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4.5 · Classify member forces as tension or compression and verify equilibrium

Learn to classify member forces as tension or compression and verify equilibrium through clear examples and targeted practice.

University of Alberta ENGG 130: Engineering Mechanics: Statics

Truss Analysis

A joint-by-joint method for deciding what a member force means and checking your answer

A truss member can pull on a joint or push on it. These actions are called tension and compression, respectively. The algebra does not label a member for you: you choose an assumed force direction, solve the equilibrium equations, and use the result’s sign to classify the member. This lesson focuses on that process at pin-connected joints. Treat each joint as a particle: the member forces and any applied loads act at the joint, and their vector sum must be zero. All examples use planar statics, ideal pins, and forces in kN.

What you will learn

  • Describe tension and compression using the direction of a member’s force on a joint.
  • Draw an isolated-joint free-body diagram and state a consistent force sign convention.
  • Use planar equilibrium equations to solve for member-force magnitudes and directions.
  • Interpret a negative assumed member force as compression when tension was assumed.
  • Check force equilibrium and explain how moment equilibrium applies at an ideal pin joint.

1. What tension and compression mean

For a pin-connected, straight member carrying force only at its two ends, the end forces act along the member. When the member is in tension, it pulls each connected joint away from the member’s other end. When it is in compression, it pushes each joint toward the other end. These terms describe the direction of the member’s force on the joint; they do not describe whether the member force is positive or negative by themselves.
To find an unknown member force, isolate one joint and draw every force acting on it. A useful default is to assume every unknown member force is tension: draw its arrow away from the joint along the member. If the solved value is positive, that assumed tension direction was correct. If it is negative, the actual force is opposite the arrow, so the member is in compression. State the magnitude and classification clearly.
Fx=0,Fy=0F_{x}=0,\quad F_{y}=0
  • Tension pulls away from the joint; compression pushes toward it.
  • A negative result reverses the assumed arrow. It is not a negative physical magnitude.

2. Isolate a joint and set up equilibrium

The system for a joint calculation is the single ideal pin and the forces applied to it. Replace each connected member by a force along that member’s axis. Include known applied loads and any known forces that act at the joint. Do not include forces exerted at other joints: they are not acting on the isolated system.
Choose horizontal and vertical axes, then resolve angled forces into components. With right and up positive, a force with magnitude FF and angle θ\theta measured counterclockwise from the positive horizontal axis has components Fx=Fcos⁡θF_x=F\cos\theta and Fy=Fsin⁡θF_y=F\sin\theta. Write one signed component equation for each axis. These equations state that the total force vector on the joint is zero.
At a pin joint, all forces in the isolated-joint diagram meet at the pin. Their moment arms about that point are zero, so the moment sum about the pin is zero as well. The joint’s moment check is therefore satisfied by the concurrent-force model; it does not replace either force equation. For a larger isolated rigid body, moment equilibrium would also need to be checked about a suitable point.
∑Fx=0,∑Fy=0,∑MO=0\sum F_x=0,\quad \sum F_y=0,\quad \sum M_O=0
  • Use one axis and sign convention throughout a calculation.
  • Resolve angled forces before adding components.
  • For an ideal isolated joint, check both force components; moments about the pin are zero because the forces are concurrent.

3. Solve, classify, and verify

Start with the equations that most directly determine the unknowns. If a joint has two unknown member forces, the two component equations often suffice. Keep unknowns signed according to the arrows in your diagram, and retain units in substitutions. If a member force comes out negative, report its positive magnitude and identify compression, rather than silently changing the equations.
Verify by substituting the signed answers into both component sums. Each must equal zero, apart from rounding. Also check the moment sum about the pin: for a particle model with concurrent forces, each force has zero perpendicular distance from the pin, so the sum is zero. A force check that fails usually signals a component, sign, or rounding error. A classification that conflicts with the arrow direction usually signals that the assumed direction was not interpreted correctly.
Fmember=Fsigned u\mathbf{F}_{\text{member}}=F_{\text{signed}}\,\mathbf{u}
  • Verify horizontal and vertical balance independently.
  • Report classification together with the magnitude and unit.
  • For a joint particle, the moment check about the pin is zero because each force acts through that point.

