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4.3 · Analyze a truss with the method of joints

Learn to analyze a truss with the method of joints through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Truss Analysis

ENGG 130 Engineering Mechanics: Statics — Study topic 4.3

A planar truss is a set of straight members joined at their ends. In the ideal truss model used here, the joints are pins, loads act at joints, and each member carries force along its own axis. Because a member is acted on only at its ends, its end forces are equal, opposite, and along the member. This lets us isolate one joint at a time and treat it as a particle. The method of joints begins with support reactions for the complete truss, then uses force balance at selected joints to find the member forces.

What you will learn

  • Model an ideal truss joint as a particle in planar equilibrium.
  • Find support reactions before solving member forces.
  • Apply the method of joints using consistent axes and signs.
  • Interpret positive and negative member-force results as tension and compression.
  • Check joint force balance and whole-truss equilibrium.

1. Model the truss and find support reactions

Start by defining the system. For reactions, the system is the complete truss: include all applied loads and the forces exerted by supports. A pin can exert horizontal and vertical reactions. A roller on a horizontal surface exerts a vertical reaction. Draw the whole-truss free-body diagram and use equilibrium to find the reactions before moving to individual joints.
Take right and upward as the positive coordinate directions. For whole-truss moments, choose a point and a sign convention; here, counterclockwise is positive. A reaction calculated as negative acts opposite to its assumed direction. Moments are especially useful for finding reactions because forces whose lines of action pass through the moment centre contribute zero moment.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Use the whole truss for external reactions, then isolate joints for member forces.
  • A member force acts along that member’s axis.
  • Keep force and moment units distinct, such as kN and kN·m.

2. Isolate a joint and solve its member forces

Choose a joint with no more than two unknown member forces. Draw that joint by itself and include every member force, applied load, and support reaction that acts there. A joint is modelled as a particle, so the forces meet at one point and the two planar force-balance equations apply.
A reliable sign convention is to assume every unknown member force pulls away from the joint. This is the tension assumption. Resolve an inclined member force into horizontal and vertical components using its geometry. If the member has horizontal and vertical offsets Δx and Δy and length L, its direction components have magnitudes |Δx|/L and |Δy|/L. Assign component signs from the direction of the assumed force.
Solve the horizontal and vertical equations together. A positive answer agrees with the assumed pull and is tension. A negative answer means the actual force is opposite the assumed pull: the member is in compression. At the other end of the same member, the force has equal magnitude and opposite direction.
∑Fx=0,∑Fy=0\sum F_x=0,\quad \sum F_y=0
  • Draw a separate free-body diagram for every joint you solve.
  • Use the member’s actual length, not just one of its offsets, to find direction ratios.
  • A negative answer is useful information; report its magnitude and identify compression.

3. Choose a useful order and check the result

Begin at a joint whose unknowns can be found from its available equations. After solving it, carry the known member forces to neighbouring joints, reversing their directions at the opposite ends. If a joint has more than two unknown member forces, use another joint first; its two force equations alone cannot determine more than two independent unknowns.
Finish by checking both local and whole-truss equilibrium. At each solved joint, add the horizontal components and the vertical components separately. For the complete truss, check external horizontal forces, vertical forces, and moments. A small nonzero remainder can result from rounding, but it should be small compared with the forces being checked.
∑FJ=0\sum \mathbf{F}_J=\mathbf{0}
  • Select joints strategically so there are at most two unknown member forces at each step.
  • Check force balance at solved joints and external force and moment balance for the truss.
  • Do not label a member force as zero unless equilibrium establishes that result.

