Trusses: Method of Joints

Introduction

The method of joints is used to determine the internal forces within members of a truss. A truss is analyzed by isolating individual joints and applying equilibrium equations to solve for the unknown members forces. Since truss members are assumed to be two-force members, each member force acts either in tension or compression along the axis of the members.

Method of Joints Procedure

  1. Solve support reactions using global equilibrium.

  2. Start at joints with the fewest unknowns.

  3. Draw a FBD of the single joint.

  4. Assume unknown member acts in tension.

  5. Apply equilibrium equations:

    1. Σ F x = 0 \Sigma F_{x} = 0

    2. Σ F y = 0 \Sigma F_{y} = 0

  6. Positive result → \rightarrow tension

  7. Negative result → \rightarrow compression

Tension vs. Compression

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Tension: member pulls away from the joint.

Compression: member pushes toward the joint.

Assume all unknown members are in tension initially.

Zero-Force Members

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If three members meet at a joint where:

  1. Two are collinear

  2. No external load or support reaction exists F D A = 0 {\ \ \ F}_{DA} = 0

then the non-collinear member is a zero-force member.

Example:
Find the force in member LK and if the member is in tension or compression.

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1) Start with Global Equilibrium to solve for Reaction Forces at points A and G:

Σ F x = 0 = A x \Sigma F_{x} = 0 = A_{x}

Σ F y = 0 = A y + G y − 10 k N − 20 k N − 10 k N \Sigma F_{y} = 0 = A_{y} + G_{y} - 10\ kN - 20\ kN - 10\ kN

A y + G y = 40 k N A_{y} + G_{y} = 40\ kN

Assuming CW as positive, Σ M G = 0 = ( A y ) ( 12 m ) − ( 10 k N ) ( 8 m ) − ( 20 k N ) ( 6 m ) − ( 10 k N ) ( 4 m ) \ \Sigma M_{G} = 0 = (A_{y})(12\ m) - (10\ kN)(8\ m) - (20\ kN)(6\ m) - (10\ kN)(4\ m)

Solving: A y = 20 k N A_{y} = 20\ kN

Substituting into A y + G y = 40 k N A_{y} + G_{y} = 40\ kN : G y = 20 k N G_{y} = 20\ kN

2) Start at Joint A:

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Σ F y = 0 = 20 k N + F A L c o s 45 ∘ \Sigma F_{y} = 0 = 20kN + F_{AL}cos45{^\circ}

F A L = − 28.28 k N F_{AL} = \ - 28.28\ kN

Σ F x = 0 = F A L s i n 45 ∘ + F A B \Sigma F_{x} = 0 = F_{AL}sin45{^\circ} + F_{AB}

F A B = 20 k N F_{AB} = 20\ kN

Since F A L F_{AL} is negative, member AL is in compression.

3) Moving on to joint B:

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Since AB and BC are collinear and there are no external reactions, member BL is a zero-force member ( F B L = 0 F_{BL} = 0 ).

Σ F x = 0 = − F A B + F B C \Sigma F_{x} = 0 = - F_{AB} + F_{BC}

F B C = 20 k N F_{BC} = 20\ kN

4) Finally, moving to Joint L:

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Σ F x = 0 = 28.28 s i n 45 ∘ + F L K s i n 45 ∘ + F L C s i n 45 ∘ \Sigma F_{x} = 0 = 28.28sin45{^\circ} + F_{LK}sin45{^\circ} + F_{LC}sin45{^\circ}

Σ F y = 0 = 28.28 c o s 45 ∘ + F L K c o s 45 ∘ − F L C s i n 45 ∘ {\Sigma F}_{y} = 0 = 28.28cos45{^\circ} + F_{LK}cos45{^\circ} - F_{LC}sin45{^\circ}

Solving System: F L K = − 28.28 k N F_{LK} = - 28.28\ kN\ ; F L C = 0 F_{LC} = 0

Since F L K F_{LK} is negative, member LK is in compression.

Final answer: F L K = 28.28 k N ( C ) F_{LK} = 28.28\ kN\ (C)