Worked example

Two members supporting a downward load

At a pin joint, member ABAB extends horizontally to the left and member ACAC extends up and right at 45∘45^\circ. A 6.00 kN6.00\,\text{kN} load acts downward. Find the force in each member and classify it. Assume tension for both unknowns.
  1. Choose axes and assumptions
    Isolate pin A. Take right and up as positive, and assume both member forces pull away from the joint. The horizontal member arrow therefore points left; the diagonal arrow points up and right.
  2. Write component balance
    The diagonal force has equal horizontal and vertical components because its angle is 45∘45^\circ. The vertical equation determines its magnitude first.
    ∑Fy=FACsin⁡45∘−6.00=0\sum F_y=F_{AC}\sin45^\circ-6.00=0
  3. Solve and classify
    Solving the vertical equation gives a positive diagonal force. Substitute it into horizontal balance to find the horizontal member force. Both values are positive under the assumed tension arrows.
    FAC=8.49 kN,FAB=6.00 kNF_{AC}=8.49\,\text{kN},\quad F_{AB}=6.00\,\text{kN}
  4. Verify equilibrium
    The vertical component of the diagonal cancels the load. Its horizontal component is balanced by the leftward force in ABAB. All forces act through A, so the moment about A is zero.
    ∑Fx=−6.00+8.49cos⁡45∘=0,∑Fy=8.49sin⁡45∘−6.00=0\sum F_x=-6.00+8.49\cos45^\circ=0,\quad \sum F_y=8.49\sin45^\circ-6.00=0
Answer: Member ACAC carries 8.49 kN8.49\,\text{kN} in tension, and member ABAB carries 6.00 kN6.00\,\text{kN} in tension.
Check: The diagonal components are approximately 6.00 kN6.00\,\text{kN} right and 6.00 kN6.00\,\text{kN} up. They balance the leftward member force and downward load, respectively.

Worked example

A negative result identifies compression

At a pin joint, two members extend up-left and up-right, each at 45∘45^\circ to the horizontal. A 10.0 kN10.0\,\text{kN} force acts upward at the joint. Assume tension for both member forces. Determine and classify the forces.
  1. Set up the equations
    With both assumed tension forces directed outward, the horizontal components oppose one another. The vertical components and the applied upward force all point upward under these assumptions.
    ∑Fx=−FBLcos⁡45∘+FBRcos⁡45∘=0\sum F_x=-F_{BL}\cos45^\circ+F_{BR}\cos45^\circ=0
  2. Use symmetry and vertical balance
    The horizontal equation gives equal signed member forces. Substituting that relation into the vertical equation shows that each assumed force must be negative. A negative value means each actual force points into the joint, so both members are in compression.
    FBL=FBR=−7.07 kNF_{BL}=F_{BR}=-7.07\,\text{kN}
  3. Verify with actual directions
    The actual compressive forces point down and inward. Their horizontal components cancel; together their downward components balance the applied upward force. Moments about B are zero because all forces pass through B.
    ∑Fx=0,∑Fy=10.0−2(7.07sin⁡45∘)=0\sum F_x=0,\quad \sum F_y=10.0-2(7.07\sin45^\circ)=0
Answer: Each member force has magnitude 7.07 kN7.07\,\text{kN} and is compressive. The negative signs indicate that both actual forces are opposite the assumed outward tension arrows.
Check: Each member contributes approximately 5.00 kN5.00\,\text{kN} downward. Together they balance the 10.0 kN10.0\,\text{kN} upward force.

