kN at C.

Draw an imaginary line that crosses exactly three members and splits the truss into a left piece and a right piece. Here the line runs between D and E through the top chord D-E, then down between B and E through B-E, and ends just to the right of B through B-C. The left piece then holds joints A, D and B and the 12 kN load. Treat the three cut members as unknown forces pulling on that piece, tension assumed.

Why a moment equation solves one member at a time

A rigid piece in a plane has three equilibrium equations: ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Three unknown cut forces would normally need all three together. The trick is to pick the pivot. If a point lies on the line of action of two unknown forces, they have no moment about it and vanish from the equation.

B-E and B-C both pass through joint B, so take moments about B, with clockwise positive. The 6 kN reaction at A pushes up 4 m to the left of B, which turns the piece clockwise. The load at B acts through the pivot and contributes nothing. D-E sits 3 m above B and, if in tension, pulls the piece to the right, which is also clockwise.

ΣM about B: 6 × 4 + N_DE × 3 = 0, so N_DE = −8.000 kN

The top chord is in compression at 8.000 kN, exactly as the joints worked out. The other two members come the same way.

B-E and D-E meet at joint E (6, 3), so moments about E leave only B-C. The reaction pushes up 6 m to the left of E, clockwise. The 12 kN load acts downward 2 m to the left of E, counterclockwise. B-C lies 3 m below E, and a tension pull to the right there turns the piece counterclockwise.

ΣM about E: 6 × 6 − 12 × 2 − N_BC × 3 = 0, so N_BC = +4.000 kN

For the diagonal, the vertical balance of the left piece does it, because D-E and B-C are horizontal and only B-E has a vertical component (3/3.606 = 0.8321 of its force):

ΣFy: 6 − 12 + N_BE × 0.8321 = 0, so N_BE = +7.211 kN

All three agree with the joints answer: top chord 8.000 kN in compression, bottom chord 4.000 kN in tension, diagonal 7.211 kN in tension. Getting the same numbers by two different routes is the best check available without a computer.

The method of sections in a long truss

The method of sections earns its place when the truss is big and the member in question is far from a support. Take the 12 m, four-panel Warren truss from the designer, with 20 kN at the centre joint C. Joint-by-joint, the centre top chord G-H would need A, F, B and G solved first.

Instead, pass a cut through G-H, G-C and B-C. The left piece holds A, F, B and G; C stays on the right with its load. G-C and B-C meet at C, so take moments about C. The reaction at A is 10 kN, 6 m to the left of C, clockwise. The top chord G-H is 3 m above C and, in tension, pulls right, also clockwise.

ΣM about C: 10 × 6 + N_GH × 3 = 0, so N_GH = −20.000 kN

The centre top chord carries 20 kN in compression, one equation, no joints. The solver gives the same figure, 20.0 kN. In a truss with ten panels instead of four the arithmetic stays one line long, whereas the joints method would have to cross half the span.

Choosing where to cut

The cut can go anywhere that crosses no more than three members whose forces are unknown, and cutting through the interior of a panel is usually enough. Two habits save time. Prefer a cut that passes through the wanted member and through two others that meet at a convenient joint, so one pivot kills both. And keep the piece that has the fewer loads and reactions on it, because there is less arithmetic to carry.

A cut that crosses four unknown members is not useless: if three of them meet at one point, moments about that point still solve the fourth. Most textbook trusses are simple enough that this is rarely needed.

The method of sections and the joints method work well together. Use sections to pick off the one or two members that matter and the joints method to fill in the neighbours, then confirm the rest in the truss analysis overview or with a solver. A Warren truss like the one above is also the layout covered on Warren truss bridge, where the same alternating push and pull appears in every panel.