Worked example

Symmetric triangular truss

A truss has supports A and B, 6 m apart, and apex C directly above the midpoint. Each sloping member is 5 m long, so C is 4 m above the base. A 10 kN downward load acts at C. A is a pin and B is a roller on a horizontal surface. Find all member forces.
  1. Find the reactions
    Use the complete truss. By symmetry, the vertical reactions are equal. The 10 kN load is centred, and there is no horizontal load, so the pin’s horizontal reaction is zero.
    Ay=By=5 kN,Ax=0A_y=B_y=5\,\mathrm{kN},\quad A_x=0
  2. Solve joint C
    Each slope has vertical direction ratio 4/5. Under the tension assumption, both vertical components at C point down. Their sum must balance the applied load, so the calculated member forces are negative and both members are in compression.
    −2(45T)−10=0,T=−6.25 kN-2\left(\frac{4}{5}T\right)-10=0,\quad T=-6.25\,\mathrm{kN}
  3. Solve joint A
    At A, the compressive member AC pushes left with horizontal component 6.25(3/5) kN. The base member balances it. The positive result under the tension assumption means AB is in tension.
    TAB=35(6.25)=3.75 kNT_{AB}=\frac{3}{5}(6.25)=3.75\,\mathrm{kN}
Answer: AC = 6.25 kN in compression; BC = 6.25 kN in compression; AB = 3.75 kN in tension.
Check: At C, each compressed slope pushes upward with component 6.25(4/5) = 5 kN, giving 10 kN upward; their horizontal components cancel. For the whole truss, the reactions total 10 kN upward, balancing the load, and the moment about A is (5 kN)(6 m) − (10 kN)(3 m) = 0.

Worked example

A horizontal load and a negative member result

A triangular truss has A at (0, 0) m, B at (4, 0) m, and C at (4, 3) m. A is a pin and B is a horizontal roller. At C, a 5 kN load acts rightward and a 10 kN load acts downward. Find the member forces. Member AC has direction ratios 4/5 horizontally and 3/5 vertically.
  1. Find the support reactions
    For the whole truss, horizontal balance gives the pin reaction. Taking moments about A includes the 4 m arm of the downward load and the 3 m arm of the horizontal load. Vertical balance then gives the reaction at A.
    Ax=−5 kN,By=13.75 kN,Ay=−3.75 kNA_x=-5\,\mathrm{kN},\quad B_y=13.75\,\mathrm{kN},\quad A_y=-3.75\,\mathrm{kN}
  2. Solve joint C
    Under the tension assumption, AC points down-left from C and CB points down. Horizontal balance gives AC. Vertical balance then gives a negative force in CB, so CB is actually in compression.
    −45TAC+5=0,−35TAC−TCB−10=0-\frac{4}{5}T_{AC}+5=0,\quad -\frac{3}{5}T_{AC}-T_{CB}-10=0
  3. Find the remaining member
    At joint A, the known forces are the reactions and AC. Their horizontal components balance without a force in AB, so AB carries no force for this loading.
    TAC=6.25 kN,TCB=−13.75 kN,TAB=0T_{AC}=6.25\,\mathrm{kN},\quad T_{CB}=-13.75\,\mathrm{kN},\quad T_{AB}=0
Answer: AC = 6.25 kN in tension; CB = 13.75 kN in compression; AB = 0 kN.
Check: At C, AC contributes 5 kN left and 3.75 kN down; compressed CB contributes 13.75 kN up. With the applied loads, both force-component sums are zero. For the whole truss, horizontal forces sum to −5 + 5 = 0 kN and vertical forces to −3.75 + 13.75 − 10 = 0 kN. Moments about A are (13.75)(4) − (10)(4) − (5)(3) = 0 kN·m.

