Engineering question: How can we calculate the internal force in each truss member using equilibrium at its joints?
Complete the modules in order. Read the worked example and sketch force arrows before revealing the practice answers.
Finish the remaining practice and worksheet outside class as assigned.
Every joint of an ideal truss is in equilibrium. If you isolate one joint, all connected members exert forces along their own axes, and those forces must balance the applied loads and support reactions at that joint. Read the reference method ↗
Initially draw an unknown member force away from the isolated joint. After solving, a positive force means tension (a pull). A negative force means compression (a push); the real force acts in the direction opposite the assumed arrow.
Rule: +F = tension under the outward-arrow convention; −F = compression.
Members are straight, weightless two-force members; ideal joints are frictionless pins; loads act at joints; and the structure is in static equilibrium. Actual manufactured bridges have connection stiffness, self-weight, imperfections and bending effects.
Critical: A calculated axial force is not a load limit or proof that a bridge will withstand that force. Real designs also require member strength, buckling, connection and stability checks.
When a member force is found, transfer its correct direction to the joint at the other end. For tension, the member pulls both joints toward itself along its axis (away from each isolated joint); for compression, it pushes outward on each joint. Check force equilibrium at a remaining joint to catch sign errors.
Geometry shortcut: For a right triangle with rise h, horizontal run r and diagonal length d, sin θ = h/d and cos θ = r/d. Units cancel in these ratios.
Start with the step-by-step Jeff Hanson lesson, then use the second explanation to reinforce how negative values identify compression. Both are linked directly if embeds are blocked by the school network.
Pause and practice: At 01:46 find support reactions; at 04:19 note how the first joint is selected; at 06:53 sketch assumed-tension member arrows.
Watch on YouTube ↗Pause and practice: Try drawing the joint FBD yourself. Decide whether a negative answer is a mistake or a compression force.
Watch on YouTube ↗Some members carry no axial force under a particular idealized load case. The on-page zero-force-member example below introduces the simplest inspection rule; identifying many zero-force members is extended in Lesson 2.9.
A triangular truss spans 4.0 m with supports A (pin) and B (roller). Apex C is centered 1.5 m above the base. An external 120 N downward point load acts at C. Neglect the truss's self-weight.
Use the whole truss, with CCW moments positive and coordinates measured from A.
Bᵧ = 60 N upward.
Aᵧ = 60 N upward; Aₓ = 0 because no horizontal external load acts.
Each diagonal has a horizontal run of 2.0 m and vertical rise of 1.5 m.
Diagonal length d = √(2² + 1.5²) = 2.5 m.
θ = arctan(1.5/2) ≈ 36.87° above the horizontal. These two ratios will simplify both joint equations.
Start with both unknown member forces assumed in tension: at joint C, draw arrows along CA and CB pointing down-left and down-right. Symmetry gives FAC = FBC = F.
Solve: −1.2F = 120 ⇒ F = −100 N.
A negative force under the assumed-tension convention means each member is actually in compression, magnitude 100 N. The diagram shows the corrected actual forces pushing upward on C.
The 60 N upward reaction is known. AC is now known to be 100 N in compression, so at joint A it acts down-left.
FAB = +80 N, so member AB is in tension.
The vertical force check at A is satisfied. The positive 80 N value is consistent with the bottom member pulling joint A to the right.
| Quantity | Signed (tension-positive) | Physical interpretation |
|---|---|---|
| Aᵧ | +60 N | Upward external reaction at A |
| Bᵧ | +60 N | Upward external reaction at B |
| AC | −100 N | Compression |
| BC | −100 N | Compression |
| AB | +80 N | Tension |
Joint B check: ΣFᵧ = +60−100(0.6) = 0; ΣFₓ = −80+100(0.8) = 0. All joint equilibrium conditions are satisfied. The signed member forces are not the same as the positive magnitudes you would use for an eventual strength calculation.
Keep the span at 4.0 m while varying apex load P and rise h. Watch what happens to the diagonal compression and bottom-tie tension. Use the stage buttons to focus on the whole truss or each joint.
What the model calculates: Aᵧ = Bᵧ = P/2; |FAC| = |FBC| = P/(2sinθ) compression; FAB = [P/(2tanθ)] tension.
Begin with the applied load and the supports. The load is centered, so each vertical reaction equals half the total downward load.
ΣFᵧ = Aᵧ + Bᵧ − P = 0
Assumptions: ideal symmetric pin-jointed triangle, load applied downward only at C, weightless members and no other loads. Display shows internal force demand, not allowable capacity or a failure prediction.
