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4.2 · Identify zero-force members

Learn to identify zero-force members through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Truss Analysis

Using equilibrium at unloaded truss joints

A truss member is a zero-force member when its axial force is zero for the loading condition being considered. Finding such members can simplify a truss analysis, but the conclusion must come from equilibrium—not from how a member looks in a drawing. The key idea is to isolate a joint and ask whether the forces at that joint can balance. This lesson focuses on two common patterns at joints with no applied load or support reaction. A zero-force member is identified for the stated loading condition; this does not mean the member can be removed from every possible truss or loading situation.

What you will learn

  • Recognize the two common unloaded-joint patterns that identify zero-force members.
  • Explain why each pattern follows from planar force equilibrium.
  • Use the rules carefully, including when an identified zero-force member allows a second joint to be examined.
  • Check force and moment equilibrium for an isolated joint.

1. Isolate the joint and apply equilibrium

A planar truss is modelled as straight members joined by pins. For this topic, isolate one pin joint at a time. Each connected member is treated as a two-force member: it pulls or pushes along its own axis. The isolated joint is therefore represented by concurrent member forces, together with any applied force or reaction at that joint.
Before using either rule, confirm that the joint has no applied load and no support reaction. A load or reaction changes the force balance and may prevent a member from being zero. Also check which members meet at the joint and whether any of their centre lines are collinear.
Choose convenient perpendicular axes, usually horizontal xx and vertical yy. Take right and up as positive. For a joint in equilibrium, the sum of the horizontal force components and the sum of the vertical force components must each be zero. Since all member forces at an ideal pin joint act through that point, their moments about the joint are zero as well.
∑Fx=0,∑Fy=0,∑MJ=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_J=0
  • A zero-force member carries zero axial force for the specified loading condition.
  • Inspect an isolated joint with no external force or support reaction before applying the common rules.
  • Member forces at a pin joint act along the member axes.

2. The two unloaded-joint rules

Rule 1: If an unloaded joint connects exactly two members, and those members are not collinear, both members are zero-force members. The two member-force directions are different. With no external force present, there is no way for two nonzero forces in non-collinear directions to sum to zero. Equilibrium therefore requires each force to be zero.
Rule 2: If an unloaded joint connects exactly three members and two are collinear, the third member is a zero-force member. Choose axes along the collinear pair and perpendicular to them. The perpendicular equilibrium equation contains only the force in the third member, so that force must be zero. The two collinear members can still carry equal and opposite forces.
These rules concern a joint’s current force balance. They do not apply unchanged when the joint has an applied force or reaction, or when the member arrangement does not match the stated pattern. A joint may have more members than the rule allows, or its apparent alignment may be misleading. In those cases, write the joint equilibrium equations rather than guessing.
Unloaded joint: 2 non-collinear members⇒F1=F2=0\text{Unloaded joint: }2\text{ non-collinear members}\Rightarrow F_1=F_2=0
  • Two non-collinear members at an unloaded two-member joint are both zero-force members.
  • At an unloaded three-member joint, if two members are collinear, the non-collinear member is zero-force.
  • A zero-force result for one member does not imply that every member connected to the same truss is zero.

3. Use the rules systematically

Start at joints that have no applied loads or support reactions and whose member layout matches one of the two patterns. Mark each identified member as zero for the loading case. Then inspect neighbouring joints again. A member already shown to have zero force contributes no force to that joint’s equilibrium equations, which can reveal a simpler pattern there.
Use a consistent sign convention when writing equations. One convenient approach is to assume each unknown member force pulls away from the joint, which is a tension assumption. If a later calculation gives a negative value, the actual force acts toward the joint instead. For the zero-force rules, the result is exactly zero, so there is no tension-versus-compression ambiguity.
The rules identify some member forces without solving the entire truss. They do not establish that all other forces are known. If a joint has two collinear members and a third non-collinear member, only the third is forced to zero by that pattern; the forces in the collinear pair may need information from elsewhere.
Fmember=0F_{\text{member}}=0
  • Recheck adjacent joints after identifying a zero-force member.
  • Use the actual joint geometry and loading, not visual intuition.
  • State which member or members the rule identifies and which forces remain undetermined.

4. Verification at a joint

A correct identification must satisfy force balance. Resolve each member force into horizontal and vertical components, then confirm that both component sums are zero. For a simple rule-based joint, this may be immediate: all member forces are zero, or the remaining two collinear forces are equal and opposite.
Moment balance is also satisfied at an ideal pin joint because the member forces and any force applied at the joint pass through that point. Their moment arms about the joint are zero. This moment check is useful as a consistency statement, although force equilibrium provides the decisive information for these rules.
Keep units consistent when known member forces are included. For example, a member force stated in kilonewtons should be balanced by forces in kilonewtons. The zero-force conclusion itself has units of force and is reported as 0 kN0\text{ kN} or 0 N0\text{ N} as appropriate.
∑Fx=0,∑Fy=0,∑MJ=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_J=0
  • Check both horizontal and vertical force balance.
  • Forces concurrent at the isolated pin joint have zero moment about that joint.
  • A zero-force result must not be confused with an absent member; the member remains part of the truss model.

Worked example

Two non-collinear members

Joint A has no applied load or support reaction. It connects only members AB and AC, which are not collinear. Identify their member forces and verify equilibrium.
  1. Define the system
    Isolate pin A. The only forces on this joint are the axial member forces in AB and AC. Assume each acts away from A; the conclusion will not depend on that assumed direction.
  2. Resolve the forces
    Take the AB direction as horizontal and let AC make an angle of 60∘60^\circ above the positive horizontal axis. With no external force, the joint must satisfy both component equations.
    FAB+FACcos⁡60∘=0,FACsin⁡60∘=0F_{AB}+F_{AC}\cos 60^\circ=0,\quad F_{AC}\sin 60^\circ=0
  3. Identify the forces
    Since sin⁡60∘\sin 60^\circ is nonzero, the vertical equation requires FAC=0F_{AC}=0. Substituting that result into the horizontal equation gives FAB=0F_{AB}=0.
    FAC=0,FAB=0F_{AC}=0,\quad F_{AB}=0
Answer: Both AB and AC are zero-force members for this loading condition.
Check: The horizontal and vertical force sums are both zero. Each force acts through A, so the moment sum about A is also zero.