Worked example

Unequal compression in a symmetric-shaped truss

Supports A and B are 6 m apart, with C 4 m above the midpoint; AC and BC are each 5 m long. A is a pin and B is a horizontal roller. At C, a 10 kN downward load and a 4 kN rightward load act. Find the reactions and member forces.
  1. Find the reactions
    For the complete truss, horizontal balance gives the pin reaction. Taking moments about A includes the downward load’s 3 m horizontal arm and the rightward load’s 4 m vertical arm; both loads create clockwise moments. Vertical balance gives the reaction at A.
    Ax=−4 kN,By=233 kN,Ay=73 kNA_x=-4\,\mathrm{kN},\quad B_y=\frac{23}{3}\,\mathrm{kN},\quad A_y=\frac{7}{3}\,\mathrm{kN}
  2. Solve joint C
    Assume both member forces pull away from C. The left member’s horizontal component is leftward, while the right member’s is rightward. Both assumed vertical components point down. The negative solutions therefore mean both members are in compression.
    −35TCA+35TCB+4=0,−45TCA−45TCB−10=0-\frac{3}{5}T_{CA}+\frac{3}{5}T_{CB}+4=0,\quad -\frac{4}{5}T_{CA}-\frac{4}{5}T_{CB}-10=0
  3. Solve joint A
    At A, the compressed AC member pushes down-left. Its horizontal component is 1.75 kN leftward, and its vertical component is 7/3 kN downward. Horizontal and vertical balance at A then determine the force in AB and confirm the vertical reaction.
    TCA=−3512 kN,TCB=−11512 kN,TAB=5.75 kNT_{CA}=-\frac{35}{12}\,\mathrm{kN},\quad T_{CB}=-\frac{115}{12}\,\mathrm{kN},\quad T_{AB}=5.75\,\mathrm{kN}
Answer: AC = 2.92 kN in compression; BC = 9.58 kN in compression; AB = 5.75 kN in tension.
Check: At C, compression in AC contributes 1.75 kN right and 7/3 kN up; compression in BC contributes 5.75 kN left and 23/3 kN up. Including the applied 4 kN rightward and 10 kN downward loads, both force sums are zero. At A, the 4 kN leftward reaction, the 1.75 kN leftward AC component, and the 5.75 kN rightward AB force balance; vertically, Ay balances AC’s downward component. Whole-truss moments about A give (23/3)(6) − 10(3) − 4(4) = 0 kN·m.

Common mistakes and how to avoid them

Using the whole truss as though it were one joint when solving member forces.
Correction: Find reactions from whole-truss equilibrium, then isolate individual joints.
Changing an assumed member-force direction without changing its sign in the equations.
Correction: Keep the tension assumption throughout; a negative result means compression.
Using a member offset as its full length when finding direction ratios.
Correction: Use the actual member length to resolve the force into components.
Checking only vertical balance or only one joint.
Correction: Check both force components at solved joints and external force and moment balance for the whole truss.

Lesson summary

  • Find support reactions using equilibrium of the complete truss.
  • Treat an isolated joint as a particle and apply horizontal and vertical force balance.
  • Assume unknown member forces pull away from the joint; positive is tension and negative is compression.
  • Choose joints with at most two unknowns, then verify joint and whole-truss equilibrium.

Check your understanding

Question 1

At a joint, an unknown force assumed to pull away from the joint is calculated as −4 kN. What does this mean?
  1. The member is in 4 kN tension.
  2. The member is in 4 kN compression.
  3. The member carries no force.
  4. The joint cannot be in equilibrium.
Show answer and explanation
The member is in 4 kN compression.
The negative result means the actual force acts opposite to the assumed outward pull, so the member is in compression.

Question 2

Why can the two particle-equilibrium equations be used to find member forces at a planar joint?
  1. They give horizontal and vertical force balance at that joint.
  2. They determine every member force in the truss at once.
  3. They remove the need to find support reactions.
  4. They apply only when all members are horizontal.
Show answer and explanation
They give horizontal and vertical force balance at that joint.
At a planar joint, the horizontal and vertical components must each sum to zero. These equations can solve up to two independent unknowns at that joint.

Question 3

For the third example, what is the horizontal component of the compressive force in AC acting on joint A?
  1. 1.75 kN to the right
  2. 1.75 kN to the left
  3. 2.92 kN to the left
  4. 5.75 kN to the right
Show answer and explanation
1.75 kN to the left
The 2.92 kN compression force acts along AC and pushes joint A away from C. Its horizontal component is (2.92)(3/5) = 1.75 kN to the left.

Key terms

Method of joints
A truss method that applies force equilibrium to isolated joints one at a time.
Two-force member
An ideal member acted on only at its two ends, with end forces along the member.
Tension
A member force that pulls away from an isolated joint.
Compression
A member force that pushes toward an isolated joint.
Support reaction
A force exerted by a support on the truss.

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