Plane trusses

Idealizing assumptions. A plane truss is modelled as an assembly of straight, weightless, two-force members connected by frictionless pins, with all external loads and support reactions applied only at the joints. Because a two-force member has forces at only its two end pins and no other load along its length, equilibrium forces those two end forces to be equal, opposite, and directed along the line joining the pins — so the member can only push (compression) or pull (tension); it carries no shear and no bending moment. This one fact is what makes truss analysis solvable with nothing but $\sum F_x=0$ and $\sum F_y=0$.

Determinacy and stability. Before solving anything, check whether the truss is even solvable with statics alone. For a plane truss with $m$ members, $r$ reaction components, and $j$ joints:

where: $m$ is the number of two-force members, $r$ is the number of independent reaction components at the supports, $j$ is the number of pin joints, and the factor $2$ counts the two scalar equilibrium equations ($\sum F_x=0$, $\sum F_y=0$) available at each joint. If $m+r<2j$ the frame has too few members for the joints available — it is a mechanism, free to collapse under load. If $m+r>2j$ it is statically indeterminate: more members than the equilibrium equations can resolve, requiring compatibility (deformation) equations beyond this course. Equality is necessary but not sufficient — the members must also be arranged so no panel can flex as a mechanism, which is why every panel of a good truss is triangulated (a triangle is the only polygon that cannot change shape without changing a side length).

Method of joints. Isolate each pin as a free particle. Because a pin transmits no moment, only two equations apply there: $\sum F_x=0$ and $\sum F_y=0$. Start by finding the external reactions from overall equilibrium of the whole truss, then work joint by joint, always choosing a joint with at most two unknown member forces so the two equations solve it directly. Assume every unknown member force is tension (pulling away from the joint); a negative answer simply means the member is in compression.

Method of sections. When you need only a few specific member forces deep inside a large truss, cut an imaginary section through no more than three members with unknown forces, isolate one side as a free body, and apply all three equilibrium equations — $\sum F_x=0$, $\sum F_y=0$, and $\sum M_O=0$ about a well-chosen point. Choosing the moment centre where two of the three unknown forces intersect eliminates them, so each unknown often falls out of a single equation, without ever visiting the joints in between.

A triangular truss has joints $A(0,0)$, $B(4,0)$, $C(2,3)$ in metres, pinned at $A$ and roller-supported (vertical reaction only) at $B$, with a downward 10 kN load at $C$. Find every member force by the method of joints, then use the method of sections on a second, larger truss to find three specific member forces directly.

Step 1 — external reactions. With no horizontal load, $\sum F_x=0$ gives $A_x=0$. Taking moments about $A$: $\sum M_A=0 \Rightarrow B_y(4) - 10(2) = 0 \Rightarrow B_y = 5$ kN. Then $\sum F_y=0 \Rightarrow A_y + 5 - 10 = 0 \Rightarrow A_y = 5$ kN.

Step 2 — joint A. Member $AC$ runs from $(0,0)$ to $(2,3)$, length $\sqrt{13}=3.606$ m, so $\cos\theta=0.5547$, $\sin\theta=0.8321$. Vertical equilibrium: $F_{AC}\sin\theta + A_y = 0 \Rightarrow F_{AC}(0.8321)=-5 \Rightarrow F_{AC}=-6.01$ kN, i.e. 6.01 kN compression. Horizontal equilibrium then gives $F_{AB} + F_{AC}\cos\theta = 0 \Rightarrow F_{AB} = 3.33$ kN tension.

Step 3 — joint B and the check at C. By the identical geometry on the other side, $F_{BC} = 6.01$ kN compression. Checking joint $C$: the vertical components of the two 6.01 kN compressive forces pushing into the joint sum to $2 \times 6.01 \times 0.8321 = 10.0$ kN, exactly balancing the 10 kN applied load, and the horizontal components cancel by symmetry — confirming $F_{AB}=3.33$ kN (T), $F_{AC}=6.01$ kN (C), $F_{BC}=6.01$ kN (C).

Competitive stretch — method of sections

A section cut through a larger truss passes through three members: top chord $DE$ (horizontal, height 3 m), diagonal $DF$, and bottom chord $CF$ (horizontal, $y=0$), with $D(4,3)$ and $F(6,0)$, so $\tan\theta=3/2$, $\theta=56.3°$. The isolated left portion carries only a 15 kN upward reaction at $x=0$. Find $F_{DE}$, $F_{DF}$, $F_{CF}$, assuming tension.

Step 1. Take moments about $D$, eliminating $F_{DE}$ (passes through $D$) and $F_{DF}$'s line does not pass through $D$ but $F_{CF}$'s moment arm is the height 3 m: $\sum M_D=0: 15(4) - F_{CF}(3) = 0 \Rightarrow F_{CF}=20$ kN tension.

Step 2. Take moments about $F$, where both $F_{DF}$ and $F_{CF}$ act along lines through $F$ and so contribute zero moment: $\sum M_F=0: 15(6) - F_{DE}(3) = 0 \Rightarrow F_{DE}=30$ kN by the assumed-tension sign convention — meaning the top chord is actually in 30 kN compression, typical for a top chord under sagging.

Step 3. Vertical equilibrium isolates $F_{DF}$: $\sum F_y=0: 15 - F_{DF}\sin56.3° = 0 \Rightarrow F_{DF} = 15/0.832 = 18.0$ kN tension.

Each of the three member forces came from a single equation — one clean cut and three moment/force choices, versus tracing joint by joint from the support all the way to $D$ and $F$.

• Forgetting a pin joint gives only two equilibrium equations. No moment equation applies at a frictionless pin — only $\sum F_x=0$ and $\sum F_y=0$.
• Sign confusion between tension and compression. Always assume tension as positive; a negative result means compression. Do not flip the assumed direction mid-calculation.
• Cutting through more than three unknown members. The method of sections needs at most three unknowns per cut, or the three equilibrium equations become simultaneous instead of one-shot.
• Treating $m+r=2j$ as sufficient for stability. The count can match exactly while a panel that isn't triangulated still flexes as a mechanism — always check the geometry, not just the arithmetic.

Ask the tutor: If you're unsure which joint to start with or want a walkthrough on spotting the fastest section cut before grinding through algebra, ask the AI tutor to work through a truss diagram with you step by step.