Interactive Lecture
Method of Joints and Method of Sections
by Dr Arnaldo Delli Carri
A truss is a structural framework composed of straight members connected at joints (nodes), designed to support loads. Trusses are commonly used in bridges, roofs, and towers because they efficiently transfer loads through axial forces in the members.
In an ideal truss, all members are two-force members experiencing only tension or compression, with no bending moments. That simplification is what makes truss analysis tractable.
For classical truss analysis, we make the following assumptions:
The Method of Joints is a technique for analysing trusses by considering the equilibrium of forces at each joint. Since each joint is in equilibrium, the sum of forces in any direction must be zero.
Equilibrium Equations at Each Joint:
\[\sum F_x = 0\]
\[\sum F_y = 0\]
Where:
The Method of Sections allows us to find forces in specific members without analysing the entire truss. We make an imaginary cut through the truss, dividing it into two sections, and apply equilibrium equations to one section.
Equilibrium Equations for a Section:
\[\sum F_x = 0\]
\[\sum F_y = 0\]
\[\sum M = 0\]
Where:
In truss analysis, we use the following sign convention:
Problem-Solving Strategy:
Click on any node to view its free body diagram and edit applied forces and constraints. Members in blue are in tension, red are in compression. The thickness represents force magnitude.
| Concept | Equation | Description |
|---|---|---|
| Joint Equilibrium (X) | \(\sum F_x = 0\) | Sum of horizontal forces at a joint |
| Joint Equilibrium (Y) | \(\sum F_y = 0\) | Sum of vertical forces at a joint |
| Force Resolution | \(F_x = F \cos\theta\), \(F_y = F \sin\theta\) | Resolve member force into components |
| Section Equilibrium | \(\sum M = 0\) | Sum of moments about any point |
| Member Count Check | \(m = 2j - 3\) | For a stable, determinate truss (m=members, j=joints) |
A triangular truss has a pin support at A(0, 0) and a horizontal roller at B(6, 0). The apex joint C is at (3, 4), where an 8 kN downward load is applied. Find the forces in all three members.
Show Solution
Solution:
Step 1: Calculate Support Reactions
By symmetry (load at centre of span): \(R_{Ay} = R_{By} = \frac{8}{2} = 4 \text{ kN}\)
\(\sum F_x = 0 \Rightarrow R_{Ax} = 0\)
Step 2: Member Geometry
Member AC: from (0,0) to (3,4), length \(L = 5\) m, \(\theta = \arctan(4/3) = 53.13°\)
Member BC: from (6,0) to (3,4), length \(L = 5\) m (symmetric)
Step 3: Method of Joints at C
Assume all forces are tension (pulling C towards the other node). By symmetry \(F_{AC} = F_{BC}\).
\[\sum F_y = 0: \quad -8 + 2F_{AC}(-\sin 53.13°) = 0\]
(Negative sine because tension in AC pulls C downward towards A.)
\[F_{AC} = \frac{-8}{2 \times 0.8} = -5 \text{ kN} \quad \Rightarrow \quad 5 \text{ kN (Compression)}\]
Step 4: Bottom Chord at Joint A
\[\sum F_x = 0: \quad F_{AB} + F_{AC}\cos(53.13°) = 0\]
\[F_{AB} = -(-5)(0.6) = 3 \text{ kN (Tension)}\]
Verification: \(\sum F_y\) at A: \(4 + (-5)(0.8) = 0\) ✓
Answer: AC = BC = 5 kN (compression), AB = 3 kN (tension).
A Warren truss has bottom chord joints at A(0, 0), B(4, 0), C(8, 0) and top chord joints at D(2, 3), E(6, 3). Pin support at A, horizontal roller at C. A 12 kN downward load is applied at the top joint D. Using the method of sections, find the force in the bottom chord member BC.
Show Solution
Solution:
Step 1: Calculate Support Reactions
Taking moments about C:
\[R_{Ay} \times 8 = 12 \times 6 \quad \Rightarrow \quad R_{Ay} = 9 \text{ kN}\]
\[R_{Cy} = 12 - 9 = 3 \text{ kN}\]
Step 2: Make a Section Cut
Cut the truss between B and E, slicing through three members: the top chord DE, diagonal BE, and bottom chord BC.
Step 3: Take Moments About E (Left Section)
Taking moments about E(6, 3) eliminates \(F_{DE}\) and \(F_{BE}\) (both pass through E), leaving only \(F_{BC}\).
The left section contains: \(R_{Ay} = 9\) kN (up) at A, and 12 kN (down) at D.
Clockwise moments about E:
\[R_{Ay} \times 6 = 9 \times 6 = 54 \text{ kN·m}\]
Counter-clockwise moments about E:
\[12 \times 4 + F_{BC} \times 3 = 48 + 3F_{BC}\]
(12 kN acts 4 m to the left of E; \(F_{BC}\) acts 3 m below E.)
Setting CW = CCW:
\[54 = 48 + 3F_{BC}\]
\[F_{BC} = 2 \text{ kN (Tension)}\]
Answer: The bottom chord member BC carries 2 kN in tension.
Identify the zero-force members in a truss where a joint has three members: two are collinear (in a straight line), and the third is at an angle. No external loads are applied at this joint.
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Solution:
Zero-Force Member Rules:
Analysis:
At the described joint, two members are collinear. For equilibrium in the direction perpendicular to these collinear members, the force in the third member must be zero.
Mathematical Proof:
Let the collinear members be along the x-axis. For vertical equilibrium:
\[\sum F_y = F_3 \sin\theta = 0\]
Since \(\sin\theta \neq 0\) (the member is at an angle), we must have \(F_3 = 0\).
Answer: The non-collinear member is a zero-force member.