Worked example

Three members with a collinear pair

An unloaded joint D connects horizontal members DE and DF in opposite directions, and a third member DG that is vertical. The force in DE is known to be 6 kN pulling away from D. Identify the force in DG and determine the force in DF.
  1. Define the system and signs
    Isolate D, which has no applied force or support reaction. Let right and up be positive. The known 6 kN force in DE points left; assume the unknown member forces point away from D.
  2. Apply vertical equilibrium
    Only DG contributes a vertical component. The vertical force sum must vanish, so DG carries no force.
    ∑Fy=FDG=0\sum F_y=F_{DG}=0
  3. Apply horizontal equilibrium
    The two collinear member forces must balance. DF must point right with magnitude 6 kN to balance the known 6 kN force to the left.
    ∑Fx=FDF−6 kN=0,FDF=6 kN\sum F_x=F_{DF}-6\text{ kN}=0,\quad F_{DF}=6\text{ kN}
Answer: DG is a zero-force member, and DF carries 6 kN pulling away from D.
Check: The horizontal sum is 6 kN−6 kN=06\text{ kN}-6\text{ kN}=0, and the vertical sum is zero. All forces act through D, so the moment sum about D is zero.

Worked example

Recheck a neighbouring joint

At unloaded joint A, members AB and AC are the only connected members and are non-collinear, so they are zero-force members. Joint B is also unloaded. It connects AB, BC (horizontal), and BD (at an angle of 45∘45^\circ above the horizontal). Use the result at A to identify the forces at B.
  1. Use the result at A
    The first joint establishes FAB=0F_{AB}=0. The force that member AB exerts at B is therefore also zero. At B, the remaining possible member forces are in BC and BD, which are not collinear.
    FBA=0F_{BA}=0
  2. Write equilibrium at B
    There is no external force at B. The vertical component equation contains only the force in BD, so it must vanish. The horizontal equation then requires the force in BC to vanish as well.
    FBDsin⁡45∘=0,FBC+FBDcos⁡45∘=0F_{BD}\sin 45^\circ=0,\quad F_{BC}+F_{BD}\cos 45^\circ=0
  3. State the result
    Since sin⁡45∘\sin 45^\circ is nonzero, FBD=0F_{BD}=0, and substitution gives FBC=0F_{BC}=0. Rechecking a joint after a zero-force result can reveal additional zero-force members.
    FBC=FBD=0F_{BC}=F_{BD}=0
Answer: AB, BC, and BD are all zero-force members for the stated loading condition.
Check: At B, both force components sum to zero after substitution. All forces act at B, so the moment sum about B is zero.

Common mistakes and how to avoid them

Applying a zero-force rule at a joint that has an applied load or support reaction.
Correction: Check for every external force and reaction first. If one is present, write equilibrium equations including it.
Calling all three members zero at an unloaded joint with two collinear members.
Correction: Only the non-collinear member is forced to zero by that rule. The collinear pair may carry equal and opposite forces.
Treating a member that looks nearly aligned as collinear.
Correction: Use the actual geometry. The rule requires the two member centre lines to lie on the same straight line.
Assuming a zero-force member can be ignored at every loading condition.
Correction: The result applies to the stated load case. A different loading condition can change the member forces.

Lesson summary

  • At an unloaded joint with exactly two non-collinear members, both member forces are zero.
  • At an unloaded joint with three members, if two are collinear, the third member has zero force.
  • After identifying a zero-force member, recheck neighbouring joints for a simpler pattern.
  • Verify horizontal and vertical force balance; concurrent forces at the joint have zero moment about it.

Check your understanding

Question 1

An unloaded joint connects three members. Two are collinear, and the third is not. What does the zero-force rule identify?
  1. All three members are zero-force members.
  2. Only the non-collinear member is a zero-force member.
  3. Only the two collinear members are zero-force members.
  4. No member can be identified without finding the reactions of the entire truss.
Show answer and explanation
Only the non-collinear member is a zero-force member.
The force balance perpendicular to the collinear pair contains only the non-collinear member force, so that force must be zero.

Question 2

An unloaded joint connects only two members that are not collinear. Which conclusion follows from equilibrium?
  1. Both member forces are zero.
  2. The members must have equal nonzero forces.
  3. Only the member with the steeper angle is zero.
  4. One force must be vertical and the other horizontal.
Show answer and explanation
Both member forces are zero.
Two non-collinear forces cannot balance each other unless each is zero; there is no external force at the joint.

Question 3

A joint has an applied force in addition to two non-collinear members. Is the two-member unloaded-joint rule directly applicable?
  1. Yes; any joint with two non-collinear members has two zero-force members.
  2. Yes, if the applied force is small.
  3. No; the applied force must be included in the equilibrium equations.
  4. No; moments must be used instead of force equilibrium.
Show answer and explanation
No; the applied force must be included in the equilibrium equations.
The rule requires an unloaded joint. An applied force changes the joint’s balance and must be included.

Key terms

Joint
A pin connection where truss members meet.
Zero-force member
A truss member whose axial force is zero for the loading condition being considered.
Collinear
Located along the same straight line.
Two-force member
A member acted on only at its two ends; in a truss model, its force acts along the member axis.

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