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Truss Method of Joints Example

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Find the force in each member of the truss. Identify tension and compression.

Concept

The method of joints analyzes trusses by isolating each joint and applying equilibrium equations. At each joint: Assume all members are in tension (forces pull away from the joint). Apply \sum F_{x} = 0 and \sum F_{y} = 0 . A negative result means the member is in compression. Start at a joint with at most two unknowns. Key formulas: \sum F_{x} = 0, \sum F_{y} = 0 Assume tension; negative result \Rightarrow compression

A simple triangular truss has pin support at A, roller support at B, and a downward load at the apex C. Find: Support reactions at A and B Force in each member (AC, BC, AB) Identify tension vs. compression

L = 6 m (span) h = 4 m (height) P = 5 kN (downward at C)

Treat the truss as a rigid body. Sum moments about A to find R_{B}, then sum vertical forces to find R_{A} . \sum M_{A} = 0 R_{B} \cdot 6 - (5 kN) (3 m) = 0 R_{B} = \frac{15}{6} = 2.5 kN = 2500 N ↑ \sum F_{y} = 0 R_{A} + R_{B} - 5 kN = 0 R_{A} = 5 - 2.5 = 2.5 kN = 2500 N ↑

At joint A we have two unknowns: F_{A C} and F_{A B} . Find the angle of member AC. θ = arctan (4/3) \approx 53.13° sin θ = 0.8, cos θ = 0.6 \sum F_{y} = 0 : R_{A} + F_{A C} sin θ = 0 F_{A C} = - \frac{R _{A}}{sin θ} = - \frac{2500}{0.8} = - 3125 N F_{A C} = 3125 N (compression) \sum F_{x} = 0 : F_{A C} cos θ + F_{A B} = 0 F_{A B} = - 0.6 F_{A C} F_{A B} = - 0.6 (- 3125) = 1875 N F_{A B} = 1875 N (tension)

By symmetry of the truss and loading, member BC carries the same force as AC. By symmetry: F_{B C} = F_{A C} F_{B C} = 3125 N (compression)

Member Force (N) Type AB 3125 Compression BC 3125 Compression CD 1875 Tension BD 0 Zero-force AD 1875 Tension

\sum F_{x} = 0, \sum F_{y} = 0 At each joint. Assume tension; negative result indicates compression.

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