Truss Analysis — Method of Joints

Truss analysis by the method of joints finds the force in each member by balancing forces at one joint at a time. In the standard statics model, the truss is planar, members are pin-connected two-force members, and external loads act only at the joints. Under those conditions, each joint must satisfy

$$\sum F_x = 0, \qquad \sum F_y = 0$$

where $\sum F_x$ is the sum of horizontal force components at a joint and $\sum F_y$ is the sum of vertical force components.

Why The Two Equations Are Enough

The method works because of what a "two-force member" means. Each truss member is pinned at both ends and loaded only at those ends, so the only way it can stay in equilibrium is to carry force purely along its own length. It cannot resist bending at the joints the way a beam or rigid frame would.

That single fact is what lets you reduce the whole structure to small problems. At a single joint in a planar truss, every member force passes through that point, so there are no moment arms and the moment equation gives nothing new. You are left with exactly two independent equations per joint. As long as a joint has at most two unknown member forces, those two equations solve it completely, which is why you always look for such a joint to start.

Formula And Symbols

For each joint you resolve angled member forces into horizontal and vertical components, then apply:

$$\sum F_x = 0, \qquad \sum F_y = 0$$

A practical convention is to assume each unknown member force is tensile at the start. If a solved force comes out negative, the member is actually in compression. Physically, a member that pushes on a joint is in compression, and one that pulls on a joint is in tension.

Worked Example: A Simple Triangular Truss

Consider a symmetric triangular truss with supports at joints $A$ and $C$, a top joint $B$, and a downward load of $10\ \mathrm{kN}$ at $B$. Let $A$ be a pin support, $C$ a roller support, and let members $AB$ and $BC$ each make a $45^\circ$ angle with the horizontal bottom member $AC$.

Because the load is centered, symmetry gives the support reactions

$$A_y = 5\ \mathrm{kN}, \qquad C_y = 5\ \mathrm{kN}$$

and $A_x = 0$.

Now start at joint $B$. Only two member forces are unknown there, and symmetry says they have the same magnitude. Call that magnitude $F$. At joint $B$, the vertical components of the two inclined members must balance the $10\ \mathrm{kN}$ downward load:

$$2F \sin 45^\circ = 10$$

so

$$F = \frac{10}{2 \sin 45^\circ} = \frac{10}{2(0.707)} \approx 7.07\ \mathrm{kN}$$

To hold up joint $B$, the inclined members must push on it, so $AB$ and $BC$ are in compression, each with magnitude $7.07\ \mathrm{kN}$.

Next move to joint $A$. The compressive force in $AB$ has a horizontal component of

$$7.07 \cos 45^\circ = 5\ \mathrm{kN}$$

At joint $A$, that horizontal component must be balanced by member $AC$, so

$$F_{AC} = 5\ \mathrm{kN}$$

Because $AC$ pulls on the joint, it is in tension. The member forces are:

$$F_{AB} = F_{BC} = 7.07\ \mathrm{kN}\ \text{(compression)}, \qquad F_{AC} = 5\ \mathrm{kN}\ \text{(tension)}$$

Practice The Same Method

Keep the same truss geometry, but change the top load from $10\ \mathrm{kN}$ to $14\ \mathrm{kN}$. Because the geometry stays the same and the model is still linear statics, each member force scales by the same factor: $F_{AB} = F_{BC} \approx 9.9\ \mathrm{kN}$ in compression and $F_{AC} = 7\ \mathrm{kN}$ in tension. Check your scaling by recomputing $2F\sin 45^\circ = 14$ directly. For a harder case, set up a truss that is not symmetric and decide which joint becomes solvable first after you compute the support reactions.

Calculation Traps To Watch

The most common mistake is starting at a joint with too many unknowns. In a planar truss, each joint gives only two independent equilibrium equations, so a joint with three unknown member forces usually cannot be solved first.

Another mistake is skipping the support reactions. If the reactions are wrong, every member force that follows will also be wrong.

Students also misread negative answers. With a consistent sign convention, a negative force usually means the member is in the opposite state from the one you assumed at the start.

The last big trap is using the method on a structure that is not modeled as a truss. The assumptions matter: the method works when joints are modeled as pins, loads and reactions are applied at joints, and the truss is in static equilibrium. Beams and rigid frames can carry bending moments, so they need a different analysis.

Where The Method Of Joints Is Used

The method of joints appears often in statics courses because it teaches how loads move through a structure. It is also useful for hand-checks of simple roof trusses, bridges, and other pin-jointed systems. In more complex engineering work, software usually analyzes the full structure, but this method still builds intuition about load paths and member-force signs.