Worked example

Resolve an angled force before classifying members

At joint C, a horizontal member extends left and a diagonal member extends up-right with direction components proportional to 33 horizontally and 44 vertically. A 2.00 kN2.00\,\text{kN} force acts right and a 6.00 kN6.00\,\text{kN} force acts down. Assume tension for both members. Find their forces and verify the joint.
  1. Write the diagonal components
    The 33-44 direction has length ratio 55, so its unit direction components are 3/53/5 horizontally and 4/54/5 vertically. This keeps the unknown force magnitude separate from its direction.
    uCD=(35,45)\mathbf{u}_{CD}=\left(\frac{3}{5},\frac{4}{5}\right)
  2. Solve vertical balance
    Only member CDCD has an upward component in the assumed directions. Its vertical component must balance the 6.00 kN6.00\,\text{kN} downward load.
    45FCD−6.00=0,FCD=7.50 kN\frac{4}{5}F_{CD}-6.00=0,\quad F_{CD}=7.50\,\text{kN}
  3. Solve horizontal balance
    The diagonal contributes 4.50 kN4.50\,\text{kN} right. Together with the applied 2.00 kN2.00\,\text{kN} right force, it is balanced by the assumed leftward force in CHCH. The positive result confirms tension.
    −FCH+35(7.50)+2.00=0,FCH=6.50 kN-F_{CH}+\frac{3}{5}(7.50)+2.00=0,\quad F_{CH}=6.50\,\text{kN}
  4. Check forces and moment
    Substitution makes both component sums zero. Each force acts through C, so every moment about C is zero as well.
    ∑Fx=−6.50+4.50+2.00=0,∑Fy=6.00−6.00=0,∑MC=0\sum F_x=-6.50+4.50+2.00=0,\quad \sum F_y=6.00-6.00=0,\quad \sum M_C=0
Answer: Member CDCD carries 7.50 kN7.50\,\text{kN} in tension, and member CHCH carries 6.50 kN6.50\,\text{kN} in tension.
Check: The force components balance independently: the horizontal total is −6.50+4.50+2.00=0 kN-6.50+4.50+2.00=0\,\text{kN}, and the vertical total is 6.00−6.00=0 kN6.00-6.00=0\,\text{kN}.

Common mistakes and how to avoid them

Calling every positive member force tension without checking the arrow assumed in the diagram.
Correction: A positive result agrees with the assumed arrow. It indicates tension only when that arrow was drawn away from the joint.
Treating a negative solved force as an impossible answer.
Correction: Keep the signed result for the equations, then report its magnitude and reverse the assumed direction. An assumed outward tension arrow that gives a negative result means compression.
Using the full angled force in a horizontal or vertical equation.
Correction: Resolve it into components using its angle or unit direction before summing forces.
Checking only one force component or claiming that the moment check replaces force balance.
Correction: Check both planar force components. At an ideal pin joint, the moment sum about the pin is zero because all joint forces are concurrent; this does not remove the need for force balance.

Lesson summary

  • Isolate the joint and draw every member force along its member.
  • Assume tension by drawing unknown member forces away from the joint.
  • Apply horizontal and vertical force equilibrium with a consistent sign convention.
  • A positive result agrees with the assumed arrow; a negative result means the actual direction is opposite.
  • Verify both force components. At an ideal joint, moments about the pin are zero because the forces pass through it.

Check your understanding

Question 1

An unknown member force is drawn away from a joint as an assumed tension force. The calculation gives −3.2 kN-3.2\,\text{kN}. What is the correct classification?
  1. Tension with magnitude 3.2 kN3.2\,\text{kN}
  2. Compression with magnitude 3.2 kN3.2\,\text{kN}
  3. No force because the answer is negative
  4. Compression with magnitude −3.2 kN-3.2\,\text{kN}
Show answer and explanation
Compression with magnitude 3.2 kN3.2\,\text{kN}
The negative sign means the actual force is opposite the assumed outward arrow. Report a positive magnitude of 3.2 kN3.2\,\text{kN} and classify the member as compression.

Question 2

For an isolated ideal pin joint, why is the moment sum about the pin zero?
  1. Every force has zero perpendicular distance from the pin.
  2. The horizontal force sum is always zero, so moments must be zero.
  3. Member forces have no direction.
  4. A joint can never have an applied force.
Show answer and explanation
Every force has zero perpendicular distance from the pin.
All forces in the particle model act through the pin, so their moment arms about that point are zero. Force equilibrium must still be checked separately.

Key terms

Member force
The force a connected member exerts on a joint, directed along the member in the ideal pin-connected model.
Tension
A member action that pulls a joint away from the member’s other end.
Compression
A member action that pushes a joint toward the member’s other end.
Equilibrium
A condition in which the total force and total moment on the chosen system are zero.
Free-body diagram
A drawing of an isolated system showing the external forces and moments acting on it.

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