Keep h = 1.50 m. Change P from 60 to 120 N. What happens to all calculated reactions and member forces?
Every calculated force doubles. This ideal static analysis is linear in P while geometry is unchanged.
Keep P constant and increase h from 1.00 m to 3.00 m. How do the diagonal compression and bottom-tie tension magnitudes change?
Both decrease. The diagonals become steeper, so each supplies more upward force per unit axial force; the bottom tie also needs less tension to balance the horizontal components. External support reactions remain P/2 each.
Use a span of 4 m, a rise of 2 m and a centered downward 100 N apex load. The two diagonals make 45° angles with the horizontal.
sin45° = cos45° ≈ 0.7071. The equal diagonal component fractions simplify the calculations.
Remember: Steeper geometry can reduce required axial forces in this particular model, but that does not by itself decide which truss design is strongest or most material-efficient.
An unloaded truss joint with three connected members, two collinear, provides a useful shortcut: the third, noncollinear member has zero axial force for that specific load case.
At joint J, member AJ and JB lie on the same horizontal line. There is no external load or support reaction applied at J.
Because JC is the only vertical force on the isolated joint, its axial force must equal 0 N.
Not a reason to remove JC from a real bridge. It may be important for bracing, buckling resistance or a different future load case.
Complete the assigned 2.8 worksheet or notebook pages. The default practice case is the 4.0 m span, 1.5 m rise triangle shown above; your teacher may assign a different force or geometry.
Use paper for free-body diagrams, equations and final results. Write force signs and whether members are in tension or compression.
Joint C has a 100 N downward load and two identical 45° members. Assume tension for both unknowns. Determine their axial force and classify them.
Symmetry gives F1 = F2 = F. From ΣFᵧ = −2F sin45°−100=0, F = −70.71 N. Both members are in compression; magnitudes are 70.71 N each.
A truss joint has three completely unknown member forces and no other information. Can you solve all three forces using the two independent equations at this single planar joint?
No. A planar joint supplies ΣFₓ = 0 and ΣFᵧ = 0: two independent scalar equations. Start at another joint with at most two unknown member forces, or solve known reactions first.
A member force of 50 N acts upward and right, along a diagonal with horizontal run 4 and rise 3. What are its signed x and y components?
Diagonal length is 5, so Fₓ = +50(4/5) = +40 N, Fᵧ = +50(3/5) = +30 N. This resolves a force, not a complete truss solution.
You drew a member force arrow away from an isolated joint and calculated −85 N. Is your answer automatically wrong? What force does the member actually exert on that joint?
The negative sign means the assumed outward tension direction was incorrect. The member has 85 N compression and pushes on the joint opposite the outward assumed arrow.
For a symmetric triangle with halfspan 2 m, determine the magnitude of compression in each diagonal and the bottom tie tension.
sinθ = 1/√5 = 0.44721, cosθ = 2/√5 = 0.89443. Each diagonal magnitude = 100/[2(0.44721)] = 111.80 N compression. Tie = 111.80(0.89443) = 100.00 N tension. Reactions are 50 N each.
At joint J, JA and JB are collinear and JC is noncollinear. No external load or support reaction is applied at J. Determine the member force JC.
Member JC = 0 N for this idealized loading case. The perpendicular component of the other two collinear forces is zero; equilibrium forces the noncollinear member force to be zero.
The diagrams and solved examples on this page are enough for the core lesson. These openly available engineering resources provide additional practice and deeper explanations.
Free, illustrated textbook explanation of joint free-body diagrams, tension assumptions, reactions and member force transfer. Includes its own truss interactive.
Open Method of Joints ↗Explains ideal-truss assumptions, counting stable arrangements and zero-force members.
Open truss introduction ↗Illustrated explanation of the method of joints and method of sections; advanced portions preview the next lesson.
Open visual truss lesson ↗Axial force: force along a member's length. Tension: member pulled outward. Compression: member pushed inward. Joint equilibrium: balance of all x and y forces acting on one isolated joint.
Complete the POE 2.8 Calculating Truss Forces classwork and homework worksheet (or an assigned notebook equivalent), including the full method-of-joints analysis. Use the instructional videos and revealed checks as study aids, not a substitute for showing your own calculations.
Due date: Check the Daily Agenda for the current deadline and follow the teacher's instructions for submitting the worksheet or notebook pages.
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